Velocity Addition: Solving Minkowski Space with Metric Techniques

[welcomeblack]

Hi all. I'm taking a course in GR and trying to get my intuition and mathematical techniques up to speed. I've been trying to derive the velocity addition formula in Minkowski space, but for some reason I can't do it. Here's what I have:

I'll use the Minkowski metric of signature (+1,-1,-1,-1). Then we have the invariant relation

$c^2d\tau^2=c^2dt^2-dz^2$

Where $\tau$ is the proper time of an observer in the metric and $t$ is the coordinate time of some omniscient being looking at the Minkowski space from afar. For observer $i$ we have the relation

$c^2d\tau_i^2=c^2dt^2-dz_i^2$

then dividing by $c^2dt^2$ we get

$(\frac{d\tau_i}{dt})^2=1-\frac{1}{c^2}(\frac{dz_i}{dt})^2=1-\frac{v_i^2}{c^2}=\frac{1}{\gamma_i^2}$

where $v_i=dz_i / dt$ is the velocity of observer $i$ according to the omniscient being. If instead I had used the chain rule on the $dz_i / dt$ factor, I would get

$\frac{1}{\gamma_i^2}=1-\frac{1}{c^2}(\frac{dz_i}{d\tau_j}\frac{d\tau_j}{dt})^2=1-\frac{v_{rel}^2}{c^2}\frac{1}{\gamma_j^2}$

where I defined the relative velocity as $v_{rel}=dz_i / d\tau_j$. Rearranging for the relative velocity I find

$v_{rel}=v_j\gamma_i$

which is certainly not the velocity addition formula. Where did I go wrong? And how should I change my thinking so it doesn't go wrong again?

 

[PAllen]

The way I've gone about this is to think in terms of dot products. (I assume you know dot product is just contraction of two vectors using the metric).

Then given two 4 velocities U1 and U2, you want to ask what would be representation of U2 in the frame of U1. All you need for velocity addition with the goal of representing the speed of U2 in the frame of U1 is to note that U1$\bullet$U2 is the timelike component of the representation of U2 in U1 frame (equivalently, the representation of U1 in U2 frame). This is all you need because this is gamma.

Thus figure the dot product, then restrict to the colinear case, then solve for v from gamma, and the velocity addition formula will fall out. It is a fun exercise.

 

[jkl71]

Hi, I’m a little confused by your notation and the way you set the problem up, e.g. why in the second equation the coordinate time for observer i doesn’t transform and why the proper time does, since the latter is an invariant under Lorentz transformations. So, I apologize if I’m not interpreting what you mean correctly. I think at a high level you may be misunderstanding the geometrical meaning of the proper time, also called the interval, in the equation. The reason is that it is frame independent and you’ve put a subscript on it.

I’ll layout an analogous problem from Euclidean geometry, it may help you see where you went wrong. Suppose I have a stick and I “boost” it to a slope s1 with respect to its original orientation (in other words I rotate it). Then I “boost” it to have a slope s2 with respect to this new baseline orientation. The question is, what is the slope s with respect to the original orientation? I know the length of the stick doesn’t change when I rotate it, but is that enough to solve the problem? I would certainly need to know if the two rotations were in the same direction or in opposite directions, e.g. if s2 = -s1, then s = 0.

You may try something like what you did, but using the infinitesimal form of Lorentz transformations for t and z instead of the interval.

I don’t think there’s any way to make sure you don’t go wrong again (except to not try, but I wouldn’t recommend that). Making mistakes is a normal part of the learning process, so don’t worry about it, just try and learn from each of them.

 

[welcomeblack]

jkl71: The time $t$ is just a coordinate on the manifold that every observer can agree on, but is not actually an observable. The proper time $\tau$ is what the observer actually measures, and is different for each observer, so I labelled them with indices according to which observer the proper time corresponds to.

PAllen: Thanks for your input. I realized that the proper method of figuring out the velocity addition formula is first finding the time dilation factor from the metric, then using your suggested invariance of the 4-velocity to pull out the required quantity. For completeness of the thread, I'll do it for the Schwarzschild metric:

$c^2d\tau^2=(1-\frac{r_s}{r})c^2dt^2-\frac{1}{1-\frac{r_s}{r}}dr^2-r^2d\theta^2-r^2\sin^2{\theta}\ d\phi^2$

Dividing through by $c^2d\tau^2$ we get

$1=(1-\frac{r_s}{r})(\frac{dt}{d\tau})^2-\frac{1}{c^2}\frac{1}{1-\frac{r_s}{r}}(\frac{dr}{d\tau})^2-\frac{r^2}{c^2}(\frac{d\theta}{d\tau})^2-\frac{r^2\sin^2{\theta}}{c^2}(\frac{d\phi}{d\tau})^2$

Assuming now a purely radial orbit along r=z, and using the chain rule on $dr/d\tau$, we get

$1=(1-\frac{r_s}{z})(\frac{dt}{d\tau})^2-\frac{1}{c^2}\frac{1}{1-\frac{r_s}{z}}(\frac{dz}{dt}\frac{dt}{d\tau})^2$

Defining $v=dz / dt$ and rearranging, we find the time dilation to be

$d\tau=\sqrt{(1-\frac{r_s}{z})-\frac{v^2}{c^2}\frac{1}{1-\frac{r_s}{z}}}dt \equiv \frac{1}{\Gamma}dt$

Note that in the limit of $r \rightarrow \infty$ we recover Minkowski space and our usual time dilation factor.

Now our four-velocity is defined as $u^\mu=dx^\mu / d\tau$, were $x^\mu=[ct\ 0\ 0\ z]$. So then by the chain rule the 4-velocity is

$u^\mu=\frac{dx^\mu}{d\tau}=\frac{dt}{d\tau}\frac{d}{dt}[ct\ 0\ 0\ z]= \Gamma [c\ 0\ 0\ v]$

Okay. Then using the tip from PAllen, we consider the four velocities of two objects in the same orbit. The inner product must be made at the same coordinates $t$ and $z$, and since this is a local inertial frame, we should expect the velocity addition formula to reduce down to that of Minkowski space. But let's show it. In the local inertial frame of object 1,

$\Delta u^2 = g_{\mu\nu}u_1^\mu u_2^\nu=\Gamma_1\Gamma_2(1-\frac{r_s}{z})c^2$

and in a frame where the two objects are moving at velocities $v_1$ and $v_2$,

$\Delta u^2 = g_{\mu\nu}u_1^{\prime\mu} u_2^{\prime\nu}=\Gamma^{\prime}_1\Gamma^{\prime}_2((1-\frac{r_s}{z})c^2-\frac{1}{1-\frac{r_s}{z}}v^{\prime}_1v^{\prime}_2)$

Matching the $\Delta u^2$ expressions we get

$\Gamma_1\Gamma_2(1-\frac{r_s}{z})c^2=\Gamma^{\prime}_1\Gamma^{\prime}_2((1-\frac{r_s}{z})c^2-\frac{1}{1-\frac{r_s}{z}}v^{\prime}_1v^{\prime}_2)$

and expanding the $\Gamma$ factors out in all their glory,

$\frac{1}{\sqrt{(1-\frac{r_s}{z})}}\frac{1}{\sqrt{(1-\frac{r_s}{z})-\frac{v_{rel}^2}{c^2}\frac{1}{1-\frac{r_s}{z}}}}(1-\frac{r_s}{z})c^2=\frac{1}{\sqrt{(1-\frac{r_s}{z})-\frac{v_1^2}{c^2}\frac{1}{1-\frac{r_s}{z}}}}\frac{1}{\sqrt{(1-\frac{r_s}{z})-\frac{v_2^2}{c^2}\frac{1}{1-\frac{r_s}{z}}}}((1-\frac{r_s}{z})c^2-\frac{1}{1-\frac{r_s}{z}}v^{\prime}_1v^{\prime}_2)$

I haven't got pen and paper with me so here's all the gory algebra. I set $c=1$ and define $\alpha=1-r_s / z$. Then

$\frac{1}{\sqrt{\alpha^2-v_{rel}^2}}\alpha=\frac{\alpha^{1/2}}{\sqrt{\alpha^2-v_1^2}}\frac{\alpha^{1/2}}{\sqrt{\alpha^2-v_2^2}}(\alpha-\alpha^{-1}v_1v_2)$

$\alpha^2-v_{rel}^2=\frac{(\alpha^2-v_1^2)(\alpha^2-v_2^2)}{(\alpha-\alpha^{-1}v_1v_2)^2}$

$v_{rel}^2=\frac{\alpha^2(\alpha-\alpha^{-1}v_1v_2)^2-(\alpha^2-v_1^2)(\alpha^2-v_2^2)}{(\alpha-\alpha^{-1}v_1v_2)^2}$

$v_{rel}^2=\frac{(\alpha^4-2\alpha^2v_1v_2+v_1^2v_2^2)-(\alpha^4-\alpha^2(v1^2+v_2^2)+v_1^2v_2^2)}{(\alpha-\alpha^{-1}v_1v_2)^2}$

$v_{rel}^2=\frac{\alpha^2(v_2-v_1)^2}{(\alpha-\alpha^{-1}v_1v_2)^2}$

so finally

$v_{rel}=\frac{v_2-v_1}{1-\alpha^{-2}v_1v_2}$

and we conclude that the Schwarzschild metric actually does have an effect on the velocity addition formula.

 

[PAllen]

Very good. Your last observation is just a question of coordinate scaling. Going back to the metric, compute what dr/dt is lightlike: your alpha. Thus, your formula is just the standard velocity addition once you've noted the scaled speed of light.

[edit: using your formula, what is the local 'speed limit'? It is your alpha. Thus this is just the units in which local light speed is expressed.]

 

[PAllen]

Actually, a better way to see that you really have the standard Minkowski formula is to suppose that v is scaled. Make your existing v variables into, e.g. u variables, let u = αv for each, and your formula turns into the standard velocity addition formula.

To see this substitution must be made, take your expression:

$\Delta u^2 = g_{\mu\nu}u_1^\mu u_2^\nu=\Gamma_1\Gamma_2(1-\frac{r_s}{z})c^2$

and note that this is the dot product of a 4-velocity and a unit vector in the t direction (which is, of course, the stationary 4-velocity in the local frame). This must be γ for the local frame. If you simplify this, you see that it is gamma if you assume velocity is really v/α. Going back earlier, you let v be defined as dr/dt, but that coordinate velocity is not necessarily the velocity expressed in local, comoving inertial frame (because r and t are both scaled compared to that frame). With this nuance, you see that you have achieved your goal.

I actually had in mind a much simpler approach, for your original question in SR with standard metric. Outlining my approach for the SC metric:

You get an expression for 4 velocity just by normalizing (1,u) per the metric (dot product with itself, square root, divide by this). Take the dot product of this with t directed unit vector ((1/√α,0) in c=1 units). This is must be gamma in a local inertial frame, and you see you should recast the general 4-velocity as the normalization of (1,vα), so v means what you think it should. Then take the dot product of two different 4-velocities (e.g. normalized (1, α v1), (1,α v2)). This will be gamma of relative velocity, and (having defined v properly) you find this dot product to be:

γ(v1)γ(v2)(1-v1v2)

and solving for v-rel from gamma-rel immediately gives you the velocity addition formula.

 

[PAllen]

Finally, note, you can use this method to easily derive the non-colinear formula from the metric:

γ(rel) = γ(v1)γ(v2)(1-v1 v2 cosθ)

from which vrel is trivially derived; θ spatial angle between 4-velocities measured in a local inertial frame in which the 4-velocities are expressed.