Stone's derivation of Thomas rotation

[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" $(1+q)^2(1-(u_1/c)^2)$?

 

[starthaus]

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?

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.

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

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 is a much better derivation (the third attachment)

 

[Rasalhague]

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.

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).

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 is a much better derivation (the third attachment)

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?

 

[starthaus]

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?

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

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).

Yes.

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

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?

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.

 

[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?

 

[starthaus]

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?

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.

 

[Rasalhague]

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.

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.

 

[Rasalhague]

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

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.

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.

 

[starthaus]

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?

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.

 

[starthaus]

But expressions containing terms with both x and y don't in general represent rotations. .

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

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

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?

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

 

[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?

You can always reorient the system of axes assocaited with the rod such that the x-axis is aligned with the rod.

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'?

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.

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?

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

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

But expressions containing terms with both x and y don't in general represent rotations.

Sure they do,

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 realize that rotation, in this context, has a more general sense: a transformation that rotates space vectors and may also change their lengths.

 

[starthaus]

Here is a very detailed treatment of the Thomas rotation.

 

[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 ], 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!

 

[starthaus]

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 ], 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?

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

 

[Rasalhague]

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

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:

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.

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:

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?

 

[starthaus]

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

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)}};

Correct

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:

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)};

Correct.

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

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?

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

 

[Rasalhague]

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

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

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.

 

[starthaus]

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

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.

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.

 

[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

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

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.

 

[starthaus]

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.

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.

 

[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!

 

[starthaus]

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.

Happens to the very best :-)

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!

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

 

[Rasalhague]

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

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.

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...

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!

 

[starthaus]

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!

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.

 

[Rasalhague]

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.

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.

 

[starthaus]

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.

Can Mathematica handle this? I am curious since I do not use it so I'd like to see if it can equal the human skills.

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.

I do not trust this type of "derivations", I much prefer the mathematical rigor. Indeed, my proof shows without doubt that there is no length contraction in the directions transverse to motion. I find Tipler extremely weak. My preferred book on relativity is "Relativity" by C.Moller, he is extremely strong on rigorous mathematical proofs.

 

[Rasalhague]

Can Mathematica handle this? I am curious since I do not use it so I'd like to see if it can equal the human skills.

There is a command Simplify[^], but it hasn't been any help in this case. Which is a shame... But I'm new to Mathematica, so there might be some other way I don't know about. If not, I'd be curious to know if there is any software that could manage it. For now though, your human skills are unmatched. Mine are another story ;-)

What I was able to do with it was to check these various formulas numerically for a range of input values. Not a rigorous proof of equality, but enough to make it unlikely they were producing the same results by chance. And it quickly showed where formulas weren't equal and which ones depended on by and bz.

 

[starthaus]

There is a command Simplify[^], but it hasn't been any help in this case. Which is a shame... But I'm new to Mathematica, so there might be some other way I don't know about. If not, I'd be curious to know if there is any software that could manage it. For now though, your human skills are unmatched. Mine are another story ;-)

What I was able to do with it was to check these various formulas numerically for a range of input values. Not a rigorous proof of equality, but enough to make it unlikely they were producing the same results by chance. And it quickly showed where formulas weren't equal and which ones depended on by and bz.

Ok, I don't think that Mathematica will "see" the obvious:

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

One day the machines will beat us but the day is not here yet :-)

 

[yuiop]

Ok, I don't think that Mathematica will "see" the obvious:

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

One day the machines will beat us but the day is not here yet :-)

Are you sure about this?

The first part:

$$2\frac{1-\gamma}{\gamma}+(\frac{1-\gamma}{\gamma})^2=\frac{1-\gamma^2}{\gamma^2}$$

seems OK, but your algebra seems to go wrong after that.

I get:

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

Maybe that is why it does not seem obvious to Mathematica?

If I am right, this means the expression x*Sqrt[1 + 2 bx^2 b^(-2) ((1 - g)/g) + bx^2 b^(-2) ((1 - g)/g)^2]
should simplify to x*Sqrt[1+bx^2/g^2].

 

[Rasalhague]

Are you sure about this?

The first part:

$$2\frac{1-\gamma}{\gamma}+(\frac{1-\gamma}{\gamma})^2=\frac{1-\gamma^2}{\gamma^2}$$

seems OK, but your algebra seems to go wrong after that.

I get:

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

I get the same as starthaus. I've been using the definition gamma = 1/Sqrt[1-beta^2], rather than gamma = Sqrt[1-beta^2].

 

[yuiop]

Therefore, perpendicular components of velocity don't contribute to length contraction.

I think you meant to say something like spatial components that are perpendicular to the velocity are not subject to length contraction. Perpendicular components of the velocity contribute to the magnitude and direction of the total velocity vector and therefore contribute to the magnitude and orientation of the length contraction.

 

[starthaus]

Are you sure about this?

The first part:

$$2\frac{1-\gamma}{\gamma}+(\frac{1-\gamma}{\gamma})^2=\frac{1-\gamma^2}{\gamma^2}$$

seems OK, but your algebra seems to go wrong after that.

Sorry, "my" algebra is correct.

I get:

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

Maybe that is why it does not seem obvious to Mathematica?

If I am right, this means the expression x*Sqrt[1 + 2 bx^2 b^(-2) ((1 - g)/g) + bx^2 b^(-2) ((1 - g)/g)^2]
should simplify to x*Sqrt[1+bx^2/g^2].

Err , you got it all wrong , try again with the correct expression for $$\gamma$$ as a function of $$\beta$$.

 

[yuiop]

I get the same as starthaus. I've been using the definition gamma = 1/Sqrt[1-beta^2], rather than gamma = Sqrt[1-beta^2].

Oops, your right. I fell into the trap of using gamma = Sqrt[1-beta^2]. Using the correct definition of gamma, the expression does simplify to -b^2 as you both got.

 

[starthaus]

I get the same as starthaus. I've been using the definition gamma = 1/Sqrt[1-beta^2], rather than gamma = Sqrt[1-beta^2].

Correct.

 

[starthaus]

I think you meant to say something like spatial components that are perpendicular to the velocity are not subject to length contraction.

I just proved that.

Perpendicular components of the velocity contribute to the magnitude and direction of the total velocity vector

Yes

and therefore contribute to the magnitude and orientation of the length contraction.

You are contradicting yourself, you got it right in the first sentence, now you are getting it wrong.