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Условие Sucessive Transformations Precession Frequency Решение

OPhO 2021, Теория — Задача 2. Thomas Precession

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2021 Online Physics Olympiad: Invitational Contest — v1.1 Theoretical Exam · 25 т.

Условие

Sucessive Transformations

In this section, we examine what happens when two successive Lorentz transformations are applied in non-parallel directions.

1. (6 pts.) Consider the three reference frames $S_1$ , $S_2$ , and $S_3$ . Events as seen from frame $S_i$ will be labeled with the space-time coordinates $(x_i,y_i,z_i,t_i)$ , for $i=1,2,3$ . All three frames coincide at $(0,0,0,0)$ . Suppose frame $S_2$ travels with velocity $\beta c$ in the $x_1$ -direction of frame $S_1$ and frame $S_3$ travels with velocity $\beta_x'c\hat{\mathbf i}_2+\beta_y'c\hat{\mathbf j}_2$ with respect to frame $S_2$ .

Perform two successive Lorentz transformations: one expressing the $S_3$ coordinates in terms of the $S_2$ coordinates and another expressing the $S_2$ coordinates in terms of the $S_1$ coordinates. Then, as the final answer, express the $S_3$ coordinates in terms of the $S_1$ coordinates.

Assume that $\beta'=(\beta_x'^2+\beta_y'^2)^{1/2}\ll\beta$ and work to first order.

2. (5 pts.) Now, find the velocity of $S_3$ in $S_1$ with the appropriate velocity addition. Perform a single Lorentz transformation to express the $S_3$ coordinates in terms of the $S_1$ coordinates. Once again, work to first order. The answer will not be the same as part 1.

3. (3 pts.) Show that your answer in Problem 1 differs from your answer in Problem 2 by a spatial rotation. In other words, two successive Lorentz transformations in non-parallel directions cannot be combined as one Lorentz transformation. Rather, they are the combination of one Lorentz transformation and one spatial rotation. Determine the magnitude and direction of this spatial rotation in the current setup.

Then argue and explain why the answer in Problem 1 and not Problem 2 is the correct transformation.

Precession Frequency

In this section, we examine the precession of the electron’s spin magnetic moment within the hydrogen atom.

Electrons possess an attribute known as spin. One can imagine the electron as being a small, spherical charged particle that is spinning on some axis. Although this mental picture is not physically correct, it is enough for us. From this mental picture, we can gather that the electron will possess a dipole moment and angular momentum intrinsic to itself and not induced by any orbital motion. Respectively, these are known as the spin magnetic moment and the spin angular momentum that behave, in our model, just as their classical counterparts do. These two vector quantities, $\boldsymbol\mu$ and $\mathbf L$ respectively, are related by

\boldsymbol\mu\simeq-\frac e{m_e}\mathbf L,

which is determined by experiment.

In this section, we will use the Bohr model of the hydrogen atom, where the electron circles the proton at some radius $r$ , pulled in by the Coulomb force.

4. (3 pts.) In the laboratory frame, the electron orbits the proton with some velocity $v$ in the $x$ - $y$ plane. Now switch to the instantaneous rest frame of the electron, where the proton moves with speed $v$ relative to the stationary electron. The moving proton will then induce some magnetic field at the electron’s location.

Suppose the spin angular momentum of the electron points in some direction other than the direction of the magnetic field. Because the electron also possesses a spin magnetic moment, it will precess as a result of the torque done on it. Find the angular frequency of the precession of the electron’s spin in terms of $r$ and whatever fundamental constants.

Ignore any relativistic effects.

5. (6 pts.) This problem will use the answer from Part 1. Once again, assume that the electron is orbiting in the $x$ - $y$ plane.

Consider the instantaneous rest frame of the electron at some time $t$ , $S_2$ . Also consider the instantaneous rest frame of the electron at some time $t+\mathrm dt$ , $S_3$ . Relative to the laboratory frame, $S_1$ , $S_2$ will have velocity $\mathbf v$ . $S_3$ will have a velocity $\mathrm d\mathbf v$ relative to $S_2$ , but $\mathbf v$ and $\mathrm d\mathbf v$ won’t be parallel.

The result from Part 1 tells us that $S_3$ will experience an infinitesimal rotation in this time $\mathrm dt$ with respect to the laboratory frame. Since the electron is continuously accelerating, its rest frame, $S_3$ , must then rotate continuously relative to the lab frame. Assume that the spin of the electron always points in the same direction in its rest frame. If the spin is not pointing in the $z$ direction, find the angular frequency of the spin’s precession in terms of $r$ and whatever fundamental constants. Ignore the effects of the previous problem.

Note: although relativistic effects cannot be ignored, you can assume that the electron’s velocity $v$ is not comparable to the speed of light in the calculation.

6. (2 pts.) Combine your answers from parts 4 and 5 and find the relativistically correct angular frequency of the precession of the electron’s spin.

Published organizer clarifications (separate source, page 2)

Quoted from the organizer clarification document:

For problem 2, you must approximate any expressions to first order given the assumption in the problem and then proceed. You do not need to use a first order approximation on $\beta$ ~~but you may use it in any other approximations.~~ (removed crossed out statement for ambiguity).
Whatever mathematical methods you use to show in part 1 does not matter as long as you are able to get the verified results.

Решение

This problem deals with a phenomenon known as Thomas precession. It is named after Llewellyn Thomas. Thomas did not discover this phenomenon, but he was the first to discover that its application was important for the electron in the hydrogen atom. He did so in one-page letter, which was printed in Nature in 1926. Thomas’s contribution was crucial for the calculation of the spin-orbit interaction in the hydrogen atom, which successfully explained the fine structure of the hydrogen spectral lines.

This entire problem is essentially an explanation of his contribution and can all be found in his one-page letter to the editor.

Part 1 — Problem 1

$(x_2,y_2,z_2,t_2)$ is related to $(x_1,y_1,z_1,t_1)$ by a simple Lorentz transformation:

\begin{aligned}ct_2&=\gamma(ct_1-\beta x_1)&&\text{(1a)}\\x_2&=\gamma(x_1-\beta ct_1)&&\text{(1b)}\\y_2&=y_1&&\text{(1c)}\\z_2&=z_1,&&\text{(1d)}\end{aligned}

where $\gamma$ is the Lorentz factor correlated with $\beta$ . To go from $S_2$ to $S_3$ we must Lorentz transform in a direction that’s not the $x$ axis. Our strategy will be to rotate $S_2$ to $S_2'$ such that $S_3$ travels along the $x_2'$ axis. Then we can Lorentz transform to find $S_3'$ and rotate back to $S_3$ .

First,

\begin{aligned}ct_2'&=ct_2&&\text{(2a)}\\x_2'&=\cos\theta x_2+\sin\theta y_2=\frac{\beta_x'}{\beta'}x_2+\frac{\beta_y'}{\beta'}y_2&&\text{(2b)}\\y_2'&=-\sin\theta x_2+\cos\theta y_2=-\frac{\beta_y'}{\beta'}x_2+\frac{\beta_x'}{\beta'}y_2&&\text{(2c)}\\z_2'&=z_2.&&\text{(2d)}\end{aligned}

If the Lorentz factor for $\beta'$ is $\gamma'$ , we can write

\begin{aligned}ct_3'&=\gamma'(ct_2'-\beta'x_2')&&\text{(3a)}\\x_3'&=\gamma'(x_2'-\beta'ct_2')&&\text{(3b)}\\y_3'&=y_2'&&\text{(3c)}\\z_3'&=z_2'.&&\text{(3d)}\end{aligned}

Then we can rotate back and find

\begin{aligned}ct_3&=ct_3'&&\text{(4a)}\\x_3&=\frac{\beta_x'}{\beta'}x_3'-\frac{\beta_y'}{\beta'}y_3'&&\text{(4b)}\\y_3&=\frac{\beta_y'}{\beta'}x_3'+\frac{\beta_x'}{\beta'}y_3'&&\text{(4c)}\\z_3&=z_3'.&&\text{(4d)}\end{aligned}

Combining all of (1), (2), (3), and (4), we can write

\begin{aligned}ct_3&=\gamma\gamma'ct_1-\gamma\gamma'\beta x_1-\gamma\gamma'\beta_x'x_1+\gamma\gamma'\beta_x'\beta ct_1-\gamma'\beta_y'y_1\\&\simeq\gamma(1+\beta_x'\beta)ct_1-\gamma(\beta+\beta_x')x_1-\beta_y'y_1\qquad\text{(5a)}\end{aligned}

\begin{aligned}x_3&=\gamma\gamma'\left(\beta\beta_x'+\frac{\beta_x'^2}{\beta'^2}\right)x_1-\gamma\gamma'\left(\beta_x'+\frac{\beta\beta_x'^2}{\beta'^2}\right)ct_1+\frac{\gamma'\beta_x'\beta_y'}{\beta'^2}y_1\\&\quad+\frac{\gamma\beta_y'^2}{\beta'^2}x_1-\frac{\gamma\beta\beta_y'^2}{\beta'^2}ct_1-\frac{\beta_x'\beta_y'}{\beta'^2}y_1\\&\simeq-\gamma(\beta+\beta_x')ct_1+\gamma(1+\beta\beta_x')x_1\qquad\text{(5b)}\end{aligned}

\begin{aligned}y_3&=\gamma\gamma'\left(\beta\beta_y'+\frac{\beta_x'\beta_y'}{\beta'^2}\right)x_1-\gamma\gamma'\left(\beta_y'+\frac{\beta_x'\beta_y'\beta}{\beta'^2}\right)ct_1+\frac{\gamma'\beta_y'^2}{\beta'^2}y_1\\&\quad-\frac{\gamma\beta_x'\beta_y'}{\beta'^2}x_1+\frac{\gamma\beta\beta_x'\beta_y'}{\beta'^2}ct_1+\frac{\beta_x'^2}{\beta'^2}y_1\\&\simeq-\gamma\beta_y'ct_1+\gamma\beta\beta_y'x_1+y_1\qquad\text{(5c)}\\z_3&=z_1.\qquad\text{(5d)}\end{aligned}

Here, everything has been done to first order, which includes writing $\gamma'\simeq1$ .

Problem 2

To transform from $S_1$ directly to $S_3$ , we need the proper velocity addition. Let the velocity of $S_3$ in $S_1$ be defined by the components $\beta_x''$ and $\beta_y''$ . Then

\beta_x''=\frac{\beta+\beta_x'}{1+\beta\beta_x'}\quad\text{and}\quad\beta_y''=\frac{\beta_y'}{\gamma(1+\beta\beta_x')}.\qquad\text{(6)}

We further define $\beta''=(\beta_x''^2+\beta_y''^2)^{1/2}$ and $\gamma''$ as the Lorentz factor corresponding to $\beta''$ .

Note that the following approximations are true to first order:

\begin{aligned}\beta_x''&\simeq\beta+\frac{\beta_x'}{\gamma^2}&&\text{(7a)}\\\gamma''&\simeq\gamma(1+\beta\beta_x')&&\text{(7b)}\\\beta''&\simeq\beta_x''.&&\text{(7c)}\end{aligned}

Once again, the Lorentz transformation needs to be made in a direction other than the $x$ direction. We employ a similar scheme. The details of the calculation are not shown here, since they are similar to what has been done previously. The final result is

\begin{aligned}ct_3'&=\gamma''ct_1-\gamma''\beta_x''x_1-\gamma''\beta_y''y_1\\&\simeq\gamma(1+\beta\beta_x')ct_1-\gamma(\beta+\beta_x')x_1-\gamma\beta_y''y_1\qquad\text{(8a)}\end{aligned}

\begin{aligned}x_3'&=\frac{\gamma''\beta_x''^2}{\beta''^2}x_1+\frac{\gamma''\beta_x''\beta_y''}{\beta''^2}y_1-\gamma''\beta''ct_1+\frac{\beta_y''^2}{\beta''^2}x_1-\frac{\beta_x''\beta_y''}{\beta''^2}y_1\qquad\text{(8b)}\\&\simeq-\gamma(\beta+\beta_x')ct_1+\gamma(1+\beta\beta_x')x_1+(\gamma-1)\frac{\beta_y''}\beta y_1\qquad\text{(8c)}\end{aligned}

\begin{aligned}y_3'&=\frac{\gamma''\beta_x''\beta_y''}{\beta''^2}x_1+\frac{\gamma''\beta_y''^2}{\beta''^2}y_1-\gamma''\beta_y''ct_1-\frac{\beta_x''\beta_y''}{\beta''^2}x_1+\frac{\beta_x''^2}{\beta''^2}y_1\\&\simeq-\gamma\beta_y''ct_1+(\gamma-1)\frac{\beta_y''}\beta x_1+y_1\qquad\text{(8d)}\\z_3'&=z_1.\qquad\text{(8e)}\end{aligned}

Note that all terms $\beta_y''^2/\beta''^2$ are second order.

Problem 3

Let us write $S_3$ in terms of $S_3'$ . Using (6), we can write (5) in terms of $\beta_y''$ instead of $\beta_y'$ . As we will see in Problem 5, $\beta_y''$ , the laboratory-observed difference in velocity between $S_2$ and $S_3$ , is more physically relevant. This is

\begin{aligned}ct_3&\simeq\gamma(1+\beta_x'\beta)ct_1-\gamma(\beta+\beta_x')x_1-\gamma\beta_y''y_1&&\text{(9a)}\\x_3&\simeq-\gamma(\beta+\beta_x')ct_1+\gamma(1+\beta\beta_x')x_1&&\text{(9b)}\\y_3&\simeq-\gamma^2\beta_y''ct_1+\gamma^2\beta\beta_y''x_1+y_1&&\text{(9c)}\\z_3&=z_1&&\text{(9d)}\end{aligned}

We can see that this is different from (8). Let us invert (8) such that we have $(x_1,y_1,z_1,t_1)$ in terms of $(x_3',y_3',z_3',t_3')$ . This is just the same Lorentz transformation in Problem 2, except with the velocities reversed. So

\begin{aligned}ct_1&\simeq\gamma(1+\beta\beta_x')ct_3'+\gamma(\beta+\beta_x')x_3'+\gamma\beta_y''y_3'&&\text{(10a)}\\x_1&\simeq\gamma(\beta+\beta_x')ct_3'+\gamma(1+\beta\beta_x')x_3'+(\gamma-1)\frac{\beta_y''}\beta y_3'&&\text{(10b)}\\y_1&\simeq\gamma\beta_y''ct_3'+(\gamma-1)\frac{\beta_y''}\beta x_3'+y_3'&&\text{(10c)}\\z_1&=z_3'.&&\text{(10d)}\end{aligned}

Plugging (10) into (9), we find

\begin{aligned}ct_3&\simeq ct_3'&&\text{(11a)}\\x_3&\simeq x_3'-(\gamma-1)\frac{\beta_y''}\beta y_3'&&\text{(11b)}\\y_3&\simeq(\gamma-1)\frac{\beta_y''}\beta x_3'+y_3'&&\text{(11c)}\\z_3&=z_3'.&&\text{(11d)}\end{aligned}

This is the equation for an infinitesimal rotation (keeping in mind that $\beta_y''/\beta\ll1$ ). Thus, to go from $S_3'$ to $S_3$ , we must rotate about the $+z$ axis an angle

-(\gamma-1)\frac{\beta_y''}\beta.

To reiterate, $\beta_y''$ is the relative velocity in the $y$ direction between frames $S_3$ and $S_2$ , as viewed from the lab frame $S_1$ .

We expect $(x_3,y_3,z_3,t_3)$ to be the correct form. The Lorentz transformation is the most general tool. The velocity addition used in problem 2 is dependent on the Lorentz transformation.

Part 2 — Problem 4

We are given

\vec\mu=-\frac e{m_e}\vec L.\qquad\text{(12)}

Suppose the electron is orbiting counter-clockwise in the $x$ - $y$ plane. In the instantaneous rest frame of the electron, the proton travels with speed $v$ a distance $r$ away from the electron. By the law of Biot and Savart, the magnetic field at the location of the electron is

B=\frac{\mu_0ev}{4\pi r^2},\qquad\text{(13)}

which points in the positive $z$ direction with our choice of coordinates.

Since we are using the Bohr model, the acceleration of the electron is given by the equation

\frac{v^2}r=\frac{e^2}{4\pi\epsilon_0r^2m_e}\qquad\text{(14)}

which gives

v=\left(\frac{e^2}{4\pi\epsilon_0rm_e}\right)^{1/2}.\qquad\text{(15)}

Now the electron will experience a torque as a result of this magnetic field because it possesses a spin magnetic moment. This torque is

\vec\tau=\vec\mu\times\vec B=-\frac e{m_e}\vec L\times\vec B.\qquad\text{(16)}

Therefore,

\left\lvert\frac{\mathrm d\vec L}{\mathrm dt}\right\rvert=\frac{eLB\sin\theta}{m_e}.\qquad\text{(17)}

$\theta$ is the angle between $\vec L$ and $\vec B$ . Thinking in analogy to $v=r\omega$ , we find that the angular velocity of the precession of $\vec L$ is

\omega_L=\frac{eB}m,\qquad\text{(18)}

which will be in the positive $z$ direction. (We have used the subscript $L$ because this precession is often called the Larmor precession.) Applying (13) and (15), we find that

\omega_L=\frac{\mu_0e^3}{4\pi m_er^2}\left(\frac1{4\pi\epsilon_0m_er}\right)^{1/2}.\qquad\text{(19)}

Problem 5

Once again, assume the electron is orbiting counter-clockwise in the $x$ - $y$ plane.

In analogy to Problem 1, $S_1$ is the lab frame. $S_2$ is the rest frame of the electron when it is travelling with velocity $\vec v$ at time $t$ . $S_3$ is the rest frame of the electron when it is travelling with velocity $\vec v+\vec a\mathrm dt$ at time $t+\mathrm dt$ . To keep the analogy, let $\vec v$ be in the $x$ direction of $S_1$ . Then $\vec a\mathrm dt$ is in the $y$ direction of $S_1$ and $S_2$ . (Since the electron is orbiting counter-clockwise, it must also be the positive $y$ direction.)

First, $\beta=v/c$ as defined in Problem 1. Second, note that $\vec a\mathrm dt$ is the relative velocity between $S_3$ and $S_2$ as measured in the lab frame $S_1$ . Therefore, $\beta_y''=a\mathrm dt/c$ . Combining these facts, the infinitesimal rotation angle between $S_2$ and $S_3$ will be

\mathrm d\theta=-(\gamma-1)\frac{a\mathrm dt}v,\qquad\text{(20)}

in the positive $z$ direction. Since we can also assume that $v/c\ll1$ , $\gamma-1$ simplifies to $v^2/2c^2$ . We can continue to repeat this process as the electron goes around in a circle. Therefore, this continuous infinitesimal rotational angle will manifest as an angular velocity: the angular frequency with which the frame $S_3$ rotates relative to the lab frame. This angular frequency is

\omega_T=-\frac{av}{2c^2}=-\frac{v^3}{2c^2r}\qquad\text{(21)}

in the positive $z$ direction. (We have used the subscript $T$ because this frequency is called the Thomas precession frequency.)

Using (15), (21) can be rewritten as

\omega_T=-\frac{e^3}{8\pi\epsilon_0c^2m_er^2}\left(\frac1{4\pi\epsilon_0m_er}\right)^{1/2}.\qquad\text{(22)}

Problem 6

The final precession frequency of the electron will be $\omega_L+\omega_t$ . Note that $c=(1/\mu_0\epsilon_0)^{1/2}$ . Thus, (22) can be rewritten as

\omega_T=-\frac{\mu_0e^3}{8\pi m_er^2}\left(\frac1{4\pi\epsilon_0m_er}\right)^{1/2},\qquad\text{(23)}

which is exactly half the result of Problem 4. Therefore, the precession frequency of the electron is

\omega=\frac{\mu_0e^3}{8\pi m_er^2}\left(\frac1{4\pi\epsilon_0m_er}\right)^{1/2}.\qquad\text{(24)}

This relativistic correction of $1/2$ because known as the Thomas half and finally matched the experimental determination of hydrogen’s fine structure with the quantum mechanical prediction. Just a few year’s later, Dirac would propose his Dirac Equation, which “came installed” with relativistic mechanics, thus eliminating the need for these kinds of calculations.

Grading Scheme

P1	For correct Lorentz transformation from $S_1$ to $S_2$	1 pt
P1	For correct Lorentz transformation from $S_2$ to $S_3$	2 pt
P1	For correct Lorentz transformation from $S_1$ to $S_3$	3 pt
P1	For not expressing the answer to first order with the given approximation. Half credit for algebraic mistakes	-2 pt
P2	For the correct velocity.	2 pt
P2	For the correct Lorentz transformation.	4 pt
P2	For not expressing to first order. Half credit for algebra mistakes.	-2 pt
P3	For correct magnitude.	2 pt
P3	For correct direction.	1 pt
P4	For the correct magnetic field at the electron’s location within its rest frame.	0.5 pt
P4	For the correct torque applied on the electron due to its spin magnetic moment.	0.5 pt
P4	For the correct precession frequency. Half credit for algebra mistakes.	2 pt
P5	For the correct analogy (assigning values for $\beta,\beta_x',\beta_y'$ , etc) to parts 1, 2, and 3.	2 pt
P5	For the correct answer. Half credit for algebra mistakes.	4 pt
P5	For not assuming $v\ll c$ .	-1 pt
P6	For the correct answer.	2 pt
P6	For not making the substitution $c=(\mu_0\epsilon_0)^{-1/2}$ .	-1 pt

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

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Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf