Stone's derivation of Thomas rotation

[yuiop]

You try (and fail) to calculate length contraction. Two very different issues. I suggest that you take a good hard look at your derivation, you will find your mistakes. There are three mistakes.

You are missing the very simple and self evident truth that if:

\| L'\| \ = L_x \sqrt{1-\beta_x^2}

is true and if:

L_x \sqrt{1-\beta_x^2} = L_x\sqrt{ sin^2\theta + \cos^2\theta(1-\beta_x^2-\beta_y^2)}

is also true, then it follows that if the LHS (your version) of the above equation is correct, then the RHS (my version) is also correct.

 

[Rasalhague]

Starthaus, in the end, I got

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

whence

\frac{1-\cos^2(\textbf{r},\pmb{\beta})(1-\frac{1}{\gamma})}{\sqrt{1-\beta^2\cos^2(\textbf{r},\pmb{\beta})}}

When you said "you both guessed wrong" were you including this guess? Is the wrongness an algebraic mistake, or a conceptual mistake? When you said, "If you want to finish the problem, you will need to finish the computations in the attached hint", which computations were you referring to? Were they the computations Kev showed in #66? These fill in the gap in your PDF that I was having trouble with. I've worked through them, and they seem okay to me. They reach the same conclusion you did. Or were you referring to the computations needed to derive a formula for transforming a general spatial vector, not necessarily aligned along the x axis. Have I made a mistake at this stage?

There are three mistakes.

It might help to say what they are. We're not short of puzzles.

 

[starthaus]

You are missing the very simple and self evident truth that if:

\| L'\| \ = L_x \sqrt{1-\beta_x^2}

is true

My derivation shows this formula to be true only if the rod is aligned with the x-axis and only if the two frames have aligned axes. It doesn't say anything about the case of arbitrary alignment. where your attempt at guessing the result fails.

and if:

L_x \sqrt{1-\beta_x^2} = L_x\sqrt{ sin^2\theta + \cos^2\theta(1-\beta_x^2-\beta_y^2)}

is also true,

But L_x\sqrt{ sin^2\theta + \cos^2\theta(1-\beta_x^2-\beta_y^2)}
isn't the correct formula for arbitrary orientation of rod and arbitrary motion between the frames of reference. You are simply copying the formula for the simpler case, when the rod is aligned with the x axis.

 

[starthaus]

When you said "you both guessed wrong"

No, I was referring to kev's two guesses. Your formula is correct but it doesn't bring you any closer to finding the answer to the question. If you complete the calculations in my hint, you'll find the correct formula.

 

[Rasalhague]

No, I was referring to kev's two guesses. Your formula is correct but it doesn't bring you any closer to finding the answer to the question. If you complete the calculations in my hint, you'll find the correct formula.

Assuming "hint" is the PDF attachment to #84, and assuming "your formula" is

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

and assuming "the question" is how to find the angle between r and r' according to the boosted-to coordimate system, you seem to be saying that my formula for this angle is wrong:

\frac{1-\cos^2(\textbf{r},\pmb{\beta})(1-1/\gamma)}{\sqrt{1-\beta^2\cos^2(\textbf{r},\pmb{\beta})}}

If so, could you eleborate, and perhaps show us the right formula for comparison.

Alternatively, if by "the question" you mean my more general question: "what is the nature of Thomas rotation, how can its effects be calculated", and that my formula does give the correct angle between these 3-vectors according to the boosted-to coordinate system, perhaps you're saying that this is not relevant to Thomas rotation. Then perhaps the "correct formula" you invite me to derive from your hint is a formula for some other quantity. If that's the case, could you elaborate a bit on why that is, and explain what quantity it is that I should be trying to find a formula for?

Alternatively, is "the question" referring to the query I began this thread with, the one about about composition of velocities?

Following your suggestion, I've worked through the derivation again, starting from the formula in your hint, and got to back to my formula for calculating r', if r and the velocity of the boost are known. Should I be looking for something else in the hint explaining why my next formula doesn't give the angle between these 3-vectors (algebraic error or conceptual error?), or why this angle is not the angle I should be concerned about? If so, I just can't see yet where this clue is. If you can't say it, could you say which part of the page it's on?

 

[yuiop]

You are missing the very simple and self evident truth that if:

\| L'\| \ = L_x \sqrt{1-\beta_x^2}

is true...

My derivation shows this formula to be true only if the rod is aligned with the x-axis and only if the two frames have aligned axes. It doesn't say anything about the case of arbitrary alignment. where your attempt at guessing the result fails.

In my earlier post I stated that:

...
Let us say we have a rod with rest length Lx aligned with the x-axis in frame S with one end at the origin of the S frame. In frame S' with relative motion (\beta_x,\beta_y), the apparent length of the rod becomes

\| L' \| \ = L_x\sqrt{ sin^2\theta + \cos^2\theta\ (1-\beta_x^2-\beta_y^2)}

where \theta = \tan^{-1}(\beta_y/\beta_x)

which makes it clear I am considering the limited case of the rod being aligned withe x-axis and arbitrary motion with respect to the x and y axes. This is exactly the same limited case that you considered in your https://www.physicsforums.com/blog.php?b=1959 .

In that attachment you state "The axis of S and S’ are presumed parallel:" and "Additionally, we can consider for simplicity that the rod is aligned with the x-axis in frame S" so your derivation has exactly the same limitations as mine, so you are either being irrational or hypercritical when you say my solution is incorrect because it does not consider a more general case than your solution. As I and others have tried to explain to you repeatedly in many other threads, a limited case is not automatically incorrect just because it is is limited. Why don't you understand that? Why, if you honestly believe that, have you considered the limited case in your blog?

In your blog and this earlier post you state the final result as:

... the proof shows clearly:

L'=L\sqrt{1-\beta_x^2}

No dependency whatsoever of \beta_y or \beta_z

This is misleading, because it implies that for any arbitrary orientation of the rod, the length contraction is a function of \beta_x only, which is obviously untrue if the rod is aligned with y or z axis. The correct way to write the expression is the way I did as:

L'=L_x\sqrt{1-\beta_x^2}

But L_x\sqrt{ sin^2\theta + \cos^2\theta(1-\beta_x^2-\beta_y^2)}
isn't the correct formula for arbitrary orientation of rod and arbitrary motion between the frames of reference. You are simply copying the formula for the simpler case, when the rod is aligned with the x axis.

I never claimed it was for arbitrary orientation of the rod and it made it clear it wasn't. It isn't, just as your solution in your blog isn't. The motion is for arbitrary motion relative to the x and y axes, and you are always free in the 3 dimensional case to rotate the axes so the motion lies in the x,y plane or for even greater simplification to align the motion with the x axis.

I note that you have not asserted that

L_x \sqrt{1-\beta_x^2} = L_x\sqrt{ sin^2\theta + \cos^2\theta(1-\beta_x^2-\beta_y^2)}

is not true, so can I presume you have checked the equality and found that it is true and that your equation and my equation are in exact agreement and are equivalent? If so, will you do the decent thing and stop claiming my equation is incorrect?

If you are unable to derive the more general case for arbitrary orientation in 3D of the rod and velocity vectors for yourself, then why don't you just say so, and I will derive it for you (when I have the time and inclination).

 

[starthaus]

Assuming "hint" is the PDF attachment to #84,

Correct

and assuming "your formula" is

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

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

How did you derive this?

and assuming "the question" is how to find the angle between r and r' according to the boosted-to coordimate system,

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

you seem to be saying that my formula for this angle is wrong:

\frac{1-\cos^2(\textbf{r},\pmb{\beta})(1-1/\gamma)}{\sqrt{1-\beta^2\cos^2(\textbf{r},\pmb{\beta})}}

No (see above)> What I am saying is that , though the angle formula is correct, it doesn't help in finding the length contraction formula.

Alternatively, if by "the question" you mean my more general question: "what is the nature of Thomas rotation, how can its effects be calculated", and that my formula does give the correct angle between these 3-vectors according to the boosted-to coordinate system, perhaps you're saying that this is not relevant to Thomas rotation. Then perhaps the "correct formula" you invite me to derive from your hint is a formula for some other quantity. If that's the case, could you elaborate a bit on why that is, and explain what quantity it is that I should be trying to find a formula for?

Sure, with pleasure,we were discussing length contraction when kev butted in. So, the problem statement is:

In frame S, the rod has projections (L_x,L_y,L_z)
Frame S" moves with speed V=(V_x,V_y,V_z) wrt S. The axes of S and S' need not be aligned.
Question: what is the formula for length contraction ? This is what we were discussing and this is what the hint answers.

 

[yuiop]

...
No, the question is finding the length contraction formula for the general case (see the hint file)
...
No (see above)> What I am saying is that , though the angle formula is correct, it doesn't help in finding the length contraction formula.
...
Sure, with pleasure,we were discussing length contraction when kev butted in.

The heading of the thread is "Stone's derivation of Thomas rotation", not "Stone's derivation of length contraction".

Might I also remind you this is an open forum and anyone is allowed to butt in.

 

[starthaus]

The heading of the thread is "Stone's derivation of Thomas rotation", not "Stone's derivation of length contraction".

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

 

[yuiop]

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

Lets look at the "testimony":

Brilliant! Thanks for that (kev). I was having no end of trouble getting through all those various powers of beta and gamma.

Thanks kev and starthaus for all your help.

Were they the computations Kev showed in #66? These fill in the gap in your PDF that I was having trouble with. I've worked through them, and they seem okay to me. They reach the same conclusion you did.

On the face of the above testimony it would seem Rasalhague has found my contributions helpful and is grateful for them.

Back on the subject of rotation, if we are allowed to discuss rotation in this rotation thread, you seem to have a major misconception in your physical understanding of rotation:

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

Your above statement suggests that the rotation is somehow not physical and just a mathematical abstraction or something like that. You do accept that Thomas rotation is a physical rotation in the normal sense, just as real as time dilation, right?

Could you just clear up that last point?

 

[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".