Velocity transformation


by dragan
Tags: transformation, velocity
dragan
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#1
Jan13-08, 08:55 AM
P: 23
Let observer O' move respect to observer O with velocity V. Suppose, that O sees an object M moving with velocity v. A standard vector formula for velocity v' of the object M seen by observer O' is attached.

My question is the following: suppose that the velocity v and v' are given. What is the relative velocity V of the two observers?

Of course in one dimension the answer is very easy :)

Thanks for your help!
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jcsd
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#2
Jan13-08, 10:02 AM
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Just consider M's frame.
dragan
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#3
Jan13-08, 10:40 AM
P: 23
It is not always possible to attribute a real reference frame to M. The simplest example - M could be a particle moving with velocity of light, or a mexican wave on a stadion, which can move with arbitrary velocity v, not necessarily slower than c.

dragan
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#4
Jan14-08, 07:31 PM
P: 23

Velocity transformation


Unfortunatelly, as it usually is, a "simple" special relativity problem turns out to be too difficult for specialists in general relativity. This always made me wonder how surprisingly difficult special relativity is. And it is always so easy when it comes to storytelling about "what is time", "black holes", etc..
jcsd
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#5
Jan14-08, 07:51 PM
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Quote Quote by dragan View Post
Unfortunatelly, as it usually is, a "simple" special relativity problem turns out to be too difficult for specialists in general relativity. This always made me wonder how surprisingly difficult special relativity is. And it is always so easy when it comes to storytelling about "what is time", "black holes", etc..
It still seems to me the best way is consider M's reference frame by assuming V<c then to extend the result.

To do it 'properly' you've got to use the fact that:

dx/dt = vx

dy/dt = vy

dz/dt = vz

dx'(x,t)/dt'(x,t) = v'x'

dy'/dt'(x,t) = dy/dt'(x,t) = v'y'

dz'/dt'(x,t) = dz/dt'(x,t) = v'z'

with the form of x'(x,t) and t'(x,t) being given the usual equations for a Lorentz boost.

There's the problem that if you allow supeluminal velocities you cannot get away from the fact that v or v' may be infinite.
dragan
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#6
Jan14-08, 07:58 PM
P: 23
Unfortunatelly when you consider 3 systems of coordinates, they cannot all be made "parallel" in general. So the transormations involed are not bare boosts, but space rotations also. This complicates the problem very much, because the rotations are parametrized by the velocity that you would like to find.
Mentz114
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#7
Jan15-08, 01:34 AM
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It is not always possible to attribute a real reference frame to M. The simplest example - M could be a particle moving with velocity of light, or a mexican wave on a stadion, which can move with arbitrary velocity v, not necessarily slower than c.
It's easy to come up with 'conundrums' if you consider fictional situations like the above.

Particles with mass are never seen travel at c from any inertial frame, so the question is meaningless. All frames of reference in SR must be related to matter.

Unfortunatelly, as it usually is, a "simple" special relativity problem turns out to be too difficult for specialists in general relativity. And it is always so easy when it comes to storytelling about "what is time", "black holes", etc..
What are you talking about ? What experts in general relativity ? You've started with a wrong premise, manipulated some equations, proved nothing, then started raving about black holes.
dragan
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#8
Jan15-08, 05:11 AM
P: 23
Mentz114,

First of all the problem is difficult to solve even for all velocities below c. I can understand that you don't see it, but once you actually try to solve it, maybe your eyes will open.

Of course I will be extremely happy if you prove me wrong and show me the right formula, but somehow I have second thoughts about it.

Other thing - "fictional situations". I understand that you never heard of photons? When you derive the formula for velocity transformation you only assume that V relates two inertial frames. v and v' can also describe a non-uniformly moving object. Who told you that they must describe inertial frames?

Jcsd,

I think you are right. I believe it is possible to start with a less general problem and to attribute inertial frame to M and then generalize it, however, as I said, the set of equations one gets seems algebraically hard to solve.
Mentz114
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#9
Jan15-08, 05:37 AM
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Other thing - "fictional situations". I understand that you never heard of photons?
What have photons got to do with this problem ? They travel at c and so are outside the scope of Lorentz transformations.

When you derive the formula for velocity transformation you only assume that V relates two inertial frames. v and v' can also describe a non-uniformly moving object. Who told you that they must describe inertial frames?.
I didn't derive the equations. Lorentz transformations are not valid between accelerating frames.

I don't know what point you are trying to make with this post. Your question about the relative velocities has been answered. You have a fundemental misunderstanding if you think

1. Particles with mass can reach light speed
2. Lorentz transformations apply between accelerating frames.
dragan
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#10
Jan15-08, 06:50 AM
P: 23
Suppose that in one inertial frame a photon moves with velocity v (of course v=c). What is the velocity of the photon v' (v'=c) in a frame that moves with velocity V? The answer is simple and given by formula from the first post.

My question is reverse - given two velocities of photon v and v' in two inertial frames derive the formula for the relative velocity of the frames V. And generalize to arbitrary case, not only photons.

Now you got it?

Do you see now, that it is not always possible to atribute an inertial system to the observed object?
jcsd
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#11
Jan15-08, 06:58 AM
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Quote Quote by dragan View Post
Unfortunatelly when you consider 3 systems of coordinates, they cannot all be made "parallel" in general. So the transormations involed are not bare boosts, but space rotations also. This complicates the problem very much, because the rotations are parametrized by the velocity that you would like to find.
Yes, but when you're considering two only there's no need to rotate. (I didn't make it clear the equations I supplky do not consider M's reference frame).
dragan
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#12
Jan15-08, 07:03 AM
P: 23
jcsd you are right, unfortunatelly it seems that considering the other frame allows one only to find the v as a function of v' and V (and the solution is trivial, you simply change the sign of V), and what I am looking for is V as a function of v and v'. This function is not symmetric, hence the difficulties.
Mentz114
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#13
Jan15-08, 08:22 AM
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Dragan:
Suppose that in one inertial frame a photon moves with velocity v (of course v=c). What is the velocity of the photon v' (v'=c) in a frame that moves with velocity V?
OK, I can see where you are going wrong. It is a postulate of special relativity that the speed of light in any frame will always be measured as c. In other words, the relative velocities of emitter and receiver are irrelevant to the observed speed of light.

Given that your observers O and O' are inertial, if either sees M with velocity c, then the other will also, and M is not an inertial frame.

If all three frames are inertial, this is a standard problem.
Ich
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#14
Jan15-08, 08:58 AM
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My question is reverse - given two velocities of photon v and v' in two inertial frames derive the formula for the relative velocity of the frames V. And generalize to arbitrary case, not only photons.
There is no single solution for v=c. You cannot determine the radial velocity. Try reversing the aberration formula if you're interested.
If v<c, jcsd already gave you the answer. Try again to understand what he's doing.
dragan
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#15
Jan15-08, 09:04 AM
P: 23
It is a postulate of special relativity that the speed of light in any frame will always be measured as c.
Einstein and Minkovski put it this way, but since 1911 (Frank and Rothe, Ann. der Phys 825 if you speak German) it is known that this is not a necessary assumption and Lorentz transform can be derived more easily with principle of democracy only. But most people don't know it, don't worry.

Given that your observers O and O' are inertial, if either sees M with velocity c, then the other will also, and M is not an inertial frame.
I am happy that you got it. But my question is still without ANY answer.

If all three frames are inertial, this is a standard problem.
Ok, I am glad that this special case seems to you a standard problem.
Please show me the right formula for V (assuming v<c, v'<c) and your PayPal account info and I promise to send you a $100 gift once I see the right answer before tomorrow.
dragan
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#16
Jan15-08, 09:14 AM
P: 23
Quote Quote by Ich View Post
There is no single solution for v=c. You cannot determine the radial velocity. Try reversing the aberration formula if you're interested.
Yes, of course there is no single solution. And for some cases there is no solution at all.

But the formula for velocity transform "works" not only for M being material particles or photons, but anything at all. It could be also superluminal phase velocity of a wave and the velocity transomation is still valid and gives proper answer. My question is to reverse it.

If v<c, jcsd already gave you the answer. Try again to understand what he's doing.
I think I understand the jcsd's answer, simply because I have already tried to solve the problem this way before :) But, if you believe the answer is easy I am ready to pay a $100 for the right formula, even for this special case of subluminal velocities v and v'.

At least till tomorrow's evening :)
Mentz114
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#17
Jan15-08, 04:10 PM
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dragan:

Here's the standard text-book answer from Prof. J. Baez.

If an observer A measures two objects B and C to be travelling at velocities u = (ux, uy, uz) and v = (vx, vy, vz) respectively, one may ask the question of what the relative speed between B and C are, or in other words at what speed w B would measure C to be travelling at, or vice versa. In Gallilean relativity the relative speed would be given by

[tex]w^2 = (u-v).(u-v) = (u_x - v_x)^2 + (u_y - v_y)^2 + (u_z - v_z)^2.[/tex]

However, in special relativity the relative speed is instead given by the formula

[tex]w^2 = \frac{(u-v).(u-v) - (uXv)^2/c^2}{(1 - (u.v)/c^2)^2 }[/tex]

where u-v = (ux - vx, uy - vy, uz - vz) is the vector difference of u and v, u.v = ux vx + uy vy + uz vz is the inner product of u and v and uXv is the vector product for which
[tex](uXv)^2 = (u.u)(v.v) - (u.v)^2[/tex]

When uy = uz = vy = vz = 0 the formula reduces to the more familiar

[tex]w = |u_x - v_x| / (1 - u_xv_x/c^2)[/tex]

References:
N. M. J. Woodhouse, "Special Relativity", Lecture Notes in Physics (m: 6), Springer Verlag, 1992.
J. D. Jackson, "Classical Electrodynamics", 2nd ed., 1975, ch 11.
P. Lounesto, "Clifford Algebras and Spinors", CUP, 1997.
Of course one assumes that A,B and C are visible to each other, or inside each others light cone. If this holds at one time, it will hold for all times, with one exception.

The general flow of the expanding universe can carry a frame outside our light-cone for instance, in which case Lorentz transformations do not apply and the formulae above become meaningless.

M
dragan
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#18
Jan15-08, 04:31 PM
P: 23
Thanks for the formula. There is no derivation given, but it is quite simple: one considers velocity four-vectors of B and C, calculates the product, which is invariant. Then one calculates the same product in an inertial frame attributed to B and equals the two. The formula for w^2 follows easily.

The only problem is that the formula gives only the magnitude, not the whole vector w (in my notation it was V) - so it is not the complete solution that I am looking for.

Thanks for your help anyway!

This is anyhow only the special case when one can attribute an inertial frame of reference with the observed object.


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