# Can photons create their own gravity?

1. May 9, 2009

### nuby

Question.. I understand photons will curve around planets in space due to their gravity (or curved space-time). But do photons create their own gravity relative to their energy? From what I can gather, gravity is more of an aspect of energy than mass.

2. May 9, 2009

### mathman

From the point of view of general relativity, mass and energy are equivalent, so photons have a gravitational effect.

3. May 9, 2009

### nuby

Thanks, Mathman.

Does this imply photons with more energy have more gravity, and will 'react' differently to a planet/star's gravity? Has this been observed?

4. May 10, 2009

### Harenil

The curvature due to photon energy would be extremely negligible, and thus, very very hard to measure. It's hard enough to measure the curvature from a massive object.

5. May 10, 2009

### meopemuk

It has been observed that photon's deflection angle does not depend on its energy/momentum (or wavelength). Otherwise, the (displaced) image of the distant star would be seen as a little rainbow. It fact, it remains point-like in spite of the light bending. The deflection angle is (approximately) $$\Delta p/p$$, where p is photon's momentum and $$\Delta p$$ is the transversal momentum change in the gravity field. From this you can conclude that $$\Delta p$$ is proportional to p. So, yes, higher energy photons are deflected more.

6. May 10, 2009

### gonegahgah

It is my understanding that a minute amount of rainbowing does occur but that this is attributed to refraction by the Sun's chromosphere and is computationally removed to produce a more accurate image. Is that right?

7. May 10, 2009

### meopemuk

I am not aware of such experimental details. On the other hand, it is plausible that the proportionality between $$\Delta p$$ and p holds only approximately, so some minor rainbowing may be present in exact theory, even if the chromosphere effect is not present.

8. May 12, 2009

### michelcolman

I thought photons just followed a "straight" path through curved space time (or whatever is closest to straight), so why would their energy make any difference? Any photon travels at the same speed and flies through the same curved space-time along the shortest path.

As for the gravity created by photons: yes, photons don't have mass but they do have energy and any kind of energy (even including gravity itself) causes gravity. In fact, the amount of gravity coming from the photons inside the sun is probably far from negligible. They have so much energy that their pressure is even important in helping to keep the sun from collapsing, so we probably feel their gravity quite noticeably too.

9. May 12, 2009

### Lasand

Last edited by a moderator: May 4, 2017
10. May 13, 2009

### tgramling

This brings up another controversial matter though, because according to Einstein energy and mass are equivalent, so saying that photons have no mass would in turn say that they have no energy, and vise versa, in saying they have energy says they have mass, so either Einstein was "wrong" or there is something about photons that we still don't know

Last edited: May 13, 2009
11. May 13, 2009

### Phrak

"The Photon" has no mass. the mass of "The Photon" can be found to be exactly equal to zero in a table of elementry particles. On the other hand, photons have mass. A culminated beam of photons has virtually no mass. A perfect planar wave has no mass. A bunch of photons distributed in a spherical box curves space exactly as you would expect from a bunch of atoms distributed in the same box.

Specifically, it is not just mass, but the stress energy tensor that curves space in 10 different ways. A perfectly culminated beam of photons curves space differently than, say, a rod of solid matter with velocity along its length. Unfortunately, I don't know the details of the difference.

Last edited: May 13, 2009
12. May 13, 2009

### granpa

I would expect light to produce the same gravitational field that an equivalent amount of matter+antimatter together would produce.

light being a wave should refract around massive objects due to gravitational time dilation. didnt some russian guy prove that it could all be accounted for by time dilation?

13. May 13, 2009

### meopemuk

The correct statement of special relativity is that energy (E), mass (m) and momentum (p) are inter-related. They are connected by formula

$$E= \sqrt{m^2c^4 + p^2c^2}$$

If a massive body is at rest (p=0), then you'll get the famous energy-mass relationship

$$E= mc^2$$

If a body has zero mass (like photons), then another famous relationship follows

$$E= pc$$

So, massless particles can have non-zero energy.

Last edited: May 14, 2009
14. May 13, 2009

### tnho

To be a bit clearer

"photon has zero rest mass, which is the mass we measured in its co-moving frame although it is not possible to do so.
In general, photon do has mass which is due to its momentum, namely $$E=pc$$.
With non-zero mass, there should have non-zero gravity created.

Here i have a thought experiment:
Suppose, in a free space, a pair of laser beams (sharp enough of course) are shot out parallel to each other with a distance apart. Will these two beams converge together eventually? it should happen, isn't it?

------------------------------------------------------
ooooops~! i was a bit slower ...

Last edited: May 13, 2009
15. May 14, 2009

### gonegahgah

May I please ask two questions? 1. When photons travel together are they at rest with respect to each other? 2. Does zero rest mass mean that photons traveling together don't feel gravity towards each other?

16. May 14, 2009

### Staff: Mentor

I think you mean "collimated."

"Culminated" is a genuine word, but it has an entirely different meaning.

17. May 14, 2009

### Phrak

Oh well. What's an ell between friends...and a u.

I object strongly. It really ought to be 'columnate'. I will rite the editors.

18. May 15, 2009

### Naty1

Photons do follow a straight path, but "straight" depends on the local curvature of spacetime which IS affected by local energies and masses....more energy means more curvature which means more deflection....

19. May 15, 2009

### Naty1

1. yes, like two cars each going along the highway at 55MPH.

2. according to post #13, it looks that way...

20. May 16, 2009

### michelcolman

1. From the point of view of a photon, everything is stationary. But that's a bit of a strange way of looking at things, because the "point of view of a photon" is not really a valid reference frame. Lengths are infinitely contracted, so the photon arrives at its destination (even if it's billions of light years away) immediately after departure (after zero time measured by the photon). The entire universe becomes two-dimensional, and the photon travels across its thickness (which is zero). If this doesn't appear to make sense, you will understand why people say this is not a "valid" reference frame. Observers can only travel slower than light.
2. No, gravity is caused by energy. Mass is just one kind of energy, but it's so huge that we used to think this was the only cause for gravity. However, other kinds of energy, like pressure for example, cause gravity too. A photon definitely has energy, so it does cause a small amount of gravity.

21. May 16, 2009

### Staff: Mentor

Calculate it for yourself using the relativistic "velocity addition" equation:

$$u^{\prime} = \frac {u - v} {1 - \frac {uv}{c^2}}$$

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html

v is the velocity of "observer B" (one of the photons) with respect to "observer A" (you), which is c, of course. u is the velocity of the other photon with respect to you (which is also c, of course). u' is the velocity of second photon with respect to the first one. What do you get for it?

And momentum. Remember in GR, the curvature of space-time is determined by the stress-energy tensor, whose components include both energy and the components of momentum.

22. May 16, 2009

### ImAnEngineer

So that's 0/0, and is undefined? How can that be explained physically?

23. May 16, 2009

### Staff: Mentor

To look at this from another "direction", take the inverse formula

$$u = \frac {u + v^{\prime}} {1 + \frac {uv^{\prime}}{c^2}}$$

and substitute u = c and v = c as before. You get the equation

$$c = \frac {c + v^{\prime}} {1 + \frac {v^{\prime}}{c}}$$

No matter what you substitute for v' (except v' = -c), it satisfies this equation. So v' can be almost anything. There is simply no well-defined answer to the question, "what is the velocity of the second photon, from the 'point of view' of the first photon?"

I think most physicists would say that this means that it doesn't make sense to talk about the "point of view" of a photon, at least in the sense of an inertial reference frame (x, y, z, t coordinate system in which Newton's First Law is valid) in which the photon is at rest.

24. May 16, 2009

### nutgeb

Here's another way to think about a photon's gravity. Think of an enormous collection of photons.

Specifically, in the first 20,000 years or so after the Big Bang the universe is believed to have been radiation dominated, meaning that any large region of the universe contained far more mass-energy in the form of free radiation than in the form of matter. The gravity of this radiation caused the original expansion of the universe to decelerate dramatically. In fact, because radiation has positive "pressure" equal to 1/3 of its mass-energy (w=1/3), free radiation causes twice as much gravity (density + 3*pressure) as would the same amount of mass-energy in the form of matter. Relative to its total mass-energy, the early universe decelerated gravitationally at twice the rate it would have if it were matter-dominated.

Each photon's individual mass-energy and pressure decay through redshift, at a rate proportional to the expansion of the universe, 1/a. Which is why the universe is no longer radiation-dominated; by comparison, the mass-energy of an atom of matter does not decay in this way.

Last edited: May 16, 2009
25. May 16, 2009

### DrGreg

Einstein's 2nd postulate implies the speed of any photon is always c in any inertial frame, therefore it's impossible to find an inertial frame in which a photon is at rest.

The fact the equations quoted early give an indeterminate answer (0/0) is the mathematics' way of telling you "you can't use that equation here".