April 2020 – CMacLaurin.com

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 book §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 book §6.4). But with the appropriate Lorentz boost map, such an identity is recovered (Jantzen+ 1992 research paper (arXiv) §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 research paper (astronomy) §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 research paper (astronomy) ; Jantzen+ 2013 book 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!]

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 book §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 research paper (astronomy) , 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.

Month: April 2020

Relative velocity in general relativity

Derivative as contraction of a 1-form and vector