# Does Hoyle C field violate conservation of energy-momentum?

1. May 17, 2015

### bcrowell

Staff Emeritus
The Einstein field equation is inconsistent unless we demand a divergence-free stress-energy tensor. This makes me think that Hoyle's steady-state cosmology is inconsistent with general relativity.

But Hawking and Ellis has this at p. 90:

I had always imagined that the C field was just some vague hand-waving by Hoyle. Is it really a field theory?

Is everything perfectly OK in classical relativity if we allow such a field?

Quantum-mechanically, is there really a reasonable field theory with such a field? I'm imagining a Dirac sea that isn't full. Not sure what state you would refer to as the vacuum. Are we talking about doing quantum mechanics with a spectrum of energy states that isn't bounded below? Don't Bad Things happen then?

If we're creating negative-energy C-field quanta in the steady-state theory, where do they all end up? Can we detect them? Don't their gravitational fields cancel out the gravitational fields of the hydrogen atoms being created?

2. May 18, 2015

### martinbn

I know almost nothing about it but my impression, from comments in books, is that the only problem within classical relativity is the violation of energy conditions.

http://arxiv.org/abs/astro-ph/0205064 (and some of the other papers by Narlikar)

On page 4 they have the stress energy tensor of the creation field

Last edited by a moderator: May 23, 2015
3. May 18, 2015

### bcrowell

Staff Emeritus
Thanks. Unfortunately, Narlikar is a kook, and that makes it more work to go through one of his papers and try to figure out what's not nonsense. Ned Wright has a nice discussion: http://www.astro.ucla.edu/~wright/stdystat.htm

4. May 18, 2015

### Staff: Mentor

The paper martinbn linked to proposes a scalar field with negative energy density. Mathematically, it works, and the covariant divergence of the resulting stress-energy tensor vanishes, so it does not violate conservation laws. Whether it's "reasonable" is a different question.

5. May 18, 2015

### martinbn

There is a little about it in Weinberg's book.

6. May 21, 2015

### bcrowell

Staff Emeritus
Do you think you could find a specific reference? I have vol I of his field theory book, but I couldn't find anything in the index.

There's a discussion in Coles and Lucchin, Cosmology, 2002, p. 57. They describe the C-field in classical terms as a modification to the Einstein field equations:

$G_{ij}+C_{ij} = 8\pi T_{ij}$

They don't mention anything about negative-energy quanta. I suppose the idea is that both $C_{ij}$ and $T_{ij}$ have nonvanishing divergences. Presumaly there is no physical significance attached to the fact that they write the C piece and the T piece separately; Hawking and Ellis consider it as just another matter field.

Coles and Lucchin say, "Hoyle suggested that $C_{ij}$ should be given by $C_{ij}=C_{;i;j}$[...]" The right-hand side is a second derivative of a scalar field C, which they say is defined as $C=-8\pi/H_0(\rho_0+p_0)t$, where $\rho_0 = 3H_0^2/8\pi$.

They don't define $p_0$. Two possibilities would be that it's numerically equal to $\rho_0$ or that it equals the actual average pressure in our universe. Since the two parameters are simply added, it seems that expressing them this way is merely for notational convenience; there is really only one adjustable parameter involved.

Apparently t is just a time coordinate. (This seems to check out in terms of dimensional analysis.) If so, then the equation assumes some preferred coordinate system, and I guess it would have to be the one defined by the Killing vector. This seems like it would clearly violate Lorentz invariance, since the local laws of physics would allow us to use local experiments in order to determine our motion in relation to the Hubble flow, without actually having to observe the Hubble flow. (I suppose the scalar field C itself is not meant to be directly observable, so its linear dependence on time wouldn't be observable.)

Last edited: May 21, 2015
7. May 21, 2015

### PAllen

I think he means Weinberg's Gravitation and Cosmology book. There does appear to be a section on this in that book.

[Chapter 14, section 8, per contents in 'look in book' on Amazon]

Last edited: May 21, 2015
8. May 21, 2015

### bcrowell

Staff Emeritus
Now that I think about it, any steady-state model of this flavor is bound to violate Lorentz invariance. For example, we could sit in a laboratory, observe a box full of vacuum, and wait for hydrogen atoms to appear. These atoms are in some state of motion, which must on the average be the motion of the Hubble flow. This makes the Hubble-flow frame a preferred frame.

9. May 21, 2015

### PAllen

But violating energy conditions rather generally allows Lorentz violations - they codify what it means for energy and momentum to be locally consistent with SR. For example, a blob of negative energy may have a spacelike world line. This fact is mentioned, for example, the the Wald-Gralla derivation of geodesic motion for test bodies from the field equations - it is necessary to either:

- assume timelike motion as an axiom (then geodesic motion can be derived)
- assume the dominant energy condition in order to derive timelike motion from the field equations

The older Geroch-Ehlers derivation has the same logical requirement: dominant energy condition is necessary and sufficient for timelike geodesic motion of test bodies.

10. May 21, 2015

### bcrowell

Staff Emeritus
I'm not following you here. You seem to be arguing that a negative-energy particle would be a tachyon, and that tachyons imply Lorentz violation. Tachyons would have imaginary mass, but are normally assumed to have real, positive energy. (For example, people speculated for a long time that neutrinos could be tachyonic, but it was known that when they deposited energy in a detector, the energy was positive.) And in any case tachyons do not imply Lorentz violation.

I also don't think it can be correct to say that energy conditions "codify what it means for energy and momentum to be locally consistent with SR." Dark energy violates various energy conditions, but it's locally consistent with SR. If violating energy conditions meant violating SR, then we could say that all the energy conditions must automatically be true. That's not the case. In fact, there are no energy conditions that are currently believed to be true in all cases: http://arxiv.org/pdf/gr-qc/0205066

It's been a while since I looked at the Ehlers-Geroch derivation, but IIRC the reason they need to assume energy conditions is that you can get effects such as an interaction between the spin of a particle and the curvature of the spacetime. In order to prove that that interaction has a negligible effect in the limit of small size, you need an energy condition.

Last edited: May 21, 2015
11. May 21, 2015

### PAllen

A negative energy particle could move ftl, but its description is not the same as a tachyon. Anyway, I see that what I meant was violating SR rather than Lorentz invariance per se. What constitutes SR violations is more a matter of opinion, but I take it to include matter of any typefollows a timelike trajectory, and that causality is observed - no influence outside the light cone.

[aside: your argument doesn't necessarily show Lorentz violation either since a frame being preferred by be able to detect motion relative to a field is not a Lorentz violation. The created atoms have a momentum determined by the field that gives rise to them. No different than the CMB picking out frames for which it is isotropic.]

12. May 21, 2015

### PAllen

No, there are papers on the necessity in Ehlers-Geroch that show spacelike motion is possible in the limit without the assumption the dominant energy condition.I am not aware of specific counter-example papers for Wald-Gralla, but their paper states the condition is necessary to ensure timelike motion.

The dominant energy condition amounts to saying that mass-energy cannot locally flow FTL. It is thus not surprising that assuming the converse allows ftl motion.

13. May 21, 2015

### bcrowell

Staff Emeritus
I don't dispute this. Certainly if you have matter that violates an energy condition, it may have spacelike world-lines.

I would agree if the field was one that followed Lorentz-invariant dynamics. But their massless, scalar C field doesn't follow the Lorentz-invariant dynamics you expect (massless Klein-Gordon equation). Instead, it simply evolves according to a prescribed equation, and that equation is coordinate-dependent. Their C field literally *is* the time coordinate (with a minus sign in front).

The CMB is different. It evolves according to Lorentz-invariant dynamics (Maxell's equations).

I wonder if their $C_{ij}$ has some simple interpretation, such as being the expansion tensor or something.

Last edited: May 21, 2015
14. May 21, 2015

### PAllen

This I have no opinion on, since I've never read a formal treatment of any steady state model. There also appear to be many of them, and Narlikar seems to change the model as soon as someone points out a flaw he can't avoid in the prior model. I found Weinberg actually has two sections on steady state models - one on Bondi-Gold, one on Hoyle's model. From 'look in book' feature, the section I mentioned above was on the Bond-Gold model.

15. May 21, 2015

### bcrowell

Staff Emeritus
What I gave in #6 is the complete classical description according to Coles and Lucchin. There's really no more to it than that. You write down a field C which is basically just a time coordinate. You take its second covariant derivative and call that a stress-energy.

16. May 21, 2015

### PAllen

I can still see a way to argue no Lorentz violation. The C field is given in some coordinates, but if you assume you get its description (locally) in other frames by Lorentz transform, then it is (arguably) Lorentz invariant.

17. May 21, 2015

### bcrowell

Staff Emeritus
Sorry, not buying it :-) That's much weaker than what Lorentz invariance really says.

18. May 21, 2015

### PAllen

Doesn't seem so bad. Consider an isolated charge. It has a uniquely simple field description (static, Coulomb) in one frame, and you can (without specifying any specific law) derive what it is in other frame's by Lorentz transform. So, the C field is specified in standard cosmological coordinates, where it takes its simplest form. As a physical field, there is, nothing, per se, wrong with being able to detect (directly or indirectly) you motion 'relative to it' [ more precisely, relative to an observer who sees its simplest form - e.g. created atoms have no momentum, for example]. Just like for an isolated charge, even an observer who only sees the field can distinguish whether they are in motion relative to the charge.

19. May 21, 2015

### bcrowell

Staff Emeritus
Your example is Lorentz-invariant because Maxwell's equations are Lorentz-invariant. The charge is a physical body, it has dynamics, and you're measuring your state of motion relative to it. That's completely different from simply prescribing the electric field to have some fixed behavior in some special coordinates.

You can make tensor-scalar theories that are Lorentz-invariant. Brans-Dicke gravity is an example. The C field isn't, at least according to the description given by Coles.

20. May 21, 2015

### Staff: Mentor

That's not necessarily the same as violating Lorentz invariance. See below.

It depends on how the $C$ field, or the $t$ quantity, transforms under local Lorentz transformations. If it transforms in such a way as to leave the laws of physics invariant, then it's Lorentz invariant. In other words, $t$ would have to be a scalar field on spacetime with the dimensions of time. This is possible AFAIK, but I don't know enough about the Coles model to know if that's what they are specifying.

If $t$ is in fact a scalar field, then the fact that it can also function as a time coordinate in a "preferred" coordinate system does not break Lorentz invariance. It just means that there is a particular coordinate chart that is adapted to the underlying symmetry of the spacetime.