Stone's derivation of Thomas rotation

[Rasalhague]

That should work now.

It does. It gives the same answer as x*Sqrt[1 - bx^2], and is unaffected by changes in by.

 

[yuiop]

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

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

And for the angle,

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

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

where, for example, $\cos(\textbf{r},\textbf{r}')$ is the angle between the original space vector $\textbf{r}$ and its boosted counterpart $\textbf{r}'$.

How does your equation compare numerically with mine?

 

[DrGreg]

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

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

I haven't been following this thread and I'm reading this post out-of-context, but if you're using standard notation, I believe you should have $\gamma$ instead of $1 / \gamma$. See Lorentz transformation - Matrix form, final equation in section.

 

[Rasalhague]

How does your equation compare numerically with mine?

In[1]:= th + ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2

Out[1]:= -0.489976

In[2]:= ArcCos[(1 - Cos[th]^2)/(1 - b^2 Cos[th])]

Out[2]:= 0.817476

This is with th = ArcTan[.3/.9] = 0.321751. Did I type it right? The angle should be getting bigger, shouldn't it?

 

[Rasalhague]

Thanks DrGreg. At the risk of speaking too soon, I think it's okay though. My equation is supposed to describe the change in a space vector (3-vector, relative vector), due to a boost, rather than the change in coordinates of a spacetime vector (4-vector). The matrix equation which is described at that link was the starting point from which this one was derived. This space vector equation has 1/gamma for the same reason as the simple, one-dimensional length-contraction equation. The intermediate steps, involving a space vector parallel to the x-axis seemed to work.

 

[yuiop]

In[1]:= th + ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2

Out[1]:= -0.489976

In[2]:= ArcCos[(1 - Cos[th]^2)/(1 - b^2 Cos[th])]

Out[2]:= 0.817476

This is with th = ArcTan[.3/.9] = 0.321751. Did I type it right? The angle should be getting bigger, shouldn't it?

Your transcription of my equation looks OK. That equation is for phi which I have defined as the rotation of the rod away from the x' axis. If you look at the bit I have highlighted in the quote from #63 below you will see that I given the total rotation of the rod away from the line of motion as -theta +phi so that you should get something like -0.321751-0.489976 which is a clockwise rotation and I think if done to enough decimal places should be close to your result give or take the sign. Which way the rod rotates depends on which side of the line of motion the rod is on, but the rotation is always away from the line of motion towards being perpendicular to the line of motion, (assuming one end of the rod is on the line of motion).

Edit: It is close..but not close enough. One of our equations is not right... probably mine.

Edit:Edit: Scrap that. I see that your equation is for the rotation of r wrt to r so it should agree with my equation for phi but they are miles apart.

The rotation $(\phi)$ of the rod relative to the x' axis in the S' frame is given by:

$$\phi = \theta + tan^{-1}\left( \frac{\cos\theta}{\sin\theta } \, \sqrt{(1-\beta_x^2-\beta_y^2)} \right)-\frac{\pi}{2}$$

This rotation is generally away from the line of motion and the orientation of the rod tends towards being orthogonal to the motion at velocities approaching c.

Now that we have the length of the rod $\| L' \|$ in S' and its angle with respect to the x' axis, it is easy enough to work out the projection (L'x) of the rod's length onto the x' axis as:

$$L'_x = \| L' \| \cos\phi$$

The angle of the rod with respect to to the line of motion is $-\theta$ in frame S and $-\theta+\phi$ in frame S'. $\phi$ has the opposite sign to $\ \theta$ and so the angle wrt the line of motion is greater in S' than in S.

 

[Rasalhague]

Your transcription of my equation looks OK. That equation is for phi which I have defined as the rotation of the rod away from the x' axis. If you look at the bit I have highlighted in the quote from #63 below you will see that I given the total rotation of the rod away from the line of motion as -theta +phi so that you should get something like -0.321751-0.489976 which is a clockwise rotation and I think if done to enough decimal places should be close to your result give or take the sign. Which way the rod rotates depends on which side of the line of motion the rod is on, but the rotation is always away from the line of motion towards being perpendicular to the line of motion, (assuming one end of the rod is on the line of motion).

Ah, I see. Curiously, for this example, they're close, but not exactly the same:

ArcCos[(1 - Cos[th]^2)/(1 - b^2 Cos[th])] = 0.817476

ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2 = ph - th = -0.811726

But in general, they're not the same. When I set by = 0, and all the velocity in the x direction, mine gives pi/2, yours indeterminate, but they should give 0 in that case, shouldn't they? I sometimes get imaginary results with mine, and it doesn't agree with the other equation I had either. Hmm. I'm going to have to leave it for now, and try again tomorrow...

 

[yuiop]

Ah, I see. Curiously, for this example, they're close, but not exactly the same:

ArcCos[(1 - Cos[th]^2)/(1 - b^2 Cos[th])] = 0.817476

ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2 = ph - th = -0.811726

But in general, they're not the same. When I set by = 0, and all the velocity in the x direction, mine gives pi/2, yours indeterminate, but they should give 0 in that case, shouldn't they? I sometimes get imaginary results with mine, and it doesn't agree with the other equation I had either. Hmm. I'm going to have to leave it for now, and try again tomorrow...

In the limit as theta goes to zero, the atan function in my expression goes to Pi/2 meaning that there is zero rotation when the by=0. The mathematical software strugglesat this extreme.

 

[Rasalhague]

This is what I did. Can you see any mistakes or wrong assumptions. I started with

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

which looks reasonable: we subtract from the original vector some multiple of its projection onto the line of motion, causing it to shrink and rotate towards the line perpendicular to the line of motion. The projection is being multiplied by a number less than one, so it should never quite reach the perpendicular line. Then I dotted the transformed vector with the original vector and divided by the lengths of the original and transformed vectors:

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

$$rr'\cos(\textbf{r},\textbf{r}')=r^2-\left ( 1-\frac{1}{\gamma} \right )\frac{r^2 \beta^2 \cos^2(\textbf{r},\pmb{\beta})}{\beta^2}$$

$$\cos(\textbf{r},\textbf{r}')=\frac{r^2-\left ( 1-\gamma^{-1} \right )r^2 \beta^2 \beta^{-2} \cos^2(\textbf{r},\pmb{\beta})}{rr'}$$

$$\cos(\textbf{r},\textbf{r}')=\frac{r^2-\left ( 1-\gamma^{-1} \right )r^2\beta^2 \beta^{-2} \cos^2(\textbf{r},\pmb{\beta})}{r\sqrt{r^2-r^2\beta^2 \cos^2(\textbf{r},\pmb{\beta})}}$$

$$\cos(\textbf{r},\textbf{r}')=\frac{1-\left ( 1-\gamma^{-1} \right ) \cos^2(\textbf{r},\pmb{\beta})}{\sqrt{1-\beta^2 \cos^2(\textbf{r},\pmb{\beta})}}$$

Well, I picked up a couple of mistakes I made when I tried to test this yesterday, but it's still not agreeing with ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2.

 

[yuiop]

$$\cos(\textbf{r},\textbf{r}')=\frac{1-\left ( 1-\gamma^{-1} \right ) \cos^2(\textbf{r},\pmb{\beta})}{\sqrt{1-\beta^2 \cos^2(\textbf{r},\pmb{\beta})}}$$

Well, I picked up a couple of mistakes I made when I tried to test this yesterday, but it's still not agreeing with ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2.

Hi. I think you might have missed a late edit to #76 where I decided that your rotation function should be the same as phi rather than the phi -theta I thought earlier.

Try comparing your function to

phi = th + ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2

Sorry, but I do not have time to check it myself at the moment.

Cheers.

 

[Rasalhague]

Hi. I think you might have missed a late edit to #76 where I decided that your rotation function should be the same as phi rather than the phi -theta I thought earlier.

Try comparing your function to

phi = th + ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2

Sorry, but I do not have time to check it myself at the moment.

Cheers.

Hey, hey:

ph = th + ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2 = -0.116569

ArcCos[(1 - Cos[th]^2 (1 - Sqrt[1 - b^2]))/Sqrt[(1 - b^2 Cos[th]^2)]] = 0.116569

!

 

[Rasalhague]

Three plots of angle(r,r') = ArcCos[(1 - Cos[th]^2 (1 - Sqrt[1 - b^2]))/Sqrt[(1 - b^2 Cos[th]^2)]]. One holding the speed fixed at 0.99, and varying the angle between the displacement vector r and the velocity vector b from 0 to 2Pi. One varying both inputs: the speed from 0 to 1 and the this angle from 0 to 2Pi. And another the same except cutting off the higher values to show more detail of the lower ones.

Thanks kev and starthaus for all your help.

 

[starthaus]

Your welcome

Oops, I had it working yesrday. It looks like another error crept in when translating my scribbled calculations from paper to latex. I should have squared the gamma factor factor as well as the trigometric functions. It should have been:

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 is the angle the velocity vector makes with the x' axis.

This is incorrect.

 

[starthaus]

Ah, I see. Curiously, for this example, they're close, but not exactly the same:

ArcCos[(1 - Cos[th]^2)/(1 - b^2 Cos[th])] = 0.817476

ArcTan[Cot[th] Sqrt[1 - (b^2)]] - Pi/2 = ph - th = -0.811726

But in general, they're not the same. When I set by = 0, and all the velocity in the x direction, mine gives pi/2, yours indeterminate, but they should give 0 in that case, shouldn't they? I sometimes get imaginary results with mine, and it doesn't agree with the other equation I had either. Hmm. I'm going to have to leave it for now, and try again tomorrow...

Charging after these angles will not provide you with the correct solution. If you want to finish the problem, you will need to finish the computations in the attached hint.

 

[yuiop]

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 is the angle the velocity vector makes with the x' axis.

This is incorrect.

That is not very helpful.

Do you agree that $$\| L'\| \ = L_x \sqrt{1-\beta_x^2}$$ is correct?

I have checked that is what the the longer equation you said is incorrect simplifies to symbolically. If you agree the short equation is correct, then my original longer equation is correct too. It also agrees numerically with the result that Rasalhague obtained by a different method. Chances are, you are in the wrong here.

 

[starthaus]

That is not very helpful.

Do you agree that $$\| L'\| \ = L_x \sqrt{1-\beta_x^2)$$ is correct.

Yes, I derived it long ago in this thread. The follow-on formulas that you tried to guess are all wrong. I gave Rasalhague a detailed blueprint as to how to derive the correct answer.

I have checked that is what the the longer equation you said is incorrect simplifies to symbolically. If you agree the short equation is correct, then my original longer equation is correct too.

No, your "long" equation is just an incorrect guess.

 

[yuiop]

No, your "long" equation is just an incorrect guess.

It is not a guess, it is derived from base principles using simple geometry.

Do you agree my "Long" equation is symbolically and numerically the same as your short equation? If so then you have no reason to say the "long" equation is incorrect. I did not use your Starthaus blueprint method, or Rasalhague's method, but we all obtained the same result. Many ways to skin a cat!

Seriously, what are the chances of guessing:

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


and the related:

$$\phi = \theta + tan^{-1}\left( \frac{\cos\theta}{\sin\theta } \, \sqrt{(1-\beta_x^2-\beta_y^2)} \right)-\frac{\pi}{2}$$

and getting both right by pure luck?

 

[starthaus]

It is not a guess, it is derived from base principles using simple geometry.

Do you agree my "Long" equation is symbolically and numerically the same as your short equation? If so then you have no reason to say the "long" equation is incorrect. I did not use your Starthaus blueprint method, or Rasalhague's method, but we all obtained the same result. Many ways to skin a cat!

Seriously, what are the chances of guessing:

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

This is incorrect.

and the related:

$$\phi = \theta + tan^{-1}\left( \frac{\cos\theta}{\sin\theta } \, \sqrt{(1-\beta_x^2-\beta_y^2)} \right)-\frac{\pi}{2}$$

and getting both right by pure luck?

Problem is, you guessed both your formulas wrong.
BTW, your formula for $$\theta$$ is also wrong.

 

[yuiop]

Problem is, you guessed both wrong.
BTW, your formula for $$\theta$$ is also wrong.

SO you are suggesting that Rasalhague, using a completely different method and presumably making a different mistake, arrived at the same (but allegedly incorrect) solution as myself?

 

[starthaus]

SO you are suggesting that Rasalhague, using a completely different method and presumably making a different mistake, arrived at the same (but allegedly incorrect) solution as myself?

Rasalhague calculates the angle between $$r'$$ and $$r$$. This angle doesn't appear in the formula for length contraction.
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.

 

[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).