Relativistic Relative Velocity Comp.: Bini, D. et al. (1995)

[More_anonymous]

TL;DR

relativistic relative velocity composition law solely in terms of relative velocities?

I'm trying to understand this paper (equation 2.16 specifically):

Bini, D., Carini, P., & Jantzen, R. T. (1995). Relative observer kinematics in general relativity. Classical and Quantum Gravity. Am I correct in reading there is no way to express the relativistic relative velocity composition law solely in terms of relative velocities? I don't think the relative boost map $B(u,u')$ can be expressed as some formula of relative velocities? At the same time I find it highly counterintuitive that special relativity would force us to use the four velocities $u$ and $u'$ as part of the composition law.

 

[PeterDonis]

Am I correct in reading there is no way to express the relativistic relative velocity composition law solely in terms of relative velocities?

No. The paper you reference has simply chosen to do everything in terms of 4-vectors instead of ordinary (3-vector) velocities. The 3-vector form of the velocity composition law can be found in countless references.

 

[PeterDonis]

I don't think the relative boost map $B(u,u')$ can be expressed as some formula of relative velocities?

Sure it can. You can also find those formulas (the Lorentz transformation formulas) in countless references, including Einstein's original 1905 paper.

 

[More_anonymous]

So by relative velocity composition law I was referring to Lorentzian version of:

$$ \vec v_{DE} + \vec v_{EF} = \vec v_{DF}$$

where $\vec v_{DE}$ is the relative velocity $D$ with respect to $E$ and $\vec v_{DE} = \vec v_D - \vec v_E$

Note this whole formula in written solely in terms of relative velocities. I was asking if $B(u,u')$ can also be expressed in terms of any of the relative velocities $\nu(u,u')$, $\nu(U,u')$ or $\nu(u,U)$ ?

Then the relativistic version of the above formula

$$ B(u',u) \nu (U,u) = \delta(U,u,u') (-\nu(u,u')+\nu(U,u') - (\gamma (u,u')^{-1}- 1)\hat \nu(u,u') \times(\hat \nu (u,u')\times \nu(U,u'))$$

would be expressible in solely in terms of relative velocity.
However, I am uncertain how to express $B(u',u)$ as a function of relative velocity?

 

[Ibix]

There are three vector composition formulae on the Wikipedia page: https://en.wikipedia.org/wiki/Velocity-addition_formula

 

[vanhees71]

See Secs. 1.5 and 1.6 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[More_anonymous]

I'm not asking for $\vec v_{DE}$ in terms of $\vec v_D$ and $\vec v_E$. I'm asking for it in terms of $\vec v_{DF}$ and $\vec v_{EF}$

 

[Sagittarius A-Star]

I'm not asking for $\vec v_{DE}$ in terms of $\vec v_D$ and $\vec v_E$. I'm asking for it in terms of $\vec v_{DF}$ and $\vec v_{EF}$

To which frame do you refer $\vec v_{DE}$?

So by relative velocity composition law I was referring to Lorentzian version of:

$$ \vec v_{DE} + \vec v_{EF} = \vec v_{DF}$$

where $\vec v_{DE}$ is the relative velocity $D$ with respect to $E$ and $\vec v_{DE} = \vec v_D - \vec v_E$

Do you only want to add closing speeds, all with reference to one inertial frame?

Example: A moving walkway $E$ in an airport $F$ moves with $v_{EFx} = 2 km/h$. A person $D$ walks on it. The speed difference $v_{DEx}$, as calculated in the airport frame, is $1 km/h$. The person gets a closing speed of $v_{DFx} = 3 km/h$ to the waiting plane.

Then the "Lorentzian version" would be equal to the "Newtonian version" and the relativistic velocity composition law, which is about transforming a velocity from one frame to another, would be irrelevant.

But if you ask, what is the backward-velocity of the moving walkway in the rest frame of the person $D$, then you could Lorentz-transform the velocity of the moving walkway from the airport frame into the rest frame of the person $D$, and requiring that the velocity of the waiting plane, with reference to this frame, must be $v_{FDx} = -3 km/h$, to reach the plane.

 

[robphy]

I was asking if $B(u,u')$ can also be expressed in terms of any of the relative velocities $\nu(u,u')$, $\nu(U,u')$ or $\nu(u,U)$ ?

Then the relativistic version of the above formula

$$ B(u',u) \nu (U,u) = \delta(U,u,u') (-\nu(u,u')+\nu(U,u') - (\gamma (u,u')^{-1}- 1)\hat \nu(u,u') \times(\hat \nu (u,u')\times \nu(U,u'))$$

would be expressible in solely in terms of relative velocity.
However, I am uncertain how to express $B(u',u)$ as a function of relative velocity?

From https://physics.stackexchange.com/questions/722890/understanding-where-equation-2-3-comes-from
and
https://physics.stackexchange.com/q...e-velocity-composition-law-equation-come-from ,
remember that it appears that (from 2.3) $B(u’,u)$ is $P(u’)P(u)/\gamma$ , which involves two spatial projection operators. By itself, it has no explicit dependence on a specific 3rd 4-velocity $U$. Given an arbitrary choice of $U$, you might be able to decompose $B(u’, u)$ in terms of $U$.

 

[PeterDonis]

So by relative velocity composition law I was referring to Lorentzian version of:

$$ \vec v_{DE} + \vec v_{EF} = \vec v_{DF}$$

where $\vec v_{DE}$ is the relative velocity $D$ with respect to $E$ and $\vec v_{DE} = \vec v_D - \vec v_E$

In the Lorentzian version it is no longer true that $\vec{v_{DE}} = \vec{v_D} - \vec{v_E}$, at least not if you want $\vec{v_{DE}}$ to be physically meaningful. That very definition presupposes the Galilean (non-relativistic) version of the velocity composition law.

Mathematically, you can of course look at quantities like $\vec{v_D} - \vec{v_E}$, but that doesn't mean they have physical meaning in a relativistic theory. They don't. So what you say you are asking for is physically meaningless in a relativistic theory. Nor does it correspond with the 4-vector equations you have been referring to. The correct Lorentzian version of "velocity composition" in terms of 3-vector velocities, the one that corresponds to the 4-vector equations you have been referring to, is the standard one that has already been referenced several times in this thread. It is true that this standard form does not make use of quantities like $\vec{v_{DE}} = \vec{v_D} - \vec{v_E}$, but that is because, as above, such quantities are physically meaningless in a relativistic theory.