Four-velocity of radially infalling gas in Schwarzschild metric

Question

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.

Answer 1

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.

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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 $y = \gamma(v) \sqrt{1-2M/r}$.

Answer 2

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.