Xjyuk

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):

$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^-1dr^2+r^2(dtheta^2+sin^2theta dphi^2)$$

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:
$$mathbfu=y(1-2M/r)^-1fracpartialpartial t-vyfracpartialpartial r$$
$$yequivmathbfucdotfracpartialpartial t=(1-2M/r)^1/2(1-v^2)^-1/2$$
$y$ 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 $v=-fracdrdt$ (minus sign because the gas is moving inward)


• The four-velocity components are usually written as $U^mu=fracdx^mudtau$, where $tau$ is the proper time


• By the chain rule, $fracdrdtau=fracdrdtfracdtdtau=-vfracdtdtau$
  


• The four-velocity as a vector is $mathbfu=U^mupartial_mu=U^mufracpartialpartial x^mu$
  

So that:
$$mathbfu=fracdtdtaufracpartialpartial t+fracdrdtaufracpartialpartial r quad\=fracdtdtauleft(fracpartialpartial t-vfracpartialpartial rright)$$

Further, since $ds^2=-dtau^2$, we have:
$$-1=-(1-2M/r)left(fracdtdtauright)^2+(1-2M/r)^-1left(fracdrdtauright)^2\
=left(fracdtdtauright)^2left(-(1-2M/r)+(1-2M/r)^-1v^2right)
$$

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

$vquad$ 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

$$mathbfe_hatt = frac1sqrt1-2M/r partial_t quad textand quad mathbfe_hatr = sqrt1 - 2M/r partial_r$$

are orthonormal, satisfying $mathbfe_hatt cdot mathbfe_hatt = -1$ and $mathbfe_hatr cdot mathbfe_hatr = 1$. In terms of these, the four-velocity is

$$mathbfu = fracysqrt1-2M/r mathbfe_hatt - fracvysqrt1-2M/r mathbfe_hatr = u^hatmu mathbfe_hatmu,$$

and we have that $v = -u^hatr/u^hatt$. Or, to put it another way, the four-velocity has the $mathbfu = (gamma, -gamma v)$ 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 $mathbfe_hatmu$ and the four-velocity $mathbfu = u^hatt mathbfe_hatt + u^hatr mathbfe_hatr$. From the special relativistic formula $mathbfu = gamma(v) mathbfe_hatt - gamma(v) v mathbfe_hatr$ (which holds since we're in an orthonormal basis), we get that $u^hatt = gamma(v)$ and $u^hatr = -gamma(v) v$. Finally, going back to the $partial_mu$ basis, we find

$$mathbfu = fracgamma(v)sqrt1-2M/r partial_t - gamma(v) v sqrt1-2M/r partial_r.$$

This is the same as the given expression if we define $y = - mathbfu cdot partial_t = - g_tt u^t$, which works out to $y = gamma(v) sqrt1-2M/r$.

Javier's answer is correct, but here's another way of seeing it: An observer's "local radial coordinate" $tilder$ and "local time coordinate" $tildet$ are defined such that the metric is (locally) $ds^2 = - dtildet^2 + dtilder^2$ (suppressing angular coordinates). It is evident that under this definition, $dtildet = dt sqrt1 - 2M/r $ and $d tilder = dr / sqrt1 - 2M/r$. Thus,
$$
-v = fracdtilderdtildet = frac11 - 2M/r fracdrdt.
$$
This implies that
$$
u^mu = fracdtdtau left( 1, fracdrdt , 0, 0 right) = fracdtdtau left( 1, -left(1 - frac2Mrright) v, 0, 0 right).
$$
Further requiring that $u^mu u_mu = -1$ then leads to
$$
fracdtdtau = sqrtfrac1(1 - 2M/r)(1 - v^2),
$$
which is equivalent to the expressions provided.