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

[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}{}' 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}{}' and \vec{v}{}' = \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!

 

[matheinste]

Hello dragan.

Just to make it clear to me, is the original question this :-

We are given a velocity (vector) in one frame and a velocity (vector) in another frame. The frames are moving relative to each other. We wish to know what is the velocity (vector) of the frames relative to each other.

Matheinste.

 

[dragan]

yes, precisely :) btw, we already have the solution, but the problem is interesting anyway. Most people usually say at first that it is a "standard" problem and it is enough to interchange the velocities (up to a minus sign) in the transformation formula.