A Coordinate Infall Time for a Vaidya Black Hole

[Markus Hanke]

TL;DR Summary

Is coordinate in-fall time (from rest at infinity to event horizon) finite for a free-fall observer in outgoing Vaidya spacetime?

Consider an observer starting a purely radial free fall from rest at infinity in outgoing Vaidya spacetime - this being a simple model for a radiating black hole. Does anyone have an explicit expression for the coordinate in-fall time (assuming purely radial motion) from infinity to event horizon in this type of spacetime? Essentially I am interested in whether this expression diverges, as it would in Schwarzschild spacetime, or whether it remains finite.

I have actually attempted to do the maths, by starting off with trying to find the equation of radial motion from the geodesic equations, but had to admit defeat very quickly as the resulting system of DE’s is very messy.

Thank you in advance!

 

[PeterDonis]

the coordinate in-fall time

What coordinates do you want to use? Note that the usual coordinates used for Vaidya metrics are not Schwarzschild coordinates, so there is no reason to expect coordinate infall times in those coordinates to behave the way Schwarzschild coordinate infall times behave in Schwarzschild spacetime.

 

[PeterDonis]

I am interested in whether this expression diverges, as it would in Schwarzschild spacetime

Since coordinate expressions are coordinate-dependent anyway, it would be better to look for an invariant expression. For example, you could ask about the redshift (or time delay) of light signals emitted radially outward by the infalling object towards the observer at rest at infinity, and how it changes as the infalling object approaches the horizon. This is an invariant quantity, and diverges in Schwarzschild spacetime, but it can be computed in any coordinates you like.

 

[Markus Hanke]

What coordinates do you want to use? Note that the usual coordinates used for Vaidya metrics are not Schwarzschild coordinates, so there is no reason to expect coordinate infall times in those coordinates to behave the way Schwarzschild coordinate infall times behave in Schwarzschild spacetime.

Thanks for replying - and I am sorry, I should have given more background.
Essentially the question arose in a discussion we were having on another forum. The thread became very convoluted, but in the end boiled down to two questions - given an observer stationary very far away, will he ever be able to detect Hawking radiation emitted from a decaying black hole (the OP argued no, because it would take an infinite amount of time on his clock to reach him)? And conversely, if the observer was to drop a test particle towards the black hole, will the particle reach the event horizon in finite time as measured on his own (far away, stationary) clock?

This isn’t Schwarzschild spacetime, so of course Schwarzschild coordinates do not apply here. Your suggestion about the invariant red shift is a good one, as that would answer both of the above questions. Do you know if this calculation has been done for Vaidya spacetime? Unfortunately I don’t have a specific mass function to supply, but this is really just a proof of concept, so any very simple exponential decay function would probably do (if necessary).

Personally I think that unlike in Schwarzschild spacetime, both of these results will remain finite. But my personal opinion doesn’t mean much without some maths to back it up (or refute it).

 

[PeterDonis]

given an observer stationary very far away, will he ever be able to detect Hawking radiation emitted from a decaying black hole

Yes. This can be seen simply by looking at the Penrose diagram for an evaporating black hole.

if the observer was to drop a test particle towards the black hole, will the particle reach the event horizon in finite time as measured on his own (far away, stationary) clock?

This question doesn't have a well-defined answer because it requires a choice of simultaneity convention, or, equivalently, a choice of coordinates. Different coordinate choices will give different answers.

An invariant question would be whether light signals emitted by the test object at the event horizon reach the distant observer at a finite time by his own clock. For an evaporating black hole, the answer to that question is yes, and that can be seen simply by looking at the Penrose diagram.

Do you know if this calculation has been done for Vaidya spacetime?

I don't know of a detailed calculation, but the answers to the questions above don't require one, as noted above.

 

[pervect]

Summary:: Is coordinate in-fall time (from rest at infinity to event horizon) finite for a free-fall observer in outgoing Vaidya spacetime?

I had a rather similar question - we are maybe thinking alike. However, I don't have any answers. But I'll give my thoughts, so far, in case they help.

I believe the geometrized (G=c=1) metric for a radiating Vaidya black hole should be given by what wiki calls the outgoing vaidya metric:

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

We can set $d\Omega=0$ for the radial case.

There aren't any obvious Killing vectors in the general metric, but my thought is to take a look at the behavior of $u(\tau)$ when M is constant. This appears to be Eddingtion Finklestein outgoing coordinates.

So I'm interested in the ingoing infalling geodesics in these outgoing EF coordinates, i,e. u($\tau$), r($\tau$). It should be workable, but so far I haven't worked it. But I'm interested if u($\tau$) remains finite in these coordinates in an infall from infinity. I have a suspicion it might not be, based on Wiki's discussion of outgoing geodesics in infalling EF coordinates, but I'm not sure it's as symmetrical.

 

[PeterDonis]

I believe the geometrized (G=c=1) metric for a radiating Vaidya black hole should be given by what wiki calls the outgoing vaidya metric

Yes, that's correct.

There aren't any obvious Killing vectors in the general metric

The Vaidya metric does not have any Killing vector fields other than the obvious spacelike ones due to spherical symmetry.

my thought is to take a look at the behavior of when M is constant. This appears to be Eddingtion Finklestein outgoing coordinates

It is, because the $M$ constant case is just Schwarzschild spacetime; if $M$ is constant then no null radiation is actually present and the spacetime is vacuum. Note that the stress-energy tensor of the null radiation in the Vaidya metric is proportional to the derivative of $M$ with respect to the null coordinate ($u$ for the outgoing Vaidya metric, $v$ for the ingoing one), so if that derivative is zero ($M$ constant), the stress-energy tensor vanishes.

 

[pervect]

Alright, then the Vaidya metric isn't static (or stationary), so it doesn't have a Killing vector.

For the case where M is constant, though, we do have a static metric which can give us some insight into the solutions. We'd need more work to confirm that it's reasonable to assume that the solution for M = constant gives some insight into the solution for when M is slowly varying, but it appears reasonable on the surface.

We are solving for $u(\tau), r(\tau)$, the u and r coordinates as a function of proper time, for a timelike geodesic.

The timelike Killing vector is $\frac{\partial}{\partial u}$. This is timelike - it's dual, du, is null.

If I haven't made any errors (and I haven't done much checking), this gives me the following as a geodesic

$$\dot{u} = -1/E \quad \dot{r} = (1-\frac{2M}{r}) \dot{u} - \frac{1}{\dot{u}}$$

*If* this is correct, it doesn't appear there is any issue with $u({\tau})$ 'blowing up' as the object approaches the event horizon. But this is pretty rushed, and there's a pretty high chance I made some errors along the way.

 

[PeterDonis]

We'd need more work to confirm that it's reasonable to assume that the solution for M = constant gives some insight into the solution for when M is slowly varying, but it appears reasonable on the surface.

I'm not so sure. The two differences--no more timelike Killing vector field, and stress-energy tensor no longer vanishing--seem quite significant to me. Note that even if $M$ is slowly varying with $u$, the calculation we are interested in is not a local one--we have to first find the ingoing timelike geodesic worldline of the infalling object, and then find the outgoing null geodesic worldline of the light ray the object emits at the horizon. Those calculations will probe the global properties of the spacetime.

 

[Markus Hanke]

An invariant question would be whether light signals emitted by the test object at the event horizon reach the distant observer at a finite time by his own clock. For an evaporating black hole, the answer to that question is yes, and that can be seen simply by looking at the Penrose diagram.

I thought so, thank you!
The trouble we were having on that other forum was that the OP rejected the validity of the Penrose diagram for Vaidya spacetime (saying it doesn’t accurately reflect the physics of this scenario), which is why I was attempting to explicitly work through the maths. But this answer will do me for now :)

 

[pervect]

A couple of other questions:

If we consider the metric

$$-\left(1-\frac{2M(u)}{r}\right)\,du^2 -2\,du\,dr+r^2d\Omega^2$$

What's the name of the surface where 2M(u)/r = 1? Would this be the "apparent horizion"?

A more ambiguous and harder to answer question. Is it "reasonable" to consider u to be a "null coordinate" and r to be a "spatial coordinate"?

Ass far as the math goes:

$\frac{\partial}{\partial u}$ is timelike, one of the things that initially confused me. The dual vector, or co-vector, the one-form du, is null, though.

There's a couple of ways of demonstrating this, considering the inverse metric:

$$g^{ab} = \begin{bmatrix} 0 & -1 & 0 & 0 \\ -1 & 1-\frac{2M(u)}{r} & 0 & 0 \\ 0 & 0 & \frac{1}{r2} & 0 \\ 0 & 0 & 0 & \frac{1}{r^2 \,\sin^2 \theta} \end{bmatrix}$$

is one of them. Similar concers apply to $\frac{\partial}{\partial r}$ and dr. So, it's another question of terminology, is it reasonable to say a coordinate is null if the one-form, or co-vector, is null, when the vector and co-vector belong to different categories?

 

[PeterDonis]

What's the name of the surface where 2M(u)/r = 1?

The event horizon. Since $M$ is a (decreasing) function of $u$, the area of the event horizon (i.e., $4 \pi r^2$, where $r$ is the radial coordinate of the horizon) is also a decreasing function of $u$.

A more ambiguous and harder to answer question. Is it "reasonable" to consider u to be a "null coordinate" and r to be a "spatial coordinate"?

That depends on how you want to define those terms. See further comments below.

$\frac{\partial}{\partial u}$ is timelike, one of the things that initially confused me.

Only outside the event horizon. On the horizon, it's null, and inside the horizon, it's spacelike.

The dual vector, or co-vector, the one-form du, is null, though.

The covector whose components are $(1, 0, 0, 0)$ is null, yes. This is a manifestation of the fact that surfaces of constant $u$ are null surfaces--more precisely, they are outgoing null surfaces, which can be thought of as expanding spheres of null dust each carrying an increment of energy away from the hole. This fact is what I understand to be the basis for referring to $u$ as a null coordinate.

However, the 1-form obtained by lowering the index of the vector $\partial / \partial u$ is not null (except on the horizon); that 1-form is $g_{\mu \nu} U^\mu = \left[ - \left( 1 - 2M(u) / r \right), -1, 0, 0 \right]$.

Note also that the vector $\partial / \partial r$, holding all other coordinates constant, is null, not spacelike, in these coordinates. Which makes the interpretation of $r$ as a "spatial" coordinate rather problematic.

 

[pervect]

The event horizon. Since M is a (decreasing) function of u, the area of the event horizon (i.e., 4πr2, where r is the radial coordinate of the horizon) is also a decreasing function of u.

A lightlike / null object, traveling radially outwards at r=2M in the coordinates under discussion, would initially stay at a constant r coordinate. But as M(r) decreased, it would escape to infinity. So I didn't think it could be the event horizion. Though I admit I'm a bit hazy on the definitions of all the horizons, which is what prompted the question.

Wiki (not too reliable) says that an event horizon is " is a boundary beyond which events cannot affect an observer. " Which is also my understanding. And the surface doesn't qualify in that regard.
[/quote]

The covector whose components are (1,0,0,0) is null, yes. This is a manifestation of the fact that surfaces of constant u are null surfaces--more precisely, they are outgoing null surfaces, which can be thought of as expanding spheres of null dust each carrying an increment of energy away from the hole. This fact is what I understand to be the basis for referring to u as a null coordinate.

I agree about the difference between gabua and du. But the next part is still puzzling me.

Something is still confusing me about using the null coordinate u, but I will have to think to ask a coherent question. I might even resort to re-introducing a time coordinate :(.

That's probably best left for a later post, the well-posed question at this point is if r=2M(u) is really an event horizon - I don't see how it can qualify if M is a decreasing function of u.

 

[PeterDonis]

A lightlike / null object, traveling radially outwards at r=2M in the coordinates under discussion, would initially stay at a constant r coordinate

It would in Schwarzschild spacetime, but this is not Schwarzschild spacetime. See further comments below.

I'm a bit hazy on the definitions of all the horizons

The definition of the event horizon is that it is the boundary of the region of spacetime that cannot send light signals to future null infinity.

the well-posed question at this point is if r=2M(u) is really an event horizon

[Edit: What follows is not entirely correct. See post #21 for an update and a reference to a better discussion.]

I thought it was, as my earlier response shows; but now I don't think so. I think it's just an apparent horizon, at least up to some value of $u$.

Consider: radially outgoing light rays with constant $u$ are carrying away energy to future null infinity. That means these radially outgoing light rays cannot originate from the event horizon (since by definition the horizon cannot send light signals to future null infinity). But that in turn means that, at any value of $u$ for which such radially outgoing light rays exist, or indeed for any value of $u$ less than that value, there cannot be an event horizon.

In other words, the entire region of spacetime up to some value of $u$ must be outside the event horizon, for all radial coordinates, all the way down to $r = 0$. The fact that $\partial / \partial u$ will become null at some positive $r$ for such values of $u$ can only show the presence of an apparent horizon (a marginally outer trapped surface, where radially outgoing light rays locally don't move outward).

What will be the maximum value of $u$ covering the region outside the event horizon? It will be that value of $u$ at which $M(u)$ reaches a minimum--either zero, if the black hole evaporates completely, or some finite positive value, if the evaporation process stops short of completion--or $u = \infty$, if the process never reaches $M = 0$ or stops at any finite positive value, but goes on forever. And that value of $u$ will mark the horizon. In other words, since radially outgoing light rays are curves of constant $u$, and the event horizon is generated by radially outgoing light rays, some constant value of $u$, call it $u_H$, must mark the event horizon.

Note that this also means the event horizon does have a fixed value of the radial coordinate $r$, namely $2M(u_H)$. If the hole evaporates completely, this value will be $r = 0$, which would seem to indicate that there isn't really a horizon (but I'm not entirely sure of that--I would have to work out what the Penrose diagram looks like for this case; it's not at all obvious to me what it should be). If the hole stops evaporating at some finite positive value of $M$, the event horizon will have $r > 0$. Or, if the hole goes on evaporating forever, as $u \rightarrow \infty$, I don't think there can be an event horizon.

So you're right, the whole question of the event horizon in outgoing Vaidya spacetime is not anything like as simple as it might appear.

 

[pervect]

So you're right, the whole question of the event horizon in outgoing Vaidya spacetime is not anything like as simple as it might appear.

Alright, thanks. I found an earlier thread that helped me with my confusion on my other issue - https://www.physicsforums.com/threa...ure-holding-constant-as-what-you-vary.516026/, one that I'd originated.

Basically when we replace t with u(t,r), we do change $\partial / \partial r$, because holding u constant is different than holding t constant, and it matters what we hold constant. So $\partial / \partial r$ changes from space-like to null. But we don't change dr, the associated one-form, by changing from the t-coordinate to the u-coordinate.

 

[PeterDonis]

when we replace t with u(t,r), we do change $\partial / \partial r$, because holding u constant is different than holding t constant, and it matters what we hold constant.

Yes. I believe this is called the "second fundamental confusion of calculus". It's bitten me plenty of times.

 

[PeterDonis]

I have done some calculations for geodesics in the outgoing Vaidya metric, but I've come up against something that doesn't make sense.

The geodesic equation for a timelike radial geodesic can be obtained most easily by the geodesic Lagrangian method. We start with the Lagrangian:

$$
L = \left( 1 - \frac{2 M (u)}{r} \right) \dot{u}^2 + 2 \dot{u} \dot{r}
$$

where the overdots are derivatives with respect to proper time (i.e., 4-velocity components). The Euler-Lagrange equations are then:

$$
\frac{d}{d\tau} \frac{\delta L}{\delta \dot{u}} = \frac{\delta L}{\delta u}
$$

$$
\frac{d}{d\tau} \frac{\delta L}{\delta \dot{r}} = \frac{\delta L}{\delta r}
$$

These give:

$$
\frac{d}{d\tau} 2 \left[ \left( 1 - \frac{2 M (u)}{r} \right) \dot{u} + \dot{r} \right] = - \frac{2}{r} \frac{dM}{du} \dot{u}^2
$$

$$
\frac{d}{d\tau} 2 \dot{u} = \frac{2 M(u)}{r^2} \dot{u}^2
$$

Which in turn give:

$$
\left[ - \frac{2}{r}\frac{dM}{du} \dot{u} + \frac{2 M (u)}{r^2} \dot{r} \right] \dot{u} + \left( 1 - \frac{2 M (u)}{r} \right) \ddot{u} + \ddot{r} = - \frac{1}{r} \frac{dM}{du} \dot{u}^2
$$

$$
\ddot{u} = \frac{M(u)}{r^2} \dot{u}^2
$$

We use the second equation to eliminate $\ddot{u}$ from the first, and rearrange terms to obtain:

$$
\ddot{r} = - \left( 1 - \frac{2 M (u)}{r} \right) \frac{M(u)}{r^2} \dot{u}^2 - \frac{2 M (u)}{r^2} \dot{u} \dot{r} + \frac{1}{r} \frac{dM}{du} \dot{u}^2
$$

For the case of constant $M$, this looks like the geodesic equation of Schwarzschild spacetime in outgoing Eddington-Finkelstein coordinates, as it should.

However, the sign of the $dM / du$ term doesn't make sense to me. Since we are talking about outgoing Vaidya spacetime, where radiation is being emitted to infinity, we must have $dM / du < 0$. But that means the term in $dM / du$ is negative--it makes the geodesic accelerate inward, just like ordinary gravity. That seems wrong to me; it seems to me that the $dM / du$ term should reflect the effects of outgoing radiation pressure, which should make the geodesic accelerate outward, not inward.

Any thoughts from anyone?

 

[PeterDonis]

As a follow-up to my previous post, we can eliminate $\dot{r}$ from the equation as well and make it even simpler by using the fact that the 4-velocity is a timelike unit vector. This means:

$$
1 = \left( 1 - \frac{2 M(u)}{r} \right) \dot{u}^2 + 2 \dot{u} \dot{r}
$$

Rearranging this gives:

$$
2 \dot{r} = \frac{1}{\dot{u}} - \left( 1 - \frac{2 M(u)}{r} \right) \dot{u}
$$

Substituting into the equation for $\ddot{r}$ leads to some nice cancellations, leaving us with:

$$
\ddot{r} = - \frac{M (u)}{r^2} + \frac{1}{r} \dot{u}^2 \frac{dM}{du}
$$

This still leaves the question of the sign of the $dM / du$ term, though.

 

[pervect]

Corrected version. I deleted the first one, it probably got out anyway, though.

I am not using the geodesic Lagrangian method - I just use GRTensor to calculate the Christoffel symbols.

If we remove $\theta$ and $\phi$ from the line element, and consider only radial motion, I get the following.

The nonzero Christoffel symbols I'm getting are:

$$\Gamma^u{}_{uu} = -\frac{M(u)}{r^2}$$

$$\Gamma^r{}_{uu} = - \frac{M'(u) r^2 - M(u) (r - 2M(u))} {r^3}$$

$$\Gamma^r{}_{ur} = \frac{M(u)}{r^2}$$

where $M'(u) = dM(u) / du$

The line element is the one I gave earlier with the angular terms removed

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

Of course, here may or may not be (more) typos and or errors in the calculation, but for what it's worth, that's what I got. Latex is also giving me a few hiccups with the double dots.

[add]
While I'm at it, I'll actually write out the geodesic equations for completenss, even though they are standard.

$$\ddot{u} + \Gamma^u{}_{uu} \dot{u}^2 = 0 \quad \ddot{r} + \Gamma^r{}_{uu} \dot{u}^2 + 2\,\Gamma^r{}_{ur} \dot{u}\dot{r} = 0$$

The symbols are the same as Peter's. These are second order differential equations, to be solved for the geodesic curve $u(\tau), r(\tau)$. A dot indicates differentiation with respect to $\tau$, a double dot indicates the second derivative with respect to $\tau$.

I believe there are some sign differences compared to Peter's result, but I'm not quite sure exactly where.

 

[PeterDonis]

The nonzero Christoffel symbols I'm getting are

This agrees with that I get from Maxima.

I believe there are some sign differences compared to Peter's result, but I'm not quite sure exactly where.

Well, let's fill in the Christoffel symbols and check. I get:

$$
\ddot{u} - \frac{M(u)}{r^2} \dot{u}^2 = 0
$$

$$
\ddot{r} + \left[ \frac{M(u)}{r^2} \left( 1 - \frac{2 M(u)}{r} \right) - \frac{M'(u)}{r} \right] \dot{u}^2 + \frac{2 M(u)}{r^2} \dot{u} \dot{r} = 0
$$

If we substitute for $2 \dot{r}$ the expression in terms of $\dot{u}$ in my previous post, we get:

$$
\ddot{r} + \left[ \frac{M(u)}{r^2} \left( 1 - \frac{2 M(u)}{r} \right) - \frac{M'(u)}{r} \right] \dot{u}^2 + \frac{M(u)}{r^2} \left[ 1 - \left( 1 - \frac{2 M(u)}{r} \right) \dot{u}^2 \right] = 0
$$

You can see that there is a nice cancellation here, as I got before, and what is left, after moving everything except $\ddot{r}$ to the RHS, is

$$
\ddot{r} = - \frac{M(u)}{r^2} + \frac{M'(u)}{r} \dot{u}^2
$$

Which is the same thing I got. The $- M/r^2$ term is the same as for Schwarzschild spacetime (except that $M$ is now a function of $u$ instead of a constant), so that part all makes sense.

The key is the sign of the term in $M'$, which, in terms of the Christoffel symbols, depends on the sign of the $M'$ term in $\Gamma^r{}_{uu}$. The sign of that term is the opposite of the sign of the $M / r^2$ term. But since $M'$ is negative, that means the effect of that term is the same as that of the $M / r^2$ term, when my physical intuition is that it should be the opposite.

It's quite possible that my physical intuition is wrong here, but if so, I would like to understand why.

 

[PeterDonis]

It's quite possible that my physical intuition is wrong here, but if so, I would like to understand why.

I have found a reference that gives a good treatment of geodesics in outgoing Vaidya spacetime:

http://www.blau.itp.unibe.ch/newlecturesGR.pdf

Chapter 41 in particular has some helpful information. Equation 41`.50 is the same equation for $\ddot{r}$ that we have obtained. The reference includes $G_N$ in the equations, which is Newton's gravitational constant; we have been using units in which $G_N = 1$. The reference also uses the notation $\mathcal{F}_0$ (I can't find the right LaTeX incantation to reproduce the font they use for this) [Edit: found it, thanks to @DrGreg] for what we are calling $- M'(u) \dot{u}^2$ (note the minus sign--since $M'(u)$ is negative, $\mathcal{F}_0$ will be positive), and notes that this quantity is the locally observed outward energy flux due to the radiation being emitted.

Writing this in terms of the outward energy flux makes it clearer what is going on: the energy flux acts as a source of gravity, and since it is spread out through the spacetime, it makes the inward "gravitational force" on a test object larger than it would otherwise be, because the "force" term due to the energy flux falls off only as $1 / r$, not $1 / r^2$, which is what we expect for radiation.

Section 41.5 also is illuminating regarding what I said in post #14. Note the title: "Future Incompleteness of Outgoing Eddington-Finkelstein Coordinates". As soon as I read this, a light bulb went on: the coordinates we are using for the outgoing Vaidya metric cannot cover the future event horizon or the region inside it, for the same reason that outgoing Eddington-Finkelstein coordinates cannot cover the corresponding region of Schwarzschild spacetime. The limit $u \rightarrow \infty$ is the event horizon in both cases, and in both cases the spacetime itself can be extended past that point, but one has to switch to some other coordinate chart that covers the future horizon and the region beyond it to analyze what happens there. (Note that there are some caveats to this, since there are edge cases where there might be a curvature singularity as $u \rightarrow \infty$. The summary referred to below discusses this.)

So my statements in post #14 were not, I think, entirely correct. The summary on p. 933 of the reference gives a better treatment of the possibilities.

 

[PeterDonis]

It's quite possible that my physical intuition is wrong here, but if so, I would like to understand why.

Just to confirm the "additional force" idea from the reference I posted previously, I've calculated the proper acceleration of an observer who is "hovering" at rest at constant $r$. Briefly, the calculation goes like this:

The observer's 4-velocity $U^\mu$ has only a $u$ component, which is $1 / \sqrt{1 - 2M(u) / r}$.

The proper acceleration 4-vector is (leaving out partial derivative terms that vanish in this case and considering only the nonzero connection coefficient terms):

$$
A^\mu = U^\mu \nabla_\mu U^\nu = \left( \Gamma^u{}_{uu} + \Gamma^r{}_{uu} \right) \left( U^u \right)^2
$$

This gives a vector (considering only the nonzero $u$ and $r$ components and leaving out some algebraic steps)

$$
A^\mu = \left( - \frac{M}{r^2} \frac{1}{1 - \frac{2M}{r}} , \frac{M}{r^2} - \frac{M'}{r} \frac{1}{1 - \frac{2M}{r}} \right)
$$

The magnitude of this vector, $\sqrt{g_{\mu \nu} A^\mu A^\nu}$, turns out to be (again leaving out some algebraic steps)

$$
A = \frac{M}{r^2} \frac{1}{\sqrt{1 - \frac{2M}{r}}} \sqrt{1 - \frac{2M' r}{M} \frac{1}{1 - \frac{2M}{r}}}
$$

If we now use the flux notation of the reference I gave in a previous post, we find that

$$
\mathcal{F}_0 = - M' \dot{u}^2 = - \frac{M'}{1 - \frac{2M}{r}}
$$

Using this to substitute for $M'$ in $A$ above gives

$$
A = \frac{M}{r^2} \frac{1}{\sqrt{1 - \frac{2M}{r}}} \sqrt{1 + \frac{2r}{M} \mathcal{F}_0}
$$

This is just the expected Schwarzschild proper acceleration (but with $M$ now a function of $u$ instead of a constant) multiplied by an extra factor due to $\mathcal{F}_0$.

 

[PeterDonis]

outgoing radiation pressure

I have realized why this intuition of mine is wrong: the Einstein tensor for this spacetime has only one nonzero component, the $uu$ component, which is the "energy density". There are no "space-space" components that would correspond to pressure. So even though the stress-energy is described as an outgoing "radiation field", it exerts no radiation pressure. (The Wikipedia article on Vaidya spacetime notes that this "radiation field" does not actually satisfy Maxwell's Equations, so its stress-energy tensor should not be expected to have all of the properties of the standard stress-energy tensor of an EM field, which does satisfy Maxwell's Equations and which does have nonzero pressure.)

 

[vanhees71]

Now the question is, what is the physics of this "radiation field". If it doesn't satisfy the Maxwell equations, it's not an electromagnetic field of course, but what physical situation is then described by this metric?

 

[DrGreg]

The reference also uses the notation $F_0$ (I can't find the right LaTeX incantation to reproduce the font they use for this)

For what it's worth, I think you wanted $\mathcal{F}_0$, obtained via \mathcal (calligraphic).

 

[PeterDonis]

For what it's worth, I think you wanted $\mathcal{F}_0$, obtained via \mathcal (calligraphic).

Yes, that's it. Thanks!

 

[PeterDonis]

what physical situation is then described by this metric?

I don't think the "radiation field" actually describes anything that is physically possible. A term that is often used to describe it in the literature is "null dust", where "dust" is physics-speak for "zero pressure". But I don't think any actual physical field can produce null radiation with zero radiation pressure.

I think the reason the Vaidya metric is studied is that it's an obvious and simple generalization of the Schwarzschild metric in Eddington-Finkelstein coordinates: just let the mass $M$ be a function of the null coordinate ($u$ for outgoing or $v$ for ingoing) instead of a constant. But the fact that that is obvious mathematically does not, as is so often the case in GR, mean that the solution so obtained mathematically is physically realistic.

It could be the case, of course, that the Vaidya metric is a reasonable approximation to some more physically realistic solution that describes radiation being emitted isotropically (or as close as possible to isotropically) by a source like an evaporating black hole. But AFAIK no such more realistic solution is known.

 

[pervect]

Now the question is, what is the physics of this "radiation field". If it doesn't satisfy the Maxwell equations, it's not an electromagnetic field of course, but what physical situation is then described by this metric?

The physics of the radiation field is that it's a "null dust". It can be described by its stress-energy tensor. Qualitatively, it's just incoherent radiation, like a flashlight. I don't know any better way of describing the physics than this. If you think about the physics of radiation from a flashlight, hopefully you will share my conclusion that it's not too productive to ask what the Faraday tensor is, it's better described by it's stress-energy tensor. If not, we could discuss it, I suppose - you could start out by writing the Faraday tensor for a flashlight beam :).

In the coordinates chosen, we can write

$$T^{ab} \propto G^{ab} = \rho v \otimes v$$

where v is a null vector. Confusingly, in these coordinates, v = $\frac{\partial}{\partial r}$. When we replace the usual time coordinate t by the coordinate u, $\frac{\partial}{\partial r}$ becomes null instead of spacelike. This had me confused earlier, as I fell victim to the second fundamental confusion of calculus in spite of knowing better. $\frac{\partial}{\partial r}$ means something different when you hold t constant than it does when you hold u constant.

But really, that's just coordinate confusion, the physical significance is that v is a null vector, and the solution is a "null dust".

The value of $\rho$ is $\frac{2}{r^2} \left| \frac{dM(u)}{du} \right|$, if it matters.

 

[PeterDonis]

The physics of the radiation field is that it's a "null dust". It can be described by its stress-energy tensor.

Yes, that is understood. The question is whether that stress-energy tensor can be obtained from any actual known physical field, the way, for example, that the standard stress-energy tensor for the EM field (which is not the same as the "null dust" stress-energy tensor) is obtained from the EM field tensor. AFAIK nobody knows of an actual physical field that can produce a "null dust" stress-energy tensor.

Qualitatively, it's just incoherent radiation, like a flashlight.

No, it isn't, because incoherent radiation, like any real EM radiation, exerts radiation pressure, and null dust has zero pressure. The standard stress-energy tensor for the EM field, which is what would describe incoherent radiation from something like a flashlight, includes radiation pressure.

 

[vanhees71]

I think "null dust" is unlikely to be realizable physically. In classical field theory a massless field should have a traceless energy-momentum tensor due to scale/conformal invariance, as it is the case for the em. field.

 

[pervect]

Yes, that is understood. The question is whether that stress-energy tensor can be obtained from any actual known physical field, the way, for example, that the standard stress-energy tensor for the EM field (which is not the same as the "null dust" stress-energy tensor) is obtained from the EM field tensor. AFAIK nobody knows of an actual physical field that can produce a "null dust" stress-energy tensor.
No, it isn't, because incoherent radiation, like any real EM radiation, exerts radiation pressure, and null dust has zero pressure. The standard stress-energy tensor for the EM field, which is what would describe incoherent radiation from something like a flashlight, includes radiation pressure.

I have to disagree with your idea that a null dust doesn't exert radiation pressure, at least as a remark with any coordinate independent content.

Given that the stress energy tensor is $\rho v \otimes v$, where v is a null vector, if we use an orthonormal basis for v, rather than a coordinate basis, we would see pressure terms as expected in the orthonormal basis.

So whatever distinction you're trying to make in your remarks about "pressure terms" is coordinate dependent.

There are some subtle differences between a null dust, which is an idealization, and a flashlight beam, however. The null dust won't diffract - it obeys purely geometric optics. This isn't a huge issue, we usually use geometric optics as an approximation for flashlight beams anyway. However, an actual flashlight beam would diffract, depending on it's component frequencies.

If we really wanted to find the Faraday tensor of a flashlight beam, the general approach that comes to mind is to decompose the field it into a weighted sum of plane waves, though some of the details of exactly how this would be accomplished I'd have to research and think about. But it'd be quite messy, and for the purposes at hand, not anything we need to do to understand the physics.

 

[PeterDonis]

Given that the stress energy tensor is $\rho v \otimes v$, where $v$ is a null vector

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.

 

[PAllen]

...
The coordinate $u$ is null because curves of constant $u$ and varying $r$ are null; but that's not the same thing.

I think you mean varying u and constant other coordinates are null, which makes u a lightlike coordinate.

 

[PeterDonis]

I think you mean varying u and constant other coordinates are null

No. Varying $u$ and constant other coordinates is described by the vector $\partial / \partial u$, which, as I have said, is not null (except on the horizon, but it's only outside the horizon that the "null dust" interpretation of the SET makes sense anyway).

To rephrase what I said before: the vector $\partial / \partial r$ (not $\partial / \partial u$) is null in the coordinates we have been using. In other words, a vector pointing in the $r$ direction in a surface of constant $u$ is null.

 

[PAllen]

No. Varying $u$ and constant other coordinates is described by the vector $\partial / \partial u$, which, as I have said, is not null (except on the horizon, but it's only outside the horizon that the "null dust" interpretation of the SET makes sense anyway).

To rephrase what I said before: the vector $\partial / \partial r$ (not $\partial / \partial u$) is null in the coordinates we have been using. In other words, a vector pointing in the $r$ direction in a surface of constant $u$ is null.

Looking back at the metric, I see you are right. But this means u is not a null coordinate. In every reference I've studied, the character of a coordinate is determined by holding other coordinates constant and varying that coordinate. In this case, the coordinate r is lightlike (or null).

For example, this is how one determines that for Schwarzschild exterior coordinates, t is timelike, while for interior coordinates it is spacelike.

 

[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.