Relative velocity in general relativity

Difficulty level:   ★ ★ ★

Suppose we have two 4-velocity vectors \mathbf u and \mathbf v at the same point in curved spacetime. (This avoids complications such as parallel transport. Intuitively, think of the two objects as not necessarily overlapping, but close enough that we can neglect curvature etc.)

Consider firstly inertial frames in Minkowski spacetime. Using coordinates (t,x,y,z) corresponding to some observer \mathbf v, the components of a different observer \mathbf u satisfy:

    \[u^\mu = \frac{dx^\mu}{d\tau} = \frac{dt}{d\tau}\frac{dx^\mu}{dt} = \gamma(1,\beta_x,\beta_y,\beta_z).\]

Here \tau is \mathbf u‘s proper time, \gamma := -\mathbf u\cdot\mathbf v := -g_{\mu\nu}u^\mu v^\nu is the Lorentz factor as I have discussed previously, and the \beta_i are the relative speeds in the coordinate directions. This calculation is inspired by Tsamparlis 2010  §6.2.

[The expression dt/d\tau = \gamma bothered me, because time-dilation is mutual, so one might argue a case for \gamma^{-1} instead. But the key point is, the derivative occurs along the direction of \mathbf u. Another way to check the expression is to write dt/d\tau = dt(\mathbf u) = -\mathbf v^\flat(\mathbf u) = \gamma. This is a contraction of the 1-form dt with the vector \mathbf u, as I explained previously. The “flat” symbol just means dt is the 1-form dual to -\mathbf v. Conversely, along \mathbf v we have d\tau/dt = -\mathbf u^\flat(\mathbf v) = \gamma, which is not a contradiction!]

With a view to generalisation, we re-express the earlier displayed formula using vectors in place of coordinate components: \mathbf u = \gamma(\mathbf v+\mathbf u_\textrm{rel}). This is also more elegant. The reader may find better notation than \mathbf u_\textrm{rel}, but this is the relative velocity of \mathbf u from \mathbf v‘s frame. Rearranging,

    \[\boxed{\mathbf u_\textrm{rel} = \gamma^{-1}\mathbf u - \mathbf v.}\]

This vector lies in the local 3-space of \mathbf v, since \mathbf v\cdot\mathbf u_\textrm{rel} = 0, so in particular \mathbf u_\textrm{rel} is spatial. It has length \beta, which is the overall relative speed, and satisfies \gamma = (1-\beta^2)^{-1/2}. If you want, there is also a decomposition \mathbf u_\textrm{rel} = \beta\hat{\mathbf n}, where \hat{\mathbf n} is a unit vector. Conversely, the relative velocity of \mathbf v with respect to \mathbf u is \mathbf v_\textrm{rel} = \gamma^{-1}\mathbf v - \mathbf u. This also has length \beta, but lies in \mathbf u‘s 3-space. However, unlike the Newtonian case, \mathbf u_\textrm{rel} \ne -\mathbf v_\textrm{rel}, unless \mathbf u = \mathbf v. See Tsamparlis (§6.4) for discussion.

All the vector formulae above transfer unchanged to curved spacetime, for 4-velocities at the same event, including worldlines with acceleration. This can be justified using local inertial coordinates. While the formulae do appear in the literature, with one example being Bini 2014  §6, the topic of observer measurements in general is not widely promoted. I recall two separate conversations with senior relativists who were unfamiliar with use of the Lorentz factor in a curved spacetime context.

For comparison, one quantity which should not be naively ported across from special relativity is acceleration. In curved spacetime, the 4-acceleration \nabla_{\mathbf u}\mathbf u requires the covariant derivative, which depends on curvature (and possibly other choices). The reader curious about relative acceleration could try Jantzen, Carini & Bini: their 1995 paper , and an unfinished book  last updated in 2013.

Derivative as contraction of a 1-form and vector

Difficulty level:   ★ ★ ★

Suppose you seek the derivative of a quantity along a curve, such as the rate of change of a scalar by proper time along a worldline: d\Phi/d\tau, or perhaps the rate of change of pressure by proper distance along a given spatial direction: dp/ds. These derivatives are conveniently expressed as a contraction between a 1-form (the gradient of the scalar) and a tangent vector to the curve. For the first example,

    \[\frac{d\Phi}{d\tau} = \frac{\partial\Phi}{\partial x^\mu} \frac{dx^\mu}{d\tau} = (d\Phi)_\mu u^\mu = d\Phi(\mathbf u),\]

where (x^\mu) is some coordinate system, d\Phi is the 1-form with components (d\Phi)_\mu = \Phi_{,\mu} = \partial\Phi/\partial x^\mu, and u^\mu = dx^\mu/d\tau is the 4-velocity. d\Phi(\mathbf u) is the contraction of the vector and 1-form, yielding a scalar. Schutz 2009  §3.3 gives this derivation.

A spacelike path can be parametrised by proper distance. Then d\Phi/ds = d\Phi(\boldsymbol\xi), where \xi^\mu := dx^\mu/ds is the unit tangent vector. An example of a paper which uses this is Gibbons 1972 , for the change in stress along a rigid cable, see the line after his Equation 4.

For a null path there is no natural parameter, at least not without additional context. But for any chosen parameter \lambda, we have d\Phi/d\lambda = d\Phi(\boldsymbol\xi) as before, where \xi^\mu := dx^\mu/d\lambda is the tangent vector. Of course this applies to the other cases as well. Note all these calculations occur within a single tangent space.