# The Gravity of Photons

1. Feb 24, 2010

### blaksheep423

How would I go about finding the gravitational force produced by a photon? They are massless, but they must warp spacetime as all energy warps spacetime. What equations could I use to find the gravitational potential energy between two (very close together) photons in terms of their energy?

2. Feb 24, 2010

### jnorman

as i understand it (keep in mind that i am always wrong), a single photon has no defined location in spacetime between the time it is emitted and the time it is absorbed. without a location, it does not have a local effect on the gravitational field.

3. Feb 24, 2010

### Nabeshin

4. Feb 24, 2010

### Staff: Mentor

We don't have a quantum theory of gravity yet, so nobody can really answer your question. However, the http://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost" [Broken] would be the spacetime produced by a very brief pulse of light.

Last edited by a moderator: May 4, 2017
5. Feb 24, 2010

### bcrowell

Staff Emeritus
Let's say you have a photon of energy E at a distance r from a material particle of mass M. Then the force between them can be found by using mass-energy equivalence, $E=mc^2$, giving $F=GME/c^2r^2$.

6. Feb 24, 2010

### blaksheep423

It doesn't seem like this should be accurate. If photons are truly massless, then I don't think you can just use E/c^2 as a replacement. Then again, I don't really know at all, which is why I asked. Do einstein's field equations work for energy in any form? Could they be used?

7. Feb 24, 2010

### nicksauce

There is no reason why you could naively use Newton's law. You would have to find the stress energy tensor for a photon and solve Einstein's equations. But then things are tricky, since it's difficult to define a "gravitational potential energy" or a "force" in GR.

8. Feb 24, 2010

### Staff: Mentor

Most definitely not!
You are correct, you cannot plug the energy into Newton's law to get the behavior of massless radiation. Einstein's field equations do apply for energy in any form, but Einstein's field equations also include terms for momentum, pressure, stress, and energy flux. When you are dealing with massless radiation the momentum terms are not negligible. The general class of spacetimes dealing with massless radiation is called http://en.wikipedia.org/wiki/Pp-wave_spacetime" [Broken] with the specific one for a brief pulse being the ultraboost that I linked to earlier.

Last edited by a moderator: May 4, 2017
9. Feb 24, 2010

### bcrowell

Staff Emeritus
Why not?

I don't claim that it's perfectly accurate, but I think it will give roughly the right result.

You have a stress-energy tensor for the photon. It has a time-time component which is equal to the energy density of the photon. It also has space-space components. It's true that the spacetime surrounding the photon won't be exactly a Schwarzschild solution (due to the space-space part of the stress-energy tensor). But you will certainly get approximately the right answer this way.

One way to see that it can't be wildly wrong is that the equivalence principle tells us that the real and virtual photons inside a piece of matter have gravitational fields exactly like the gravitational fields contributed by the rest mass of the material particles. In matter, the space-space part of the stress-energy tensor averages out to zero by symmetry. The lack of this cancellation for a single photon will change the result somewhat, but not by an order of magnitude.

If the photon was bouncing back and forth between two mirrors, then the e.p. argument would become exact, and the photon's gravitational force would be given by exactly the expression I gave earlier.

Yet another way to see that it can't be wildly wrong is that there is simply no other way to put together the variables in the problem to give an answer with the right units and the right dependence on the variables (e.g., proportionality to E to the first power).

And I would claim that this would give the same result as the equation I gave, to a reasonable approximation.

The lack of a quantum theory of gravity is completely irrelevant here. It would only be relevant if there were some scale in the problem that was at the Planck scale.

Last edited by a moderator: May 4, 2017
10. Feb 24, 2010

### atyy

11. Feb 25, 2010

### Mute

$E = mc^2$ does not apply to massless particles. You cannot justifiably use it to get your result, even if it somehow happened to be correct. It may be approximately the correct result on dimensional grounds, but that's not a good enough reason to use $E = mc^2$ for the derivation when it's not correct.

12. Feb 25, 2010

### atyy

13. Feb 25, 2010

### Staff: Mentor

I already explained why not, the momentum terms are not negligible. Newton's formula is an approximation to GR only in a specific limit, namely the limit as M->0 and the limit as v<<c. Obviously for a photon v=c so the approximation is simply not valid.

14. Feb 25, 2010

### Ich

Hi bcrowell,

I'm with DaleSpam here, the Photon's gravity seems to be more like a Dirac-pulse. I mean, maybe you can get to a similar result if you include the finite propagation speed of gravity, and model some "sonic bang". But that's far from the formula you gave.
No, that's not right. The stress-energy tensor still includes pressure, wich is a space-space component (only the time-space components vanish). And pressure contributes to gravity: the source term for a perfect fluid like this is the trace of T, not the tt-component. Thus, a photon gas has double gravitation compared to pressureless dust of the same mass.
To get some closed system to push around and test the EP, you'll have to keep the pressure in some containment, which experiences tension and so cancels the effect of pressure.

15. Feb 25, 2010

### jnorman

excuse me, but let me repeat something i stated above.

a single photon does not a defined location in spacetime. with no location, it cannot have a gravitational effect.

if i am wrong about that, i really need someone to clarify for me. thanks.

16. Feb 25, 2010

### Ich

It has a defined location. You can't define a frame in which the photon is at rest, but in every other frame it has a well defined position at each time.

17. Feb 25, 2010

### DrGreg

The problem is, we do not currently have any theory that combines general relativistic effects with quantum effects, so the question for a single isolated photon, or a pair, can't be answered from a quantum theoretical point of view. (In quantum theory a photon's location cannot be measured with perfect precision. In fact, if you don't measure it, it can be in several places at once.)

So the best we can do is consider a "classical photon" (if that's not a contradiction) which behaves as a non-quantum particle.

The question is less problematic (from the quantum view) if we consider instead a collection of, say, a trillion photons whose collective behaviour can be modelled more accurately.

18. Feb 25, 2010

### George Jones

Staff Emeritus
There have been candidates put forward (by Margaret Hawton and others), but I do not think there is an accepted position operator for the photon in quantum theory.

19. Feb 25, 2010

### Redbelly98

Staff Emeritus
Are you referring to the fact that there is uncertainty in the photon's location? That is true of anything, including the Earth (which definitely has a gravitational effect.)

20. Feb 26, 2010

### Dmitry67

If matter annihilates with antimatter, the created photon gas *MUST* has exactly the same gravity as matter before, otherwise GR would be violated. Of course, rest mass is zero, so gravity is created by the pressure components of the tensor.

21. Feb 26, 2010

### Ich

Agreed to all, but here's the relativity forum. Please understand my statements as referring to a wave packet with negligible spatial extension, not zero extension. I think that's what is meant by a "photon" in a classical relativistic thought experiment (a null geodesic).

22. Feb 28, 2010

### jimgraber

To the OP or whoever did the calculations:
What actually happens has been determined by the famous observation of light deflection, and also gravitational lensing. According to GR, in the weak field limit, light is bent twice as much as in the Newtonian calculation, so your force will also be off by a factor of two if you are not careful. On the other hand, it will have the right dimensions and be of the right order of magnitude.

If you are interested to pursue this further, use the slightly more sophisticated formula

E^2 = m^2 c^4 + p^2 c^2

to get a more defensible result.

Best,
Jim Graber

23. Mar 1, 2010

### jnorman

all - no, i was not talking about uncertainty (HUP) in a photon's position (though that does apply to a degree, since we know the exact momentum of a photon, it's location is completely unknown). i am saying that a single photon does not have a location until it is detected/absorbed. for the responder who mentioned a "wave packet with negligible extension" - that is simply not the nature of a photon as i understand it. if a photon had negligible extension, you would not see the effects we see when we perform a dual slit experiment. until/unless a photon is detected/absorbed, it simply has no location - it only exists as a probability function. it is moving by "all possible paths" to whever it may be finally detected/absorbed.

further, since time does not exist in the reference frame of a photon (if a photon can be said to have a ref frame), distance also does not exist for a photon - implying that at the moment a photon is emitted, it is essentially everywhere in the universe at once - there is no distance between it and anything else.

again, if i am mis-stating something here, please straighten me out...

24. Mar 1, 2010

### Dmitry67

1 No measurement apparatus can measure the exact moment (frequncy) of photon.

2 Just point a laser beam somewhere. You can calculate the position of the beam. For example, about 1/2 seconds after it is emitted it will be about halfway between Earth and Moon (if you point it to the Moon)

3 Only if you dont focus the light it is going in all directions. Think again about the laser.

25. Mar 1, 2010

### jnorman

dmitry - from my understanding, given any particular atom emitting a photon, we know the exact energy required for an electron to drop to the next lowest level when it emits a photon, and thus we do know its frequency. using your laser analogy, "A photon of red-orange light from a HeNe laser has a wavelength of 632.8 nm. Using the equation gives a frequency of 4.738X1014 Hz or about 474 trillion cycle per second." so, we do know exactly the frequency of that photon, and thus cannot know anything about it's position, as per HUP.

however, a wiki article states this:
"Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate , and, thus, the normal Heisenberg uncertainty principle ΔxΔp > h / 2 does not pertain to photons."

here is a quote from an article on the copenhagen interpretation:
"It's more than simply saying we don't know which slit the photon passes through. The photon doesn't pass through just one slit at all. In other words, as the photon passes through the slits, not only don't we know it's location, it doesn't even have a location. It doesn't have a location until we observe it on the film. This paradox is the heart of what has come to be called the Copenhagen interpretation of quantum physics."

source - http://webs.morningside.edu/slaven/physics/uncertainty/uncertainty7.html [Broken]

also, "the photon is a bit of a problem, because it has turned out to be impossible to identify a position operator for it....These findings tell us two things: first, unlike electrons, photons really can't be localized to an arbitrary precision, and second, a position operator is meaningless because there really is no position to operate on. "

source - APS - http://pra.aps.org/abstract/PRA/v79/i3/e032112

Last edited by a moderator: May 4, 2017