How did L.H. Thomas derive his 1927 expressions for an electron with an axis?

Question

I'm looking at the 1927 paper of Thomas, The Kinematics of an Electron with an Axis, where he shows that the instantaneous co-moving frame of an accelerating electron rotates and moves with some infinitesimal velocity. He states:

At $t=t_0$ let the electron have position $\mathbf{r}_0$ and velocity $\mathbf{v}_0$, with $\beta_0=(1-{\mathbf{v}_0}^2/c^2)^{-\frac 1 2}$, in $(\mathbf{r}, t)$. Then, by (2.1), that definite system of coordinates $(\mathbf{R}_0, T_0)$ in which the electron is instantaneously at rest at the origin and which is obtained from $(\mathbf{r},t)$ by a translation and a Lorentz transformation without rotation is gven by $$\begin{align*}\mathbf{R}_0 &= \mathbf{r} - \mathbf{r}_0 + (\beta_0 - 1)\frac{ (\mathbf{r}-\mathbf{r}_0)\cdot \mathbf{v}_0}{{\mathbf{v}_0}^2}\mathbf{v}_0-\beta_0 \mathbf{v}_0(t-t_0)\tag{3.1a}\\ T_0 &= \beta_0\left( t - t_0 - \frac {(\mathbf{r} - \mathbf{r}_0)\cdot \mathbf{v}_0}{c^2}\right)\tag{3.1b}\end{align*}$$

By eliminating $(\mathbf{r},t)$ from equations (3.1) and the similar equations for $(\mathbf{R}_1, T_1)$,$$\begin{align*}\mathbf{R}_1 &= \mathbf{R}_0 + \frac{(\beta_0 - 1)} {{\mathbf{v}_0}^2}(\mathbf{R}_0\times (\mathbf{v}_0\times \mathbf{dv}_0)) - \beta_0 T_0(\mathbf{dv}_0 + (\beta_0 - 1)\frac{(\mathbf{v}_0\cdot \mathbf{dv}_0)}{{\mathbf{v}_0}^2}\mathbf{v}_0)\tag{a}\\ T_1 &= T_0 - \frac {\beta_0} c^2((\mathbf{R}_0\cdot(\mathbf{dv}_0 + (\beta_0 - 1)\frac{(\mathbf{v}_0\cdot \mathbf{dv}_0)}{{\mathbf{v}_0}^2}\mathbf{v}_0))) - d\tau_0\tag{3.3b} \end{align*}$$

What are the steps to get from (3.1) to (3.3)?

Answer 1

I will reference the original paper ( http://www.clifford.org/drbill/csueb/4250/topics/thomas_papers/Thomas1927.pdf ). One key to the Thomas Procession is understanding infinitesimal Lorentz transformations, since one needs to understand the effects of transforming from the lab frame into the "rotating" rest frame of the electron and vice versa.

First, you need to understand how to get from 2.1 to 2.2. Equation 2.1 is the Lorentz transformation for the motion of a frame with an arbitrary direction. Look here for a good discussion of arbitrary Lorentz transformations and Thomas precession in a modern context using only vector notation at first starting in section 15.2 and the discussion continues on your topic through section 17.

To derive 2.2, we need to understand how successive Lorentz transformations behave. I think the approach in Thomas is pretty good. What you need to do is compute the transformation with velocity -v, followed by a transformation with v+dv. It is more clearly seen if you do this for motion along one axis at first, the algebra is easier to follow. Then you will see that equation 2.2 collapses to the result for 1D motion and this is also done explicitly in the above reference.

Equation 3.1 is the full NON-rotational Lorentz transformation, but that therein lies the problem. We need to transform from a rotating frame back into the lab frame. Lets look at the time transformation in equation 3.23. Tau1 is T1 and Tau0 is T0. Substitute dt0/beta0 for ds0. Thomas states that n0 is the linear acceleration which is equal to dv0/dt0. Substitute and cancel dt0 (scalar). The transformation in 3.23 assumes the electron is stationary but rotating, so rho0 is R0-- essentially the angular position indicated by the eularian angles can be decoupled from the relative position.

After substitution note that the time equation in 3.23 has the same form as the time equation in 3.1 except we now have a dv0 term instead of v0. Thomas states "infinitesimal...must be of the form" 3.23, which makes sense and come from classical mechanics. But what is dv0? It is the infinitesimal velocity given in 2.2. Substitute this infinitesimal velocity into 3.23 and the time transformation in 3.3 follows.

The R0 term in 3.1 transforms in the way, except the algebra is a bit messier. The form for the infinitesimal transformation for R0 given by 3.23 is a consequence of classical mechanics (cf Coriolis effect). A position vector in a non–rotating frame can be expressed in a rotating (for instance the electrons rest frame) frame by adding the vectors time rate of change in direction due to the rotation of the frame. This is also discussed in the reference.

Answer 2

Thomas is using suboptimal language in his paper. I will answer the specific question you asked, but it is best to do it in a more modern way, which has more feeling for the geometry of Lorentz space. I will digress to do this first.

Geometrical precession

To understand the effect, it is best to start with the analogous effect in geometry. Suppose I have a curve parametrized by arclength x(s), where x is a vector in 3-space. Consider sliding a frame of three vectors along the curve, so that the z-axis is always parallel to the tangent of the curve, and at any time, you go forward along the curve by tilting the frame so that the tilting does no rotation of the x-y plane at any instant.

Suppose the curve starts and ends parallel to the z-axis. What is the total net x-y rotation of the frame beginning to end?

The answer is not zero. As you go around the curve, the tangent vector is making a loop on the two sphere. It starts out at the north pole, wanders around, then comes back to the north pole. As this is happening, the x-y plane is parallel transported to stay tangent to the sphere, and the total turn angle you get is equal to the curvature of the sphere times the area of the loop.

That this is true is easy to see (it's one of Gauss's theorems)--- the turning angle is additive over loops end to end, and for an infinitesimal square, it is the definition of the intrinsic curvature.

This gives the rotation angle for a loop. The corresponds to a curve which ends with the same tangential direction as where it started. You can't define a frame everywhere on the tangent plane of the sphere at once and ask what angle you make with respect to this frame, because you can't comb the hair on a sphere.

Things are only this simple in 3d, because the 2d rotation group is abelian.

Thomas Precession in 2+1 dimensions

This is just as simple Now you have a curve in space-time x(s) parametrized by relativistic arc-length. The tangent vector to the curve is making a path on the unit hyperbola in Minkowksi space $t^2 -x^2 - y^2 = 1$. This curve is constant curvature inside Minkowkski space, because any point can be moved to the origin by a boost.

The curvature of this space is one, as can be seen from the commutator of two infinitesimal Lorentz transformations.

So if you start the electron at rest, and you return at rest, the angle of rotation is equal to the area cut out by the path of the tangent vector on the unit hyperbola. This solves the problem, except for the area of the hyperbola. This area can be worked out by noting that for any two vectors v and w, the area they bound is:

$$|A| = \int_{(x,y)\in A} {1\over 1+x^2 + y^2} dx dy$$

If the electron doesn't wind up at rest, you can still work out the angle relative to a frame at every point of the hyperbola, because you can comb the hair on a hyperbola. The way you do this is to boost from the origin to velocity v, and define the result of boosting the x,y,z,t vectors to be unrotated. This is what Thomas does at each point of the velocity hyperbola. This then allows him to define the precession amount when you go from any vector to a nearby one.

Precession in 3+1 dimensions

For this, all you need to note is that at any time, the velocity, the acceleration and time make a 2+1 dimensional space. At any one time, the amount of precession is just given by the 2 dimensional precession from above.

Thomas's paper

Thomas first writes down the Lorentz transformations to boost the time axis to have slope v. I set c to 1, capitalized (r,t) to (R,T) because these are frame variables and this really should be consistent, and got rid of the bolding on the vectors:

$$ R_0 = R - r_0 + (\beta_0 - 1){ (R-r_0)\cdot v_0 \over v_0^2} v_0- \beta_0 v_0(T-t_0)$$ $$ T_0 = \beta_0 ( T - t_0 - (R - r_0)\cdot v_0)$$

That is copied from your question, to get the notation straight. All the capital R's and T's, are frame variables--- as they vary, they describe the whole space, and their only purpose is place-holders for describing the Lorentz transformation involved. He then writes down the Lorentz transformation for another time by copy-paste, changing 0 to 1:

$$ R_1 = R - r_1 + (\beta_1 - 1){ (R-r_1)\cdot v_1 \over v_1^2} v_0- \beta_1 v_1(T-t_1)$$ $$ T_1 = \beta_1 ( T - t_1 - (R - r_1)\cdot v_0)$$

Then he notes that $v_1$ is only infinitesimally different from $v_0$, $v_1=v_0 + dv_0$ (you should have said that in the question), and expands the above to first order in dv.

Next he solves for $(R,T)$ in terms of $(R_1,T_1)$, getting that they are related by boost of -v. He then substitutes the R,T solution into the equation for $R_1,T_1$ to determine what Lorentz transformation has occurred between frame $R_1,T_1$ and $R_0,T_0$. He gets:

$$ R_1 = R_0 + {(\beta_0 - 1)\over v_0^2}(R_0\times (v_0\times dv_0)) - \beta_0$$ $$ T_0(dv_0 + (\beta_0 - 1){(v_0\cdot dv_0)\over v_0^2}v_0)$$

$$ T_1 = T_0 - \beta_0(R_0\cdot dv_0 + (\beta_0 - 1){v_0\cdot dv_0)\over v_0^2}v_0))) - d\tau_0 $$

This is just an inane way to compose Lorentz transformations. The factor of $dv_0 + (\beta_0 - 1){(v_0\cdot dv_0)\over v_0^2}$ is the infinitesimal velocity of the particle at 2 when viewed in frame 1. The result is a translation, an infinitesimal boost, and an infinitesimal rotation. That the rotation is significant is because it is relative to the combing of the hyperbola defined before.

Checking Thomas's work

Thomas almost certainly didn't do the steps above in his personal notebook--- that's just what he wrote in the published paper. Nobody should check using the steps he gives in his paper, it would be silly.

The way you check his work is to ignore the translations, just look at Lorentz boosts between the frames. Then you choose your x axis in the direction of $v_0$, which I will call v below, and you choose the y-axis so the acceleration dv is in the x-y plane.

To boost from zero velocity to velocity "v" in the x-direction in three dimensions (which is the general case) you do:

$$ x_0 = \beta(x - v t) = x+ (\beta-1)x - \beta vt $$ $$ y_0 = y $$ $$ t_0 = \beta(t - vx) $$

The reason for separating out $\beta-1$ is to write the Lorentz transformation in vector form. From the above, you conclude that the general Lorentz boost is:

$$ r = r + (\beta -1) {v\cdot r\over v^2} v - \beta vt$$ $$ t = \beta (t - v\cdot r ) $$

Where the dot product times v over v^2 is the vector way of selecting a component in the direction of v. This is correct, because it is in vector language, so it is rotationally invariant, and reduces to the equations above it when you choose the x-axis along v. This is what Thomas means by a "Lorentz boost without rotation", a Lorentz boost of this type. These boosts are not a group, if you compose them, you get rotations. That's all that Thomas is doing.

Going back to the case of v in the x direction, write the reverse transformation, which tells you x,y,t in terms of $x_0,y_0,z_0$:

$$ x = \beta( x_0 + vt_0)$$ $$ y = y_0$$ $$ t = \beta(t_0 + vx_0)$$

This is what Thomas means by "solve for $(r,t)$ in terms of $(R_0,T_0)$".

Next you need a boost which takes zero velocity to a velocity v in the x direction and dv in the y direction (I am assuming that the acceleration dv is all in the y direction, this turns out the be the general case, the x component doesn't do any rotation). This doesn't change $\beta$ at all to leading order in dv. Using the vectorial form for Lorentz transformations

$$ x_1 = x + (\beta-1)(x + y{dv \over v}) - \beta vt = \beta(x-vt) + (\beta-1) {dv\over v}y $$ $$ y_1 = y - (\beta-1) {dv\over v} x - \beta dv t $$ $$ t_1 = \beta (t - v x) - \beta dv y$$

Now you plug in the formulas for $x,y,t$ in terms of $x_0,y_0,t_0$ to get

$$ x_1 = x_0 + (\beta-1){dv \over v} y_0 $$ $$ y_1 = y_0 - (\beta-1){dv \over v} x_0 - \beta^2 dv t_0$$ $$ t_1 = t_0 - \beta^2 dv y_0$$

This is a superposition of an infinitesimal rotation in the x-y plane of magnitude $(\beta_0 -1) {dv\over v}$ and an infinitesimal boost in the y direction of some magnitude you don't care about. The answer is linear in dv, because it is to linear order. If $dv$ were in the x direction, the answer would have been just a boost, because the 1-d Lorentz boosts form a group just by themselves, and who cares.

From the above, you conclude that the amount of rotation is given by the component of dv perpendicular to v times the magnitude of v divided by v^2, so that the $\omega$ vector is

$$\omega = (\beta-1){v\times dv \over v^2}$$

This verifies that the rotation part of Thomas's paper is accurate.

When reading old papers which use vector notation, it is important to read between the lines like this, because this is almost certainly what Thomas did privately before publishing. When reproducing the work, you can't just fill in the intermediate steps.