Stone's derivation of Thomas rotation

[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&#039; \ = \|L&#039;\| = \sqrt{L&#039;_x^2 + L&#039;_y^2 + L&#039;_z^2} \,<br /> <br /> = \sqrt{L_x^2(1-\beta_x^2) + L_y^2(1-\beta_y^2) + L_z^2(1-\beta_z^2)} \,<br /> <br /> = \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&#039;=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&#039;\| = \|L\|\sqrt{1-\beta_x^2}

which is wrong. It should be either:

\|L&#039;\| = \sqrt{\|L\|^2 -L_x^2\beta_x^2 -L_y^2\beta_y^2-L_z^2\beta_z^2}

or:

L&#039;_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&#039; \ = \|L&#039;\| = \sqrt{L&#039;_x^2 + L&#039;_y^2 + L&#039;_z^2} \,<br /> <br /> = \sqrt{L_x^2(1-\beta_x^2) + L_y^2(1-\beta_y^2) + L_z^2(1-\beta_z^2)} \,<br /> <br /> = \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&#039;_x^2 + L&#039;_y^2 + L&#039;_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&#039;\| = \|L\|\sqrt{1-\beta_x^2}

which is wrong. It should be either:

\|L&#039;\| = \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&#039; \| \ = 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&#039;\| \ = 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&#039; \| 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&#039;_x = \| L&#039; \| \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&#039; \| \ = 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}&#039;=\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&#039;\\y&#039;\\z&#039; \end{bmatrix}=\textbf{p}=\textbf{r}&#039;=\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 ] =<br /> \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&#039;\\y&#039;\\z&#039; \end{bmatrix}=<br /> \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&#039; \| = 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&#039; \| = 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&#039; \| = 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&#039; \| = 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&#039; \| = Lx\sqrt{ 1+2b_x^2\ b^{-2}(g-1) + b_x^2\ b^{-2} (g-1)^2 }

\|L&#039; \| = 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&#039; \| = Lx\sqrt{ 1 - b_x^2\ b^{-2} + b_x^2\ b^{-2}g^2 }

\|L&#039; \| = Lx\sqrt{ 1 - b_x^2\ b^{-2} (1-g^2) }

\|L&#039; \| = 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&#039; \| \ = 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&#039; \| \ = 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}&#039;=\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}&#039;)=\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}&#039;) is the angle between the original space vector \textbf{r} and its boosted counterpart \textbf{r}&#039;.

 

[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}&#039;=\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}&#039;)=\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}&#039;) is the angle between the original space vector \textbf{r} and its boosted counterpart \textbf{r}&#039;.

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}&#039;=\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&#039; \| 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&#039;_x = \| L&#039; \| \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}&#039;=\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}&#039;=\textbf{r}\cdot \left [ \textbf{r}-\left ( 1-\frac{1}{\gamma} \right )\frac{\textbf{r}\cdot \pmb{\beta}}{\beta^2} \pmb{\beta}\right ]

rr&#039;\cos(\textbf{r},\textbf{r}&#039;)=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}&#039;)=\frac{r^2-\left ( 1-\gamma^{-1} \right )r^2 \beta^2 \beta^{-2} \cos^2(\textbf{r},\pmb{\beta})}{rr&#039;}

\cos(\textbf{r},\textbf{r}&#039;)=\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}&#039;)=\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}&#039;)=\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&#039; \| \ = 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&#039; \| \ = 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&#039;\| \ = 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&#039;\| \ = 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 &#039;\| \ = 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 &#039;\| \ = 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&#039; 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.