# Can something with zero rest mass (photon) curve spacetime?

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1. Jun 22, 2013

### 49ers2013Champ

My understanding is that the presence of energy and matter curve spacetime.

Is a photon considered energy?

If so, how can it curve spacetime while having zero rest mass?

2. Jun 22, 2013

### tiny-tim

hi 49ers2013Champ!
correct
yes
it doesn't have zero rest mass, it doesn't have a rest mass

3. Jun 22, 2013

### Staff: Mentor

It does have zero invariant mass, though, which mathematically is the same thing as "rest mass" for objects that can be at rest; but calling it invariant mass makes it clear that the concept also applies to photons.

4. Jun 22, 2013

### 49ers2013Champ

Tiny-tim:

Can we calculate the degrees of curvature created by photons? Certainly they are much less influential than the ones created by, say, the sun and earth, right?

5. Jun 22, 2013

### bobc2

Interesting question you bring up, 49er (what a season!). Keep in mind that the photon is not at rest. And yes, a photon has energy equal to (Planck's constant) x (frequency of the photon).

I think Planck's constant is approximately 6 x 10^-27 erg-sec, a quite small number, so in the scheme of things the curvature is quite negligible.

6. Jun 22, 2013

### WannabeNewton

Rest mass (if present) isn't the only thing that contributes to space-time curvature. The Einstein equations (with vanishing cosmological constant) read $R_{ab} - \frac{1}{2}g_{ab}R = 8\pi T_{ab}$ where $R_{ab}$ is the Ricci curvature tensor, $R$ is the Ricci scalar curvature, and $T_{ab}$ is of course the energy-momentum tensor. The point is that the space-time curvature couples not only to rest mass (if present) but also to energy density, momentum density, and stresses from shear forces and pressure, as codified by $T_{ab}$.

An electromagnetic field has no rest mass in the classical sense but it still carries energy and momentum and contributes to space-time curvature; conversely the electromagnetic field itself couples to the space-time geometry (the metric tensor).

7. Jun 22, 2013

### 49ers2013Champ

But, Wannabe, you agree with Bob that the curvature is very small, right? That's what I'm trying to understand right now.

The large curvatures in spacetime come from things with large rest mass (Earth, Sun, Jupiter, Cornell's running back, etc.), right?

Last edited: Jun 22, 2013
8. Jun 22, 2013

### Staff: Mentor

6.626 069 57 x 10-34 J s according to NIST:

http://physics.nist.gov/cgi-bin/cuu/Value?h

One J s = 10^7 erg-sec, so yes, 6 x 10^27 erg-sec is the right order of magnitude.

For any photon frequency we have observed, yes. The most energetic gamma rays ever observed have frequencies of about 10^27 cycles/sec, which equates to an energy of about 1 erg or 10^-7 J. That's the equivalent of a rest mass of about 10^-20 grams.

9. Jun 22, 2013

### WannabeNewton

Not necessarily. There is quite a bit of freedom in the field equations with regards to what kind of space-time geometries can be generated by mass-energy distributions. Only in the Newtonian limit does the rest energy-density unequivocally dominate the field equations.

The problem is that there is no trivial way to quantify the "strength" of space-time curvature in the manner of which you speak and it is even less trivial to compare space-time curvature of different solutions (I'm not even sure if this would have any physical meaning). The relevant solution is the pp-wave metric: http://en.wikipedia.org/wiki/Pp-wave_spacetime

10. Jun 22, 2013

### Bill_K

I do not agree that the curvature produced by a photon is very small. The question is meaningless, for the following reasons. A photon has no rest frame. The energy of a photon depends on which rest frame you choose. The same photon may be infrared in one frame and ultraviolet in another. A photon's energy is one component of a tensor, which is another way of saying the same thing - it's frame dependent. Photons detected in cosmic rays have been observed with incredibly high energies, above 1018 eV.

Spacetime curvature is also described by a tensor, the Riemann tensor. Objects with rest mass, and objects with no rest mass produce curvature of different types. The curvature tensor of a Schwarzschild particle is called Type D, while the curvature produced by a massless object like a photon is called Type N. There is no frame-invariant way to say how large or small a Type N curvature is. So the claim that a photon produces a "small" curvature is meaningless. It's "not even wrong".

11. Jun 22, 2013

### 49ers2013Champ

So Peter and Bob agree that the curvature created by the presence of a photon is negligible, but WN and Bill are saying it's not necassarily the case?

Peter, please come clean this up. It's what you do!

12. Jun 22, 2013

### Staff: Mentor

Agreed. All the statements I made about energies of photons were relative to Earth's rest frame.

Yes, but as I noted, that's still a small amount of energy by ordinary standards. (And as you note, it's frame-dependent anyway.)

I assume this is because there are no non-vanishing scalar curvature invariants for Type N spacetimes? If so, I agree that makes the issue less straightforward than it is with ordinary massive objects.

However, there are still ways of approaching the question in a practical sense. For example: consider the effective stress-energy tensor that is used in cosmology to model the universe as a whole. How large is the contribution of photons to this effective SET? The answer is, very small, much smaller than any of the other contributions we know of (ordinary matter, dark matter, and dark energy). It's true that there was a period, the radiation-dominated era, in which the photon contribution was the largest, but that period ended, IIRC, a few hundred thousand years after the Big Bang.

Another example: what is the contribution of photons in the Sun to the Sun's observed mass? I haven't seen an actual computation of this, but my educated guess is, again, very small, much smaller than the contribution of the Sun's hydrogen and helium. (An interesting question is how large the photon contribution is compared to the contribution of ordinary kinetic pressure inside the Sun. I haven't seen this calculated either; if I have time I may try to do a back of the envelope estimate if nobody can give a link to a computation already done.)

So while I agree that, in principle, there is no way to invariantly quantify the amount of curvature produced by a photon, I think there are ways of quantifying, in practical terms, the contribution of photons to overall curvature in scenarios of interest.

13. Jun 22, 2013

### 49ers2013Champ

Peter, as always, thanks for contributing to a thread of mine. Bill, thanks for your insight. I look forward to reading your response to Peter's notes, if you have one.

14. Jun 22, 2013

### pervect

Staff Emeritus
The idea that "mass" causes gravity is a holdover idea from Newtonian mechanics. In GR, it's the stress energy tensor that "causes gravity", in the sense of being on the right hand side of Einstein's equation. (This is perhaps not as strong a notion as it might first appear, but frankly this post is already getting complex enough, I don't want to deal with anything more at the moment, so will stick with this particular notion of cause for now.)

Mass turns out to be a slippery quantity to even define in GR.

If you stick with the stress energy tensor, though you can easily say that the photon has a nonzero one. More specifically, you can say it has energy, momentum, and pressure, all of which are components of the stress energy tensor, and all of which contribute to the gravitational field.

The idea that momentum, and even pressure, contribute to gravity is part of Einstein's theory that's NOT part of Newton's theory.

15. Jun 23, 2013

### 49ers2013Champ

Pervect,

The strength of the stress-energy tensor is dictated by how much energy or matter is present, right?

Addtionally, are you saying that the mass of an object has nothing to do with the curvature of spacetime?

(Bob, thanks for the 49er shoutout. I have us going undefeated this year--sorry, Seahawk fans.)

16. Jun 24, 2013

### BruceW

Exactly. At one point, the 'standard model' of cosmology says that the stress-energy tensor was dominated by radiation (with zero rest mass). So in that case, normal matter produced a negligible curvature to spacetime.

you were asking Pervect, but I thought I'd jump in. Yes, in a sense invariant mass has nothing to do with curvature of spacetime, since the invariant mass does not enter into the equation for the stress-energy tensor. Only the relativistic mass (a.k.a. energy), momentum, stress and pressure.

17. Jun 24, 2013

### pervect

Staff Emeritus
More or less - the stress energy tensor has more than one component. The density of energy (as seen by a specific observer) is one component of the tensor. But there are other components as well. Terms like momentum, and pressure.

You may or may not be aware that momentum and energy in special relativity "intermix". Similar things happen with the different components of the stress energy tensor.

The really rigorous answer is that we don't have a good definition of mass in general relativity. People assume that such a fundamental concept must have a good definition, but so far nobody's found one. While we don't have a good defintion of mass, we DO have a good defintion of the stress-energy tensor. That's a major reason why people use it.

The less rigorous answer is that while we don't have a perfect defintion of mass, we've got some defintions that work in particular cases, even if they don't work in general. But even in such cases where we can define mass, it does not tell the whole story. Certainly it contributes to curvature - but so do things that are not mass, such as momentum, and pressure.

18. Jun 24, 2013

### PAllen

"Nothing to do with" is a little strong. It isn't a direct source term, however it approximates ADM mass for certain simple systems (these also have the feature that Bondi mass = ADM mass). Roughly:

- if Bondi mass = ADM mass (radiation is insignificant, including GW), and self gravitation effects are insignificant (thus dust with relativistically moving particles; clusters of comets; but not planets or stars), then it turns out that ADM mass is given by invariant mass (of the system) to a very good approximation. ADM mass is total curvature measured at asymptotic infinity. Oh, that's another limitation - the approximate equivalence is only good for asymptotically flat spacetime.

This fact relates to why two co-moving bodies (no matter how fast they move per some observer) have no extra tendency to form a BH (obviously, because an observer change of motion can't make something else more likely to become a BH). However, two bodies with approaching each other at a close flyby trajectory at near c can form a BH without touching (initially). This would happen when their invariant mass at closest approach exceeds the threshold set by the hoop conjecture (which I assume to be true for this argument).

[edit: a clarification on the flyby example: if the invariant mass computed for the objects as a system, far away from each other, significantly exceeds the hoop conjecture criterion for their closest approach computed per SR, then they are likely to capture each other and form a BH. The reason excess is needed is that significant GW will be emitted as they approach.]

Last edited: Jun 24, 2013
19. Jun 24, 2013

### 49ers2013Champ

Are you gentlemen just saying that the numbers that are plugged into the GR equations can vary from observer to observer, that there is no definite set of numbers (or quantitites...whatever the correct mathematical word is--I'm talking about mathematical language that represents measurements of momentum, energy, and pressure) that we can use to perfectly solve his field equations?

Is that perhaps what is meant by the quote from Walter Isaacson's book: "...the general relativity of all forms of motion...."?

20. Jun 24, 2013

### WannabeNewton

Yes the components of the tensors involved are not frame independent, only the geometric objects (the tensors) themselves are.