Stone's derivation of Thomas rotation

1. Jul 30, 2010

Rasalhague

Has anyone here read and understood this introduction to Thomas rotation/precession?

http://homepage.ntlworld.com/stone-catend/ThomRotn.pdf

I'm stuck on section 4: General composition of velocities.

Letting $\textbf{u}$ be the velocity of frame F' wrt F, and $\textbf{v}$ the velocity of a point P wrt F', Stone uses the notation $\textbf{u}_1=\textbf{u} \oplus \textbf{v}$ for the velocity of P wrt F. What then does he mean by $\textbf{u}_2=\textbf{v} \oplus \textbf{u}$? That's to say, what is the significance of changing the order of the terms?

And later in this section, what does he mean by "calculate" $(1+q)^2(1-(u_1/c)^2)$?

Last edited by a moderator: Apr 25, 2017
2. Jul 30, 2010

starthaus

He's telling you later on in the page that "$u_2$ is obtained from $u_1$ by interchanging the roles of u and v". He's trying to prove one of the properties of Thomas precession: order independence. The whole derivation isn't very good.

This is even worse. He's attempting to calculate an expression that is well known in SR and he's making it more complicated is than necessary. https://www.physicsforums.com/blog.php?b=1959 [Broken] is a much better derivation (the third attachment)

Last edited by a moderator: May 4, 2017
3. Jul 30, 2010

Rasalhague

In your derivation, you show that "Thomas precession is order dependent". I guess you meant to write that Stone is trying to prove that something else--something relating to composition of velocities--is order independent?

By "interchanging the roles of u and v" do you mean he's saying that if (instead of denoting the velocity of F' in F) the letter u denoted the velocity of a point P measured in F', and if (instead of denoting the velocity of a point P measured in F') the letter v denoted the velocity of F' in F, and if these newly defined variables had the same numerical values as they did when they had their original meanings, then the numerical value of this new velocity composition calculation would be the same as the original calculation? (The original calculation being the one where u denotes the velocity of F' in F, and v the velocity of a point P measured in F'.) In other words, switching the numerical value of the inputs (arguments), while holding the significance of their order fixed, has no effect on the output (value of the function).

Thanks. Your algebra is nice and clear. I just have just a couple of questions about the simplifying assumptions, and where to go from there.

All three frames have the same event as their origin. I take it the axes of your second frame, F', are aligned with those of your first, F, and the axes of F'' are aligned with F', but the point of Thomas rotation is that the axes of F'' are not then aligned with F. So alignment of axes isn't transitive. Is that right? Given an orthonormal basis, one boost in the x direction changes the length of the unit vector in the x direction, while preserving spatial orthogonality among the (originally) spacelike basis vectors, and a subsequent boost in a different direction destroys the spatial orthogonality of the basis.

How can we tell that the term $y+\gamma x v v' c^{-2}$ represents a pure rotation and how can we work out the angle?

Last edited by a moderator: May 4, 2017
4. Jul 31, 2010

starthaus

Yes. In my opinion, this part of his calculation has nothing to do with the Thomas rotation.

Yes.

F' is a boost in the x direction wrt F.
F" is a boost in the y direction wrt F'.

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.

5. Jul 31, 2010

Rasalhague

By the way, in the first attachment, time dilation and length contraction for a boost in an arbitrary direction, how is it that we can ignore the y' and z' components?

6. Jul 31, 2010

starthaus

You can always reorient the system of axes assocaited with the rod such that the x axis is aligned with the rod. $$\Delta y = \Delta z =0$$ is meant that the rod has no component along the y and z axis.

7. Jul 31, 2010

Rasalhague

So is this what you did: you boosted in an arbitrary direction from an inertial frame F, where the rod lies along the x axis, to another inertial frame F', then reoriented by rotating from F' to another frame, call it F'', moving with the same velocity as F', such that the rod lies along the x'' axis? And you used a single prime symbol for values with respect to this rotated frame that I've called F'' as well as for values with respect to the merely boosted frame F'? If so, does delta x' in eq. (2) represent a different quantity from the same symbol in eq. (3)?

So far I've got

$$\textbf{r}'=\beta_x \Delta x\left ( \frac{\gamma-1}{\beta^2}-\gamma \right ) \pmb{\beta}+\begin{bmatrix}\Delta x\\0\\0\end{bmatrix}$$

Is the next step to rotate this space vector by some angle about an arbitrary spatial axis:

$$R_xR_yR_z\textbf{r}'=\textbf{r}''=\begin{bmatrix}\Delta x''\\0\\0\end{bmatrix}$$

How do we solve for the three angles, the arguments of $R_x$ etc., or is there another way to get the rotation matrix?

$$R_xR_yR_z=\begin{bmatrix} \textbf{i}'' \cdot \textbf{i}' & \textbf{i}'' \cdot \textbf{j}' & \textbf{i}'' \cdot \textbf{k}'\\ \textbf{j}'' \cdot \textbf{i}' & \textbf{j}'' \cdot \textbf{j}' & \textbf{j}'' \cdot \textbf{k}' \\ \textbf{k}'' \cdot \textbf{i}' & \textbf{k}'' \cdot \textbf{j}' & \textbf{k}'' \cdot \textbf{k}' \end{bmatrix}$$

where i'' etc. are the basis vectors of F'', and i' etc. those of F', both sets expressed wrt F', I suppose. I'm not sure where to go from here.

8. Jul 31, 2010

Rasalhague

But expressions containing terms with both x and y don't in general represent rotations. Is it possible to see from your derivation that this one does, or is that something which isn't obvious and would need a more detailed proof?

By pure, I meant a rotation only with no dilation: a transformation that doesn't change the lengths of space vectors.

9. Jul 31, 2010

starthaus

No, absolutely not. Don't mix the two files together, they are not connected at all.
One is about the length contraction/time dilation under one arbitrary boost.
The other is about the Thomas rotation under two consecutive boosts, one in x, the second one in y.

I simply put the attachments together , under the same caption, because this forum disallows opening as many subjects as you want. The two files aren't connected.

10. Jul 31, 2010

starthaus

Sure they do, here are the equations of rotation in 2D:

$$x'=x cos(\phi)+ysin (\phi)$$
$$y'=-x sin(\phi)+ycos (\phi)$$

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

11. Jul 31, 2010

Rasalhague

If not by rotating the axes, in what sense did you mean "reorient the system"? Notice that my wrong guess at what you meant does only involve one boost (from F to F'), followed by a rotation (from F' to F''). Is my first formula in #7 right, as far as it goes? If so, how do I get from that to your much simpler formula?

Ah, I see now. Thanks for your patience. (I thought you were objecting to the terminology, but you were actually objecting to the statement.)

Okay, I wrote this thinking that rotation was being used synonymously with pure rotation (as in your example here), a proper orthogonal (i.e. orthonormal) transformation, and as in the Wikipedia and Mathworld entries "Rotation matrix". I didn't realise that rotation, in this context, has a more general sense: a transformation that rotates space vectors and may also change their lengths.

Last edited: Jul 31, 2010
12. Jul 31, 2010

starthaus

Here is a very detailed treatment of the Thomas rotation.

13. Aug 1, 2010

Rasalhague

Thanks, starthaus. I'm reading it now. Meanwhile, just to clarify my question about your length-contraction attachment, I applied the general boost matrix to a spacetime vector with all components 0 except the t and x components. Then I imposed the condition that t' (the time component that results from the boost) be 0, and wrote the resulting space vector as

$$\textbf{r}'=\beta_x \Delta x\left ( \frac{\gamma-1}{\beta^2}-\gamma \right ) \pmb{\beta}+\begin{bmatrix}\Delta x\\0\\0\end{bmatrix}$$

where bold beta is the velocity of the boost, in units where c = 1, with respect to the frame boosted from, call that frame F. Beta subscript x is the x component of the velocity. And beta squared is the square of its magnitude. Gamma, as usual, is 1/sqrt(1-beta^2). Delta x is the x component of the original spacetime vector, and represents the length of a rod parallel to the x axis in frame F.

Unless I've made an arithmetic mistake--which is quite possible--this seems to show that the rod doesn't lie along the x' axis in frame F', but rather has nonzero x', y' and z' components in F'. At first glance, at least, this seems to agree with my sketchy and, as yet, only qualitative understanding of Thomas rotation; after all, if a boost along the x axis rotates an arbitrarily aligned rod, shouldn't a boost in an arbitrary direction rotate a rod aligned along the x axis? Aren't these just different ways of approaching the same question?

I would expect the length of the rod to be the magnitude of this space vector, r'. The calculation looks rather complicated though, and I haven't yet found a way to simplify it.

But I fear I may have completely misunderstood the concept because in your attachment on length contraction [ https://www.physicsforums.com/blog.php?b=1959 [Broken] ], in equation (3), you offer a much simpler-looking formula for the length of the rod in F', namely

$$\sum \Delta x'^2=\left ( 1+\frac{1-\gamma^2}{\gamma^2} \frac{\beta_x^2}{\beta^2} \right ) \Delta x$$

where the length is delta x'. But what is the significance of the summation sign on the left? Isn't there only one delta x'? Was I right to think there are other nonzero components? If not, can you see where I went wrong, and what is wrong with my reasoning above that there could, in general, be other nonzero components, by analogy with the case of an arbitrarily aligned rod and a boost in the x direction? If the boost does result in other nonzero components for r' (delta y' and delta z'), how were you able to ignore them; wouldn't they also contribute to the length of the rod?

You answered this last question by saying that we "can always reorient the system of axes associated with the rod such that the x axis is aligned with the rod" (i.e. in this case, the x' axis). By reorient, I thought you meant switch from F' to another frame moving at the same velocity and differing only by a rotation of axes. You said this was "absolutely not" what you meant. If you have time, could you try again to explain what you did mean? As I can't think what reorienting could mean here apart from rotating the axes, and as you went on to insist that there was only one boost involved (as if I'd suggested a second boost, rather than a boost followed by a rotation), I get the impression we may have been talking at cross purposes. But I'm pretty confused, so I could be way off the mark... Sorry to pester you with all these questions!

Last edited by a moderator: May 4, 2017
14. Aug 1, 2010

starthaus

$$\sum \Delta x'^2=\Delta x'^2+\Delta y'^2+\Delta z'^2$$

Last edited by a moderator: May 4, 2017
15. Aug 1, 2010

Rasalhague

Okay, but how did you simplify it? I've just had a go at checking my result in mathematica:

Code (Text):

In[1]:= L = {{g, -bx*g, -by*g, -bz*g}, {-bx*g,
1 + (g - 1) bx^2 b^(-2), (g - 1) bx*by*b^(-2), (g - 1) bx*bz*
b^(-2)}, {-by*g, (g - 1) by*bx*b^(-2),
1 + (g - 1) by^2*b^(-2), (g - 1) by*bz*b^(-2)}, {-bz*g, (g - 1) bz*
bx*b^(-2), (g - 1) bz*by*b^(-2), (g - 1) bz^2*b^(-2)}};

In[2]:= r = {t, x, 0, 0};

In[3]:= L.r

Out[3]:= {g t - bx g x, -bx g t + (1 + (bx^2 (-1 + g))/b^2) x, -by g t + (
bx by (-1 + g) x)/b^2, -bz g t + (bx bz (-1 + g) x)/b^2}

Setting the first component of this result to 0, I made the (I hope) appropriate substitutions to write the 3-vector consisting of the space components of L.r:

Code (Text):

In[4]:= p = {-bx^2*g*x + (1 + bx^2 (g - 1)*b^(-2))*x, -by*g*bx*x +
x*bx*by (g - 1)*b^(-2), -bz*g*bx*x + bx*bz (g - 1)*x*b^(-2)};

Examining this with the Simplify[^] command, e.g. Simplify[(p - {x, 0, 0})/(bx*x)], it looks like what I got by hand.

Code (Text):

In[5]:= Norm[p]

Out[5]:= \[Sqrt](Abs[(1 + (bx^2 (-1 + g))/b^2) x - bx^2 g x]^2 +
Abs[(bx by (-1 + g) x)/b^2 - bx by g x]^2 +
Abs[(bx bz (-1 + g) x)/b^2 - bx bz g x]^2)

In[6]:= Simplify[Norm[p]]

Out[6]:= \[Sqrt](Abs[(bx by (-1 + g - b^2 g) x)/b^2]^2 +
Abs[(bx bz (-1 + g - b^2 g) x)/b^2]^2 +
Abs[(1 + (bx^2 (-1 + g))/b^2 - bx^2 g) x]^2)

Is this equivalent to your much simpler looking result with no beta_y and beta_z terms? Apparently not:

Code (Text):

In[7]:= g = Sqrt[1 - b^2]; b = Sqrt[bx^2 + by^2 + bz^2]; b<1; b>0; TrueQ[
Norm[p]^2 == (x (1 + (1 - g^2)*g^(-2)*bx^2*b^(-2)))^2]

Out[7]:= False

...but I'm not sure if that TrueQ test works. When I just type the equation with the == sign, it just prints out the input without offering any truth value. I wonder if that's because of the possibility that one of the denominators might be zero.

Can you see if I'm making any false assumptions? Are there any extra assumptions that I should be making?

Last edited: Aug 1, 2010
16. Aug 1, 2010

starthaus

Correct

Correct.

Are you using the fact that $$\beta_y^2+\beta_z^2=\beta^2-\beta_x^2$$?

17. Aug 2, 2010

Rasalhague

Yes, in input line 7. As I suspect the truth-testing commands may not be working here, I tried it with a numerical example:

Code (Text):

In[8]:= bx = .5; by = .5; bz = .5; x = 1; p

Out[8]:= {0.708333, -0.291667, -0.291667}

In[9]:= Norm[p]

Out[9]:= 0.81968

In[10]:= Norm[p] == (x (1 + (1 - g^2)*g^(-2)*bx^2*b^(-2)))^2

Out[10]:= False

In[11]:= Sqrt[bx^2 + by^2 + bz^2] - b

Out[11]:= 0

In[12]:= (x (1 + (1 - g^2)*g^(-2)*bx^2*b^(-2)))^2

Out[12]: 4

Barring typos, or some fundamental misunderstanding on my part, I think this shows that your equation is not equivalent to the more complicated one I got.

Last edited: Aug 2, 2010
18. Aug 2, 2010

starthaus

You are right, I made an error in the last step in the reduction of terms (I do not use software packages like Mathematica, I do everything by hand). Must have been wishful thinking, I inadvertently replaced $$\gamma$$ with $$\gamma^2$$ in front of the last term and I got a simpler than correct expression. I have uploaded the corrected file.

19. Aug 2, 2010

Rasalhague

Looking at it now, I see in lines 10 and 12 I miscopied the expression on the RHS of your original eq. 3, and forgot to take the square root. Sorry about that. What I should have been comparing to Norm[p] is this

Code (Text):

In[13]:= x*Sqrt[1 + (1 - g^2)*g^(-2)*bx^2*b^(-2)]

Out[14]:= 1.41421

Recall from line 9 that, in this example, Norm[p] = 0.81968.

Your new version of eq. 3 gives

Code (Text):

In[14]:= x*Sqrt[1 + g^2 bx^2*b^(-2) (b^4 - bx^2) - g*bx^2*b^(-2) (b^2 - bx^2)]

Out[14]:= 0.970932

which is at least a contraction, but I still don't understand why you have no terms involving the y and z components of beta in your revised length contraction equation. When I try it in Mathematica with different values of by or bz, Norm[p] gives a different result.

Last edited: Aug 2, 2010
20. Aug 2, 2010

starthaus

Because you get a term in $$\beta_y^2+\beta_z^2=\beta^2-\beta_x^2$$, so the terms in y and z must go away.

Last edited: Aug 2, 2010
21. Aug 2, 2010

Rasalhague

Phew, I tracked down what the problem was with my numerical example, eventually. It was that I'd accidentally defined g as Sqrt[1-b^2] (as, weirdly, does Kevin Brown on that page about Thomas precession), rather than 1/Sqrt[1-b^2] to be consistent with the rest of the derivation. I also simplified my equation a bit by hand. Now, with p = x ({1, 0, 0} + bx*(Sqrt[1 - b^2] - 1)*b^(-2) {bx, by, bz}),

$$\textbf{p}=\textbf{r}'=\Delta x \left [ \begin{pmatrix}1\\0 \\0 \end{pmatrix} + \frac{\beta_x(-1+\sqrt{1-\beta^2})}{\beta^2} \begin{pmatrix}\beta_x\\ \beta_y \\ \beta_z \end{pmatrix} \right ]$$

it gives answers that only depend on the x component of beta, which--come to think of it--makes sense, since any component of velocity perpendicular to x should have no effect on the length of the rod, as shown by all those textbook thought experiments with rulers and paintbrushes and objects fitting through each other. I must have forgotten them in my confusion at Thomas rotation. Anyway, thanks again for all your help!

22. Aug 2, 2010

starthaus

Happens to the very best :-)

You are more than welcome. Try calculating the norm and compare it with mine.

23. Aug 2, 2010

Rasalhague

Here's my original version, called p, and my simplified version, called q. They agree with each other on the norm and all components. The norm doesn't depend on by or bz. (I see what you mean now about being able to replace by^2 + bz^2 with b^2 - bx^2.) Also, for all values of Sqrt[bx^2 + by^2 + bz^2] < 1 that I've tried, the norm is less than x.

Code (Text):

In[15]:= bx = .9; by = 0.3; bz = .2; x = 1; b = Sqrt[bx^2 + by^2 + bz^2]; q =
x ({1, 0, 0} + bx*(Sqrt[1 - b^2] - 1)*b^(-2) {bx, by, bz})

Out[15]:= {0.349371, -0.216876, -0.144584}

In[16]:= Norm[q]

Out[16]:= 0.43589

In[17]:= g = 1/Sqrt[
1 - b^2]; p = {-bx^2*g*x + (1 + bx^2 (g - 1)*b^(-2))*x, -by*g*bx*
x + x*bx*by (g - 1)*b^(-2), -bz*g*bx*x + bx*bz (g - 1)*x*b^(-2)}

Out[17]:= {0.349371, -0.216876, -0.144584}

In[18]:= Norm[p]

Out[18]:= 0.43589

Your current version, unless I've mistyped it...

Code (Text):

In[19]:= x*Sqrt[1 + g^2 bx^2 b^(-2) (b^4 - bx^2) - g*bx^2 b^(-2) (b^2 - bx^2)]

Out[19]:= 1.26479

This can give outputs greater than 1, as in this example. It doesn't agree with p and q. And it varies when I change by and bz, as my earlier attempts did.

I also tried deriving a simple expression for the norm by hand, but I haven't yet managed to simplify it much. I tried inputting a couple of versions of the still fairly complicated version I had, and got different and unrealistic answers each time. Conclusion: time for bed!

24. Aug 2, 2010

starthaus

The original answer I gave you was correct all along. The bottom line is that the formula simplifies to :

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

I'll go back to the original answer and show you how it was derived. I reverted to the original file with a few additional explanations.

Last edited: Aug 2, 2010
25. Aug 3, 2010

Rasalhague

Yeah, now that I've corrected the problem with g, your expression x*Sqrt[1 + 2 bx^2 b^(-2) ((1 - g)/g) + bx^2 b^(-2) ((1 - g)/g)^2] = x*Sqrt[1 - bx^2] = Norm[p] = Norm[q]. Sorry to make all this trouble! I haven't yet managed to work through the whole simplification the long way by hand without any errors, but this must be right.

This is the logic I was forgetting: Tipler & Mosca in Physics for Scientists and Engineers illustrate this with two rulers. One has two marker pens attached at certain points so that if the sticks are brought together at rest wrt each other, the markers on the one ruler would mark the other ruler at the corresponding points along its length. While at rest wrt each other, the rulers are placed parallel to each other and their ends lined up with each other. Then, without changing anything else about their relative position or alignment, the rulers move past each other at a constant velocity perpendicular to their length. If this perpendicular movement contributed to length contraction of one ruler in the rest frame of the other, it would, by symmetry, contribute the same amount of length contraction to the other in the rest frame of the first. But this would lead to a contradiction, since from the perspective of the rest frame of the ruler-with-pens, the other ruler wouldn't be marked if it was contracted enough to pass between the pens without touching them, whereas according to the rest frame of the ruler-without-pens, it would be marked, since the ruler-with-pens would be contracted. The marking of the rulers is a spacetime coincidence: either it happens or it doesn't, and it can't both happen and not happen. Therefore, perpendicular components of velocity don't contribute to length contraction.