TP_TO_physics_en_COMP.tsv

id	question	solution	final_answer	context	image	modality	difficulty	is_multiple_answer	unit	answer_type	error	question_type	subfield	subject	language
0	"(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$"	['We can write down the equations of motion, multiplying the second one by $i$ :\n\n$$\n\\begin{aligned}\n& 0=\\ddot{x}+2 \\Omega \\dot{y}-\\Omega^{2} x \\\\\n& 0=\\ddot{y}-2 \\Omega i \\dot{x}-\\Omega^{2} i y\n\\end{aligned}\n\\tag{19}\n$$\n\nWe can add these equations together to obtain\n\n$$\n0=(\\ddot{x}+i \\ddot{y})+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2}(x+i y)=\\ddot{\\eta}+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2} \\eta\n\\tag{20}\n$$\n\nWe note that\n\n$$\n\\dot{y}-i \\dot{x}=-i(\\dot{x}+i \\dot{y})=-i \\eta\n\\tag{21}\n$$\n\nThen\n\n$$\n0=\\ddot{\\eta}-2 i \\Omega \\dot{\\eta}-\\Omega^{2} \\eta\n\\tag{22}\n$$']		"4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$"	[]	Text-only	Competition	False				Theorem proof	Mechanics	Physics	English
1	(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.	['We can multiply both sides of the answer to part (e) by $\\hbar$ and use $E=\\hbar \\omega$ and $p=\\hbar k$, we have\n\n$$\nE=\\frac{p^{2}}{2 m}\n\\tag{33}\n$$\n\nA classical momentum has $p=m v$, so this gives the classical energy for a free particle (i.e., one without a potential):\n\n$$\nE=\\frac{1}{2} m v^{2}\n\\tag{34}\n$$']		"4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$
Context question:
(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$
Context answer:
\boxed{证明题}


Context question:
(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.

Plugging in this guess, what must $\lambda$ be?
Context answer:
\boxed{$\lambda=i \Omega$}


Context question:
(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.

This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.
Context answer:
\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}


Context question:
(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by

$$
\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{5}
$$

A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation.""
Context answer:
\boxed{$\omega=-i k^{2} a$}


Context question:
(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by

$$
i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{6}
$$

Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?
Context answer:
\boxed{$\omega=\frac{\hbar k^{2}}{2 m}$}
"	[]	Text-only	Competition	False				Theorem proof	Mechanics	Physics	English
2	"(b) Laplace's equation is a second order differential equation

$$
\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
\tag{8}
$$

Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.

Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.

Hint: Consider $\phi=\phi_{1}-\phi_{2}$."	"[""Suppose that we have two potentials satisfying the Laplace's equation and the boundary conditions. $\\nabla^{2} \\phi_{1}=0, \\nabla^{2} \\phi_{2}=0$ and $\\phi_{1}=\\phi_{2}$ on the boundary of $V$. Define $\\phi_{3}=\\phi_{1}-\\phi_{2}$. By the linearity of Laplace's equation, $\\nabla^{2} \\phi_{2}=0$ and $\\phi_{3}=0$ on the surface. By the maximum principle, all extrema occurs at the surface. Thus, both the minimum and the maximum of $\\phi_{3}$ is 0 . Therefore, $\\phi_{3}=0$. This concludes the proof that $\\phi_{1}=\\phi_{2}$ identically.""]"		"5. Polarization and Oscillation 

In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.

Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
Context question:
(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
Context answer:
开放性回答
"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
3	(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)	"[""The homogeneity of space implies the following:\n\n$$\nX\\left(x_{2}+h, t, v\\right)-X\\left(x_{2}, t, v\\right)=X\\left(x_{1}+h, t, v\\right)-X\\left(x_{1}, t, v\\right)\n$$\n\n\n\n<img_4548>\n\nFigure 3: The shaded area is the ruler's trajectory in spacetime. Any measurement of its length must intersect the shaded area.\n\n\n\nnow dividing by $h$ and sending $h \\rightarrow 0$ is the partial derivative, therefore\n\n$$\n\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{2}}=\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{1}}\n$$\n\nThe same method is repeated for the other variables.""]"		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
4	(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)	['The isotropy of space implies\n\n$$\n\\begin{aligned}\nX(-x, t,-v) & =-x^{\\prime}=-X(x, t, v) \\\\\nT(-x, t,-v) & =T(x, t, v)\n\\end{aligned}\n$$\n\nand then plugging into (9) we see that\n\n$$\n\\begin{aligned}\n& A(-v)=A(v) \\\\\n& B(-v)=-B(v) \\\\\n& C(-v)=-C(v) \\\\\n& D(-v)=D(v)\n\\end{aligned}\n$$']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
5	"(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$."	['The principle of relativity implies that the coordinate transformation can be inverted such that\n\n$$\n\\begin{aligned}\nX\\left(x^{\\prime}, t^{\\prime},-v\\right) & =x \\\\\nT\\left(x^{\\prime}, t^{\\prime},-v\\right) & =t\n\\end{aligned}\n$$\n\ntherefore, because $X, T$ are linear, we can write the generalized boost in matrix form, and then using the even/oddness of $A, B, C, D$ derived previously\n\n$$\n\\left(\\begin{array}{ll}\nA(v) & B(v) \\\\\nC(v) & D(v)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -B(v) \\\\\n-C(v) & D(v)\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n1 & 0 \\\\\n0 & 1\n\\end{array}\\right)\n$$\n\nwhich is precisely the system of equations listed.']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
6	"(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$"	"[""The $F^{\\prime}$ frame is defined by $x^{\\prime}=0$, therefore $A(v) x+B(v) t=A(v) v t+B(v) t$ plugging in the equation $x=v t$. This implies that $A v+B=0$. This let's all the undetermined functions $A, B, C, D$ to be solved in terms of $A$.""]"		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
7	"(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that

$$
\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .
$$

for an arbitrary constant $\kappa$."	['Consider a frame $F^{\\prime \\prime}$ related to $F^{\\prime}$ by a boost in the $x^{\\prime}$-direction with relative velocity $u$. Therefore, the composition of these two boosts results in a boost $F \\rightarrow F^{\\prime \\prime}$ given by\n\n$$\n\\Lambda_{F \\rightarrow F^{\\prime \\prime}}=\\left(\\begin{array}{cc}\nA(u) & -u A(u) \\\\\n-\\frac{A(u)^{2}-1}{u A(u)} & A(u)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -v A(v) \\\\\n-\\frac{A(v)^{2}-1}{v A(v)} & A(v)\n\\end{array}\\right)\n$$\n\nMultiplying out these matrices and noting that we have shown that the diagonal terms must be equal, we find that\n\n$$\n\\frac{A(v)^{2}-1}{v^{2} A(v)^{2}}=\\frac{A(u)^{2}-1}{u^{2} A(u)^{2}}\n$$\n\nas the left hand side is a function of $v$ only and the right hand side is a function of $u$ only, they must both be equal to some constant $\\kappa$.']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}


Context question:
(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
8	"(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)
\end{aligned}
$$"	['Solve for $A$,\n\n$$\nA(v)=\\frac{1}{\\sqrt{1-\\kappa v^{2}}}\n$$\n\nand the dimensions of $\\kappa$ are determined by the restriction that $\\kappa v^{2}$ is being added to a dimensionless number. Substituting this form of $A(v)$ into\n\n$$\n\\begin{aligned}\nx^{\\prime} & =A(v) x-v A(v) t \\\\\nt^{\\prime} & =-\\left(\\frac{A(v)^{2}-1}{v A(v)}\\right) x+A(v) t\n\\end{aligned}\n$$\n\nyields the answer.']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}


Context question:
(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$
Context answer:
\boxed{证明题}


Context question:
(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that

$$
\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .
$$

for an arbitrary constant $\kappa$.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
9	(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.	['If $v$ is unbounded and $\\kappa \\neq 0$, it may be large enough so that the square root gives an imaginary number, and as $x^{\\prime}, t^{\\prime}$ cannot be imaginary, it must be that $\\kappa=0$. There is nothing stopping a particle from traveling arbitrarily fast under a Galilean structure of spacetime, as given any particle we may Galilean boost to an inertial frame moving at $v$ arbitrarily fast, in which the particle is then moving at $-v$. By the principle of relativity, there is nothing wrong with doing physics in this frame, so it must be that it is acceptable to have particles move arbitrarily fast']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}


Context question:
(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$
Context answer:
\boxed{证明题}


Context question:
(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that

$$
\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .
$$

for an arbitrary constant $\kappa$.
Context answer:
\boxed{证明题}


Context question:
(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)
\end{aligned}
$$
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
10	(h) Assume that $v$ must be smaller than a finite value. Show that $1 / \sqrt{\kappa}$ is the maximum allowable speed, and that this speed is frame invariant, i.e., $\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x}{d t}$ for something moving at speed $1 / \sqrt{\kappa}$. Experiment has shown that this speed is $c$, the speed of light. Setting $\kappa=1 / c^{2}$, show that you recover the Lorentz boost.	['Again by requiring that $x^{\\prime}, t^{\\prime}$ are real, we can find the desired bound on $v$ from $1-\\kappa v^{2}>0$. One way to show that the speed is frame invariant is by deriving the relativistic velocity addition formula as follows\n\n$$\n\\begin{aligned}\nd x^{\\prime} & =\\gamma(d x-v d t) \\\\\nd t^{\\prime} & =\\gamma\\left(d t-v / c^{2} d x\\right)\n\\end{aligned}\n$$\n\nand dividing to yield\n\n$$\n\\frac{d x^{\\prime}}{d t^{\\prime}}=\\frac{\\frac{d x}{d t}-v}{1-\\frac{v}{c^{2}} \\frac{d x}{d t}}\n$$\n\nLet $w=d x^{\\prime} / d t^{\\prime}, u=d x / d t$, and, as often makes relativity problems easier to deal with, set $c=1$ (this is equivalent to choosing a new system of units). Then, we have\n\n$$\nw=\\frac{u-v}{1-v u}\n$$\n\nnow if $w=c=1$, we can solve the above equation to show that $u=1$, in other words frame $F$ and $F^{\\prime}$ both agree on what velocities move at $c$.']		"4. Lorentz Boost

In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts:

$$
\begin{aligned}
x^{\prime} & =x-v t \\
t^{\prime} & =t
\end{aligned}
$$

In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts:

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right)
\end{aligned}
$$

In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$.

The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$,

$$
\begin{aligned}
x^{\prime} & =X(x, t, v) \\
t^{\prime} & =T(x, t, v)
\end{aligned}
$$

which we will refer to as the generalized boost.
Context question:
(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form

$$
\begin{aligned}
X(x, t, v) & =A(v) x+B(v) t \\
T(x, t, v) & =C(v) x+D(v) t
\end{aligned}
$$

where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.
Context question:
(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)
Context answer:
\boxed{证明题}


Context question:
(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold:

$$
\begin{aligned}
A(v)^{2}-B(v) C(v) & =1 \\
D(v)^{2}-B(v) C(v) & =1 \\
C(v)(A(v)-D(v)) & =0 \\
B(v)(A(v)-D(v)) & =0 .
\end{aligned}
$$

(Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.
Context answer:
\boxed{证明题}


Context question:
(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form:

$$
\begin{aligned}
x^{\prime} & =A(v) x-v A(v) t \\
t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t
\end{aligned}
$$
Context answer:
\boxed{证明题}


Context question:
(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that

$$
\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa .
$$

for an arbitrary constant $\kappa$.
Context answer:
\boxed{证明题}


Context question:
(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form

$$
\begin{aligned}
x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\
t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x)
\end{aligned}
$$
Context answer:
\boxed{证明题}


Context question:
(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
11	"(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$."	['A periodic wave on interval $[0, L]$ must fit an integer number of wavelengths $\\lambda$ into the length $L$,\n\n$$\nn \\lambda=L\n\\tag{1}\n$$\n\nTherefore $k_{n}=\\frac{2 \\pi}{\\lambda_{n}}=\\frac{2 \\pi n}{L}$.\n\nFor the next part, use $c^{\\prime}=\\frac{\\omega}{k}$. Therefore,\n\n$$\n\\omega_{n}=c^{\\prime} k_{n}=\\frac{2 \\pi c^{\\prime} n}{L} \\Rightarrow d \\omega_{n}=\\frac{2 \\pi c^{\\prime}}{L} d n \\Rightarrow \\frac{d n}{d \\omega_{n}}=\\frac{L}{2 \\pi c^{\\prime}}\n\\tag{2}\n$$']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$."	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
12	"(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$."	['Use the given Taylor expansion,\n\n$$\n\\left\\langle N_{n}\\right\\rangle=\\frac{1}{\\exp \\frac{\\hbar \\omega_{n}}{k T}-1} \\approx \\frac{1}{\\left(1+\\frac{\\hbar \\omega_{n}}{k T}\\right)-1}=\\frac{k T}{\\hbar \\omega_{n}}\n\\tag{3}\n$$']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
13	(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.	['The electromagnetic modes are excitations of the electromagnetic field like photons, so they are also massless. The energy of a photon is $E=h f=\\hbar \\omega$.\n\nAlternatively, you can start from $E^{2}=\\left(m c^{2}\\right)^{2}+(p c)^{2}$ and use $m=0, p=h / \\lambda$.']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
14	"(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is

$$
P[f, f+d f] \approx k T d f .
\tag{6}
$$

Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)"	['Power equals energy per time. The average energy delivered by one boson to the resistor is $\\hbar \\omega_{n}$.\n\nFor each state $n$, the number of bosons which either enter or exit the resistor per time is equal to their population divided by the time taken to travel the length of the resistor, $t=L / c^{\\prime}$ :\n\n$$\n\\frac{d\\left\\langle N_{n}\\right\\rangle}{d t}=\\frac{\\left\\langle N_{n}\\right\\rangle}{L / c^{\\prime}}=\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\n\\tag{4}\n$$\n\n\n\nFor a frequency interval $d \\omega_{n}$, the number of states is $d n=\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}$ (from part (a)). Therefore, the energy delivered per time, per frequency interval is\n\n$$\nd P\\left[\\omega_{n}, \\omega_{n}+d \\omega_{n}\\right]=\\left(\\hbar \\omega_{n}\\right) \\times\\left(\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\\right) \\times\\left(\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}\\right)=k T \\frac{d \\omega_{n}}{2 \\pi}\n\\tag{5}\n$$\n\nUsing $f=\\frac{\\omega}{2 \\pi}$,\n\n$$\nd P[f, f+d f]=k T d f\n\\tag{6}\n$$']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Context question:
(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
15	"(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is

$$
P_{\max }=\frac{V^{2}}{4 R} .
\tag{7}
$$

Give the optimal value of $r$ in terms of $R$ and $V$."	['The total resistance in this circuit is $R+r$. The power dissipated in $r$ is therefore\n\n$$\nP=r I^{2}=r\\left(\\frac{V}{R+r}\\right)^{2}\n\\tag{7}\n$$\n\nThe maximization of $\\frac{r}{(R+r)^{2}}$ is equivalent to the minimization of $\\phi(r)=\\frac{(R+r)^{2}}{r}=\\frac{R^{2}}{r}+2 R+r$. Setting $\\frac{d \\phi}{d r}=0$, for example, gives the solution $r=R$. Substituting $r=R$ into 7 yields\n\n$$\nP=\\frac{V^{2}}{4 R}\n\\tag{8}\n$$']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Context question:
(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
Context answer:
\boxed{证明题}


Context question:
(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is

$$
P[f, f+d f] \approx k T d f .
\tag{6}
$$

Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Nyquist equivalent noisy voltage source 

The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$."	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
16	(b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$.	['Differentiate the result of the previous part, and apply a time-expectation $\\langle\\rangle: d\\left\\langle V^{2}\\right\\rangle=4 R d\\langle P\\rangle=$ $k T d f$. Therefore $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$.']		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Context question:
(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
Context answer:
\boxed{证明题}


Context question:
(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is

$$
P[f, f+d f] \approx k T d f .
\tag{6}
$$

Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Nyquist equivalent noisy voltage source 

The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.
Context question:
(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is

$$
P_{\max }=\frac{V^{2}}{4 R} .
\tag{7}
$$

Give the optimal value of $r$ in terms of $R$ and $V$.
Context answer:
证明题
"	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
17	(a) Explain why no Johnson-Nyquist noise is produced by ideal inductors or capacitors. There are multiple explanations; any explanation will be accepted. (Hint: the impedance of an ideal inductor or capacitor is purely imaginary.)	"[""For example, the impedance of an ideal inductor or capacitor is purely imaginary. Replacing $R \\rightarrow i \\omega L, \\frac{1}{i \\omega C}$ in the formula $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$ would give an imaginary squared-voltage, which doesn't make sense.\n\nMore physically, the current and voltage in a pure inductor or capacitor are always orthogonal (out-of-phase), so no power is dissipated.""]"		"2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Context question:
(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
Context answer:
\boxed{证明题}


Context question:
(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is

$$
P[f, f+d f] \approx k T d f .
\tag{6}
$$

Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Nyquist equivalent noisy voltage source 

The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.
Context question:
(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is

$$
P_{\max }=\frac{V^{2}}{4 R} .
\tag{7}
$$

Give the optimal value of $r$ in terms of $R$ and $V$.
Context answer:
证明题


Context question:
(b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Other circuit elements

We derived the Johnson-Nyquist noise due to a resistor, $R$. It turns out the equation $\frac{d\left\langle V^{2}\right\rangle}{d f}=$ $4 k T R$ is not generalizable to inductors or capacitors."	[]	Text-only	Competition	False				Theorem proof	Electromagnetism	Physics	English
18	4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.	"[""Consider the following systems, a frame $S^{\\prime}$ is moving with respect to another frame $S$, with velocity $u$ in the $x$ direction. If a particle is moving in the $\\mathrm{S}^{\\prime}$ frame with velocity $v^{\\prime}$ also in $x$ direction, then the particle velocity in the $\\mathrm{S}$ frame is given by\n\n$$\nv=\\frac{u+v^{\\prime}}{1+\\frac{u v^{\\prime}}{c^{2}}}\n\\tag{7}\n$$\n\n\n\nIf the particles velocity changes with respect to the $S$ ' frame, then the velocity in the $S$ frame is also change according to\n\n$$\n\\begin{aligned}\nd v & =\\frac{d v^{\\prime}}{1+\\frac{u v^{\\prime}}{c^{2}}}-\\frac{u+v^{\\prime}}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{2}} \\frac{u d v^{\\prime}}{c^{2}} \\\\\nd v & =\\frac{1}{\\gamma^{2}} \\frac{d v^{\\prime}}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{2}}\n\\end{aligned}\n\\tag{8}\n$$\n\nThe time in the $\\mathrm{S}^{\\prime}$ frame is $t^{\\prime}$, so the time in the $\\mathrm{S}$ frame is given by\n\n$$\nt=\\gamma\\left(t^{\\prime}+\\frac{u x^{\\prime}}{c^{2}}\\right)\n\\tag{9}\n$$\n\nso the time change in the $S^{\\prime}$ frame will give a time change in the $\\mathrm{S}$ frame as follow\n\n$$\nd t=\\gamma d t^{\\prime}\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)\n\\tag{10}\n$$\n\nThe acceleration in the $\\mathrm{S}$ frame is given by\n\n$$\na=\\frac{d v}{d t}=\\frac{a^{\\prime}}{\\gamma^{3}} \\frac{1}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{3}}\n\\tag{11}\n$$\n\nIf the $\\mathrm{S}^{\\prime}$ frame is the proper frame, then by definition the velocity $v^{\\prime}=0$. Substitute this to the last equation, we get\n\n$$\na=\\frac{a^{\\prime}}{\\gamma^{3}}\n\\tag{12}\n$$\n\nCombining Eq.(3) and Eq.(12), we get\n\n$$\na^{\\prime}=\\frac{F}{m} \\equiv g\n\\tag{13}\n$$""]"		"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.

First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).

By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.

Some mathematics formulas that might be useful

- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$



- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$


Part A. Single Accelerated Particle 

Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}


Context question:
2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}


Context question:
3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
19	1.  At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?	"[""When the clock in the origin time is equal to $t_{0}$, it emits a signal that contain the information of its time. This signal will arrive at the particle at time $t$, while the particle position is at $x(t)$. We have\n\n$$\nc\\left(t-t_{0}\\right) =x(t)\n\\tag{20}\n$$\n$$\nt-t_{0}=\\frac{c}{g}\\left(\\sqrt{1+\\left(\\frac{g t}{c}\\right)^{2}}-1\\right)\n$$\n$$\nt =\\frac{t_{0}}{2} \\frac{2-\\frac{g t_{0}}{c}}{1-\\frac{g t_{0}}{c}} .\n\\tag{21}\n$$\n\nWhen the information arrive at the particle, the particle's clock has a reading according to eq.(19). So we get\n\n$$\n\\begin{aligned}\n\\frac{c}{g} \\sinh \\frac{g \\tau}{c} & =\\frac{t_{0}}{2} \\frac{2-\\frac{g t_{0}}{c}}{1-\\frac{g t_{0}}{c}} \\\\\n0 & =\\frac{1}{2}\\left(\\frac{g t_{0}}{c}\\right)^{2}-\\frac{g t_{0}}{c}\\left(1+\\sinh \\frac{g \\tau}{c}\\right)+\\sinh \\frac{g \\tau}{c} .\n\\end{aligned}\n$$\n\n$$\n\\frac{g t_{0}}{c} =1+\\sinh \\frac{g \\tau}{c} \\pm \\cosh \\frac{g \\tau}{c}\n\\tag{22}\n$$\n\nUsing initial condition $t=0$ when $\\tau=0$, we choose the negative sign\n\n$$\n\\begin{aligned}\n\\frac{g t_{0}}{c} & =1+\\sinh \\frac{g \\tau}{c}-\\cosh \\frac{g \\tau}{c} \\\\\nt_{0} & =\\frac{c}{g}\\left(1-e^{-\\frac{g \\tau}{c}}\\right) .\n\\end{aligned}\n\\tag{23}\n$$\n\nAs $\\tau \\rightarrow \\infty, t_{0}=\\frac{c}{g}$. So the clock reading will freeze at this value.""]"		"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.

First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).

By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.

Some mathematics formulas that might be useful

- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$



- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$


Part A. Single Accelerated Particle 

Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}


Context question:
2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}


Context question:
3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}


Context question:
4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
Context answer:
\boxed{证明题}


Context question:
5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
Context answer:
\boxed{$\beta=\tanh \frac{g \tau}{c}$}


Context question:
6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.
Context answer:
\boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$}


Extra Supplementary Reading Materials:

Part B. Flight Time 

The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
20	2.  Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value?	"[""When the particles clock has a reading $\\tau_{0}$, its position is given by eq.(6), and the time $t_{0}$ is given by eq.(19). Combining this two equation, we get\n\n$$\nx=\\frac{c^{2}}{g}\\left(\\sqrt{1+\\sinh ^{2} \\frac{g \\tau_{0}}{c}}-1\\right)\n\\tag{24}\n$$\n\nThe particle's clock reading is then sent to the observer at the origin. The total time needed for the information to arrive is given by\n\n$$\nt =\\frac{c}{g} \\sinh \\frac{g \\tau_{0}}{c}+\\frac{x}{c}\n\\tag{25}\n$$\n\n$$\n=\\frac{c}{g}\\left(\\sinh \\frac{g \\tau_{0}}{c}+\\cosh \\frac{g \\tau_{0}}{c}-1\\right)\n$$\n\n$$\nt =\\frac{c}{g}\\left(e^{\\frac{g \\tau_{0}}{c}}-1\\right)\n\\tag{26}\n$$\n\n$$\n\\tau_{0} =\\frac{c}{g} \\ln \\left(\\frac{g t}{c}+1\\right)\n\\tag{27}\n$$\n\nThe time will not freeze.""]"		"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.

First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).

By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.

Some mathematics formulas that might be useful

- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$



- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$


Part A. Single Accelerated Particle 

Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
Context question:
1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
Context answer:
\boxed{$a=\frac{F}{\gamma^{3} m}$}


Context question:
2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
Context answer:
\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}


Context question:
3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
Context answer:
\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}


Context question:
4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
Context answer:
\boxed{证明题}


Context question:
5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
Context answer:
\boxed{$\beta=\tanh \frac{g \tau}{c}$}


Context question:
6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.
Context answer:
\boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$}


Extra Supplementary Reading Materials:

Part B. Flight Time 

The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A.
Context question:
1.  At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?
Context answer:
\boxed{证明题}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
21	"1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)"	['From the two equations given in the text, we obtain the relation\n\n$$\n\\frac{d \\boldsymbol{\\mu}}{d t}=-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}\n\\tag{1}\n$$\n\nTaking the dot product of eq (1). with $\\boldsymbol{\\mu}$, we can prove that\n\n$$\n\\begin{aligned}\n\\boldsymbol{\\mu} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\boldsymbol{\\mu} \\cdot(\\boldsymbol{\\mu} \\times \\mathbf{B}) \\\\\n\\frac{d|\\boldsymbol{\\mu}|^{2}}{d t} & =0 \\\\\n\\mu=|\\boldsymbol{\\mu}| & =\\text { const. }\n\\end{aligned}\n\\tag{2}\n$$\n\nTaking the dot product of eq. (1) with $\\mathbf{B}$, we also prove that\n\n$$\n\\begin{aligned}\n\\mathbf{B} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\mathbf{B} \\cdot(\\boldsymbol{\\mu} \\times \\mathbf{B}), \\\\\n\\mathbf{B} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =0, \\\\\n\\mathbf{B} \\cdot \\boldsymbol{\\mu} & =\\text { const. }\n\\end{aligned}\n\\tag{3}\n$$\n\nAn acute reader will notice that our master equation in (1) is identical to the equation of motion for a charged particle in a magnetic field\n\n$$\n\\frac{d \\mathbf{v}}{d t}=\\frac{q}{m} \\mathbf{v} \\times \\mathbf{B}\n\\tag{4}\n$$\n\nHence, the same argument for a charged particle in magnetic field can be applied in this case.']		"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
22	"1.  Show that the time evolution of the magnetic moment follows the equation

$$
\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
$$

where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field."	['Using the relation given in the text, it is easily shown that\n\n$$\n\\begin{aligned}\n\\left(\\frac{d \\boldsymbol{\\mu}}{d t}\\right)_{\\mathrm{rot}} & =\\left(\\frac{d \\boldsymbol{\\mu}}{d t}\\right)_{\\mathrm{lab}}-\\boldsymbol{\\omega} \\times \\boldsymbol{\\mu} \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}-\\omega \\mathbf{k}^{\\prime} \\times \\boldsymbol{\\mu} \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times\\left(\\mathbf{B}-\\frac{\\omega}{\\gamma} \\mathbf{k}^{\\prime}\\right) \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}_{\\mathrm{eff}} .\n\\end{aligned}\n\\tag{6}\n$$\n\nNote that $\\mathbf{k}$ is equal to $\\mathbf{k}^{\\prime}$ as observed in the rotating frame.']		"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession
Context question:
1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)
Context answer:
\boxed{证明题}


Context question:
2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
Context answer:
\boxed{$\omega_{0}=\gamma B_{0}$}


Extra Supplementary Reading Materials:

Part B. Rotating frame

In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes

$$
\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
$$



$$
\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
$$

where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$."	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
23	"3.  Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is

$$
\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}}
$$"	['Since the magnetic field as viewed in the rotating frame is $\\mathbf{B}=B_{0} \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}$,\n\n$$\n\\mathbf{B}_{\\mathrm{eff}}=\\mathbf{B}-\\omega / \\gamma \\mathbf{k}^{\\prime}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\Omega & =\\gamma\\left|\\mathbf{B}_{\\mathrm{eff}}\\right| \\\\\n& =\\gamma \\sqrt{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}}\n\\end{aligned}\n\\tag{8}\n$$']		"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession
Context question:
1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)
Context answer:
\boxed{证明题}


Context question:
2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
Context answer:
\boxed{$\omega_{0}=\gamma B_{0}$}


Extra Supplementary Reading Materials:

Part B. Rotating frame

In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes

$$
\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
$$



$$
\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
$$

where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
Context question:
1.  Show that the time evolution of the magnetic moment follows the equation

$$
\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
$$

where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
Context answer:
\boxed{证明题}


Context question:
2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?
Context answer:
\boxed{$\Delta =\gamma B_{0}-\omega$}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English
24	"2.  Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as

$$
M(t)=N \mu(\cos \Omega t) .
$$"	['Since the angle $\\alpha$ that $\\boldsymbol{\\mu}$ makes with $\\mathbf{B}_{\\text {eff }}$ stays constant and $\\boldsymbol{\\mu}$ is initially oriented along the $z$ axis, $\\alpha$ is also the angle between $\\mathbf{B}_{\\text {eff }}$ and the $z$ axis which is\n\n$$\n\\tan \\alpha=\\frac{b}{B_{0}-\\frac{\\omega}{\\gamma}}\n\\tag{11}\n$$\n\n<img_4496>\n\nFrom the geometry of the system, we can show that $\\left(\\cos \\theta=\\mu_{z} / \\mu\\right)$ :\n\n$$\n\\begin{aligned}\n2 \\mu \\sin \\frac{\\theta}{2} & =2 \\mu \\sin \\alpha \\sin \\frac{\\Omega t}{2} \\\\\n\\sin ^{2} \\frac{\\theta}{2} & =\\sin ^{2} \\alpha \\sin ^{2} \\frac{\\Omega t}{2} \\\\\n\\frac{1-\\cos \\theta}{2} & =\\sin ^{2} \\alpha \\frac{1-\\cos \\Omega t}{2} \\\\\n\\cos \\theta & =1-\\sin ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t \\\\\n\\cos \\theta & =\\cos ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t .\n\\end{aligned}\n$$\n\nSo, the projected magnetic moment along the $z$ axis is $\\mu_{z}(t)=\\mu \\cos \\theta$ and the magnetization is\n\n$$\nM=N \\mu_{z}=N \\mu\\left(\\cos ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t\\right) .\n$$\n\nNote that the magnetization does not depend on the reference frame $S$ or $S^{\\prime}$ ( $\\mu_{z}$ has the same value viewed in both frames).\n\nTaking $\\omega=\\omega_{0}=\\gamma B_{0}$, the angle $\\alpha$ is $90^{\\circ}$ and $M=N \\mu \\cos \\Omega t$.']		"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.

The classical torque equation of spin is given by

$$
\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
$$

In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation

$$
\boldsymbol{\mu}=-\gamma \boldsymbol{L}
$$

where $\gamma$ is the gyromagnetic ratio.

In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.

Part A. Larmor precession
Context question:
1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.

(Hint: You can use properties of vector products.)
Context answer:
\boxed{证明题}


Context question:
2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
Context answer:
\boxed{$\omega_{0}=\gamma B_{0}$}


Extra Supplementary Reading Materials:

Part B. Rotating frame

In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes

$$
\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
$$



$$
\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
$$

where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
Context question:
1.  Show that the time evolution of the magnetic moment follows the equation

$$
\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
$$

where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
Context answer:
\boxed{证明题}


Context question:
2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?
Context answer:
\boxed{$\Delta =\gamma B_{0}-\omega$}


Context question:
3.  Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is

$$
\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}}
$$
Context answer:
\boxed{证明题}


Context question:
4.  Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?
Context answer:
\boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$}


Extra Supplementary Reading Materials:

Part C. Rabi oscillation  

For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation

$$
N_{\uparrow}+N_{\downarrow}=N
$$

The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis:

$$
M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} .
$$

In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by

$$
\boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} .
$$
Context question:
1.  In the rotating frame $S^{\prime}$, show that the effective field can be approximated by

$$
\boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime},
$$

which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?
Context answer:
\boxed{$\Omega=\gamma b$}
"	[]	Text-only	Competition	False				Theorem proof	Modern Physics	Physics	English