# A Paper Spelunking: "Particle Creation by Black Holes" - S.H.

1. Mar 28, 2017

### wraith

Hey all! I'm new to Physics Forums. I'm going through Hawking's famous 1975 paper "Particle Creation by Black Holes" which presents the phenomenon later known as Hawking radiation. I'm going through it for the sake of my own learning and benefit since the paper is a bit ahead of my edge as a physics student. And, since I have lots of questions about it, I figured I'd leave them here to see if I can get some answers and share the knowledge.

Q1: Equation 1.1. Yep, the first one. It says if there's a field $\phi$ that satisfies
$$\phi_{; \mu\nu} \eta^{\mu\nu} = 0$$
then one can express $\phi$ as
$$\phi = \sum_i f_i a_i + \bar{f}_i a^\dagger_i$$

where $\eta^{\mu\nu}$ is the Minkowski metric, the $a_i^\dagger ,a_i$ are the usual creation and annihilation operators and $f,\bar{f}$ are the complete, complex, and orthonormal family of solutions to the wave equation $f_{i ; ab} \eta^{ab} = 0$

I'm leaving out a few subscripts to declutter. Now, here's the questions:
• How is $f_{i ; ab} \eta^{ab} = 0$ a wave equation? Can we compare to E&M's $\partial_\mu\partial^\mu A^{\nu} = 0$?
• Isn't $\phi = \sum_i f_i a_i + \bar{f}_i a^\dagger_i$ something that comes from QM to express a state as a linear expansion of basis states? Why does Hawking put the solution in this form?
• Last, and maybe most important part: why is Hawking considering fields that satisfy $\phi_{\mu\nu} \eta^{\mu\nu} = 0$ in first place?

EDIT 3/29: Fixed the things LeandroMdO pointed out.

Last edited: Mar 29, 2017
2. Mar 29, 2017

### LeandroMdO

You left out a semicolon in Hawking's wave equation. It wasn't a typo: it really was supposed to be there. It denotes covariant differentiation, so that really is a wave equation. See here: https://en.wikipedia.org/wiki/Inhomogeneous_electromagnetic_wave_equation#Curved_spacetime

Your expression for the flat space E&M wave equation is incorrect, by the way. It should be \partial_mu\partial^\mu A^\nu = 0. A is a four-vector, so it only has one index, and the wave equation has two derivatives.

That expression is an example of what's called a Bogoliubov transformation. Phi is not a state, it's an operator. Hawking's objective is to write down the linear transformation that takes the initial state, as described by observers in the far past,before the black hole forms, to the final state, as described by observers in the far future. Generically speaking, if you know how a transformation acts on a basis, you know how it acts on any vector. The vectors here are states of a quantum field, so you need to learn a bit about field quantization before understanding this. The short version is that (in flat space) each Fourier component gets treated as an independent harmonic oscillator, and each has an occupation number. In flat space we say there is "a particle" in that mode, but in curved space there are subtleties and the particle language isn't very useful.

But in any case, roughly speaking, the Bogoliubov transformation takes initial states, described for example in the occupation number basis, into final states, which will in general have different occupation numbers. We say that particles were created or destroyed in the process. In particular, the vacuum state (a state where all occupation numbers are zero) is taken by the Bogoliubov transformation into a state where there are particles. What looks to the observer in the past as a perfect vacuum is filled with particles to the observer in the future. This we call Hawking radiation.

Anyway, that is the rough idea, but to really understand this you need to learn at least some quantum field theory and GR.

3. Mar 29, 2017

### wraith

Awesome! Thanks for clearing that up. So is it equivalent to writing $$\nabla_{a}\nabla_{b} f_i \eta^{ab}$$
Thanks, that was a horrible error on my part. I was getting two EM things confused while rushing to post. Fixed the things you mentioned in an edit.

I will look into the Bogoliubov transformation! I'm relying on my notes from undergrad GR and currently going through Peskin & Schroeder's QFT on my own to help me through this.

4. Mar 30, 2017

### Demystifier

Contrary to what @LeandroMdO said, this is not a Bogoliubov transformation. It is a general solution of the wave equation. It has the form of a linear expansion because the wave equation is linear. All linear equations (not only those appearing in quantum mechanics) have general solutions that can be written in the form of a linear expansion.

For a beginner, the Hawking's paper is rather difficult to read. To learn that stuff I would suggest you to first read the pedagogic review by Jacobson:
http://lanl.arxiv.org/abs/gr-qc/0308048

5. Mar 30, 2017

### LeandroMdO

You're right of course. I was trying to decipher the TeX without compiling, so that, together with distraction, made my brain fill in the blanks with the knowledge of what Hawking's paper says.

Last edited: Mar 30, 2017
6. Mar 30, 2017

### Demystifier

It seems that wikipedia is wrong. There should be no term in the wave equation for EM field that depends on the curvature. At least not in the minimal coupling.

7. Mar 30, 2017

### dextercioby

Wikipedia is right. The field equations for the E_M field in the presence of charges in curved space-time read:

$\nabla^{\mu} F_{\mu\nu} = -\kappa J_{\nu}$ (reference Dirac, page 57, eqn. 29.8, see my comment in post #14 below)

$\nabla^{\mu}\nabla_{\mu} A_{\nu} - g^{\mu\alpha} \nabla_{\alpha}\nabla_{\nu} A_{\mu} = -\kappa J_{\nu}$ (by using non-metricity=0)

$\nabla^{\mu}\nabla_{\mu} A_{\nu} - g^{\mu\alpha} R_{\alpha\nu |\mu}{}^{\sigma} A_{\sigma} = -\kappa J_{\nu}$ (by using torsion=0 and the curved-spacetime Lorenz gauge)

$\nabla^{\mu}\nabla_{\mu} A_{\nu} - R_{\nu}{}^{\sigma} A_{\sigma} = -\kappa J_{\nu}$ (the contracted Riemann tensor is the Ricci tensor)

Last edited: Mar 31, 2017
8. Mar 30, 2017

### Demystifier

I see, it's only for the scalar field that the wave equation does not depend on curvature, basically because only for the scalar field $\nabla_{\mu}\phi=\partial_{\mu}\phi$. For other fields, a second covariant derivative of the field necessarily involves a first derivative of the Christoffel connection $\Gamma$. The connection can always be made vanishing at a given point by an appropriate coordinate transformation (that's the equivalence principle), but the first derivative of the connection cannot be made vanishing. That's where the dependence on the curvature comes from.

But then again, in the case of EM field it seems to be a Lorenz gauge artefact. Since
$$F_{\mu\nu}=\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$
in any gauge, the quantity $\nabla^{\mu}F_{\mu\nu}$ does not really involve derivatives of the connection. The connection itself can always be made vanishing at a given point, so local physics of EM fields does not really depend on curvature.

This can also be seen as follows. As shown e.g. in the book N. Straumann, General Relativity, we have
$$\nabla_{\nu}F^{\mu\nu}=\frac{1}{\sqrt{-g}} \partial_{\nu}(\sqrt{-g}F^{\mu\nu})$$
This shows that only first derivatives (not the second ones) of the metric appear, which can always be made vanishing at a given point, so we see again that curvature is not relevant for local physics of EM fields.

Last edited: Mar 30, 2017
9. Mar 30, 2017

### dextercioby

What do you mean, there's the second derivative on the metric in there, once you put potentials. The "curvy" Lorenz gauge only simplifies the field equations, but the curvature term globally cannot be ignored, if potentials are used. In field theory, the potentials ("connections in the gauge bundle", if I were to use the language of @samalkhaiat) are the field variables, not the fields (!).

10. Mar 30, 2017

### Demystifier

Using
$$\nabla_{\alpha}A_{\beta}-\nabla_{\beta}A_{\alpha}=\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}$$
we see that
$$F^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}=g^{\mu\alpha}g^{\nu\beta}(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})$$
does not contain any derivatives of the metric. Consequently
$$\partial_{\nu}(\sqrt{-g}F^{\mu\nu})$$
does not contain second derivatives of the metric, even when the potentials are put in.

The second derivative of the metric appears if we take $A_{\alpha}=g_{\alpha\gamma}A^{\gamma}$. But this means that the wave equation can depend on Ricci tensor only when the wave equation is written as an equation for $A^{\mu}$, and not $A_{\mu}$. (Does it mean that your last equation could actually be wrong? You should check out!) Indeed, it is consistent with Straumann's "General Relativity", Eq. (2.63), which is an equation for $A^{\mu}$ similar to your last equation for $A_{\mu}$. If you don't have Straumann, you can also check out the free lectures by Blau, Eq. (7.110), which is also a wave equation (with Ricci tensor) for $A^{\mu}$ (not $A_{\mu}$). Wikipedia also has $A^{\mu}$ (not $A_{\mu}$).

Is there a deeper message of all this? I think the deeper message is that in curved coordinates the fundamental quantity is $A_{\mu}$ and not $A^{\mu}$, which is related to the fact that in curved coordinates the spacetime coordinate is $x^{\mu}$ and not $x_{\mu}$. Indeed, the natural gauge transformation is
$$A_{\mu}\rightarrow A_{\mu} +\partial_{\mu}\Lambda$$
while
$$A^{\mu}\rightarrow A^{\mu} +\partial^{\mu}\Lambda=A^{\mu} +g^{\mu\nu}\partial_{\nu}\Lambda$$
looks rather clumsy.

Last edited: Mar 30, 2017
11. Mar 31, 2017

### Demystifier

After some more thought, now I understand much better why the wave equation for $A_{\mu}$ cannot depend on Ricci curvature. Suppose that we have a differential manifold which is not equipped with metric and affine connection. Without metric and affine connection, we cannot define curvature. Can we define sourceless electrodynamics on a such a manifold?

The answer is - yes (see e.g. the book B.F. Schutz, Geometrical Methods of Mathematical Physics) - by formulating electrodynamics in terms of differential forms. Geometrically, the EM potential is a 1-form
$$A=A_{\mu}dx^{\mu}$$
The corresponding EM field is a 2-form
$$F=d\wedge A = F_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$$
where
$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$
It is invariant under the gauge transformation
$$A\rightarrow A +\partial_{\mu}\Lambda dx^{\mu}= A+d\Lambda$$
The sourceless Maxwell equations are
$$dF=0$$
and
$$d\tilde{F}=0$$
where $\tilde{F}$ is dual to $F$. Clearly, all the equations above are independent of metric and affine connection, so they cannot depend on Ricci curvature.

Note however that all equations are expressed in terms of the 1-form $A_{\mu}dx^{\mu}$ and not in terms of a 1-vector $A^{\mu}\partial_{\mu}$. To define $A^{\mu}$ we would need a metric tensor. This is the reason why Maxwell equations, when expressed in terms of $A^{\mu}$, can receive a dependence on Ricci curvature.

12. Mar 31, 2017

### rubi

What you call $\tilde F$ is the Hodge dual of $F$ and it is defined in terms of the metric. You can't write down Maxwell's equations without a background metric.

13. Mar 31, 2017

### Demystifier

Damn, you are right! Thanks for the correction! But it is still true that, geometrically, the 1-form with components $A_{\mu}$ is more fundamental than the 1-vector with components $A^{\mu}$, do you agree?

14. Mar 31, 2017

### dextercioby

From a geometrical perspective, writing equations in terms of $A^{\mu}$ is incorrect, for the true geometrical interpretation of relativistic electrodynamics on arbitrary Lorentzian spacetimes defines the e-m potential as a 1-form. Field theorists working with Standard Model on the flat Minkowski spacetime don't really care about the geometrical meanings behind, for they have the metric tensor which can easily convert co-vectors into vectors (component-wise). Only when one uses curved spacetimes, thus in a GR-compatible formulation of electrodynamics, one needs to separate forms from vectors. Dirac was not a geometer, but a field theorist, in his GR book, the field equations of electromagnetism are written using his semi-colon (I call it awful) notation as.

I rewrote them by switching the covariant derivative to a contravariant one and lowering indices to account for the 1-form character of the field. I will check my calculations over the weekend, but I think the lowering makes sense because the non-metricity is zero.

Last edited: Mar 31, 2017
15. Mar 31, 2017

### rubi

Yes, I would say so, because $A$ ist most naturally thought of as the pull-back of a connection form on some principal fiber bundle. (It is in principle possible not to use the connection formalism and still have a geometric formulation, but that seems less natural.) However, this won't make the dependence of the equations of motion on the background metric go away. An example of a gauge theory without dependence on a background metric would be Chern-Simons theory.

16. Mar 31, 2017

### Demystifier

Fine, but the crucial question is: Will the dependence on Ricci curvature go away?

17. Mar 31, 2017

### rubi

Well, the Ricci curvature is computed from the metric, so everything that depends on the metric also depends on the Ricci curvature and possibly other information about the metric that is not contained in the Ricci curvature. That's the case for the equation on the Wikipedia page: There is still a covariant derivative, which contains the Christoffel symbols. But that doesn't make the equation incorrect. The Wikipedia page just points out that the correct generalization of the wave equation to curved spacetime needs the curvature term in addition to the substitution $\partial\rightarrow\nabla$.

18. Mar 31, 2017

### Demystifier

The question is whether the Maxwell equation for $A_{\mu}$ contains second derivatives of the metric. I argue in post #10 that it does not. Physically that's important because the dependence on second and higher derivatives of the metric indicates violation of the equivalence principle.

19. Mar 31, 2017

### vanhees71

You don't need the Hodge dual, because you already work with $A_{\mu}$, i.e., you implicitly already fulfill the homogeneous Maxwell equations, which you don't need to state anymore. You can get pretty far with "premetric electromagnetism":

https://arxiv.org/abs/physics/0610221

20. Mar 31, 2017

### rubi

Of course $*j = \mathrm d * F = \mathrm d * \mathrm d A$ doesn't depend on the second derivatives of the metric. But if you write the equation in terms of covariant derivatives instead of exterior derivatives as on the Wikipedia page, you get the additional curvature term. It's just a different way to state the equations, which is closer to the ordinary textbook formulation of electromagnetism on flat spacetime. The equivalence principle is really independent of the number of derivatives of the metric, however we have different reasons for allowing only first derivatives in the Lagrangian (such as Ostrogradsky instability).

The Maxwell equations in terms of differential forms are $\mathrm d F = 0$ and $\mathrm d * F = *j$. From the homogeneous equation, we find $F=\mathrm d A$ (on $U$ with $H^2_\text{dR}(U)=0$). But that doesn't remove the Hode star from the second equation. All you can do is insert $A$ into the equation: $\mathrm d * \mathrm d A = * j$, but the dependence of the metric stays there of course.

Premetric electromagnetism is a different theory, which is not subject of this discussion.