Relative velocity in general relativity

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. Physically, think of the two objects as not necessarily overlapping, but close enough that we can neglect curvature etc.) We can calculate the relative velocity each determines of the other, which is not \mathbf u-\mathbf v.

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 2019  §6.2.

With a view to generalisation, we re-express the displayed formula above using vectors in place of coordinate components: \mathbf u = \gamma(\mathbf v+\mathbf u_\textrm{rel}). This is more elegant, and explicitly tensorial. 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, for \beta\ne 0 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. Why is \mathbf u_\textrm{rel} \ne -\mathbf v_\textrm{rel} unlike in Newtonian physics, aside from the trivial case \mathbf u = \mathbf v? They are in different frames (Tsamparlis §6.4). But with the appropriate Lorentz boost map, such an identity is recovered (Jantzen+ 1992  §4).

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 (Jantzen+ 1992; Bini 2014  §6, etc), 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.

In contrast, 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 of a particle requires the covariant derivative, which depends on curvature. Relative acceleration between observers is more complicated, as it depends on one’s choice of affine connection, for which there are various natural options. For instance, Fermi-Walker transport, or co-rotation with an observer’s frame, see e.g. Jantzen, Carini & Bini (Jantzen+ 1992; Jantzen+ 1995 ; Jantzen+ 2013  draft).

[Finally, the expression dt/d\tau = \gamma given near the start bothered me at first, because time-dilation is mutual, so one might equally argue a case for \gamma^{-1}. But the key point is, the derivative occurs along the direction of \mathbf u, not \mathbf v. 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, and this is not a contradiction!]

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