What is the relative velocity of two observers with given object velocities?

[dragan]

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!

 

[jcsd]

Just consider M's frame.

 

[dragan]

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]

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]

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 = v^x

dy/dt = v^y

dz/dt = v^z

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]

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]

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]

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]

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 fundamental misunderstanding if you think

1. Particles with mass can reach light speed
2. Lorentz transformations apply between accelerating frames.

 

[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? 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]

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]

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]

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]

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]

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]

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]

dragan:

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

If an observer A measures two objects B and C to be traveling 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 traveling at, or vice versa. In Gallilean relativity the relative speed would be given by

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

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

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

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
(uXv)^2 = (u.u)(v.v) - (u.v)^2

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

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

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]

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.

 

[yuiop]

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.

Standard solution from http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

wx = (ux + vx) / (1 + ux vx / c2)
wy = uy / [(1 + ux vx / c2) gamma(vx)]
wz = uz / [(1 + ux vx / c2) gamma(vx)].

gamma(vx) = 1/sqrt(1 - vx2 / c2).

Thus the velocity w = (wx, wy, wz) of C with respect to A is given by the above three formulae, assuming that B is orientated in the standard way with respect to A. Note that if uy=uz=0 then this reduces to the simpler velocity addition formula given before.

References: "Essential Relativity", W. Rindler, Second Edition. Springer-Verlag 1977.


Claims "prize"

 

[Mentz114]

Hey, Kev, I beat you to it ! J. Baez gets the prize.

Dragan,

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.

As you've specified the problem, there's not enough information to get the components of V.

M

 

[dragan]

Ok, guys :)

Kev, what you have shown is the first thing that comes to one's mind, but the formulas are not right. To find out, just plug them into the equation that I have given. Your formula are not solution of this equation. They only apply to v and v', not V.

The reason they are wrong is because there is an extra spatial rotation in one of the reference frames in respect to the other. And it has to be taken into the account.

Mentz114: I also realized that Baez' answer also do not apply to the velocity V - it only applies to v and v'. Also, the easies way to find out is to put it into the equation and see that it is not consistent. The same reason - an extra spatial rotation.

In conclusion: there is no formula even for the magnitude of V.

Sorry guys :)

Right now, I am just reversing the equations and it seems that it is possible to determine V this way.

 

[George Jones]

When dragan says that a rotation is involved, what he means is something like the following.

The set of all boosts doesn't form a group.

The identity is a boost, every boost B has an inverse that is a boost, composition (multiplication) of boosts is associative, but the set of boosts in not closed under composition. However, any (general, i.e, linear transformation that preserves the Minkowski metric) Lorentz transformation can be written as the product of a boost (often called a Lorentz transformation in elementary treatments that don't mention what a Lorentz transformation really is) and a rotation or a rotation and a boost. In particular, the product of two boosts, like we have here, can be written in these forms:

B_1B_2 = RB&#039; = BR.

Roughly, boost first, and then rotate to align axes, or first rotate first such that after then boosting, axes end up aligned. Note that the rotations are the same, and the boosts are different (but related).

This is the source of Thomas precession.

Many years ago, I used to do a lot of calculations of this type. I haven't tried this calculation, partly because, right now, I don't have any free time, and partly for other reasons.

 

[dragan]

George,

Of course you are right in all that you said. This is exactly the reason for Thomas precession. The good news is that I think I already know how to determine V, but the calculations are surprisingly painful.

What is also so funny, that the problem seems usually "obvious" to everyone, and it makes some tiime to make people realize that it actually isn't :)

And I can keep my $100, unless someone gets the right results faster ;)

 

[Mentz114]

So, is the problem posed then purely a mathematical one ? I.e. rearranging the formula given by dragan ? If so, then there are no physical implications at all and the whole thing has been a waste of time.

What is also so funny, that the problem seems usually "obvious" to everyone, and it makes some time to make people realize that it actually isn't :)

You've got a twisted sense of humour.

 

[dragan]

So, is the problem posed then purely a mathematical one ? I.e. rearranging the formula given by dragan ? If so, then there are no physical implications at all and the whole thing has been a waste of time.

I would say that you just don't see physical implications. Of course they are, only I did not discuss them. Solution to the problem I posted is a small piece of a very general question which, is from the physical point of view very interesting.

I just asked the question I couldn't answer hoping, that more clever guys can find a tricky solution.

Anyway, I am sorry if you feel that you have wasted your time.

You've got a twisted sense of humour.

Thanks!

 

[jcsd]

There's a point that hasn't been made, but as you've got the answer now it seems moot. The velocity vectors shouldn't be expressed in terms of each other as they do not all belong . The answer should be expressed in terms of components.

 

[dragan]

jcsd: All vectors v, v' and V always lie in one plane (you can show it by taking a vector product v'xV). Therefore it is reasonable to look for V in the form V = av + bv'. At least that is the way I tried :) I don't have the answer, but I am close, I believe.

Btw, I was actually looking for a case, when v has infinite magnitude and a given direction. Physically, this corresponds to something happening along a line at one instant (for example apples falling simultaneously at a straight line). In other frame of reference an observer notices that apples fall not simultaneously at the line. Suppose, that apples fall along this line with velocity v' (which must be naturally larger than c). My question was: what is the velocity V of the other observer.

This question is a little bit easier to answer, but I was curious if anyone can generalize it to arbitrary v and v'.

Cheers!

 

[yuiop]

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!

In your message you mention 3 variables:

1) Small bold v ( v )
2) Capital bold v ( V )
3) Primed small bold v ( v' )

In your attached formula there appears to be 4 or 5 variables:

1) Small bold v ( v )
2) Capital bold v ( V )
3) Primed small bold v ( v' )
4) Large capital bold v ( V )
5) Capital italic v ( V )

If your intention is to actually solve this problem and not just confuse everyone you should define all the variables clearly.

If I rename variables 1 to 5 as v,w,v',V and u the formula can be written as :

v\prime = \frac{\sqrt{1-u^2/c^2}.(v-v.w.V/u^2)-(V-v.w.V/u^2)}{(1-v.w/c^2)}

or assuming that variable 4 is actually the same thing as variable 2 then the equation would be written as :

v\prime = \frac{\sqrt{1-u^2/c^2}.(v-v.w^2/u^2)-(w-v.w^2/u^2)}{(1-v.w/c^2)}

Is that what you meant? If so, could you define variable u?

P.S. To use TEX simply click on the formulas and copy/ paste the text from the pop-up window. It's fairly self explanatory and easy to edit. A cheat sheet is here: http://www.stdout.org/~winston/latex/latexsheet.pdf

 

[dragan]

Kev,

I have been teaching Relativity for several years, the formula is part of a textbook I've written on the topic and no student has ever complained

Here is the formula in tex. c=1 for simplicity:

v\prime = \frac{\sqrt{1-V^2}\left(v-\frac{v\cdot V}{V\cdot V}V\right)-V+\frac{v\cdot V}{V\cdot V}V}{1-v\cdot V}

where V is relative velocity of two observers, v and v' are velocities of some object measured by the two observers. All velocities are vectors.

I hope that this helps :)

 

[yuiop]

Kev,

I have been teaching Relativity for several years, the formula is part of a textbook I've written on the topic and no student has ever complained

Here is the formula in tex. c=1 for simplicity:

v\prime = \frac{\sqrt{1-V^2}\left(v-\frac{v\cdot V}{V\cdot V}V\right)-V+\frac{v\cdot V}{V\cdot V}V}{1-v\cdot V}

where V is relative velocity of two observers, v and v' are velocities of some object measured by the two observers. All velocities are vectors.

I hope that this helps :)

OK, the term v\cdot V is the scalar or dot product of the vectors v and V.
My bad

 

[jcsd]

jcsd: All vectors v, v' and V always lie in one plane (you can show it by taking a vector product v'xV). Therefore it is reasonable to look for V in the form V = av + bv'. At least that is the way I tried :) I don't have the answer, but I am close, I believe.

The cross product is defined on two vectors in a 3-D Euclidian space, but in this case the two vectors are part of different vector spaces and as such you cannot take their cross product, you cannot take their dot product, you cannot add them together.

This point would simply be nitpicking if there was a natural isomorphism between the two spaces, but whilst they are certainly isomorphic there isn't a natural isomorphism between them.

In this context there is a natural mapping from the space of v to the space of v', but this isn't an isomorphism. It's this mapping you're interested in.

So firtsly you decide on a basis in each space, this gives you a way of expressing v in terms of v'.

It is in some ways quibbling, but it's important to understand what's going mathematically.

 

[George Jones]

Ok, I have an answer.

Instead of trying to invert the formula in post#1, I went back and did the calculation from scratch.

To get my answer, interchange \vec{v}{}&#039; and \vec{V} in post #1.

But there is more going on than just this interchange, i.e., the the points that jcsd is making are good ones. I hope I have taken them into account correctly.

Roughly, the formula in post #1 is with respect to the basis vectors for the frame of O', while the formula I derived is with respect to the basis vectors for the frame of O.

I hope to revisit this in a few days to: 1) make sure what I did makes sense; 2) make pedagogical post(s). The way things are going right now, I'm not sure that this will happen.

Note that the Galilean view has symmetry: \vec{V} = \vec{v} - \vec{v}{}&#039; and \vec{v}{}&#039; = \vec{v} - \vec{V}.

 

[dragan]

jcds,

Of course this is an issue, but it can be sorted out easily. All the velocities are defined in two inertial frames. V and v are defined in 3D vector space of observer O and v' is defined in 3D vector space of the observer O'. We have freedom of choice of both bases, so we simply choose the same basis for both observers.

For two observers it is always possible to choose a common system of 3D coordinates. For three observers it is not possible (due to the fact that Boosts do not form a closed group, as someone stated before).

 

[dragan]

George,

Let us denote v' = f(V,v). From your suggestion it follows that the function f is symmetric and this is not true. This is obviously true for Galilean boost, but not for Lorentz transformation.

The easiest way to find out that the transformation cannot be symmetric is choosing V and v to be orthogonal.

The transformation is symmetric only in a special case, when all velocities are parallel. Therefore in general your answer cannot be right. I am afraid it is not so easy :)

 

[George Jones]

George,

Let us denote v' = f(V,v). form your suggestion it follows that the function f is symmetric

Maybe, but I don't think so.

I'm done with this for a while. My wife is already angry that I'm going to be two hours late for supper!

 

[Mentz114]

Dragan:

I would say that you just don't see physical implications. Of course they are, only I did not discuss them. Solution to the problem I posted is a small piece of a very general question which, is from the physical point of view very interesting.

What physical implications? The three observer case in SR is completely handled and understood. If anyone of the three is outside the light cone of the others there's no point in inverting your formula. Until you can tell me what great discovery you will make with your manipulations, I maintain this is a purely mathematical problem.

 

[dragan]

Maybe, but I don't think so.

Well, here is a little proof :) From the definition:

v' = f(V,v) (1)
v = f(-V,v') (2)

Function f must also have the property that changing the sign of its arguments changes its sign:
f(-x,-y) = -f(x,y) (3)

Now, let us assume that what you said is right:
V = f(v',v) (4)

From the definition it follows that also:
v = f(-v',V) = (5)

Comparing (2) and (5) and using (3) you get:
f(-v',V) = f(-V,v') = -f(V,-v') (5)

You can easily check the formula - it is not antisymmetric (nor symmetric), so your answer (4) is not correct. I hope your wife is not very angry :)

 

[jcsd]

jcds,

Of course this is an issue, but it can be sorted out easily. All the velocities are defined in two inertial frames. V and v are defined in 3D vector space of observer O and v' is defined in 3D vector space of the observer O'. We have freedom of choice of both bases, so we simply choose the same basis for both observers.

For two observers it is always possible to choose a common system of 3D coordinates. For three observers it is not possible (due to the fact that Boosts do not form a closed group, as someone stated before).

The basis vectors also belong to different spaces so you cannot choose the same basis.

Notice I have only been talking about two observers only.

Using the dot product is not only incorrect, but it's obscuring to yourself what's actually going on. You need a function that relates the components of v in the 3-space basis we have choosen for O and the components of v' in the 3-space basis with have choosen for O'. Note V decides how we choose the bases for O and O' and V will take the form of (V,0,0) in the basis we have choosen for O, so the answer you want was psoted a while back.

 

[dragan]

You need a function that relates the components of v in the 3-space basis we have choosen for O and the components of v' in the 3-space basis with have choosen for O'.

Yes, precisely. I choose this function to be identity, so I define the problem in such a way, that O and O' use the same basis. Therefore the dot product makes perfect sense. Maybe I should have clarified this.

Please note that the formula I shown is just a vector form of the usual formulas:

v_x\prime=\frac{v_x-v}{1-v_x V}

v_y\prime=\frac{v_y\sqrt{1-V^2}}{1-v_x V}


If these formulas make sense, then of course also dot product makes sense. Using the vector form alows one to get free of a particular choice of basis provided, that both observes use the same basis of the 3D space.

And this is the basis that I want to use to express V. The answer given above gives the anser in a different basis, and the relation between these two bases is the problem that needs has not been solved. And this is what I was asking.

 

[yuiop]

Hi Dragan,

If you combine the Baez equations for the components of w in the obvious way so that w= \sqrt{w_x^2+w_y^2+w_z^2} you find that if u=c for observer O' then w=c for observer O, whatever the relative velocities of the two observers, which is what you would expect. Unfortunately I cannot get your equation to behave as nicely.

Using the combined Baez equations we get:

W= \sqrt{\frac{(U_X+V)^2+U_Y^2(1-V^2)}{(1+U_X V)^2}}

where W is the velocity of object M as measured by observer O, V is the relative velocity of the two observers and U_X and U_Y are the components of the velocity (U) of M as measured by O'. I have ignored the z component as we can choose axes such that all the vectors lie in one plane (as you mentioned).

The combined Baez equation can be reversed to give the relative velocities of the two observers:

V=U_X\frac{2-2W^2\pm\sqrt{(2W^2-2)^2-4B(W^2-U_X^2-U_Y^2)/U_X^2}}{2B}

where B = (W^2 U_X^2+U_Y^2-1).

Substituting U_X = U cos(\theta) and U_Y = U sin(\theta) gives the relative velocity purely in terms of U and W and the angle of vector U to V.

The forward and reverse equations work for all values of V and U (even superluminal) except for the case when M is a photon (U=c) when there are an infinite number of possible answers. Note that due to the square root term having a positive or negative answer there are two possible solutions in all other cases. The negative root gives the correct answer when the velocity (U) of M is less than c and the positive root gives the correct answer when U>c.

 

[yuiop]

Kev,

I have been teaching Relativity for several years, the formula is part of a textbook I've written on the topic and no student has ever complained

Here is the formula in tex. c=1 for simplicity:

v\prime = \frac{\sqrt{1-V^2}\left(v-\frac{v\cdot V}{V\cdot V}V\right)-V+\frac{v\cdot V}{V\cdot V}V}{1-v\cdot V}

where V is relative velocity of two observers, v and v' are velocities of some object measured by the two observers. All velocities are vectors.

I hope that this helps :)

It has gone strangely quiet here since I mentioned in my last post that your formula does not appear to meet the most basic requirement that when v=c then v'=c.

I entered your formula in a spreadsheet making these assumptions and simplifications:

The angle between vector v and vector V is \theta in radians.
v \cdot V = the dot product = v*cos(a)*V
V \cdot V = V*cos(0)*V = V^2

Your formula can then be written in a form that can be entered in a spreadsheet as:

v\prime = \frac{\sqrt{1-V^2}\left(v-v cos(\theta)\right)-V+v cos(\theta)}{1-v cos(\theta) V}

The Baez formula that you claim to be incorrect does at least meet the basic requirement that when v=c then v'=c.

The baez formula written in similar terms to your formula for comparison is:

v\prime= \sqrt{\frac{(v cos(\theta)+V)^2+(v sin(\theta))^2(1-V^2)}{(1+v cos(\theta) V)^2}}

It may well be that I am interpreting your formula incorrectly. Maybe you are using a transformed value for \theta? Could you post your formula in a form that can entered in a spreadsheet?

 

[yuiop]

$100?

 

[George Jones]

<br /> \vec{V}=\frac{\vec{v} - \vec{v&#039;}\cdot\hat{v}\hat{v} - \gamma_v^{-1} \left( \vec{v&#039;} - \vec{v&#039;}\cdot\hat{v}\hat{v}\right)}{1 - \vec{v&#039;}\cdot\vec{v}}<br />

 

[dragan]

kev, I offered the prize for a limited time, please check it up :) I already found a good answer for V, but it demanded going through SL(2,C) group. I also managed to revert the equation in a special case when one of the velocities has infinite magnitude. Both sulutions agree and I am happy :)

According to your question about the initial formula - it is exactly the standard formula that Baez gives, but written without the assumption that V is along x axis, but for arbitrary V (to check it, please assume V = Vx, where x is the x-axis versor). It also guarantees that |v'| = c when |v| = c. Please remember that all quantities in the given formula are vectors. If you get the wrong answer, I will write the calculations here in tex.

To verify that just calculate the dot product v'.v'.

If you are not sure of its correctness, just look at the Lorentz transform in a vector form (for example here http://en.wikipedia.org/wiki/Lorentz_transformation) and take a time derivative.

Cheers!

PS: argh! where is the message about the dog? :) It floored me :)

 

[dragan]

Dear George, you have written the same answer again (with an extra minus sign?) and I am sure again that it is wrong again. This is, however, the first answer that comes to one's mind. The right formula should be symmetric in respect to interchange of the arguments and the given formula is not. To make sure that it is not, just take v and v' to be orthogonal. You obtain:

<br /> \vec{V}=\vec{v} - \gamma_v^{-1} \vec{v&#039;}<br />

The result is not symmetric.

 

[yuiop]

<br /> \vec{V}=\frac{\vec{v} - \vec{v&#039;}\cdot\hat{v}\hat{v} - \gamma_v^{-1} \left( \vec{v&#039;} - \vec{v&#039;}\cdot\hat{v}\hat{v}\right)}{1 - \vec{v&#039;}\cdot\vec{v}}<br />

George, you have defined vector V on the left of the equation in terms of vector V on the right. This is confusing to a lay person like me. Any possibility of posting your equation in a form that is more accessable to lay persons without all the unit hat vector stuff so that we might learn something?

Dragan, thanks for the response. I'll try and absorb what you have written later ;)

 

[George Jones]

George, you have defined vector V on the left of the equation in terms of vector V on the right.

I don't see it. If I did, it's a typo. I corrected a bunch of typos (maybe while you were typing).

\hat{v} is a vector of unit length in the same direction as \vec{v}, so \hat{v} = \vec{v}/v, \vec{v} = v\hat{v}, and

\vec{v}\cdot\hat{V}\hat{V} = \vec{v}\cdot\vec{V}\vec{V}/V^2.

I hope to make a much longer post in a few hours.

 

[tim_lou]

I don't know if a full formula has been proposed. However, there is a fairly simple geometric construction of \mathbf{V} from \mathbf{v} and \mathbf{v}&#039;

now, let a=\gamma \mathbf{v}, similarly for for a'. These are components of the four velocity vector. let v and v' be the four velocity vector constructed from the given velocity.

Firstly, one knows that \mathbf{v}=\Lambda \mathbf{v}&#039; where Lambda is a general lorentz transformation (with rotation). The idea is to choose a coordinate such that Lambda is simple.

consider a and a', if we can find a coordinate system such that a and a' only differs by one component, then the lorentz transformation in this coordinate would be simple, i.e. the boost is in the direction in which the coordinates of a and a' differ.

This can be accomplished by noting that:
1. choose e2 as the unit vector in the direction of axa' (cross product), a dot e2 = a' dot e2 =0 trivially.

2. draw a triangle by putting a and a' at the origin, and then connect the tip of a and a', denote the vector a-a' (or a'-a, either way) by b, now, draw a vector orthogonal to b and e2, normalize it, and call it e3 (it is effectively in the same direction as the latitude constructed from the base b). notice a dot e3 = a' dot e3 by geometry.

3. let e1= e2 x e3. This is the direction of V, notice e1 is parallel (or anti-parallel) to b. (the direction of the lorentz boost)

Secondly, we decompose a and a' in this coordinate system, we only need to consider the component in the e1 direction. The problem reduces to a two dimensional (let the e1 direction be x, and together with the ct coordinate).

we see that (c=1)
(a \cdot e_1, 1/\sqrt{1-v^2}) = \Lambda (a&#039; \cdot e_1, 1/\sqrt{1-v&#039;^2})

find the magnitude of V (with sign) would be fairly simple then.

 

[tim_lou]

after a little algebra... one can obtain a formula,

if one let c=1 and change v'=u for simplicity, and define

\gamma_v=1/\sqrt{1-v^2}

\mathbf{e}=\frac{\gamma_v\mathbf{v}-\gamma_u \mathbf{u}}{||\gamma_v\mathbf{v}-\gamma_u \mathbf{u}||}

then
\mathbf{V}=\frac{(\mathbf{v}-\mathbf{u})\cdot\mathbf{e}}{1-(\mathbf{v}\cdot\mathbf{e})(\mathbf{u}\cdot\mathbf{e})}\,\mathbf{e}

I don't know if I have made any mistake, but the formula looks decent.
checking limiting cases:
1. if v=u, V isn't really well defined, since e is not well defined. But as v approaches u, we see that V approaches zero.
2. if u=0, V should equal v, this is true from the formula.
3. as u approaches c=1, gamma u goes to infinity and e is in the direction of -u, e= -u approximately, so that (v-u) dot e looks like v dot -u + u dot u = 1-v dot u. The top cancels out with the bottom and V approaches e which goes to -u, as expected.
4. if v and u are collinear, the formula reduces to the simple one dimensional case.

 

[dragan]

Dear Tim,

I think your answer is right. I got something a little different, and our formulas are probably equivalent (my formula also reduces to right answer in the special cases that you have mentioned). The only problem is how to use the formula when the given velocities are superluminal. It is probably obvious, but I had to check if I am right by assuming a tachyonic four-vectors as v and v' (using your notation). In this case it is very easy to make a mistake and one has, first, to derive the proper formulas for superluminal "four-velocities" because in literature it is usually only a one-dimensional case begin considered and, for example in famous Feirabend paper its sign seems wrong (in general).

How did I check that I got the right answer? Simply, by reversing my initial equation algebraically in the special case when one of the velocities is infinite. This demanded some clever observations, but turned out to be quite easy. Both formulas agreed, so the approach seems correct.

Thanks for your help and congratulations for getting the answer!