Coordinate Infall Time for a Vaidya Black Hole

[pervect]

It isn't. The stress-energy tensor has only one nonzero component in the coordinates we've been using, the $uu$ component, so the SET is $\rho \, \partial / \partial u \otimes \partial / \partial u$. And $\partial / \partial u$ is not a null vector--more precisely, it's not null except on the horizon $r = 2M(u)$; outside the horizon, which is the area where the "null dust" interpretation makes sense, it's timelike. The coordinate $u$ is null because curves of constant $u$ and varying $r$ are null; but that's not the same thing.

I disagree.

Let's compute the Einstein tensor from the metric to be sure. (Using GRTensor). For the line element

$$-(1-2M(u)/r)\, du^2 -2\,du\,dr+r^2 d\theta^2 + r^2\sin^2 \theta d\phi^2$$

the components of the Einstein tensor $G^{ab}$ are

$$G^{ab} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & -2M'(u)/r^2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Or, more concisely, in index free notation

$$G^{ab} = -(2M'(u) / r^2 ) \, \frac{\partial}{\partial r} \otimes \frac{\partial}{\partial r}$$

And has been noted $\frac{\partial}{\partial r}$ is a null vector . So the Einstein and stress energy have the expected form of a null dust solution.

By definition, the Einstein tensor of a null dust solution has the form

$$G^{ab} = 8 \pi \Phi k^a k^b$$

where $k^a$ is a null vector field.

I'll agree that it has only one nonzero component in the coordinates we are using, but as I pointed out, if you use a local orthonormal basis, you'll see two nonzero components in the stress energy tensor. So physically, there is nothing special about the radiation "lacking pressure".

It may or may not be helpful to note that $G_{ab}$ has the form

$$G_{ab} = -2M'(u)/r^2 \, du \otimes du$$

where $du$ is a null one-form or covector. This is somewhat similar to what you wrote, but there is a significant difference.

See post #11 for why the co-vector du has zero length.

 

[PeterDonis]

$$G^{ab} = -(2M'(u) / r^2 ) \, \frac{\partial}{\partial r} \otimes \frac{\partial}{\partial r}$$

...

$$G_{ab} = -2M'(u)/r^2 \, du \otimes du$$

Ah, yes, I see where the disconnect was--I was looking at the lower-index form of the Einstein tensor, but you are looking at the upper-index form.

However, I don't think this changes the "zero pressure" conclusion. See below.

if you use a local orthonormal basis, you'll see two nonzero components in the stress energy tensor

There's a simpler way to do it: just use the general perfect fluid stress-energy tensor formula, which is (with the - + + + metric signature convention we are using)

$$
T^{ab} = \left( \rho + p \right) u^a u^b + p g^{ab}
$$

Here $u^a = \partial / \partial r$ (the tangent vector to the fluid flow worldlines), so the $ur$ component only contains the pressure; we have $T^{ur} = p g^{ur} = - p$ since $g^{ur} = - 1$. But $T^{ur} = 0$, so we must have $p = 0$.

 

[PeterDonis]

There's a simpler way to do it

Hm, looking over this again, I see another possible disconnect. The standard interpretation of $\rho$ and $p$ in the general stress-energy tensor formula is the energy density and pressure in the fluid rest frame. But a null fluid has no rest frame, so that interpretation does not work for a null fluid.

 

[PAllen]

On the other hand, if null dust lives up to its name, it should have no pressure. "dust" encapsulated absence of pressure and tension in GR solutions.

 

[PeterDonis]

if null dust lives up to its name, it should have no pressure.

Yes, that was my original thought, and the fact that $p = 0$ in the standard perfect fluid stress-energy tensor formula when applied to the null dust in the Vaidya metric appears to bear this out.

However, I do think that the fact that a null fluid has no rest frame makes the interpretation of $p = 0$ as meaning "zero pressure" in the usual sense at least potentially problematic.

 

[pervect]

It really depends on how you define "pressure". But it's not in principle any different than other radiation pressure.

Dust had no pressure in it's rest frame, but - as has been noted - there isn't any rest frame for a null dust. The point I'm making is that it has just as much "pressure" as any other form of radiation.

We can formally introduce an orthonormal basis of 1-forms (w1,w2,w3,w4) to get some physical insight.

$$w1 = du \, \sqrt{1-2M(u)/r} + \frac{dr} { \sqrt{1-2M(u)/r}} \quad w2 = \frac{dr} { \sqrt{1-2M(u)/r}} \quad w3 = rd\theta \quad w4 = r\sin \theta d\phi$$

so that the line element becomes $-w1^2 + w2^2 + w3^2 + w4^2$.

Letting $T^{\hat{a}\hat{b}}$ denote the stress-energy tensor in the orthonormal basis, we have.

[fixed- sign error]

$$T^{\hat{a}\hat{b}} = -\frac{2M'(u)}{r^2(1-2M(u)/r)} \begin{bmatrix} 1 & -1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

As before M'(u) = dM/du

 

[PAllen]

Here’s a thought on this. Ordinary dust in GR will clearly exert a force on a body with motion relative to the dust. We would typically call this pressure on the body, even if the dust has no internal pressure. The thing about null dust is that it cannot be at rest in relation to any body, so it invariably exerts pressure on any body.

 

[PeterDonis]

Ordinary dust in GR will clearly exert a force on a body with motion relative to the dust. We would typically call this pressure on the body, even if the dust has no internal pressure. The thing about null dust is that it cannot be at rest in relation to any body, so it invariably exerts pressure on any body.

I think this is consistent with the stress-energy tensor in the orthonormal basis that @pervect gave: the off-diagonal components $T^{01}$ and $T^{10}$ indicate momentum relative to the observer, and the component $T^{11}$ can be viewed as dynamic pressure due to that relative motion.

 

[PeterDonis]

Letting $T^{\hat{a}\hat{b}}$ denote the stress-energy tensor in the orthonormal basis, we have.

[fixed- sign error]

$$T^{\hat{a}\hat{b}} = -\frac{2M'(u)}{r^2(1-2M(u)/r)} \begin{bmatrix} 1 & -1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

It's worth noting that locally, this is identical to the SET of EM radiation as derived from the standard SET for the EM field, in an orthonormal basis. So you are correct that locally, this does "look like" EM radiation and has similar local properties.

The difference between this outgoing null dust and actual EM radiation is global: the null dust is spherically symmetric, and it's impossible to have actual EM radiation (i.e., a global solution of the source-free Maxwell's Equations) that is spherically symmetric. The lowest order actual EM radiation is dipole, not monopole.

 

[PeterDonis]

[fixed- sign error]

I actually think that, if we are looking at the SET in terms of upper indexes, the original form, with $+1$ for the off-diagonal terms in the matrix, is correct. The $-1$ for the off-diagonal terms, I think, is correct for the SET in terms of lower indexes.

To put this another way, in terms of basis vectors, the tangent vector to the outgoing null dust (i.e., the vector $\partial / \partial r$ that gets tensored with itself) is (neglecting a coefficient in front that depends on global properties) $\hat{e}_0 + \hat{e}_1$ (note the plus sign). That means the off-diagonal terms in the SET with upper indexes in the orthonormal basis, $T^{\hat{a} \hat{b}}$, denoting the energy flux/momentum density, should be positive.

But in terms of basis 1-forms, the 1-form that describes the outgoing null dust (i.e., the 1-form $du$ that gets tensored with itself) is (again neglecting a coefficient in front) $w^0 - w^1$ (note the minus sign). That means the off-diagonal terms in the SET with lower indexes in the orthonormal basis, $T_{\hat{a} \hat{b}}$, should be negative.

 

[vanhees71]

Hm, looking over this again, I see another possible disconnect. The standard interpretation of $\rho$ and $p$ in the general stress-energy tensor formula is the energy density and pressure in the fluid rest frame. But a null fluid has no rest frame, so that interpretation does not work for a null fluid.

I'm still puzzled about how to intepret this specific matter model. I don't think it's physically interpretable very well. In which physics context is this metric discussed? Or is it just a mathematically interesting example with no physics interpretation?

The reason is that for a classical massless field theory, where no dimensionful quantities occur in the Lagrangian, this Lagrangian should be scale invariant and then Noether's theorem tells you that the trace of the energy-momentum tensor vanishes. An important example is the free electromagnetic field. The trace of the em. tensor vanishes, and the equation of state of black-body radiation indeed is $u-3P=0$ ($u$ energy density, $P$ pressure). Also pressure has a well-defined meaning for massless fields. Adding matter in the usual "minimal-coupling" way, you see taking the momentum-balance equation that it indeed describes the pressure of the black-body radiation on the container walls. For a free em. field it leads to the usual Lorentz-force density as well as to the Maxwell stress tensor in continuum mechanics.

 

[PeterDonis]

In which physics context is this metric discussed?

The outgoing Vaidya metric is the simplest spacetime I know of that can be used (at least as an approximation) to model an evaporating black hole. That is the main physics context in which I have seen it used.

the trace of the energy-momentum tensor vanishes

The trace of the SET for the Vaidya metric does vanish. This is easily seen in both the original coordinates used in this thread (since the trace would be either $g^{uu} T_{uu}$ or $g_{rr} T^{rr}$, and in both cases the metric coefficient appearing in the trace vanishes) and in the orthonormal basis @pervect gave (since the metric is just the Minkowski metric in that basis and so the two $1$'s on the main diagonal cancel each other).

As I said in post #44, locally, the SET of the Vaidya metric is identical to the SET of EM radiation--a more precise way of putting it would be that it is identical to the SET of a null EM field. The difference is global; as an exact solution, the Vaidya metric is physically unrealistic globally since it is spherically symmetric, and it is impossible to have a spherically symmetric null EM field as a solution to Maxwell's Equations. However, the Vaidya metric can still be a useful approximation.