Stone's derivation of Thomas rotation

[Rasalhague]

Sure but we were discussing length contraction. Testimony is that your only contribution has been guesses about length contraction.

Not so. In #66 Kev showed me how to get through the fiddly algebra of a step you left out of your original attachment on TD and LC. While I'd got a formula that agreed numerically with yours, my attempts at filling in the missing step in the algebra had always gone astray. It was also very helpful to see Kev's formula for the angle. When I mistakenly applied it, he explained to me where I was going wrong: which angles in his formula corresponded to which in mine. It was good to see that, when correctly applied, the two gave the same results.

 

[Rasalhague]

No, the question is finding the length contraction formula for the general case (see the hint file).

That'd just be the norm of r', wouldn't it? When all the cancelling's done: r' = (r²-(r.b)²)^1/2. I used it to get my formula for the angle.

 

[starthaus]

That'd just be the norm of r', wouldn't it? When all the cancelling's done: r' = (r²-(r.b)²)^1/2. I used it to get my formula for the angle.

Correct, so it does not depend on the angle r,r' it depends on the angle r,V. You would arrive to the same answer if you finished the calculations in the hint.
What troubles me in your derivation is that the correct starting point is:

\textbf{r'}=\textbf{r}+\textbf{V}(\frac{\gamma-1}{||V||^2} \textbf{r.V}-\gamma t)

How did you arrive to the final formula? Your
\textbf{r}'=\textbf{r}-\pmb{\beta} \frac{\textbf{r}\cdot \pmb{\beta}}{\beta^2}\left ( 1-\frac{1}{\gamma} \right )

comes out of nowhere, how did you derive it from:

\textbf{r'}=\textbf{r}+\textbf{V}(\frac{\gamma-1}{||V||^2} \textbf{r.V}-\gamma t)?

 

[starthaus]

Your above statement suggests that the rotation is somehow not physical

No, what gives you this misconception? Here is the exact statement.

 

[starthaus]

For the effect of a general boost on a general space vector, I get:

\textbf{r}'=\textbf{r}-\frac{\textbf{r}\cdot \pmb{\beta}}{\beta^2}\pmb{\beta}\left ( 1-\frac{1}{\gamma(\beta)} \right )

How do you get this? I am not saying that it is incorrect but it appears out of nowhere.
BTW, the transformation for vectors is time dependent:

\textbf{r'}=\textbf{r}+\textbf{V}(\frac{\gamma-1}{||V||^2} \textbf{r.V}-\gamma t)

(see C.Moller "The Theory of Relativity",p.47, for example).

 

[Rasalhague]

How did you arrive to the final formula?

\begin{pmatrix}<br /> \gamma &amp; -\gamma \pmb{\beta}\\ <br /> -\gamma \pmb{\beta} &amp; I+\frac{\gamma-1}{\beta^2} \pmb{\beta\beta}<br /> \end{pmatrix}\begin{pmatrix}<br /> t\\\textbf{r}\end{pmatrix}

=\begin{pmatrix}<br /> \gamma t - \gamma \pmb{\beta}\cdot \textbf{r}\\ -\gamma t \pmb{\beta}+\textbf{r}+\frac{\gamma-1}{\beta^2} (\textbf{r}\cdot \pmb{\beta})\pmb{\beta}<br /> \end{pmatrix}

And t&#039; = 0:

\gamma t - \gamma \pmb{\beta}\cdot \textbf{r}=0

so

t = \pmb{\beta}\cdot \textbf{r}.

Substituting for \pmb{\beta}\cdot \textbf{r} for t in the spatial part:

\textbf{r}&#039;=-\gamma (\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}+\textbf{r}+\frac{\gamma-1}{\beta^2} (\textbf{r}\cdot \pmb{\beta})\pmb{\beta}

which simplifies like this:

\textbf{r}&#039;=\textbf{r}+(\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}\left ( \frac{\gamma-1}{\beta^2} -\frac{\gamma\beta^2}{\beta^2}\right )

\textbf{r}&#039;=\textbf{r}+\frac{(\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}}{\beta^2}\left ( \gamma-1 -\gamma\beta^2 \right )

\textbf{r}&#039;=\textbf{r}-\frac{(\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}}{\beta^2}\left ( -\gamma+1 +\gamma\beta^2 \right )

\textbf{r}&#039;=\textbf{r}-\frac{(\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}}{\beta^2}\left ( 1-\gamma (1-\beta^2) \right )

\textbf{r}&#039;=\textbf{r}-\frac{(\textbf{r}\cdot \pmb{\beta}) \pmb{\beta}}{\beta^2}\left ( 1-\frac{1}{\gamma} \right )

since

\gamma(1-\beta^2)=\frac{\gamma}{\gamma^2}.

 

[yuiop]

No, what gives you this misconception? Here is the exact statement.

Your exact statement is the one I quoted, i.e.:

I have alread y answered that, it isn't a "pure" rotation. It is simply called a rotation , by abuse of language.

which is a variation on your earlier statement:

It isn't a pure rotation, it is called "rotation". Generally, it 2D proper rotations contain terms mixed in both x and y, this is where the name "Thomas rotation" originated from. This is the way the effect is associated with its name.

O.K. so we are in agreement that the Thomas rotation is a real physical rotation of a rod when it boosted in a direction that is not exactly parallel or orthogonal to its length?

Basically you are saying that a pure rotation combined with a pure translation (which is what happens in the case of Thomas rotation) should not be called a rotation and so it should be called Thomas-pure-rotation-combined-with-pure-translation which is not very catchy and besides, the length contraction part was already well known and should not be attributed to Thomas.

 

[starthaus]

\begin{pmatrix}<br /> \gamma &amp; -\gamma \pmb{\beta}\\ <br /> -\gamma \pmb{\beta} &amp; I+\frac{\gamma-1}{\beta^2} \pmb{\beta\beta}<br /> \end{pmatrix}\begin{pmatrix}<br /> t\\\textbf{r}\end{pmatrix}

=\begin{pmatrix}<br /> \gamma t - \gamma \pmb{\beta}\cdot \textbf{r}\\ -\gamma t \pmb{\beta}+\textbf{r}+\frac{\gamma-1}{\beta^2} (\textbf{r}\cdot \pmb{\beta})\pmb{\beta}<br /> \end{pmatrix}

And t&#039; = 0:

The above gives you the correct formula with an incorrect derivation. There is no reason to set t&#039;=0. What you need to remember is that we are marking the endpoints of the rod simultaneously in frame S'. So, you need to transform the above in its differential form and to set \Delta t&#039;=0. This makes the above derviation somewhat more complicated since you need to write in differential form but makes it rigorous:

\begin{pmatrix}<br /> \gamma &amp; -\gamma \pmb{\beta}\\ <br /> -\gamma \pmb{\beta} &amp; I+\frac{\gamma-1}{\beta^2} \pmb{\beta\beta}<br /> \end{pmatrix}\begin{pmatrix}<br /> dt\\\textbf{dr}\end{pmatrix}
and dt&#039; = 0

It has the advantage that , if you get really ambitious, you can derive length contraction in accelerating frames by differentiating \textbf{v} (i.e. \beta) as well in the above.

 

[Rasalhague]

The above gives you the correct formula with an incorrect derivation. There is no reason to set t&#039;=0. What you need to remember is that we are marking the endpoints of the rod simultaneously in frame S'. So, you need to transform the above in its differential form and to set \Delta t&#039;=0. This makes the above derviation somewhat more complicated since you need to write in differential form but makes it rigorous:

That was my intended meaning, but I should have made it clear, and maybe used a different letter to r. By r and r' I meant displacement 3-vectors. I referred to r as such in #82, but it's been a long thread... The 4-vectors of which they're the spatial parts I took to be displacement vectors in flat spacetime.

Lucky the extra complication is no worse than inserting the letter d or delta, or is there a catch?

 

[starthaus]

Lucky the extra complication is no worse than inserting the letter d or delta, or is there a catch?

Correct, there is no further catch.
Note the observation from post 103.
It was nice interacting with you, we're done with this subject.

 

[Rasalhague]

Correct, there is no further catch.
Note the observation from post 103.
It was nice interacting with you, we're done with this subject.

Thanks.

 

[yuiop]

Seeing as how Starthaus is done with this thread, this is a question for Rasalhague.

Do your equations agree numerically with the following simple trigonometric analysis?

Consider two frames S and S' that have relative velocity v/c=\beta and axes aligned with each other when there origins coincide and S' is moving along the x-axis of S. (We are always free to align the coordinates frames in such a way that this is true.) The rod is at rest in S with proper length L and one end at the origin and the rod is orientated in at an angle (theta) in the x,y plane. (Again we are always free to align the frames this way.)

In frame S' the angle (theta') of the rod relative to the x' axis is by simple trigonometry and allowing for length contraction is:

\theta&#039; = \tan^{-1}\left( \frac{L_y&#039;}{L_x&#039;} \right) = \tan^{-1}\left(\frac{L_y}{L_x \sqrt{1-\beta^2}}\right) = \tan^{-1}\left(\frac{\tan{\theta}}{ \sqrt{1-\beta^2}} \right)

The difference between the two angles (phi) is the "rotation" of the rod (The Thomas rotation) and is given by:

\phi = (\theta - \theta&#039;) = \theta - \tan^{-1}\left(\frac{\tan(\theta)}{ \sqrt{1-\beta^2}}\right) \qquad \qquad (Eq1)

This rotation angle is independent of the orientation of the axes with respect to the motion or the rod, and only depends on the angle of the rod with respect to the motion.

In the earlier example I gave the equation for the rotation as

\phi = (\theta - \theta&#039;) = \theta + \tan^{-1}(\cot(\theta)\sqrt{1-\beta^2}) - \pi/2

when the rod was aligned with x-axis and the motion was at angle (theta) wrt the x axis. It is easy to see that the two equations are equivalent because of the truth of the simple trigonometric equality (at least when 0<x<=1):

-\tan^{-1}\left(\frac{\tan(\theta)}{x}\right) = \tan^{-1}(\cot(\theta)x) - \pi/2

Now for the length contraction aspect of the problem.

In frame S' the length of the moving rod is given by:

\| L &#039; \| = \sqrt{L_Y&#039;^2 +L_x&#039;^2} = \sqrt{L_y^2 + L_x^2 (1-\beta^2)}

In the rest frame of the rod L_y = \tan(\theta)L_x so the above equation can restated as:

\| L &#039; \| = \sqrt{\tan^2(\theta) L_x^2 + L_x^2 (1-\beta^2)} <br /> = L_x \sqrt{\tan^2(\theta) + (1-\beta^2)}

It is also true that in the rest frame the x component of the proper rod length \| L \| is given by L_x = \cos(\theta) \| L \| so the equation can be further restated as:

\| L &#039; \| = \| L \| \sqrt{\sin^2(\theta) + \cos^2{\theta}(1-\beta^2)} \qquad \qquad (Eq2)

Again, this equation is independent of the orientation of the frames wrt the motion or the rod. It is easy to check that when the rod is parallel to the motion and theta=0, the equation reduces to the familiar \| L &#039; \| = \| L \| \sqrt{(1-\beta^2)} and when the rod is exactly orthogonal to the motion and theta=pi/2 the equation reduces to the expected \| L &#039; \| = \| L \|.

In summary, the rotation (Eq1) and the length contraction (Eq2) is only a function of the angle of the rod wrt to the motion and is independent of the orientation of the axes wrt the motion and independent of the orientation of the rod wrt the axes. The only limitation is that the equations require that the two frames have their axes orientated in the "standard way" (See http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html). In practice this means in our case the z axes remain parallel to each other and the x,y planes remain parallel to each other. This slightly awkward definition comes about because when motion is not parallel to the x or y-axis the x and y axes themselves appear to rotate from the point of view of frame S'.

 

[starthaus]

Seeing as how Starthaus is done with this thread, this is a question for Rasalhague.

Do your equations agree numerically with the following simple trigonometric analysis?

Consider two frames S and S' that have relative velocity v/c=\beta and axes aligned with each other when there origins coincide and S' is moving along the x-axis of S. (We are always free to align the coordinates frames in such a way that this is true.) The rod is at rest in S with proper length L and one end at the origin and the rod is orientated in at an angle (theta) in the x,y plane. (Again we are always free to align the frames this way.)

In frame S' the angle (theta') of the rod relative to the x' axis is by simple trigonometry and allowing for length contraction is:

\theta&#039; = \tan^{-1}\left( \frac{L_y&#039;}{L_x&#039;} \right) = \tan^{-1}\left(\frac{L_y}{L_x \sqrt{1-\beta^2}}\right) = \tan^{-1}\left(\frac{\tan{\theta}}{ \sqrt{1-\beta^2}} \right)

The difference between the two angles (phi) is the "rotation" of the rod (The Thomas rotation) and is given by:

\phi = (\theta - \theta&#039;) = \theta - \tan^{-1}\left(\frac{\tan(\theta)}{ \sqrt{1-\beta^2}}\right) \qquad \qquad (Eq1)

This rotation angle is independent of the orientation of the axes with respect to the motion or the rod, and only depends on the angle of the rod with respect to the motion.

In the earlier example I gave the equation for the rotation as

\phi = (\theta - \theta&#039;) = \theta + \tan^{-1}(\cot(\theta)\sqrt{1-\beta^2}) - \pi/2

when the rod was aligned with x-axis and the motion was at angle (theta) wrt the x axis. It is easy to see that the two equations are equivalent because of the truth of the simple trigonometric equality (at least when 0<x<=1):

-\tan^{-1}\left(\frac{\tan(\theta)}{x}\right) = \tan^{-1}(\cot(\theta)x) - \pi/2

This is not correct since the Thomas rotation is a function of the change in the angle of consecutive boosts. Your formula contains no such angle.

 

[yuiop]

This is not correct since the Thomas rotation is a function of the change in the angle of consecutive boosts. Your formula contains no such angle.

You are wrong. While it is true that the rotation can be analysed in terms of two successive boosts, e.g one parallel to the rod \beta_x followed by an orthogonal boost \beta_y, it is equally valid to analyse the rotation in terms of a single boost at an angle \theta = atan(\beta_y/\beta_x).

In terms of successive boosts, it can be shown that after the initial parallel boost, the clocks at either end of the rod are no longer simultaneous from the the point of view of the original frame and the next boost does not occur simultaneously at either end of the rod in the original frame. This asynchronous second boost can be thought of as what that brings about the rotation of the rod in the original frame. Usually when analysed like this, the order of the boosts is reversed to shown that the order of the two successive boosts is not important and from there it is implied that two successive boosts is equivalent to a single boost at an angle that is a function of the two velocity components.

I am fairly sure (barring typos) that if you compare my single boost equations, numerically with the two boost methods, you will see that they are equivalent.

 

[starthaus]

You are wrong. While it is true that the rotation can be analysed in terms of two successive boosts, e.g one parallel to the rod \beta_x followed by an orthogonal boost \beta_y, it is equally valid to analyse the rotation in terms of a single boost at an angle \theta = atan(\beta_y/\beta_x).

I don't think so, you are missing a basic element, the change in direction of the successive boosts.

I am fairly sure (barring typos) that if you compare my single boost equations, numerically with the two boost methods, you will see that they are equivalent.

Try proving it with math, not with words.

 

[Rasalhague]

Do your equations agree numerically with the following simple trigonometric analysis?

Yes, with the proviso below, and except sometimes for a difference in sign. E.g.

In[1]:= \[Beta] = {.7, 0, 0}; x = {89, 1501, 0}; \[Theta] = 
 ArcTan[x[[2]]/x[[1]]]; \[Iota] = 
 ArcTan[Tan[\[Theta]]/
   Sqrt[1 - \[Beta].\[Beta]]]; \[Phi] = \[Theta] - \[Iota]; \[Phi]

Out[1]:= -0.0169055

In[2]:= ArcCos[(1 - Cos[\[Theta]]^2 (1 - Sqrt[1 - \[Beta].\[Beta]]))/
  Sqrt[(1 - \[Beta].\[Beta]*Cos[\[Theta]]^2)]]

Out[2]:= 0.0169055

In:[3]:= ArcCos[(x.x - (\[Beta].\[Beta])^(-1) (x.\[Beta])^2 (1 - 
       Sqrt[1 - \[Beta].\[Beta]]))/(Norm[x] Sqrt[
     x.x - (x.\[Beta])^2])]

Out:[3]:= 0.0169055

In the extreme cases: expression 3 gives 0 when the rod is aligned with the y axis, while expressions 1 and 2 are indeterminate due to the way we defined the angle theta for them; when the rod is aligned along the x axis, expressions 1 and 2 give 0, while expression 3 either gives zero or a certain very small number, 2.10734 * 10^(-8) or 2.10734 * 10^(-8) i, depending on speed and length.

 

[Rasalhague]

Given the 3-velocity, \mathbf{v}, of an object in one orthonormal spatial basis, what is is its velocity, \textbf{v}&#039;, in another orthonormal spatial basis derived from the first by a boost of velocity \pmb{\beta}, the latter velocity wrt to the first basis. From the general boost formula, I get the following expression:

\textbf{v}&#039;=\textbf{v}\ominus \pmb{\beta}\equiv \frac{\textbf{v}+\pmb{\beta}\left [ (\gamma-1)(\textbf{v}\cdot \pmb{\beta})\beta^{-2}-\gamma \right ]}{\gamma(1-\textbf{v}\cdot \pmb{\beta})}

I've just used the "ominus" symbol as a convenient abbreviation to represent the function on the far RHS. Can two boosts, in general, be expressed as one? I think Starthaus, you're is saying no, and Kev yes. Is that right? If it was possible, how would we compose boosts? For example, suppose we had a boost with velocity \pmb{\beta} relative to some spacetime basis S, in the xy plane of S. Call this boost L(\pmb{\beta}). Now suppose we try to decompose it into two boosts, the first entirely along the x axis, L(\pmb{\beta}_x), which transforms from S to another spacetime basis S'. What boost should we try to compose with this? I guessed that a natural choice might be L((\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta}), that is L(\pmb{\beta}_y \ominus \pmb{\beta}).

So would you expect the following equation to be generally true if the boost on the left of the equation lies in the xy plane

L(\pmb{\beta})=L((\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta})\circ L(\pmb{\beta}_x) ?

My first attempts at testing it numerically suggest it isn't, but from past experience, it's all too possible that I've made mistakes. But what of the concept; is this the right question to be asking? If not, and if you think the composition (in some sense) of boosts is always a pure boost, what would be the right way to compose them?

 

[starthaus]

if you think the composition (in some sense) of boosts is always a pure boost

It isn't, I showed it when I introduced the notion of https://www.physicsforums.com/blog.php?b=1959 .

what would be the right way to compose them?

Easy, multiply:

\begin{pmatrix}<br /> \gamma_1 &amp; -\gamma_1 \pmb{\beta_1}\\ <br /> -\gamma_1 \pmb{\beta_1} &amp; I+\frac{\gamma_1-1}{\beta_1^2} \pmb{\beta_1\beta_1}<br /> \end{pmatrix}

by:

\begin{pmatrix}<br /> \gamma_2 &amp; -\gamma_2 \pmb{\beta_2}\\ <br /> -\gamma_2 \pmb{\beta_2} &amp; I+\frac{\gamma_2-1}{\beta_2^2} \pmb{\beta_2\beta_2}<br /> \end{pmatrix}

When you do that, you get the most general expression for the Thomas rotation.

 

[Rasalhague]

This is not correct since the Thomas rotation is a function of the change in the angle of consecutive boosts. Your formula contains no such angle.

Assuming "this" means "to call this phenomenon Thomas rotation" (rather than "these equations don't correctly give the angle they claim to do"), the above statement accords with the Wikipedia definition, which explicitly mentions two boosts: "The composition of two Lorentz boosts which are non-colinear, results in a Lorentz transformation that is not a pure boost but is the product of a boost and a rotation. This rotation is called Thomas rotation, [...]"

It seems then that Thomas rotation is something that happens to 4-vectors due to multiple boosts, so anything that happens to 3-vectors or is not due to more than one boost must be some other kind of rotation. (EDIT: On second thoughts, I suppose a pure rotation--with no boost component--of a 4-vector would correspond to a pure rotation--with no stretch component--of its relative spatial part.) But it's easy to see how one could get the impression Thomas rotation referred to the rotation of a 3-vector due to a single boost not parellel to it, given than introductions to the topic such as Kevin Brown's [ http://www.mathpages.com/rr/s2-11/2-11.htm ] and the similar one in Spacetime Physics take this phenomenon as their starting point.

I suppose another way of putting it would be: boosts don't comprise the underlying set of a subgroup of the Lorentz group. Is the following statement right? Any composition of any number of boosts can be decomposed into two Lorentz transformations: one a pure boost, the other a pure rotation, the latter called a Thomas rotation.

\pmb{\beta}_2 in #118 corresponds to \textbf{v}&#039; = (\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta} = \pmb{\beta}_y \ominus \pmb{\beta} in #117, with the "ominus" symbol as defined in #117.

I've just corrected a typo in the final equation in #117, the one with the question mark. I take it, Starthaus, your answer to the corrected version is also no, in general

L(\pmb{\beta}) \neq L((\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta})\circ L(\pmb{\beta}_x)

 

[starthaus]

\pmb{\beta}_2 in #118 corresponds to \textbf{v}&#039; = (\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta} = \pmb{\beta}_y \ominus \pmb{\beta} in #117, with the "ominus" symbol as defined in #117.

\pmb{\beta}_2=\textbf{v}_2/c

I've just corrected a typo in the final equation in #117, the one with the question mark. I take it, Starthaus, your answer to the corrected version is also no, in general

L(\pmb{\beta}) \neq L((\pmb{\beta}-\pmb{\beta}_x)\ominus \pmb{\beta})\circ L(\pmb{\beta}_x)

Correct, the anser is "no".

 

[Rasalhague]

I should make it clear that I intended all my velocities (and all other length and time quantities) to be in units where c = 1. That is, I didn't mean the letter v to suggest I'd begun using a different system of units, or that there was any relationship between v and beta, except that, for the sake of example, I put them both in the xy-plane. By v I only meant a different velocity to beta.

 

[starthaus]

I should make it clear that I intended all my velocities (and all other length and time quantities) to be in units where c = 1. That is, I didn't mean the letter v to suggest I'd begun using a different system of units, or that there was any relationship between v and beta, except that, for the sake of example, I put them both in the xy-plane. By v I only meant a different velocity to beta.

Yes, I understood that. In #118 there is absolutely no correlation between \textbf{v}_1 and \textbf{v}_2, so there is no correlation between the "betas".

 

[yuiop]

Yes, with the proviso below, and except sometimes for a difference in sign. E.g.

My bad. My equation should have been Phi = Iota - Omega if the signs are to be consistent (e.g positive angle means anticlockwise rotation) and if Iota and Omega are measured wrt the motion vector, rather than the Phi = Omega - Iota that I orginally gave.

 

[Rasalhague]

My bad. My equation should have been Phi = Iota - Omega if the signs are to be consistent (e.g positive angle means anticlockwise rotation) and if Iota and Omega are measured wrt the motion vector, rather than the Phi = Omega - Iota that I orginally gave.

I see, yes, now your formula gives the correctly signed angle according to which quadrant the displacement vector is pointing, while mine both just give the absolute value of the angle.

 

[Rasalhague]

It seems as if this kind of rotation, whatever it's called, that we've been talking about is the same phenomenon that both Brown and Taylor & Wheeler begin their discussions of Thomas rotation with, due to the relativity of simultaneity, but that Thomas rotation is something else, only associated with multiple boosts in different directions, albeit presumably related to this single-boost phenomenon in some way. If this rotation due to a single boost was all there was to Thomas rotation, and multiple sucessive pure boosts could be composed into a single pure boost, then I don't see how there could be any net precession due to motion in a complete circle which ends with the starting velocity, since all the intermediate boosts could be composed into a single identity boost.

 

[jason12345]

It seems as if this kind of rotation, whatever it's called, that we've been talking about is the same phenomenon that both Brown and Taylor & Wheeler begin their discussions of Thomas rotation with, due to the relativity of simultaneity, but that Thomas rotation is something else, only associated with multiple boosts in different directions, albeit presumably related to this single-boost phenomenon in some way. If this rotation due to a single boost was all there was to Thomas rotation, and multiple sucessive pure boosts could be composed into a single pure boost, then I don't see how there could be any net precession due to motion in a complete circle which ends with the starting velocity, since all the intermediate boosts could be composed into a single identity boost.

If you do a Google search on Wigner rotation, Thomas precession, you'll find that there seems to be some disagreement on how it manifests it self, which is why you should have a look at his original paper if you can.

In Thomas's original paper, he defined an infinite number of inertial frames in the laboratory frame all moving with a velocity that the electron acquires for t > 0. He then gives the LT from the laboratory frame to one of these inertal frames instantaneously co-moving with the electron centred at the origin. The Thomas Rotation is the rotation of the instantaneously co-moving frame at tau+dtau relative to the comoving frame at tau.

 

[starthaus]

If you do a Google search on Wigner rotation, Thomas precession, you'll find that there seems to be some disagreement on how it manifests it self, which is why you should have a look at his original paper if you can.

In Thomas's original paper, he defined an infinite number of inertial frames in the laboratory frame all moving with a velocity that the electron acquires for t > 0. He then gives the LT from the laboratory frame to one of these inertal frames instantaneously co-moving with the electron centred at the origin. The Thomas Rotation is the rotation of the instantaneously co-moving frame at tau+dtau relative to the comoving frame at tau.

This is explained fairly well in Taylor and Wheeler's "Spacetime Physics" but it is explained even better in Moller's "The Theory of Relativity". You need 3 frames in order to understand what is going on:

-the lab frame \Sigma
-a frame S boosted by an arbitrary speed u wrt \Sigma
-a second frame S' boosted by an arbitrary speed u&#039; wrt \Sigma (where, in "Spacetime Physics". u and u&#039; describe the sides of an n-sided polygon, so they make an angle \alpha=2\pi/n).

The frames S and S' are necessary in order to calculate the precession effect due to a particle "jumping" from frame S to frame S' when it "turns the corner" at each vertex of the n-sided polygon. The net effect is that a vector (for example the classical spin of the particle) that has a fixed orientation wrt the axis of S and S' precesses (i.e. "jumps angle") from the perspective of the lab frame \Sigma every time a corner is being rounded. So, when the respective vector has made a ful turn around the polygon , traveling with the succession of inertial frames S, S', S",...back to S, its orientation has changed! This exact computation isn't for the faint of heart and is given in pages 124-125 in Moller's book.

 

[Rasalhague]

My plan: work out what happens, then worry about why.

Outline. Begin with a 4-vector, s_0, representing the spin of a gyroscope, at some event (the starting event) on a spatially circular path which the gyroscope is to travel at constant speed, this vector having no time component in an orthonormal basis whose time axis is parallel to the 4-velocity of the gyroscope. The gyroscrope's spin after it's completed one orbit of the circle:

(1) \; \lim_{n\rightarrow \infty}\circ_{i=1}^{n} L(-\pmb{\beta}_i) \circ R(s_{i-1})

where the circle represents a composition of n boosts and

(2) \; \pmb{\beta}_i=\textbf{v}_i\ominus \pmb{\beta}_{i-1}\equiv \frac{\textbf{v}_i+\pmb{\beta}_{i-1}\left [ (\gamma-1)(\textbf{v}_i\cdot \pmb{\beta}_{i-1})\beta^{-2}_{i-1}-\gamma \right ]}{\gamma(1-\textbf{v}_i\cdot \pmb{\beta}_{i-1})}

where the gammas are a function of beta sub i-1, and \textbf{v}_i is the velocity of each incremental boost according to one constant orthonormal basis field, call it "the lab frame", in which the circlular trajectory is defined, and \pmb{\beta}_i is the velocity of this same boost in the previous comoving basis of the polygonal approximation, and R_i is a function which derives a 3-vector from the previous value of s_i according to the formula derived in #106, then creates from this a 4-vector having the same spatial components as this 3-vector and time component 0.

Finally, transform the coordinates of the resulting 4-vector and the starting 4-vector into the lab frame and compare the angle between them in that frame. Does this make sense? Would it work? Would it measure the right thing? Can you see any conceptual flaws in this plan?

 

[Rasalhague]

If you do a Google search on Wigner rotation, Thomas precession, you'll find that there seems to be some disagreement on how it manifests it self, which is why you should have a look at his original paper if you can.

In Thomas's original paper, he defined an infinite number of inertial frames in the laboratory frame all moving with a velocity that the electron acquires for t > 0. He then gives the LT from the laboratory frame to one of these inertal frames instantaneously co-moving with the electron centred at the origin. The Thomas Rotation is the rotation of the instantaneously co-moving frame at tau+dtau relative to the comoving frame at tau.

So are you saying Thomas's Thomas rotation is what Kev called "Thomas rotation" in #112 after all? I'm just guessing here, but could it be that what Starthaus calls Thomas rotation in #113 is what those who call Kev's Thomas rotation "Thomas rotation" would call Thomas precession?

 

[yuiop]

So are you saying Thomas's Thomas rotation is what Kev called "Thomas rotation" in #112 after all? I'm just guessing here, but could it be that what Starthaus calls Thomas rotation in #113 is what those who call Kev's Thomas rotation "Thomas rotation" would call Thomas precession?

My guess is that the Thomas precession is an accumulation of instantaneous Thomas rotations. For example the total rotation of a gyroscope (that is onboard a satellite) after (the satellite completes) one orbit, relative to the orientation of the satellite that has its orientation "locked on" a distant fixed star. This precession continues to accumulate, precessing by an additional amount for each full orbit.

 

[jason12345]

So are you saying Thomas's Thomas rotation is what Kev called "Thomas rotation" in #112 after all? I'm just guessing here, but could it be that what Starthaus calls Thomas rotation in #113 is what those who call Kev's Thomas rotation "Thomas rotation" would call Thomas precession?

Kev's analysis doesn't make sense to me because his rod isn't being accelerated from one frame to another and the rod isn't rotating in the frame comoving with it. He hasn't shown how his analysis is equivalent to the usual way of showing the Thomas rotation.

At time t1, the electron in the lab frame has position x1,y1,z1 and velocity v1. In frame S' comoving with the electron at the origin, the electron is stationary at time tau. When the electron returns to that comoving frame, the frame attached to it will have rotated relative to it, hence the subtlety.

My guess is that the Thomas precession is an accumulation of instantaneous Thomas rotations. For example the total rotation of a gyroscope after completing one orbit, relative to the orientation of a satellite that has its orientation "locked on" a distant fixed star. This precession continues to accumulate, precessing by an additional amount for each full orbit.

Correct.

 

[Rasalhague]

At time t1, the electron in the lab frame has position x1,y1,z1 and velocity v1. In frame S' comoving with the electron at the origin, the electron is stationary at time tau. When the electron returns to that comoving frame,

i.e. returns to a state of rest in S'?

the frame attached to it will have rotated relative to it, hence the subtlety.

Doesn't this depend on how "the frame attached to it" is defined. There must be something about the physics of what happens that suggests a certain definition as natural, so natural that someone familiar with the concept can overlook other possibilities. Robert Littlejohn's account talks about a rule for generating a vector field along a curve, which rule he calls Fermi-Walker transport, although it emerged in this thread, that his FWT is a only a special case of FWT. In that thread, bcrowell mentioned the idea of Fermi coordinates. Is this what you mean by "the frame attached to" the electron, a (noninertial) frame in which its spin or angular momentum is constant, a frame which can itself be defined by the motion of gyroscopes? The coordinate bases associated with this coordinate system, Littlejohn's \left \{ f_\alpha \right \}(\tau) are a kind of FW-transported basis field. One defining characteristic seems to be that the spin vector at each point along the world line is orthogonal to the gyroscope's 4-velocity. Why is that?

 

[Rasalhague]

Brown begins his description of Thomas precession with a simple example of how a change of coordinates (a coordinate boost) can result in a change of angle, due to the relativity of simultaneity. Likewise Taylor & Wheeler (## 5 and 6). Isn't it the same phenomenon, a change of angle resulting from a coordinate boost, that Kev described in #112 of this thread? Also in #127, Starthaus writes, "You need 3 frames in order to understand what is going on".

But if this is a physical effect due to the acceleration of a gyroscope, shouldn't it be possible to describe it in anyone frame that includes the whole journey from "starting event", where the gyroscope begins its journey along a circular path in space, to "finishing event" where the gyroscope returns to the same spatial coordinates? I think the prediction is: if we send a gyroscope on a circular path in space (in flat spacetime, spinning freely, not subject to any force except that causing its circular motion which is exerted through its centre of mass) and let another follow an inertial world line from starting event to finishing event, and the gyroscopes have the same spin at the start, they'll differ by some angle when they meet at the finish. Wouldn't they differ in any meaningful coordinate system that included the whole journey?

 

[Rasalhague]

Just a guess, but by analogy with the rotation of a Euclidean vector in a fixed basis, which is described by the inverse of the function that gives the coordinates of a fixed Eucludean vector when the basis is rotated, I was wondering... could it be that the change undergone by a 3-vector representing the spin of a gyroscope at one corner of a polygonal path is described, with respect to the constant coordinate basis field associated with one fixed inertial coordinate system, by the same function I derived in #106 for the change undergone by a 3-vector representing a displacement due to a coordinate boost, but using the inverse of that boost... which results in the same 3-vector, since there are an even number of betas in the final equation, so the minuses would cancel out.

Inverting the boost by changing beta to -beta is what I had in mind when I put the minus before the beta in eq. (1) of #128, where L represents a general coordinate boost, the general "boost in any arbitrary direction" as described here in the section "Matrix form", so that L(-beta) should be the inverse.

 

[yuiop]

Kev's analysis doesn't make sense to me because his rod isn't being accelerated from one frame to another and the rod isn't rotating in the frame co-moving with it. He hasn't shown how his analysis is equivalent to the usual way of showing the Thomas rotation.

I do not actually have to accelerate the rod. For example if I have a rod of length 1.0x that is at rest wrt my frame and then the rod is physically accelerated to 0.8c relative to me in the x direction, I will now consider the rod's length to be 0.6x. If on the other hand the rod remains at rest wrt myself and is viewed by an observer with a relative velocity of 0.8c then the other observer will consider the rod's length to be 0.6x even though the rod has not actually undergone any physical acceleration. The two measurements are equivalent and that is why my equations are numerically the same as Raselhague's. In the rotating rod case, rather than accelerating the rod I just consider the point of view of an observer with relative motion at some angle to the rod.

Now consider the rotation of the rod in terms of successive boosts. In frame S the rod is initially at rest and enclosed in a lab. Now the rod and lab are boosted in the x direction towards some fixed star. The rod and lab are now at rest in a new frame S'. Now the lab is accelerated sideways in the y direction by two lateral rockets at the head and tail of the lab that are fired simultaneously as measured in frame S'. According to an observer that remains at rest in frame S, the two lateral rockets do not fire simultaneously and the lab and rod rotate to an angle that is no longer pointing at the fixed star. Call this final frame S''. Observers inside the lab are unaware of any rotation of the rod because the lab and rod have rotated by the same amount. Observers that remain in frame S' (the intermediate frame) still consider the rod and lab to be parallel to their orientation before the second boost, so observers in frame S' do not see any rotation of the rod or the lab. The only observers that see any rotation are the original observers that remained at rest in frame S.

The only way that observers in a "comoving lab frame" will see a rotation of the rod relative to their own frame, is if the lab is artificially accelerated to compensate for the Thomas rotation so that the lab remains pointing at the distant fixed star. This is of course very like the Gravity Probe B experiment. In that experiment the gyroscopes are free to precess and the satellite is artificially accelerated to keep pointing at the distant star.

 

[Rasalhague]

Now consider the rotation of the rod in terms of successive boosts. In frame S the rod is initially at rest and enclosed in a lab. Now the rod and lab are boosted in the x direction towards some fixed star. The rod and lab are now at rest in a new frame S'. Now the lab is accelerated sideways in the y direction by two lateral rockets at the head and tail of the lab that are fired simultaneously as measured in frame S'. According to an observer that remains at rest in frame S, the two lateral rockets do not fire simultaneously and the lab and rod rotate to an angle that is no longer pointing at the fixed star. Call this final frame S''. Observers inside the lab are unaware of any rotation of the rod because the lab and rod have rotated by the same amount. Observers that remain in frame S' (the intermediate frame) still consider the rod and lab to be parallel to their orientation before the second boost, so observers in frame S' do not see any rotation of the rod or the lab. The only observers that see any rotation are the original observers that remained at rest in frame S.

Interesting. Would observers in S'' see rotation (at the first boost)? I'm guessing so, by symmetry, since we could imagine the whole scenario in reverse.

 

[yuiop]

Interesting. Would observers in S'' see rotation (at the first boost)? I'm guessing so, by symmetry, since we could imagine the whole scenario in reverse.

If I understand you correctly, then I think I agree.

I have done a little sketch that illustrates the apparent relative motions and orientations of 3 rods A, B and C that are at rest in frames S, S' and S'' respectively. Is this what you have in mind?

It is clear from the diagrams that "parallel" is a relative concept. You could call it the relativity of parallelism.

Rod A remains at rest in S. Rods B and C are initially boosted in the x direction towards the distant star. Finally rod C is boosted in the y direction. The red lines are very long refrence rods that are at rest and parallel in frame S.

The orientations are what observers would measure in their respective frames and not what they would "see" due to aberration and light travel times. Only the observer in frame S' considers all 3 rods to be parallel to each other.

 

[Rasalhague]

It makes sense to me that C would rotate as shown from S to S', and that A would rotate as shown from S'' to S', according to the relativity of simultaneity. But at first sight, there seems to be a contradiction between (1) our earlier formulas and statements in textbooks which suggested there would be no rotation of a 3-vector due to a parallel or perpendicular boost and (2) the boost from S' to S'', where a boost perpendicular to A rotates A, and the boost from S' to S, where a boost parallel to C rotates C.

 

[yuiop]

But at first sight, there seems to be a contradiction between (1) our earlier formulas and statements in textbooks which suggested there would be no rotation of a 3-vector due to a parallel or perpendicular boost

This is true for single boosts of a rod that is initially at rest in the observer's rest frame. However a combination of boosts such as a parallel boost followed by a perpendicular boost (or vice versa) is equivalent to a single boost at an angle that is neither parallel of perpendicular.

and (2) the boost from S' to S'', where a boost perpendicular to A rotates A, and the boost from S' to S, where a boost parallel to C rotates C.

Let us say the second boost is perpendicular. This second boost is a boost of a rod that is not at rest in the original frame and the statement that a perpendicular boost does not cause a rotation is only valid for a rod that no parallel motion. Similarly, the statement that a parallel boost does not cause a rotation is only true for a rod that has no perpendicular motion.

 

[Rasalhague]

This is true for single boosts of a rod that is initially at rest in the observer's rest frame. However a combination of boosts such as a parallel boost followed by a perpendicular boost (or vice versa) is equivalent to a single boost at an angle that is neither parallel of perpendicular.

What about the single coordinate boost from S' to S'', and the single boost from S' to S, each considered in isolation? Surely these boosts have no memory: they can't know if they're part of a sequence, they don't know what other coordinate systems we may have thought about before or after. All we tell them is what the coordinates are of events according to S'. Can we describe, just from the perspective of the coordinate boost from S' to S'', how the relativity of simultaneity relates to the rotation of A? In S', neither end of A is further north than the other at any time, so the "intuitive" idea that events to the north happen sooner in S'' doesn't seem to explain it. Won't two events, one at each end of A, that are simultaneous in S' also be simultaneous in S''?

I suppose the difference between this scenario and my earlier formula must be that it assumed the 3-vector to be transformed represented the length and alignment of a rod at rest in the input frame, whereas here, for example, neither A nor C are at rest in S'. So we have to take account of that movement in some way, and maybe a displacement 3-vector isn't the best way to represent that, or maybe I'd need to define it differently.

In the derivation of length contraction, it was possible to calculate the rod's length in a frame where it's moving by measuring its ends at the same instant in that frame. So I guess here we'd have to measure the position of its ends at the same instant in the frame that results from the coordinate boost. Maybe I need to think of it from the other side: given that A has a certain angle in S'', what how will it end up when we boost back to S', or what angle would it have to have, given that a coordinate boost of a certain velocity leaves it perpendicular to the direction of the boost. Hey, if A was at rest in S'', it would be rotating in the opposite direction when we boost coordinates to S', wouldn't it? Wow, even more complicated! Would it be possible for the rotation due purely to a rod's alignment wrt the boost to be canceled out by the rotation due to the direction of its velocity?

Let us say the second boost is perpendicular. This second boost is a boost of a rod that is not at rest in the original frame and the statement that a perpendicular boost does not cause a rotation is only valid for a rod that no parallel motion. Similarly, the statement that a parallel boost does not cause a rotation is only true for a rod that has no perpendicular motion.

Hmm... Suppose we begin by analysing the scenario in S', then we boost coordinates to S''. Beforehand we see three rods parallel to each other. We boost coordinates by a velocity perpendicular to the rods. One rotates as we change coordinates, two don't rotate. What was different about the one that rotated: before and after the boost, it's velocity is not parallel to the boost. Likewise for the coordinate boost from S' to S.

In boosting coordinates from S' to S'', A has no motion parallel to the boost in S', and it rotates, whereas the rods whose motion was parallel to the boost didn't rotate. In boosting coordinates from S' to S, C has no motion parallel to the boost in S', and it rotates from S' to S.

What would the scenario look like in a frame with the same velocity as S'' has in S, but obtained directly from S with one pure boost? How would this frame differ from S''?

 

[yuiop]

Hi Rasalhague.

There are far too many questions in you last post!

I feel confident I can answer them, but it would be too time consuming. Could you pick one scenario or issue you most want addressing and describe it in careful unambiguous detail?

 

[Rasalhague]

Okay, fair enough. It was a very rambly post! A lot of that was me just thinking aloud, trying to blunder onto solid ground. With a bit more brain-racking, I hope I'll be able to home in on a more-or-less coherent question. But for now, how about this? It refers to your post #137 and the attached diagram, and deals with the basic issue of whether there are two different kinds of rotation associated with a single boost, one due to the alignment of a rod or displacement vector, the other due to the direction of a rod's velocity wrt that of the boost.

If A was at rest in S'', would it rotate in the opposite direction (clockwise) to that shown in your diagram (counterclockwise) when we boost coordinates to S'? And if so, does this mean that it would be possible, given an appropriate choice of alignments and velocities, for the rotation due purely to a rod's alignment wrt the boost (the rotation for which we each found a formula earlier in this thread) to be canceled out by the rotation due to the direction of the rod's velocity wrt that of the boost?

 

[yuiop]

If A was at rest in S'', would it rotate in the opposite direction (clockwise) to that shown in your diagram (counterclockwise) when we boost coordinates to S'? And if so, does this mean that it would be possible, given an appropriate choice of alignments and velocities, for the rotation due purely to a rod's alignment wrt the boost (the rotation for which we each found a formula earlier in this thread) to be canceled out by the rotation due to the direction of the rod's velocity wrt that of the boost?

Hi Raselhague. I started answering your question, but while working on some examples I ran into some paradoxical situations, so it is quite possible that some of your concerns are valid and that Jason's comments about requiring actual acceleration rather than just a transformation boost to another frame is required. I will have to give it some more thought. Bear with me!