# Four-velocity of radially infalling gas in Schwarzschild metric

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

• 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? – Simon G. Aug 5 '19 at 20:19
• @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. – Javier Aug 5 '19 at 20:20
• 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? – Simon G. Aug 5 '19 at 20:28
• @SimonG. I've updated my answer. – Javier Aug 5 '19 at 20:41
• Thank you, I did not know that this connection between orthonormal bases and special relativity could be made. It makes sense! – 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.

• 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! – Javier Aug 5 '19 at 20:22
• 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? – Simon G. Aug 5 '19 at 20:43
• @Michael. Just a small correction, $dt/d\tau = (1-2M/r)^{-1/2}(1-v^2)^{-1/2}$. – Simon G. Aug 5 '19 at 21:34
• 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? – DilithiumMatrix Oct 24 '19 at 3:22
• @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. – Michael Seifert Oct 24 '19 at 11:23