Trouble with 2 step velocity transformation in SR

[MikeLizzi]

Homework Statement

Given:
An object at rest with respect to an inertial reference frame S.
2 other inertial reference frames S' and S''.
S' has velocity (vx, vy) = (-.6c, 0) with respect to S.
S'' has velocity (vx, vy) = (-.6c, +.6c) with respect to S.

Assumptions:
If I transform my observations from S to S', the object's velocity with respect to S' is (vx', vy') = (+.6c, 0).

If instead I transform my observations from S to S'', the object's velocity with respect to S'' is (vx'', vy'') = (+.6c, -.6c).

Task:
Determine the object's velocity with respect to S'' by means of 2 sequential reference frame transformations, S to S' and then S' to S''.

Homework Equations

For velocity transformations when more than one dimension is involved, I do a velocity 4-vector transformation.
$$\begin{bmatrix}
\gamma_{u'} \\
\gamma_{u'} u'_x \\
\gamma_{u'} u'_y \\
\gamma_{u'} u'_z
\end{bmatrix}
=
\begin{bmatrix}
\gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\
-\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\
-\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\
-\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2}
\end{bmatrix}
\begin{bmatrix}
\gamma_{u} \\
\gamma_{u} u_x \\
\gamma_{u} u_y \\
\gamma_{u} u_z
\end{bmatrix}$$
Where v is the velocity of reference frame S' with respect to reference frame S and
$$\beta_x = v_x/c$$
$$\beta_y = v_y/c$$
$$\beta_y = v_y/c$$
$$\beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2$$
$$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$
$$\gamma_u = \frac{1}{\sqrt{1-\beta_u^2}}$$
$$\gamma_{u'} = \frac{1}{\sqrt{1-\beta_{u'}^2}}$$

The Attempt at a Solution

For S to S' I already have the velocity for the object as stated previously (vx', vy') = (+.6c, 0).

Now I want to transform from S' to S''. I need the velocity of S'' with respect to S'. It is the relativistic velocity difference between the two. That’s not (0, +.6c). Using the formulas presented in Latex above I calculate it as (0, +.75c).

Then the velocity of the object with respect to S'' becomes the relativistic velocity difference between (+.6c, 0) and (0, +.75c) Using the formulas presented in Latex above I calculate that to be (+.34c, -.75c).

So the velocity of the object is (+.6c, -.6c), magnitude = .848 with respect to S'' using a one step process
and (+.34c, -.75c) magnitude .848 with respect to S'' using a two step process.

If the calculations are correct then I am forced to conclude that the reference frame I labeled S'' in the one step process is not the same reference frame as the one I labeled S'' in the 2 step process. They have the same origin, are at rest with respect to each other, but one is rotated with respect to the other.

So here are my questions.
Even after spending hours checking my work, do I have a mistake in my calculations? Have I misunderstood the meaning of the calculations? Was gibt? (That's German for “What gives?”)

P.S.
I am working a more complicated version of this problem in another forum and a poster has suggested I have run into Thomas Precession.

 

[Charles Link]

My expertise with working problems in special relativity is quite limited, but if you can write the matrix equations, $S'=(Q)S$ and $S^{''}=(R)S$, you should be able to find the matrix $U$ such that $S^{''}=(U)S'$. That matrix is $U=(R)(Q^{-1})$.

 

[MikeLizzi]

My expertise with working problems in special relativity is quite limited, but if you can write the matrix equations, $S'=(Q)S$ and $S^{''}=(R)S$, you should be able to find the matrix $U$ such that $S^{''}=(U)S'$. That matrix is $U=(R)(Q^{-1})$.

I'm sorry I don't understand what you are saying. I provided the relevant transform matrix. It's a straightforward process to populate it for a specific transformation. Do you want to see them?

 

[PeroK]

Even after spending hours checking my work, do I have a mistake in my calculations? Have I misunderstood the meaning of the calculations? Was gibt? (That's German for “What gives?”)

Perhaps a better translation is "what's going on". But, anyway, it's not clear to me what you have calculated.

1) You know that the answer should be $(0.6c, -0.6c)$.

2) If you use the 2D Lorentz transformation, you must get symmetry in $x$ and $y$ from the data. It's not clear what you have done there. Also, that is an alternative one-step calculation.

PS although that gives you the four-velocity, which you would have to convert to a three-velocity.

3) You haven't shown your working for the two-step calculation, which should involve a Lorentz boost in the x-direction followed by one in the y-direction, using the appropriate velocity transformation formulas.

 

[MikeLizzi]

Perhaps a better translation is "what's going on". But, anyway, it's not clear to me what you have calculated.

1) You know that the answer should be $(0.6c, -0.6c)$.

2) If you use the 2D Lorentz transformation, you must get symmetry in $x$ and $y$ from the data. It's not clear what you have done there. Also, that is an alternative one-step calculation.

PS although that gives you the four-velocity, which you would have to convert to a three-velocity.

3) You haven't shown your working for the two-step calculation, which should involve a Lorentz boost in the x-direction followed by one in the y-direction, using the appropriate velocity transformation formulas.

Looks like people are asking me to show the transformation formula with the actual numbers not just variables. Makes sense if you believe there is an error. I will try to post and update with actual numbers tomorrow.

 

[PeroK]

For velocity transformations when more than one dimension is involved, I do a velocity 4-vector transformation.
$$\begin{bmatrix}
\gamma_{u'} \\
\gamma_{u'} u'_x \\
\gamma_{u'} u'_y \\
\gamma_{u'} u'_z
\end{bmatrix}
=
\begin{bmatrix}
\gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\
-\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\
-\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\
-\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2}
\end{bmatrix}
\begin{bmatrix}
\gamma_{u} \\
\gamma_{u} u_x \\
\gamma_{u} u_y \\
\gamma_{u} u_z
\end{bmatrix}$$
Where v is the velocity of reference frame S' with respect to reference frame S and

Looks like people are asking me to show the transformation formula with the actual numbers not just variables. Makes sense if you believe there is an error. I will try to post and update with actual numbers tomorrow.

If you put the same value for $v_x$ and $v_y$ into that formula, you must get the same output in $x$ and $y$. Did you put different values for $v_x$ and $v_y$ into that?

PS And, the four velocity going in should be $(1, 0, 0, 0)$ as this is the four-velocity of the particle in the S frame.

Or did you apply that matrix twice?

You don't need to post all your working, but you need to explain what you actually calculated. What numbers went in where.

 

[Charles Link]

I worked the problem in more detail using velocity transformation equations, and I did get a result that was somewhat paradoxical but also had some consistency. If you consider the three frames as A, B, and C, then C has velocity relative to B that is .75 c. (In the y direction). Working from C and observing what he sees, I got that B thereby moves relative to C with speed $v_y=.75c$. This is also the speed of A in that direction(vertical=y) relative to C. The odd result was that A as seen by C moves at a speed in the horizontal=x direction of $v_x=.6c(\sqrt{1-((1.25)(.6))^2}$. This result is consistent in that $v_x^2+v_y^2=2(.6c)^2$, so that the speed that A observes C to move at agrees with the speed that C observes A to move at. $\\$ In any case I used $u_x'=(u_x-v)/(1-(u_xv/c^2))$ and $u_y'=u_y/(\gamma (1-u_xv/c^2))$ as found in an Oxford publication on Lorentz transformations of velocity 4 vectors. I also found the identical result in a publication on Special Relativity by J.H. Smith that I had from my college days (1975). $\\$ The result I got is paradoxical enough that I am not certain that it is correct, but I worked through the velocity transformation equations quite carefully. I'd be interested in seeing if the OP and/or anyone else can concur with my result. I anticipate the Lorentz contraction along the direction of motion would resolve the apparent paradox. Editing... Meanwhile if you go to the rest frame of B, he sees C moving away from him=B at .75c and A moving away from him=B at .6c. Both A and C report slower motion in these directions when observing each other. If we had both parties moving at .6c relative to B as observed by B, then they would agree (by symmetry) on each others velocities that they observe in these directions.

 

[PeroK]

I worked the problem in more detail using velocity transformation equations, and I did get a result that was somewhat paradoxical but also had some consistency. If you consider the three frames as A, B, and C, then C has velocity relative to B that is .75 c. (In the y direction). Working from C and observing what he sees, I got that B thereby moves relative to C with speed $v_y=.75c$. This is also the speed of A in that direction(vertical=y) relative to C. The odd result was that A as seen by C moves at a speed in the horizontal=x direction of $v_x=.6c(\sqrt{1-((1.25)(.6))^2}$. This result is consistent in that $v_x^2+v_y^2=2(.6c)^2$, so that the speed that A observes C to move at agrees with the speed that C observes A to move at. $\\$ In any case I used $u_x'=(u_x-v)/(1-(u_xv/c^2))$ and $u_y'=u_y/(\gamma (1-u_xv/c^2))$ as found in an Oxford publication on Lorentz transformations of velocity 4 vectors. I also found the identical result in a publication on Special Relativity by J.H. Smith that I had from my college days (1975). $\\$ The result I got is paradoxical enough that I am not certain that it is correct, but I worked through the velocity transformation equations quite carefully. I'd be interested in seeing if the OP and/or anyone else can concur with my result. I anticipate the Lorentz contraction along the direction of motion would resolve the apparent paradox. Editing... Meanwhile if you go to the rest frame of B, he sees C moving away from him=B at .75c and A moving away from him=B at .6c. Both A and C report slower motion in these directions when observing each other. If we had both parties moving at .6c relative to B as observed by B, then they would agree (by symmetry) on each others velocities that they observe in these directions.

I must admit I haven't seen this before. The problem must be that A, B and C cannot share a set of x-y axes. If you imagine that A and C agree on a set of axes, one along their relative separation and the other perpendicular to this, then to B these axes are not the x-y axes. Or, if A and C agree that they are moving at 45° to their agreed x-y axes, then these axes will not be B's x-y axes.

If we apply two velocity transformations, then the second (involving B and C) will not be defined along A's x-y axes, but some other pair of axes (in A's frame).

I haven't worked out the details. Perhaps someone else already knows where this is documented?

 

[PeroK]

I took a closer look at this. Because A and C are moving in two directions relative to each other, C's axes are not perpendicular in A's reference frame, but obtuse. And vice versa.

If motion is in either the x or the y direction, then the axes remain perpendicular. But not for 2D motion. The two velocity transformations, I believe, calculate the velocity components along a mixture of x and y- axes. Not along C's x and y axes.

The components of A's velocity you end up with are along some intermediate axes (which appear to be perpendicular as you get the right total velocity squaring them). But they are not the components of A's velocity along C's coordinate axes.

 

[PeroK]

I worked the problem in more detail using velocity transformation equations, and I did get a result that was somewhat paradoxical but also had some consistency. If you consider the three frames as A, B, and C, then C has velocity relative to B that is .75 c. (In the y direction). Working from C and observing what he sees, I got that B thereby moves relative to C with speed $v_y=.75c$. This is also the speed of A in that direction(vertical=y) relative to C. The odd result was that A as seen by C moves at a speed in the horizontal=x direction of $v_x=.6c(\sqrt{1-((1.25)(.6))^2}$. This result is consistent in that $v_x^2+v_y^2=2(.6c)^2$, so that the speed that A observes C to move at agrees with the speed that C observes A to move at. $\\$ In any case I used $u_x'=(u_x-v)/(1-(u_xv/c^2))$ and $u_y'=u_y/(\gamma (1-u_xv/c^2))$ as found in an Oxford publication on Lorentz transformations of velocity 4 vectors. I also found the identical result in a publication on Special Relativity by J.H. Smith that I had from my college days (1975). $\\$ The result I got is paradoxical enough that I am not certain that it is correct, but I worked through the velocity transformation equations quite carefully. I'd be interested in seeing if the OP and/or anyone else can concur with my result. I anticipate the Lorentz contraction along the direction of motion would resolve the apparent paradox. Editing... Meanwhile if you go to the rest frame of B, he sees C moving away from him=B at .75c and A moving away from him=B at .6c. Both A and C report slower motion in these directions when observing each other. If we had both parties moving at .6c relative to B as observed by B, then they would agree (by symmetry) on each others velocities that they observe in these directions.

In general, if C has velocity $(v_x, v_y)$, then in the B frame moving with $v_x$ in the x-direction, C has a velocity: $(0, \gamma_x v_y)$. Where $\gamma_x$ is associated with the velocity $v_x$

Then, in the C frame, A has a velocity $v = (-\frac{v_x}{\gamma}, -\gamma_x v_y)$, where $\gamma$ is associated with the velocity $\gamma_x v_y$.

This gives the correct speed for A:

$v^2 = v_x^2(1- \gamma_x^2 v_y^2) + \gamma_x^2 v_y^2 = v_x^2 + \gamma_x^2 v_y^2(1-v_x^2) = v_x^2 + v_y^2$

But these components are along a pair or orthogonal axes in C's frame that do not necessarily coincide with C's x-y axes.

 

[Charles Link]

In general, if C has velocity $(v_x, v_y)$, then in the B frame moving with $v_x$ in the x-direction, C has a velocity: $(0, \gamma_x v_y)$. Where $\gamma_x$ is associated with the velocity $v_x$

Then, in the C frame, A has a velocity $v = (-\frac{v_x}{\gamma}, -\gamma_x v_y)$, where $\gamma$ is associated with the velocity $\gamma_x v_y$.

This gives the correct speed for A:

$v^2 = v_x^2(1- \gamma_x^2 v_y^2) + \gamma_x^2 v_y^2 = v_x^2 + \gamma_x^2 v_y^2(1-v_x^2) = v_x^2 + v_y^2$

But these components are along a pair or orthogonal axes in C's frame that do not necessarily coincide with C's x-y axes.

I don't yet have a really good answer for this paradox. Because of the Lorentz contraction in the direction of motion, perhaps they won't agree on the directions that each other are travelling.

 

[Charles Link]

In general, if C has velocity $(v_x, v_y)$, then in the B frame moving with $v_x$ in the x-direction, C has a velocity: $(0, \gamma_x v_y)$. Where $\gamma_x$ is associated with the velocity $v_x$

Then, in the C frame, A has a velocity $v = (-\frac{v_x}{\gamma}, -\gamma_x v_y)$, where $\gamma$ is associated with the velocity $\gamma_x v_y$.

This gives the correct speed for A:

$v^2 = v_x^2(1- \gamma_x^2 v_y^2) + \gamma_x^2 v_y^2 = v_x^2 + \gamma_x^2 v_y^2(1-v_x^2) = v_x^2 + v_y^2$

But these components are along a pair or orthogonal axes in C's frame that do not necessarily coincide with C's x-y axes.

I think my results concur with this, but in the transformation equations, the x and y were clearly retained. There was no ambiguity or rotation that might result with these equations.

 

[PeroK]

I think my results concur with this, but in the transformation equations, the x and y were clearly retained. There was no ambiguity or rotation that might result with these equations.

Yes, your results do concur.

To see the issue, consider metre sticks at the origin of all three frames. Assume that we have $v_x = v_y = v$ and the combined velocity of C in A's frame is $w^2 = 2v^2$.

Consider A's first: In A's frame, at $t=0$ the ends are at $(1, 0)$ and $(0,1)$.

In B's frame, at $t' = 0$, these are at $(\frac{1}{\gamma_v}, 0)$ and $(0, 1)$.

Similarly, for each pair of frames, they can agree a common set of x-y axes.

In C's frame, however, the ends of A's metre sticks are, at $t'' = 0$ at:

$\frac12(\frac{1}{\gamma_w} + 1, \frac{1}{\gamma_w} - 1)$ and $\frac12(\frac{1}{\gamma_w} - 1, \frac{1}{\gamma_w} + 1)$

Similarly, the end's of C's metre sticks in A' frame, at time $t=0$, at the same points:

$\frac12(\frac{1}{\gamma_w} + 1, \frac{1}{\gamma_w} - 1)$ and $\frac12(\frac{1}{\gamma_w} - 1, \frac{1}{\gamma_w} + 1)$

The full Lorentz transformation takes account of all this, so if you transform the four-velocities, you automatically transform to C's normal x-y axes. The two separate velocity transformations, however, end up giving the components along a different set of axes in C's frame - not the ones you might expect. Although, I've not been able to work out what these axes are!

 

[Charles Link]

Yes, your results do concur. Consider metre sticks at the origin of all three frames. Assume that we have $v_x = v_y = v$ and the combined velocity of C in A's frame is $w^2 = 2v^2$.

Consider A's first: In A's frame, at $t=0$ the ends are at $(1, 0)$ and $(0,1)$.

In B's frame, at $t' = 0$, these are at $(\frac{1}{\gamma_v}, 0)$ and $(0, 1)$.

Similarly, for each pair of frames, they can agree a common set of x-y axes.

In C's frame, however, the ends of A's metre sticks are, at $t'' = 0$ at:

$\frac12(\frac{1}{\gamma_w} + 1, \frac{1}{\gamma_w} - 1)$ and $\frac12(\frac{1}{\gamma_w} - 1, \frac{1}{\gamma_w} + 1)$

Similarly, the end's of C's metre sticks in A' frame, at time $t=0$, at the same points:

$\frac12(\frac{1}{\gamma_w} + 1, \frac{1}{\gamma_w} - 1)$ and $\frac12(\frac{1}{\gamma_w} - 1, \frac{1}{\gamma_w} + 1)$

The full Lorentz transformation takes account of all this, so if you transform the four-velocities, you automatically transform to C's normal x-y axes. The two separate velocity transformations, however, end up giving the components along a different set of axes in C's frame - not the ones you might expect. Although, I've not been able to work out what these axes are!

The result seems to make sense if you go to the rest frame of B and consider the case where C and A move away from B in perpendicular directions at equal velocities. The problem then has a symmetry to it, and both A and C will say that the velocity of each other along these directions is less than that that B would measure or that their counterpart would measure. Perhaps the Lorentz contraction in the direction of motion accounts for the difference. Even if they don't agree on the direction that each other are heading, surely they must agree on the path that gets traversed. $\\$ Editing... Additional comment: In our coursework, we normally did addition of velocities in one dimension. The case with 2-D relative motion is new to me as well. It's an interesting case that the OP @MikeLizzi presented. @PeterDonis Special Relativity seems to be one of your areas of specialization. What might the answer be?

 

[TSny]

If the calculations are correct then I am forced to conclude that the reference frame I labeled S'' in the one step process is not the same reference frame as the one I labeled S'' in the 2 step process. They have the same origin, are at rest with respect to each other, but one is rotated with respect to the other.
...
P.S.
I am working a more complicated version of this problem in another forum and a poster has suggested I have run into Thomas Precession.

Yes. You can read more about this here
https://en.wikipedia.org/wiki/Wigner_rotation

 

[PeroK]

PS if anyone is interested, I calculated the angle for a pair of boosts in the x, followed by the y direction. The velocity of A in the $S''$ frame is:

$v''_A = (-\gamma_x v_x \sqrt{1- v_x^2 - v_y^2}, \ -\gamma_x v_y)$

C's frame is, therefore, rotated clockwise by an angle, $\theta$, such that:

$\cos \theta = \frac{\gamma_x v_x^2 \sqrt{1- v_x^2 - v_y^2} \ + \ \gamma_x v_y^2}{v_x^2 + v_y^2}$

In the case where $v_x = v_y = v$ this reduces to:

$\cos \theta = \frac{\sqrt{1- 2v^2} \ + \ 1}{2\sqrt{1- v^2}}$

Note that this means that A's velocity in C's frame is rotated anti-clockwise by the same angle.

 

[MikeLizzi]

Does this mean I can stop building Latex formulas with the variables filled in with values? I'm only halfway through the effort. Transcribing computer code into Latex is a good exercise but posting every step of calculations in Latex is still days away.

After reviewing the posts of TSny, Charles Link and PeroK it looks like everyone is in agreement. My results are correct. Just to reassure everybody, I have performed this multi transformation in problems containing many 3D shapes with many different relative velocities. What I get as I switch reference frames is a consistent world, the velocities (and geometric contractions) of all the shapes behave as they should except that the entire universe displays on my computer monitor rotated.

 

[PeroK]

Does this mean I can stop building Latex formulas with the variables filled in with values? I'm only halfway through the effort. Transcribing computer code into Latex is a good exercise but posting every step of calculations in Latex is still days away.

After reviewing the posts of TSny, Charles Link and PeroK it looks like everyone is in agreement. My results are correct. Just to reassure everybody, I have performed this multi transformation in problems containing many 3D shapes with many different relative velocities. What I get as I switch reference frames is a consistent world, the velocities (and geometric contractions) of all the shapes behave as they should except that the entire universe displays on my computer monitor rotated.

Yes, two boosts lead to a rotation. The Wikipedia page gives the general case, and I've posted the angle for the special case of a boost in the x-direction followed by one in the y-direction.

It's not a question of the universe being rotated, it's a question of two boosts breaking the simple geometric assumption that all three frames can be linked by a common set of x-y axes.

It's certainly something of a paradox. But, I think when you look at the analysis in post #13 of the metre sticks along the coordinate axes, you can see that the axes between frames A and C do not have a common direction.

Very intertesting, in any case.