I am reading this paper by Thorne, Flammang & Zytkow (1981), which discusses the dynamics of spherical accretion onto a black hole in the Schwarzschild metric ($c=G=1$ units):

$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$$

For a gas with inward velocity $v$ (as measured by an observer at rest in this metric), equations 2-3 of the paper give this expression for the four-velocity: $$\mathbf{u}=y(1-2M/r)^{-1}\frac{\partial}{\partial t}-vy\frac{\partial}{\partial r}$$ $$y\equiv\mathbf{u}\cdot\frac{\partial}{\partial t}=(1-2M/r)^{1/2}(1-v^2)^{-1/2}$$ $y$ is the so-called energy parameter, and it would be a constant if the trajectory was a geodesic.

I am having trouble understanding how to obtain this expression. Here is what I tried so far:

  • The velocity is $v=-\frac{dr}{dt}$ (minus sign because the gas is moving inward)
  • The four-velocity components are usually written as $U^{\mu}=\frac{dx^\mu}{d\tau}$, where $\tau$ is the proper time
  • By the chain rule, $\frac{dr}{d\tau}=\frac{dr}{dt}\frac{dt}{d\tau}=-v\frac{dt}{d\tau}$
  • The four-velocity as a vector is $\mathbf{u}=U^\mu\partial_\mu=U^\mu\frac{\partial}{\partial x^\mu}$

So that: $$\mathbf{u}=\frac{dt}{d\tau}\frac{\partial}{\partial t}+\frac{dr}{d\tau}\frac{\partial}{\partial r} \quad\\=\frac{dt}{d\tau}\left(\frac{\partial}{\partial t}-v\frac{\partial}{\partial r}\right)$$

Further, since $ds^2=-d\tau^2$, we have: $$-1=-(1-2M/r)\left(\frac{dt}{d\tau}\right)^2+(1-2M/r)^{-1}\left(\frac{dr}{d\tau}\right)^2\\ =\left(\frac{dt}{d\tau}\right)^2\left(-(1-2M/r)+(1-2M/r)^{-1}v^2\right) $$

I don't see how this can lead to the expression for the four-velocity from the paper, so I must be making a mistake somewhere. Any help would be appreciated.


The velocity $v$ is not $dr/dt$. What the paper says is

$v\quad$ inward velocity of gas as measured by an observer at fixed $r$ in his local proper reference frame

This means that you need to build an orthonormal frame from the $\{\partial/\partial^\mu\}$ vectors at a given spacetime point. Well, you can check that the vectors

$$\mathbf{e}_\hat{t} = \frac{1}{\sqrt{1-2M/r}} \partial_t \quad \text{and} \quad \mathbf{e}_\hat{r} = \sqrt{1 - 2M/r} \partial_r$$

are orthonormal, satisfying $\mathbf{e}_\hat{t} \cdot \mathbf{e}_\hat{t} = -1$ and $\mathbf{e}_\hat{r} \cdot \mathbf{e}_\hat{r} = 1$. In terms of these, the four-velocity is

$$\mathbf{u} = \frac{y}{\sqrt{1-2M/r}} \mathbf{e}_\hat{t} - \frac{vy}{\sqrt{1-2M/r}} \mathbf{e}_\hat{r} = u^{\hat{\mu}} \mathbf{e}_\hat{\mu},$$

and we have that $v = -u^\hat{r}/u^\hat{t}$. Or, to put it another way, the four-velocity has the $\mathbf{u} = (\gamma, -\gamma v)$ structure we expect from special relativity.

What if we want to go the other way and write down an expression for the four-velocity? We start with our expressions for the orthonormal basis vectors $\mathbf{e}_\hat{\mu}$ and the four-velocity $\mathbf{u} = u^\hat{t} \mathbf{e}_\hat{t} + u^\hat{r} \mathbf{e}_\hat{r}$. From the special relativistic formula $\mathbf{u} = \gamma(v) \mathbf{e}_\hat{t} - \gamma(v) v \mathbf{e}_\hat{r}$ (which holds since we're in an orthonormal basis), we get that $u^\hat{t} = \gamma(v)$ and $u^\hat{r} = -\gamma(v) v$. Finally, going back to the $\partial_\mu$ basis, we find

$$\mathbf{u} = \frac{\gamma(v)}{\sqrt{1-2M/r}} \partial_t - \gamma(v) v \sqrt{1-2M/r} \partial_r.$$

This is the same as the given expression if we define $y = - \mathbf{u} \cdot \partial_t = - g_{tt} u^t$, which works out to $y = \gamma(v) \sqrt{1-2M/r}$.

| cite | improve this answer | |
  • $\begingroup$ Thanks, I did have a misconception about which velocity v is. I see how this leads to the expression in the paper, but could you explain how to get the u components in this basis? $\endgroup$ – Simon G. Aug 5 '19 at 20:19
  • $\begingroup$ @SimonG. I just replaced the $\partial_\mu$ vectors in terms of the $\mathbf{e}_\hat{\mu}$, using that $\partial_t = \sqrt{1-2M/r} \mathbf{e}_t$ and so on. $\endgroup$ – Javier Aug 5 '19 at 20:20
  • $\begingroup$ But how would I get the expression for the four-velocity in the first place? Where do $y$ and $v$ come from if I'm starting from this orthonormal basis? $\endgroup$ – Simon G. Aug 5 '19 at 20:28
  • $\begingroup$ @SimonG. I've updated my answer. $\endgroup$ – Javier Aug 5 '19 at 20:41
  • $\begingroup$ Thank you, I did not know that this connection between orthonormal bases and special relativity could be made. It makes sense! $\endgroup$ – Simon G. Aug 5 '19 at 20:54

Javier's answer is correct, but here's another way of seeing it: An observer's "local radial coordinate" $\tilde{r}$ and "local time coordinate" $\tilde{t}$ are defined such that the metric is (locally) $ds^2 = - d\tilde{t}^2 + d\tilde{r}^2$ (suppressing angular coordinates). It is evident that under this definition, $d\tilde{t} = dt \sqrt{1 - 2M/r} $ and $d \tilde{r} = dr / \sqrt{1 - 2M/r}$. Thus, $$ -v = \frac{d\tilde{r}}{d\tilde{t}} = \frac{1}{1 - 2M/r} \frac{dr}{dt}. $$ This implies that $$ u^\mu = \frac{dt}{d\tau} \left( 1, \frac{dr}{dt} , 0, 0 \right) = \frac{dt}{d\tau} \left( 1, -\left(1 - \frac{2M}{r}\right) v, 0, 0 \right). $$ Further requiring that $u^\mu u_\mu = -1$ then leads to $$ \frac{dt}{d\tau} = \sqrt{\frac{1}{(1 - 2M/r)(1 - v^2)}}, $$ which is equivalent to the expressions provided.

| cite | improve this answer | |
  • 1
    $\begingroup$ Although this is correct, one has to be careful, because the equations with the differentials don't necessarily define actual coordinate functions $\tilde{t}$ and $\tilde{r}$. The quotation marks around "local coordinates" are very important! $\endgroup$ – Javier Aug 5 '19 at 20:22
  • $\begingroup$ This answer is more intuitive to me. The fact that local coordinates are equivalent to flat spacetime is only true exactly at point $p$ where the observer is, correct? $\endgroup$ – Simon G. Aug 5 '19 at 20:43
  • $\begingroup$ @Michael. Just a small correction, $dt/d\tau = (1-2M/r)^{-1/2}(1-v^2)^{-1/2}$. $\endgroup$ – Simon G. Aug 5 '19 at 21:34
  • $\begingroup$ Maybe I'm missing something, but it seems like this gives the reverse transformation that @Javier gets in his answer. This answer gives $d\hat{t} = dt (1 - 2M/r)^{1/2}$, while Javier's answer below gives, $e_\hat{t} = e_t (1 - 2M/r)^{-1/2}$. Am I misunderstanding the notation? $\endgroup$ – DilithiumMatrix Oct 24 '19 at 3:22
  • 1
    $\begingroup$ @DilithiumMatrix: I believe that the difference is that my answer effectively deals with coordinate one-forms, while Javier's uses basis vectors. So it's not surprising that they have "opposite" transformation rules. $\endgroup$ – Michael Seifert Oct 24 '19 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.