An apparent paradox with couple of frames

[Ookke]

Please see the pictures. In lab frame, we have rockets A and B initially at rest and clocks in sync. When clocks reach certain time T, both A and B accelerate at 45 degrees to up-right direction. There are inertial observers X and Y, which match the velocity x- and y-components that the rockets are going to have.

In X's frame, the rockets are initally moving left with some length contraction, but the clocks are not in sync. B's clock is ahead, so B reaches time T first, when it stops in x-direction and accelerates in y-direction. Soon after, A reaches T also and does the same, but B has gained advantage in y-direction that seems permanent.

In Y's frame, the rockets are initially moving down and the clocks are in sync. Both reach T simultaneously, stop in y-direction and accelerate in x-direction. After stopping in y-direction, the setup is practically identical with Bells's spaceship paradox (without a rope, though). A and B do not have distance in y-direction in this frame.

After the accelerations are done, A and B will end up into their final common rest frame. Strangely enough, things in this frame seem quite different depending on our point of view. If we start from X's frame and jump into final rest frame, we can expect that A and B do have some distance in y-direction. But if we start from Y's frame and jump into final rest frame, we can expect to have somewhat increased distance in x-direction (like Bell's) but no distance in y-direction. This doesn't seem right. Things should not be that relative.

 

[Dale]

It doesn't seem like you have an actual paradox, just a vague hunch that "things shouldn't be that relative".

The best approach for something like this is:
1) explicitly write down the world line of the objects in one frame such as the lab frame
2) boost into any other frame of interest
3) algebraically remove any references to the original coordinates

 

[Ookke]

I need to try something like this. The point will be that is there or is there not distance in y-direction between A and B, in their common final rest frame.

 

[jartsa]

When are the rockets turned off?

 

[Ookke]

jartsa, the rocket engines operate only briefly just to give the acceleration. After that, rockets drift free. And you don't need to think rockets if you don't like: any "object" with clock would do, and you can imagine that someone gives a bump to A and B when their clock reach T.

 

[jartsa]

jartsa, the rocket engines operate only briefly just to give the acceleration. After that, rockets drift free. And you don't need to think rockets if you don't like: any "object" with clock would do, and you can imagine that someone gives a bump to A and B when their clock reach T.

I see. Let me think ... The formation rotates according to X, that is the problem. Each rocket also rotates according to X, that's the solution.

Let's say a rocket consists of two rocket motors strapped onto a fuel tank. One motor starts early according to X, and the rocket turns according to X.

Right?

 

[Ookke]

Not sure if I was able to explain it clear enough. 45 degrees acceleration to up-right in lab frame needs to be achieved (see the picture). Each rocket could have one engine at bottom, one at left side. Each rocket uses both engines for brief time. The rockets accelerate to up-right, but do not turn.

The rockets accelerate simultaneously in Y-frame, but not in X-frame. This results that A and B are at the same plane (no distance in y-direction between them) in Y-frame, but not that in X-frame, which I find problematic.

 

[Dale]

The first step is to write an expression for the world line of each rocket. Do you know how to do that?

 

[Ookke]

No, I don't, sorry. If that is the only way to analyze this, I need to study it. Hopefully already tomorrow.

 

[pervect]

Perhaps we can skip ahead a bit to the fact that the composition of two Lorentz boosts generates a rotation, such that a boost in the X direction followed by a boost in the Y direction generates an additional rotation.

See for instance http://en.wikipedia.org/wiki/Lorentz_transformation#Composition_of_two_boosts
though I'm not familiar with some important details, just the overview.

$B(u)B(v) = B(u \oplus v) Gyr[u,v]$

B(u) and B(v) are both Lorentz boosts. The circled plus represents the SR relativistic velocity addition rule.

If this isn't sufficient, then we have to wade through a more complete problem specification, as Dale suggests. If Gyr is antisymmetric, as it appears, then the order of the boosts generates a different rotation in the result. I have to run now, no time to delve more into the nature of Gyr.

 

[jartsa]

Not sure if I was able to explain it clear enough. 45 degrees acceleration to up-right in lab frame needs to be achieved (see the picture). Each rocket could have one engine at bottom, one at left side. Each rocket uses both engines for brief time. The rockets accelerate to up-right, but do not turn.

The rockets accelerate simultaneously in Y-frame, but not in X-frame. This results that A and B are at the same plane (no distance in y-direction between them) in Y-frame, but not that in X-frame, which I find problematic.

Draw into a space-time diagram the world-tube of a short and wide rocket that is accelerating. Then transform that diagram to other frame. I predict that you will see that the rocket is turning in the other diagram.

EDIT: Stepwise acceleration might be a good idea here, or one sudden acceleration.

 

[Dale]

No, I don't, sorry. If that is the only way to analyze this, I need to study it. Hopefully already tomorrow.

I am not sure that it is the only way to analyze it, there is usually more than one way to do something. However, I do think that it is the easiest and best way to correctly analyze it.

A worldline is a parametric equation in 4D. For basics about parametric equations see:
http://en.wikipedia.org/wiki/Parametric_equation

Basically you want to write a set of four functions of one variable (the parameter) which describes the path in spacetime of one of the rockets. You would write this something like $(t,x,y,z)=(f_t (\lambda), f_x (\lambda), f_y (\lambda), f_z (\lambda))$ where $\lambda$ is the parameter.

For example, an object moving at v in the y direction would be $(\lambda,0,v\lambda,0)$ where $\lambda=t$ is the parameter.

Given that, can you write the worldline for the rockets?

 

[Ookke]

Let's try. In lab frame, set x=0 and y=0 at rocket A's initial position, d is the distance between rockets and v is the rocket speed in x- or y-direction (the same) then

$(t,x,y,z) = (t,vt,vt,0)$ for A
$(t,x,y,z) = (t,d+vt,vt,0)$ for B

I'm still a bit obsessed with the setup and jumping between frames, and trying to make my point more clear, but we can work on this path also.

 

[Ookke]

Draw into a space-time diagram the world-tube of a short and wide rocket that is accelerating. Then transform that diagram to other frame. I predict that you will see that the rocket is turning in the other diagram.

EDIT: Stepwise acceleration might be a good idea here, or one sudden acceleration.

We can use sudden acceleration, because this is a thought experiment and duration of the acceleration is really not the point. I need to also think about the rotation issue that has been brought up. It's somewhat surprising to me.

 

[Dale]

Let's try. In lab frame, set x=0 and y=0 at rocket A's initial position, d is the distance between rockets and v is the rocket speed in x- or y-direction (the same) then

$(t,x,y,z) = (t,vt,vt,0)$ for A
$(t,x,y,z) = (t,d+vt,vt,0)$ for B

I'm still a bit obsessed with the setup and jumping between frames, and trying to make my point more clear, but we can work on this path also.

Excellent job, particularly if you have never done this before.

One important thing is that the equations you wrote are the parametric equations of a straight line. In other words, they are for inertial objects. You will need to make v into a function of t, for example by using the Heaviside step function, to use it for a non inertial rocket.

http://en.m.wikipedia.org/wiki/Heaviside_step_function

 

[jartsa]

We can use sudden acceleration, because this is a thought experiment and duration of the acceleration is really not the point. I need to also think about the rotation issue that has been brought up. It's somewhat surprising to me.

Well here's my updated understanding of the rotation:

|________|
. /\./\./\./\
<--------------
X <- observer X

First X sees the rocket moving to the left, then he sees the engines firing, starting from the right (there are 4 engines). The right side of the rocket is lifted first. The floor of the rocket turns, or tilts, or rotates. It's not correct to say that the rocket turns, because the vertical walls stay vertical.

(I didn't want to consider the 45 degree motion right now)

 

[Ookke]

Excellent job, particularly if you have never done this before.

One important thing is that the equations you wrote are the parametric equations of a straight line. In other words, they are for inertial objects. You will need to make v into a function of t, for example by using the Heaviside step function, to use it for a non inertial rocket.

http://en.m.wikipedia.org/wiki/Heaviside_step_function

Maybe world lines sound scarier than they are :) Of course this was just the first step.

The rockets are inertial almost all the time, apart from instantaneous acceleration when their clocks read T. So, the velocity function could be
$v(t)=0, t < T$
$v(t)=v, t > T$
where the constant $v$ is some fixed "relativistic" speed.

 

[Ookke]

Meanwhile, I need to follow my obsession by pushing some images... But not pushing these any more. I just wanted to the idea clear, since no one has explicitly shown it wrong yet.

The rotation part that came up is tricky and rockets may well experience rotation due to combined boosts, but since X and Y are inertial observers that accelerate into one direction only, I suppose they do not experience rotation.

 

[Ookke]

First X sees the rocket moving to the left, then he sees the engines firing, starting from the right (there are 4 engines). The right side of the rocket is lifted first. The floor of the rocket turns, or tilts, or rotates. It's not correct to say that the rocket turns, because the vertical walls stay vertical.

"Tail" (i.e the right side in this case) is lifted first, due to relativity of simultaneity. But there are no combined boosts in that case.

 

[Dale]

Maybe world lines sound scarier than they are :) Of course this was just the first step.

The rockets are inertial almost all the time, apart from instantaneous acceleration when their clocks read T. So, the velocity function could be
$v(t)=0, t < T$
$v(t)=v, t > T$
where the constant $v$ is some fixed "relativistic" speed.

Good. So the next step is to use the Lorentz transform to calculate the world line in the moving frame.

 

[Ookke]

Good. So the next step is to use the Lorentz transform to calculate the world line in the moving frame.

This is much harder. I found some material on the net, but didn't understand it very well and may have applied it wrong. But let's try anyway:

X-frame spacetime point presented in terms of lab frame coordinates would be
$(t',x',y',z') = (\gamma(t-vx), \gamma(x-vt), y, 0)$
and vice versa
$(t,x,y,z) = (\gamma(t'+vx'), \gamma(x'+vt'), y', 0)$

The same for Y-frame
$(t',x',y',z') = (\gamma(t-vy), x, \gamma(y-vt), 0)$
and
$(t,x,y,z) = (\gamma(t'+vy'), x', \gamma(y'+vt'), 0)$

A's world line in lab frame coordinates is $(t,vt,vt,0)$, so applying the transformations (in X-frame) would result $(\gamma(t'+vx'),v\gamma(t'+vx'),v\gamma(t'+vx'),0)$ which looks quite complicated.

B's world line in lab frame is $(t,d+vt,vt,0)$. I wonder how $d$ should be handled, but distance is shorter in X-frame, so I would guess that $d=\gamma d'$. That would make the world line $(\gamma(t'+vx'),\gamma d' + v\gamma(t'+vx'), v\gamma(t'+vx'), 0)$.

 

[jartsa]

How about this solution:

If the speeds of X and Y are large, over 0.71 c, then the scenario is unphysical, rockets can not move that fast in x-direction and in y-direction at the same time.

If the speeds of X and Y are not large, the scenario is still unphysical, If X says the rocket has correct x-velocity, then X says the rocket has incorrect y-velocity.

The reason for the latter thing is the aberration of rocket's path.

EDIT: If the launchpad personnel says the rockets were launched correctly, then both X and Y say the rockets were launched incorrectly: "too little speed to my direction, too much speed to the other direction"

 

[Dale]

This is much harder. I found some material on the net, but didn't understand it very well and may have applied it wrong. But let's try anyway:

X-frame spacetime point presented in terms of lab frame coordinates would be
$(t',x',y',z') = (\gamma(t-vx), \gamma(x-vt), y, 0)$
and vice versa
$(t,x,y,z) = (\gamma(t'+vx'), \gamma(x'+vt'), y', 0)$

You are right, this next step is considerably harder than the first.

One of the possible problems is confusion regarding $v$. There are two completely separate quantities that both have the symbol $v$ here in this scenario. The first is the $v(t)$ which is part of the expression for the non-inertial worldline of the rocket. It is a function of t because the worldline it is describing is non-inertial, and it applies specifically to the rocket's worldline.

The other is the $v$ which is the relative velocity of the primed and unprimed frames. This is a constant, and can be considered simply a parameter of the Lorentz transform. It does not change over time (the primed and unprimed frames are both inertial), and it applies to all coordinates, not only the ones on the worldline.

I would recommend to separate the two in order to avoid confusion. I would use $u$ to refer to the constant velocity between the primed and unprimed frame, and I would use $v(t)$ to refer to the non-inertial worldline of the rocket. I would not use $v$ at all, but make sure that every occurrence of $v$ in any formula is identified as either $u$ or $v(t)$.

I will try to write more this evening.

 

[Ookke]

I would not use v at all, but make sure that every occurrence of v in any formula is identified as either u or v(t).

This is reasonable, let's do that.

 

[Ookke]

If the speeds of X and Y are not large, the scenario is still unphysical, If X says the rocket has correct x-velocity, then X says the rocket has incorrect y-velocity.

In lab frame, A and B travel at 45 degrees angle with total speed that is less than $c$. Speed components in x- and y-direction are equally large, they can be calculated if we know the total speed.

In X-frame, rockets stop in x-direction and y-component is not the same than it's in the lab frame, but still it's the same for A and B. Likewise, in Y-frame rockets stop in y-direction and x-component differs from x-component in lab frame, but it's still the same for both A and B.

 

[Dale]

X-frame spacetime point presented in terms of lab frame coordinates would be
(t′,x′,y′,z′)=(γ(t−vx),γ(x−vt),y,0)(t',x',y',z') = (\gamma(t-vx), \gamma(x-vt), y, 0)

OK, so as we discussed, to clarify the notation we have:
$(t',x',y',z') = (\gamma(t-ux), \gamma(x-ut), y, z)$ where $\gamma=1/\sqrt{1-u^2}$ in units where c=1.

Now, from your earlier efforts we also have:

(t,x,y,z)=(t,d+vt,vt,0)(t,x,y,z) = (t,d+vt,vt,0) for B

$(t,x,y,z)=(t,d+t~v(t),t~v(t),0)$

So now comes the conceptually easy but algebraically difficult part. All we have to do is to substitute this expression into the above expression, eliminate any of the "leftover" coordinates, and simplify.

Substituting gives us:
$(\gamma (-d u+t (-u) v(t)+t),\gamma (d-t u+t v(t)),t v(t),0)$

We want to solve for t in terms of t', but $t'=\gamma (-d u+t (-u) v(t)+t)$ cannot be solved for t analytically. So, rather than solve it numerically, let's get rid of $v(t)$ where t is the coordinate time in the lab frame, and replace it with $v(\tau)$ where $\tau$ is the frame-invariant proper time.

Without loss of generality we will set T=0 and then frame invariant proper time is given by $t=\tau$ for $t=\tau<0$ and $t \sqrt{1-2u^2} = \tau$ for $t=\tau>0$. So then we have:
$(\gamma (-d u+t (-u) v(\tau)+t),\gamma (d-t u+t v(\tau)),t v(\tau),0)$

Solving $t'=\gamma (-d u+t (-u) v(t)+t)$ for t gives:
$$t=\frac{t'+d u \gamma}{\gamma(1-u v(\tau))}$$

Substituting in and simplifying gives:
$$(t',x',y',z')=\left( t', \frac{d-t'u\gamma+t'\gamma v(\tau)}{\gamma(1-uv(\tau))}, \frac{(t'+du\gamma)v(\tau)}{\gamma(1-uv(\tau))}, 0 \right)$$

 

[Ookke]

Substituting in and simplifying gives:
$$(t',x',y',z')=\left( t', \frac{d-t'u\gamma+t'\gamma v(\tau)}{\gamma(1-uv(\tau))}, \frac{(t'+du\gamma)v(\tau)}{\gamma(1-uv(\tau))}, 0 \right)$$

Yikes! And worse is probably coming. :)

There are few things to digest, but I think I'm getting the idea. One question that came up was that do we have to substitute world line into coordinate transfrom (like you did) or can we do other way around i.e. substitute coordinate transform into world line (something I was trying to do in #21), or is it just a matter of personal taste. Another difficult part is using and justifying the use of proper time. We don't necessarily need to get deeper into these, but I just wanted to tell about the learning experience in case it's useful information for someone.

Thanks for your help so far. I'm curious to complete this excercise in time, to see world lines in action and get the problem resolved. I might stop by here within couple of days, if the thread is active, but after the weekend seems more likely.

 

[Ookke]

Perhaps we can skip ahead a bit to the fact that the composition of two Lorentz boosts generates a rotation, such that a boost in the X direction followed by a boost in the Y direction generates an additional rotation.

See for instance http://en.wikipedia.org/wiki/Lorentz_transformation#Composition_of_two_boosts
though I'm not familiar with some important details, just the overview.

$B(u)B(v) = B(u \oplus v) Gyr[u,v]$

B(u) and B(v) are both Lorentz boosts. The circled plus represents the SR relativistic velocity addition rule.

If this isn't sufficient, then we have to wade through a more complete problem specification, as Dale suggests. If Gyr is antisymmetric, as it appears, then the order of the boosts generates a different rotation in the result. I have to run now, no time to delve more into the nature of Gyr.

I did some preliminary study for the rotation, which kept bothering me, by imagining a stationary grid in the lab frame and checking how it would behave when we jump from frame to another. I didn't do the math, but the result seems to be consistent with pervect's suggestion that observers X and Y indeed rotate due to composition of two boost. So the differing observations are not caused by A and B, but the rotation of X and Y itself. But if this is a solution, it also raises some questions.

I suppose that the rotation is something absolute that can be measured by gyroscope. If the rotation is more about coordinates than physics, and does not show in gyroscope, it's hard to explain how X and Y finally end up into same rest frame with different angle.

Also I find it strange that single acceleration in one direction, that X and Y experience, would cause any rotation. If that is the case, should every accelerating object rotate also? Any acceleration is composition of two boosts, if we look it from moving frame. Moreover, X and Y accelerate at the same magnitude, but experience different rotation. The only thing that differs is the direction. Is the space not same in all directions?