Stone's derivation of Thomas rotation

[Rasalhague]

In[20]:= 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}); Norm[q]

Out:= 0.43589

In[21]:= 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)}; Norm[p]

Out[21]:= 0.43589

In[22]:= x*Sqrt[1 - bx^2]

Out[22]:= 0.43589

In[23]:= x*Sqrt[1 + 2 bx^2 b^(-2) ((1 - g)/g) + bx^2 b^(-2) ((1 - g)/g)^2]

Out[23]:= 0.43589

But

In[24]:= x*Sqrt[1 + bx^2/g^2]

Out[24]:= 1.02401

 

[Rasalhague]

You must have $$b_x^2+b_y^2+b_z^2=1$$, you don't have that in the above.

Surely the magnitude of beta has to be less than 1 = c, doesn't it? In this case, it's 0.969536.

 

[starthaus]

Surely the magnitude of beta has to be less than 1 = c, doesn't it? In this case, it's 0.969536.

Yes, my mistake. You still have something wrong in the Mathematica expression.

 

[Rasalhague]

You still have something wrong in the Mathematica expression.

What?

 

[starthaus]

What?

Your formula is fine, you are simply stopping before getting rid of $$\beta_y$$ and $$\beta_z$$. If you read the attachment, you will see how I did that.

 

[yuiop]

I just proved that.

I must have missed your proof. In all sincerity I would like to see it.

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

I think it is a question of semantics. I am saying that perpendicalur components of the velocity indirectly affect the length contraction. For example an object with Vx=0.6, Vy=0, Vz=0 will be length contracted by a factor of 0.8 in the x direction. An object with Vx=0.6, Vy=0.6, Vz=0 will be length contracted by a factor of about 0.529 and the orientation of the length contraction will no longer be parallel to the x axis. I suppose technically you could say there is no such thing as the "perpedicular components of the velocity" if we mean components perpendicular to the total resultant velocity vector, because by definition these components are always zero.

 

[starthaus]

I must have missed your proof. In all sincerity I would like to see it.

See my blog https://www.physicsforums.com/blog.php?b=1959 .

I think it is a question of semantics. I am saying that perpendicalur components of the velocity indirectly affect the length contraction.

This is incorrect since the proof shows clearly:

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

No dependency whatsoever of $$\beta_y$$ or $$\beta_z$$

This is not self-evident, it required some heavy lifting to prove.

For example an object with Vx=0.6, Vy=0, Vz=0 will be length contracted by a factor of 0.6 in the x direction.

This is not correct, the contraction is 0.8.

An object with Vx=0.6, Vy=0.6, Vz=0 will be length contracted by a factor of about 0.529 and the orientation of the length contraction will no longer be parallel to the x axis.

This is false as well, the correct contraction is 0.8. See the correct formula I gave above.

 

[Rasalhague]

The correct multiplication factor is :

$$\frac{1-\gamma}{\gamma}$$

Is this what you mean: s = x ({1, 0, 0} + g^(-1)*(1 - g) {bx, by, bz}) ? It doesn't work. Norm is generally different from that of the other expressions, including x*Sqrt[1 - bx^2], and the norm of s changes when I change the values of by or bz.

As you saw, Norm[q] gave the same answers as all of these other expressions, including x*Sqrt[1 - bx^2], and changing by or bz had no effect on it.

 

[starthaus]

Is this what you mean: s = x ({1, 0, 0} + g^(-1)*(1 - g) {bx, by, bz}) ?

No, this is not what I mean. You can find the correct expression in the blog attachment, I inserted a step between expression (2) and (3) specifically for your benefit. It isn't bx*by+by*bz+bz*bx, as you input into Mathematica, it is : bx^2+bx(by+bz)

It doesn't work. Norm is generally different from that of the other expressions, and the norm of s changes when I change the values of by or bz.

As you saw, Norm[q] gave the same answers as all of these other expressions, and changing by or bz had no effect on it.

Try working on the symbolic expressions, your approach using Mathematica is wrong. I am tired of figuring out what goes wrong in your derivation.

 

[Rasalhague]

This is false as well, the correct contraction is 0.8. See the correct formula I gave above.

With velocity all in x direction, magnitude 0.6, I get a contractrion of 0.8. Wolfram Alpha agrees. This with my supposedly wrong method which so far has always agreed exactly with your formula x*Sqrt[1 - bx^2]!

 

[starthaus]

With velocity all in x direction, magnitude 0.6, I get a contractrion of 0.8. Wolfram Alpha agrees. This with my supposedly wrong method which so far has always agreed exactly with your formula x*Sqrt[1 - bx^2]!

Work on the symbolic formulas. You have all the information in the attachment. I don't trust your verification via Mathematica because it suffers from errors, if you input the incorrect thing, don't be surprised to get the wrong output.

 

[Rasalhague]

Work on the symbolic formulas. You have all the information in the attachment. I don't trust your verification via Mathematica because it suffers from errors, if you input the incorrect thing, don't be surprised to get the wrong output.

Your claim that it's wrong would be more convincing if you could come up with an example where Norm[q] gives a different answer from Norm[p] and x*Sqrt[1 - bx^2]. With their current definitions, so far, they've all agreed exactly. While that doesn't prove they're identical, it does make it unlikely that they're not. All of them are independent of by and bz. It seems that like Mathematica, we humans are also struggling to establish the identity of these expressions symbolically.

 

[yuiop]

This is incorrect since the proof shows clearly:

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

No dependency whatsoever of $$\beta_y$$ or $$\beta_z$$

This is not self-evident, it required some heavy lifting to prove.

This is only true if $$\beta_y=0$$ and $$\beta_z=0$$. This is easy to demonstrate. If $$\beta_x=0$$ and $$\beta_y=0.8$$ then the length contraction is 0.6 and dependent on $$\beta_y$$ and is orientated in the direction of the y axis.

This is not correct, the contraction is 0.8.

That was a typo that did not really affect the argument. I started with vx=0.8 and length contraction =0.6 and later changed vx to 0.6 because the resulant velocity of vx=0.8, vy=0.8 was greater than 1.0 so I changed the velocities to 0.6 and overlooked changing the contraction to 0.8. This does not change the fact that the all the velocity components contribute to the direction and magnitude of the total velocity and therefore they all contribute to the magnitude and direction of the length contraction.

 

[Rasalhague]

starthaus's example only deals with the contraction of a space vector having no y or z component. Could that be the source of the confusion? If instead of a rod aligned along the x axis, imagined as having no thickness, we had an object with spatial extent in all directions, then (if I've understood this) velocity components along the y and z axes would affect the extent of the object along those axes. I think that's the situation you're describing, kev, isn't it?

 

[yuiop]

This is incorrect since the proof shows clearly:

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

No dependency whatsoever of $$\beta_y$$ or $$\beta_z$$

This is not self-evident, it required some heavy lifting to prove.

I think in the end you will find the general solution for the total length contraction is given by:

$$L' \ = \ L\sqrt{1-(\beta_x^2+\beta_y^2+\beta_z^2)}\ =\ L\sqrt{1-\beta^2}$$

 

[yuiop]

starthaus's example only deals with the contraction of a space vector having no y or z component. Could that be the source of the confusion? If instead of a rod aligned along the x axis, imagined as having no thickness, we had an object with spatial extent in all directions, then (if I've understood this) velocity components along the y and z axes would affect the extent of the object along those axes. I think that's the situation you're describing, kev, isn't it?

Yes, I was talking about the more general solution, which is what I thought you guys were looking for.

If you wish to break the total length contraction down into its components then you get:

$$L'_x \ =\ L_x \sqrt{1-\beta^2_x}$$

$$L'_y \ =\ L_y \sqrt{1-\beta^2_y}$$

$$L'_z \ =\ L_z \sqrt{1-\beta^2_z}$$

and the total length contraction is:

$$L' \ = \sqrt{L'_x^2 + L'_y^2 + L'_z^2} = \sqrt{(L_x^2 + L_y^2 + L_z^2)(1-\beta_x^2-\beta_y^2-\beta_z^2)} =\ L \sqrt{1-\beta^2}$$

so yes, maybe we are at cross purposes and maybe my fault for not reading all the thread.

As for the proof that spatial components orthogonal to the motion are not length contracted, Starthaus starts with the Lorentz transformations which explicitly assume that in the first place. There are other possible formulations of the transformations that allow length contraction in the transverse plane that are consistent with a constant speed of light and MM experiment etc. and they had to be ruled out using the logic of rulers and markers that you gave or considering rings moving past each other. If the radius of a moving orthogonal ring changed, then you could have have ring A passing inside ring B in one frame and ring B passing inside ring A in another frame which is physically impossible.

 

[starthaus]

This is only true if $$\beta_y=0$$ and $$\beta_z=0$$.

This is false, you have not read the proof, it assumes non null $$v_y$$ and non-null $$v_z$$

 

[yuiop]

This is false, you have not read the proof, it assumes non null $$v_y$$ and non-null $$v_z$$

I think where we differ is that I am defining

$$L' \ = \sqrt{L'_x^2 + L'_y^2 + L'_z^2}$$

while you are defining L' as $L'_x$ and the confusion comes about because you have not made your definition of L' clear.

 

[starthaus]

Yes, I was talking about the more general solution, which is what I thought you guys were looking for.

If you wish to break the total length contraction down into its components then you get:

$$L'_x \ =\ L_x \sqrt{1-\beta^2_x}$$

$$L'_y \ =\ L_y \sqrt{1-\beta^2_y}$$

$$L'_z \ =\ L_z \sqrt{1-\beta^2_z}$$

and the total length contraction is:

$$L' \ = \sqrt{L'_x^2 + L'_y^2 + L'_z^2} = \sqrt{(L_x^2 + L_y^2 + L_z^2)(1-\beta_x^2-\beta_y^2-\beta_z^2)} =\ L \sqrt{1-\beta^2}$$

so yes, maybe we are at cross purposes and maybe my fault for not reading all the thread.

Err, the correct math would say that your final formula is incorrect. The error is just glaring.

 

[Rasalhague]

I think in the end you will find the general solution for the total length contraction is given by:

$$L' \ = \ L\sqrt{1-(\beta_x^2+\beta_y^2+\beta_z^2)}\ =\ L\sqrt{1-\beta^2}$$

Let $\textbf{r}=\left ( x,y,z \right )$, where y and z are not necessarily equal to zero. Then I get

$$\left \| \textbf{r}' \right \|=\sqrt{\textbf{r} \cdot \textbf{r}-(\textbf{r}\cdot\pmb{\beta})^2}$$

$$=\sqrt{\left \| \textbf{r} \right \|^2-x^2\beta_x^2-y^2 \beta_y^2-z^2\beta_z^2}$$

This agrees with Norm[p] and Norm[q], as previously defined, but gives a different answers to Norm[r]*Sqrt[1 - bx^2 - by^2 - bz^2] = Norm[r]*Sqrt[1 - b^2] in the special case where y = z = 0, and, in this case, kev's formula Norm[r]*Sqrt[1 - b^2] does depend on by and bz, which I think we agree would lead to contradictions.

 

[starthaus]

In[20]:= 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}); Norm[q]

Out:= 0.43589

In[21]:= 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)}; Norm[p]

Out[21]:= 0.43589

In[22]:= x*Sqrt[1 - bx^2]

Out[22]:= 0.43589

In[23]:= x*Sqrt[1 + 2 bx^2 b^(-2) ((1 - g)/g) + bx^2 b^(-2) ((1 - g)/g)^2]

Out[23]:= 0.43589

But

In[24]:= x*Sqrt[1 + bx^2/g^2]

Out[24]:= 1.02401

I re-read this post, it makes no sense. Where do you think I use anything remotely similar to:

x*Sqrt[1 + bx^2/g^2]

?

The correct formula is :

x*Sqrt[1-bx^2]

No wonder you don't get the outputs to agree.

 

[starthaus]

Let $\textbf{r}=\left ( x,y,z \right )$, where y and z are not necessarily equal to zero. Then I get

$$\left \| \textbf{r}' \right \|=\sqrt{\textbf{r} \cdot \textbf{r}-(\textbf{r}\cdot\pmb{\beta})^2}$$

$$=\sqrt{\left \| \textbf{r} \right \|^2-x^2\beta_x^2-y^2 \beta_y^2-z^2\beta_z^2}$$

This agrees with Norm[p] and Norm[q], as previously defined, but gives a different answers to Norm[r]*Sqrt[1 - bx^2 - by^2 - bz^2] = Norm[r]*Sqrt[1 - b^2] in the special case where y = z = 0, and, in this case, kev's formula Norm[r]*Sqrt[1 - b^2] does depend on by and bz, which I think we agree would lead to contradictions.

Your formula is correct while kev's is obviously marred by an elementary algebraic mistake.

Now, having said that, there is absolutely no reason for attempting to add up the contracted dimensions the way kev did. So he compunded the algebraic mistake with a physics mistake. The length contraction of a rod with arbitrary orientation needs to be derived from basic principles. This is not what kev did.

 

[starthaus]

As for the proof that spatial components orthogonal to the motion are not length contracted, Starthaus starts with the Lorentz transformations which explicitly assume that in the first place.

This is also false. The general Lorentz transforms do not "assume" any such thing.

 

[Rasalhague]

I re-read this post, it makes no sense. Where do you think I use anything remotely similar to:

x*Sqrt[1 + bx^2/g^2]

?

That was kev's erroneous formula from #29.

 

[starthaus]

That was kev's erroneous formula from #29.

Arrgh, I see. So he made both of us waste time. We are good.

 

[yuiop]

That was kev's erroneous formula from #29.

I had already admited and corrected that error in #33, so I am not sure why you were still harping on about it in #36. Looking back through the thread I see you made a very similar error earlier.

If you wish to break the total length contraction down into its components then you get:

$$L'_x \ =\ L_x \sqrt{1-\beta^2_x}$$

$$L'_y \ =\ L_y \sqrt{1-\beta^2_y}$$

$$L'_z \ =\ L_z \sqrt{1-\beta^2_z}$$

and the total length contraction is:

$$L' \ = \sqrt{L'_x^2 + L'_y^2 + L'_z^2} = \sqrt{(L_x^2 + L_y^2 + L_z^2)(1-\beta_x^2-\beta_y^2-\beta_z^2)} =\ L \sqrt{1-\beta^2}$$

OK, I have to admit another error in the result above. The final expression is only true if the one dimensional rod is orientated parallel to the motion. Obviously not my week. It should have been:

$$L' \ = \|L'\| = \sqrt{L'_x^2 + L'_y^2 + L'_z^2} \, = \sqrt{L_x^2(1-\beta_x^2) + L_y^2(1-\beta_y^2) + L_z^2(1-\beta_z^2)} \, = \sqrt{\|L\|^2 -L_x^2\beta_x^2 -L_y^2\beta_y^2-L_z^2\beta_z^2}$$

which is the equation given by Rasalhague earlier.

Back to the main subject of the thread, the above general equation shows that if a rod is not parallel to the motion, the length contraction of only the components parallel to the motion causes an effective rotation of the rod orientation in the boosted frame relative to the rest frame. For example, if the rod has Lx=1, Ly=1, Lz=0 in the rest frame so that it is orientated at 45 degrees to the y-axis in frame S and if it is boosted in the y direction by 0.8c, then it will have Lx'=1, Ly'=0.6, Lz=0 and will be orientated at 90-atan(0.6)*180/pi = aprox 59 degrees to the y' axis in frame S'.

This is incorrect since the proof shows clearly:

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

No dependency whatsoever of $$\beta_y$$ or $$\beta_z$$

I still contend that the above equation is at best misleading.

It implies that:

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

which is wrong. It should be either:

$$\|L'\| = \sqrt{\|L\|^2 -L_x^2\beta_x^2 -L_y^2\beta_y^2-L_z^2\beta_z^2}$$

or:

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

 

[starthaus]

I had already admited and corrected that error in #33, so I am not sure why you were still harping on about it in #36. Looking back through the thread I see you made a very similar error earlier.
OK, I have to admit another error in the result above. The final expression is only true if the one dimensional rod is orientated parallel to the motion. Obviously not my week. It should have been:

$$L' \ = \|L'\| = \sqrt{L'_x^2 + L'_y^2 + L'_z^2} \, = \sqrt{L_x^2(1-\beta_x^2) + L_y^2(1-\beta_y^2) + L_z^2(1-\beta_z^2)} \, = \sqrt{\|L\| -L_x^2\beta_x^2 -L_y^2\beta_y^2-L_z^2\beta_z^2}$$

which is the equation given by Rasalhague earlier.

While mathematically correct, the above is not the correct answer to the problem, you cannot blindly add up $$L'_x^2 + L'_y^2 + L'_z^2$$. You need to figure out how to solve this problem correctly. The attachment in my blog gives you the blueprint how to get the correct solution, you need to go back, read it and understand it. Ask questions and I'll give you hints how to solve the more complicated situation when $$\Delta y \ne 0$$ and $$\Delta z \ne 0$$. The way you are trying to hack it is not correct.

I still contend that the above equation is at best misleading.

It implies that:

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

which is wrong. It should be either:

$$\|L'\| = \sqrt{\|L\| -L_x^2\beta_x^2 -L_y^2\beta_y^2-L_z^2\beta_z^2}$$

Are you guessing again? Because , if you are, you are guessing wrong.

 

[yuiop]

... if a rod is not parallel to the motion, the length contraction of only the components parallel to the motion causes an effective rotation of the rod orientation in the boosted frame relative to the rest frame.

The above statement is as far as I can tell, an undeniable physical fact. Unfortunately it does not show up in any of the equations we have produced so far. I think one problem is that when two frames S' and S are moving relative to each other in a direction that is not parallel to one of the main axes, then the axes themselves ar no longer parallel to each other and rotate in the same way as the rod. Imagine that the observers in frame S construct a large physical grid made up up of rods welded at right angles to each other that label x, y and z. When another observer in frame S' is moving in a direction not parallel to x, y or z then the angles of the physical grid S do not appear to appear to be at right angles to each other according to frame S'.

This rotation of the rod due to motion not parallel or orthogonal to the rod can be alternatively explained by two sequential boosts at right angles to each other where the rotation comes about due to a difference in simultaneity during the second boost. I suspect the two explanations are just two facets of the same phenomena. I prefer the original one because it is does not require a two stage boost, when in nature both boosts might occur simultaneously.

The calculations are bit involved but this is my initial attempt, considering motion only in the x and y directions.

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

This length contraction of the length of the rod itself (without regard to any particular axis) and it turns out that the above equation is equivalent to:

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

which is probably what Starthaus was trying to say, but what he did not realize is that the rod is no longer parallel to the x' axis in the S' frame.

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.

I have done these calculations off the top of my head, so they might well contain errors. Does anyone know if they look like anything in any of the references?

 

[Rasalhague]

I had already admited and corrected that error in #33, so I am not sure why you were still harping on about it in #36. Looking back through the thread I see you made a very similar error earlier.

No harping intended, sorry if it sounded that way! I suspect I hadn't seen #33 yet when I posted #36. Even if I had, given that all three of us have made mistakes with this (although starthaus's mistake was to believe one of mine!), my instinct now is to test with numerical examples: both what we think is right and what we think is wrong. And yes, indeed, I did make the same mistake earlier.

 

[yuiop]

No harping intended, sorry if it sounded that way! I suspect I hadn't seen #33 yet when I posted #36. Even if I had, given that all three of us have made mistakes with this (although starthaus's mistake was to believe one of mine!), my instinct now is to test with numerical examples: both what we think is right and what we think is wrong. And yes, indeed, I did make the same mistake earlier.

No problem I am interested what you think about my new approach in the last post. Promising, or just a whole new can of worms?

Basically, I am splitting the rod into components parallel and orthogonal to the motion and only length contracting the component parallel to the motion by a gamma factor that is a function of the Euclidean velocity norm and then piecing it all back together again.

P.S. Yes, agree about numerical testing. I should do it more often. However, my claim that the trigometric expression for the total length contraction is the same as the $$\| L' \| \ = L_x \sqrt{1-\beta_x^2}$$ is based on numerical testing and not symbolically derived, so there is room for error there too.

 

[yuiop]

... 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...

Here is how to proceed.

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

For By=0 and Bz=0 the above reduces to the familiar:

$$\begin{bmatrix}x'\\y'\\z' \end{bmatrix}= \Delta x \begin{bmatrix} \sqrt{1-\beta^2} \\ 0\\ 0 \end{bmatrix} \right ]$$

but for non zero values of By and Bz, the y' and z' coordinates are not zero and the rod is no longer aligned with the x' axis.
This the rotation effect.

With a bit of luck it should agree with the trigometric rotation I gave in the earlier post but I have not checked it yet.

Now the Euclidean norm is obtained in the normal way from the squared coordinates and for simplicity one end of the rod is considered to be at the origin of S.

$$\|L' \| = Lx\sqrt{ \left[1+b_x^2\ b^{-2} (-1+\sqrt{1-b^2})\right]^2 + \left[b_x\ b_y\ b^{-2}(-1+\sqrt{1-b^2})\right]^2 + \left[b_x\ b_z\ b^{-2} (-1+\sqrt{1-b^2})\right]^2 }$$

Now defining g = (1-b^2) = 1/gamma

$$\|L' \| = Lx\sqrt{ \left[1+b_x^2\ b^{-2} (g-1)\right]^2 + \left[b_x\ b_y\ b^{-2}(g-1)\right]^2 + \left[b_x\ b_z\ b^{-2} (g-1)\right]^2 }$$

$$\|L' \| = Lx\sqrt{ 1+2b_x^2\ b^{-2}(g-1) + b_x^4\ b^{-4} (g-1)^2 + b_x^2\ b_y^2\ b^{-4}(g-1)^2 + b_x^2\ b_z^2\ b^{-4} (g-1)^2 }$$

$$\|L' \| = Lx\sqrt{ 1+2b_x^2\ b^{-2}(g-1) + b_x^2\ b^{-4} ( b_x^2+ b_y^2+ b_z^2)(g-1)^2 }$$

$$\|L' \| = Lx\sqrt{ 1+2b_x^2\ b^{-2}(g-1) + b_x^2\ b^{-2} (g-1)^2 }$$

$$\|L' \| = Lx\sqrt{ 1+2b_x^2\ b^{-2}g - 2b_x^2\ b^{-2} + b_x^2\ b^{-2}g^2- 2 b_x^2\ b^{-2}g + b_x^2\ b^{-2} }$$

$$\|L' \| = Lx\sqrt{ 1 - b_x^2\ b^{-2} + b_x^2\ b^{-2}g^2 }$$

$$\|L' \| = Lx\sqrt{ 1 - b_x^2\ b^{-2} (1-g^2) }$$

$$\|L' \| = Lx\sqrt{ 1 - b_x^2}$$

 

[Rasalhague]

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

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

where

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

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

This length contraction of the length of the rod itself (without regard to any particular axis) and it turns out that the above equation is equivalent to:

$$\| L\| \ = L_x \sqrt{1-\beta_x^2)$$

In[1]:= bx = .9; by = .3; x = 1; x*Sqrt[1 - bx^2]

Out[1]:= 0.43589

In[2]: th = ArcTan[by/bx]; x*
 Sqrt[Sin[th]^2 + Cos[th]^2*Sqrt[1 - bx^2 - by^2]]

Out[2]:= 0.620165

In[3]:= by = .1; Sqrt[1 - bx^2]

Out[3]:= 0.43589

In[4]:= th = ArcTan[by/bx]; x*
 Sqrt[Sin[th]^2 + Cos[th]^2*Sqrt[1 - bx^2 - by^2]]

Out[4]:= 0.656723

 

[Rasalhague]

Here is how to proceed.

Brilliant! Thanks for that. I was having no end of trouble getting through all those various powers of beta and gamma. I must have copied at least one thing wrong every time I tried it.

 

[yuiop]

Brilliant! Thanks for that. I was having no end of trouble getting through all those various powers of beta and gamma. I must have copied at least one thing wrong every time I tried it.

Your welcome

In[1]:= bx = .9; by = .3; x = 1; x*Sqrt[1 - bx^2]

Out[1]:= 0.43589

In[2]: th = ArcTan[by/bx]; x*
 Sqrt[Sin[th]^2 + Cos[th]^2*Sqrt[1 - bx^2 - by^2]]

Out[2]:= 0.620165

In[3]:= by = .1; Sqrt[1 - bx^2]

Out[3]:= 0.43589

In[4]:= th = ArcTan[by/bx]; x*
 Sqrt[Sin[th]^2 + Cos[th]^2*Sqrt[1 - bx^2 - by^2]]

Out[4]:= 0.656723

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.

That should work now. I will have to check the error has not propagated elsewhere.

P.S. I have taken the liberty of editing and correcting the original equation in #63 to reflect your correction.

 

[Rasalhague]

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 $\textbf{r}'$.