# Are photons affected differently by gravity?

1. Apr 2, 2010

### Nethral

I haven't found any source confirming my thoughts so I guess I'm wrong, but I was hoping someone could explain it to me :) This is how I'm thinking...

Mass <=> energy, and therefore you could say that energy is also affected by gravity (light, for example). Then more energetic light should be more affected by gravity than less energetic light, like in the way a heavier object is more affected by the earth than a lighter object.

Or am I wrong? For example, is it wrong to say that "energy is affected by gravity"? I guess the common explanation is that light just follows a straight path through universe, thereby being "affected" by bent space-time/gravity...

Hope someone can explain this to me. Thanks in advance!

2. Apr 2, 2010

### bcrowell

Staff Emeritus
There's nothing wrong with this.

Example #1: A 1-joule flash of light passes by the earth.

Example #2: A 2-joule flash of light passes by the earth, at the same distance of closest approach.

The change in the earth's momentum is twice as much in example #2. An observer who sees this change in momentum will infer that the earth was subjected to twice as much force.

This is just a different way of describing the same thing.

3. Apr 2, 2010

### Nethral

Thank you, bcrowell!

I kind of understand your examples but also kind of... not. For example, I have learned that photons are affected by gravity by gravitational redshift and bending. Would then blue light be more bent than red light? Or would blue light have a greater gravitational redshift/energy loss?

4. Apr 2, 2010

### bcrowell

Staff Emeritus
No. General relativity is a geometrical theory. Think of the axiom in Euclidean geometry that says that a line segment can always be extended in a unique way. The same is true in GR. In GR, the "lines" are geodesics, which are interpreted as the world-lines of particles that aren't subjected to any nongravitational forces. Since a geodesic is uniquely determined by any initial segment, you can't have different geodesics for red and blue light.

No. The fractional energy change $\Delta E/E$ is the same regardless of initial color. One way to see this is that you can interpret the frequency change as a time dilation effect, and time dilation is universal in the sense that it doesn't matter what clock you use to measure it.

5. Apr 2, 2010

### Altabeh

Energy of photon does not have any interference with the bending of its trajectory in a strong gravitational field. This is because the geodesic equations have no inclusion of mass, but only the components of the proper 4-velocity, metric and its first derivatives.

AB

6. Apr 3, 2010

### cragar

would blue photons creates a stronger gravitational field then red photons ?

7. Apr 3, 2010

### bcrowell

Staff Emeritus
Yes, for a fixed number of photons, because the energy would be greater.

8. Apr 3, 2010

### utesfan100

Suppose we have a "purple" photon make of a red photon at 700 hertz and a blue photon at 400 hertz moving at the same spot in the same direction.

Would this pair of photons stay coherent in a gravitational lens? The observation of star light shows no angular defect due to gravitational lenses, so experimentally the answer is not as far as we can measure.

The force of gravity should be higher for the blue photon than for the red photon by a factor of 7/4.

The momentum for the blue photon will also be 7/4 that of the red photon. Force is dP/dt.

The bending is caused by the force in the direction perpendicular to the motion. This is P*sin(angle of approach). Since gravitational force is proportional to momentum for a photon its energy/frequency does not alter its path. This is like the mass not changing the path of an object in classical mechanics, as the mass of the orbiting object cancels out of the equations of motion.

9. Apr 3, 2010

### Altabeh

How are photons able to produce gravitational field with a vanishing rest-mass? Let alone the energy of the field. The gravitational correction to the Minkowski spacetime in the case of a photon is zero due to its zero mass!

AB

10. Apr 3, 2010

### Frame Dragger

I feel as though this is yet another place where The Parable of The Apple from MTW can do some good for the OP. Granted, it's a bit of a step back, but I think we may have lost the OP there.

@Nethral: From Misner, Thorne, and Wheeler's "Gravitiation:
@Altabeh: They may have no (or miniscule) 'rest-mass', but energy/stress/momentum deforms spacetime. I don't see why that would be different for light; we already know it's SUBJECT to gravitational fields, and I was under the impression that was always (excepting theoretical constructs) a mutual effect. Light still follows a geodesic informed and deformed by local geometry, or so it seems, and if we accept photons as discrete quanta of energy, then the SET has to kick in.

11. Apr 3, 2010

### atyy

How? Classically, the electromagnetic field should curve spacetime as a solution of Einstein-Maxwell equations. But does frequency enter the classical energy-momentum tensor? Photon energy being proportional to frequency seems to be from quantum field theory. But in QFT on curved spacetime, the field doesn't contribute to spacetime curvature.

12. Apr 3, 2010

### bcrowell

Staff Emeritus
No, frequency doesn't. But for a fixed number of photons, the energy is proportional to the frequency.

13. Apr 3, 2010

### Frame Dragger

@bcrowell: Do you know of any good papers on this? I'd love to read some experimental and theoretical thinking on this.

14. Apr 3, 2010

### bcrowell

Staff Emeritus
Isn't this all pretty standard textbook stuff?

15. Apr 3, 2010

### atyy

What does the formalism look like? Is it QED + Donoghue's GR as an effective QFT?

16. Apr 3, 2010

### bcrowell

Staff Emeritus
I'm claiming that there's nothing in any of this beyond freshman relativity. If you keep the number of photons constant while increasing the frequency, you get more mass-energy. A greater mass-energy means proportionately stronger gravitational interactions.

17. Apr 3, 2010

### Frame Dragger

Ah, that's not in line with my (albiet limited) understanding of QED, nor is it "textbook".

So, if Bob sends a radio signal to Alice, and Alice and Bob are converging (so, blue-shift), its participation in the gravitational interaction becomes stronger as it shifts from radio, through visible, into UV and beyond?

18. Apr 3, 2010

### bcrowell

Staff Emeritus
This seems like a much less conceptually straightforward case to me. In your example, you have a whole bunch of other factors that are changing, not just the energy of the radio beam.

The original example seems very simple and straightforward to me. If you want a slightly more detailed version of the argument in #16, here you go. In the limit of weak fields, GR is to an excellent approximation a linear theory, so that we can think in terms of superposing the gravitational field of a ray of light on top of some background field. The light's field is some (non-Schwarzschild) metric, and its $g-\eta$ superposes with the $g-\eta$ of the background field, where $\eta$ is the Minkowski metric. The light beam's $g-\eta$ is (in the limit of weak-field GR) linearly proportional to the stress-energy tensor, and the stress-energy tensor is in turn linearly proportional to the energy of the beam of light.

19. Apr 3, 2010

### Frame Dragger

Well, that does seem very straightforward and "freshman". Thanks bcrowell, sometimes it's tough sorting out the Classical from the Semi-Classical from the Quantum, etc... etc...

20. Apr 3, 2010

### yuiop

It is probably better to think of it as the Earth being more affected (accelerated) by the heavier object than the lighter object. The heavier object on the other hand is affected exactly the same as the lighter object by the Earth and both are accelerated towards the Earth at exactly the same rate. The rate that a particle is accelerated towards the Earth is independent of its rest mass and so even a particle with zero rest mass (such as a photon) is accelerated like any other particle.

The amount of gravitational deflection is independent of the rest mass of the test particle but it does depend on the horizontal velocity of the particle. All photons have the same velocity, so we can confidently say the red photon and the blue photon will deflect to exactly the same extent and so there is no chromatic or prismatic aberration in a gravitational lens unlike an uncorrected glass lens and so no galactic rainbows due to gravitational lensing.