Are photons affected differently by gravity?

[bcrowell]

I mean this is probably crazy to bring this up , But wouldn't a heavier object distort space more so the force between that object and the Earth would be a little greater , I mean the effect would be very small and negligible , but I am just wondering , There was like a thread a month ago about , shouldn't heavier objects fall faster based on this idea. I mean this would obviously make sense with 2 large bodies coming together.

GR is a nonlinear theory.

In the limit where we consider a test particle of very small mass, the test particle moves along a geodesic. This is required by the correspondence principle, since universality of trajectories for low-mass objects has already been observed in Eotvos experiments and verified in the context of Newton's laws and Newton's law of gravity.

When you're no longer in the limit of a low-mass object, it's no longer true that the trajectory is a geodesic; for instance, you can get effects like gravitational radiation.

 

[Frame Dragger]

I don't think you need any quantum-mechanical ideas at all in order to understand why the vacuum is nondispersive. Does my explanation in #4 work for you?

There really is no luminiferous aether is there papa?!

 

[atyy]

I don't think you need any quantum-mechanical ideas at all in order to understand why the vacuum is nondispersive. Does my explanation in #4 work for you?

No, because if light of different wavelengths bend spacetime differently, then the curvature would be different, so the geodesics would be different.

 

[yuiop]

No, because if light of different wavelengths bend spacetime differently, then the curvature would be different, so the geodesics would be different.

The equivalence principle tells you that the paths of light with different wavelengths in a gravitational field must be exactly the same.

In the falling elevator thought experiment, two horizontal laser beams of different wavelength, must appear to follow the same straight path as they would in a "stationary" elevator in flat space.

In an accelerating rocket in flat space thought experiment, two laser beams of different wavelength directed orthogonally to the acceleration of the rocket must follow the same path, because they have no reason to do anything else in flat space. In this case both beams must be intercepted by the rear of the accelerating rocket at exactly the same time so they must appear to fall at exactly the same rate in the accelerating rocket and by the EP they must fall at the same rate in a gravitational field.

 

[atyy]

The equivalence principle tells you that the paths of light with different wavelengths in a gravitational field must be exactly the same.

In the falling elevator thought experiment, two horizontal laser beams of different wavelength, must appear to follow the same straight path as they would in a "stationary" elevator in flat space.

In an accelerating rocket in flat space thought experiment, two laser beams of different wavelength directed orthogonally to the acceleration of the rocket must follow the same path, because they have no reason to do anything else in flat space. In this case both beams must be intercepted by the rear of the accelerating rocket at exactly the same time so they must appear to fall at exactly the same rate in the accelerating rocket and by the EP they must fall at the same rate in a gravitational field.

Yes, I'm aware of that. Certainly this is true in the formalism GR + geodesic equation for test particles such as test photons. But in that formalism, the test photons do not contribute to spacetime curvature. I'm trying to understand if it is true that light of different frequencies bend spacetime differently, because from semiclassical old QM, we expect E=hf, and heuristically, E contributes to spacetime curvature.

 

[yuiop]

Yes, I'm aware of that. Certainly this is true in the formalism GR + geodesic equation for test particles such as test photons. But in that formalism, the test photons do not contribute to spacetime curvature. I'm trying to understand if it is true that light of different frequencies bend spacetime differently, because from semiclassical old QM, we expect E=hf, and heuristically, E contributes to spacetime curvature.

Well, as I implied in #28, if photons of different wavelength or frequency travel through space at different times, there will be a slight difference in the paths, but if they travel alongside each other at the same time they will follow exactly the same path as each other.

Yes, I'm aware of that. ...

I am sure you were. The post was not directed directly at you. Most of the stuff I post is for the general benefit of less advanced members, basically because that is what I am too. No offence intended.

 

[Jonathan Scott]

Note that there is no hard distinction between photons and objects with rest mass when it comes to gravity. Something massive moving at very very nearly c will follow essentially the same path as a photon.

Any object (with or without rest mass) which is small enough to be considered a test object compared with the object generating the gravitational field is accelerated in the same way when traveling with approximately the same velocity, regardless of its total energy.

 

[starthaus]

No, because if light of different wavelengths bend spacetime differently, then the curvature would be different, so the geodesics would be different.

Math gives us the precise answer to your question.
The trajectory of a non-charged test particle in a non-rotating gravitational field is given by:

d^2u/dphi^2+u=m/h^2+3mu^2

where h=angular momentum/unit of rest mass and m=GM/c^2 is related to the Schwarzschild radius

For ANY frequency photon, rest mass=0 so h=infinity

The equation becomes:

d^2u/dphi^2+u=3mu^2

Neither the equation nor the solution depend on the nature of the photons. This is confirmed experimentally by the absence of any difraction ("rainbowing") in any of the experiments measuring the starlight deflection by massive bodies. After all, starlight is white light, made up of all frequencies photons.

 

[bcrowell]

No, because if light of different wavelengths bend spacetime differently, then the curvature would be different, so the geodesics would be different.

The only reason we talk about geodesics is that GR is a geometrical theory, and the only reason Einstein chose a geometrical formulation for GR is because different test particles follow the same trajectories.

Yes, I'm aware of that. Certainly this is true in the formalism GR + geodesic equation for test particles such as test photons. But in that formalism, the test photons do not contribute to spacetime curvature. I'm trying to understand if it is true that light of different frequencies bend spacetime differently, because from semiclassical old QM, we expect E=hf, and heuristically, E contributes to spacetime curvature.

There is nothing inherently quantum-mechanical about this issue. Anything with a sufficiently high mass-energy will follow a trajectory that differs significantly from a geodesic. "Sufficiently high" means really, really high, e.g., like one of the members of the Hulse-Taylor pulsar, which don't follow geodesics because they're radiating gravitational waves. (If you replace one of the members of that system with a body having half the mass, it would have a different trajectory, because it would radiate at a much lower rate.) A single photon has such a low mass-energy that there's no way you'd ever be able to detect its deviation from a geodesic.

 

[Frame Dragger]

For what it's worth kev, I learn every day from the to and fro that goes on here. I learn when people are right, wrong, and especially when they just can't agree. Thanks for not forgetting the peanut gallery.

@starthaus: So, lack of diffraction in an intervening medium (vacuum, etc) = constant for all photons. regardless of energy. That does make sense.

I realize this is only tangentially related, but if the Photon has a vanishingly small rest mass, so that h \neq \infty there would be SOME "rainbowing" as you call it? At what point does a photon with rest-mass start to interfere with observational data such as this?

 

[bcrowell]

I realize this is only tangentially related, but if the Photon has a vanishingly small rest mass, so that h \neq \infty there would be SOME "rainbowing" as you call it?

Yes.

At what point does a photon with rest-mass start to interfere with observational data such as this?

The best upper limits on the photon mass come from torsion balance experiments: R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters , 1998, 80, 1826-1829; download from his web page at http://silver.neep.wisc.edu/~lakes/mu.html

Here is a test using time of flight: "Testing Einstein's special relativity with Fermi's short hard gamma-ray burst GRB090510," http://arxiv.org/abs/0908.1832 The motivation here is more to do with testing possible predictions of loop quantum gravity (which later turned out not to be actual predictions of LQG).

Since times are much easier to measure with ultrahigh precision than directions on the celestial sphere, I suspect that there is no measurement of angular dispersion that comes anywhere near the precision in the two experiments above.

 

[Frame Dragger]

Yes.


The best upper limits on the photon mass come from torsion balance experiments: R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters , 1998, 80, 1826-1829; download from his web page at http://silver.neep.wisc.edu/~lakes/mu.html

Here is a test using time of flight: "Testing Einstein's special relativity with Fermi's short hard gamma-ray burst GRB090510," http://arxiv.org/abs/0908.1832 The motivation here is more to do with testing possible predictions of loop quantum gravity (which later turned out not to be actual predictions of LQG).

Since times are much easier to measure with ultrahigh precision than directions on the celestial sphere, I suspect that there is no measurement of angular dispersion that comes anywhere near the precision in the two experiments above.

Ahh, thank you very much!

 

[starthaus]

Yes.

I would have to disagree with your answer. "Rainbowing" would appear if the rest mass of the photon depended on its frequency. In this case, you would have "h" depending on frequency and the solutions of the differential equations would also depend on frequency.
If the rest mass of the photon were merely nonvanishing (but constant wrt frequency) then, the effects would be exactly the same as explained, the photons would describe trajectories that are independent of their respecive frequencies.

The best upper limits on the photon mass come from torsion balance experiments: R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters , 1998, 80, 1826-1829; download from his web page at http://silver.neep.wisc.edu/~lakes/mu.html

Yes, this is a very good paper on strong limits on the photon rest mass.

 

[atyy]

There is nothing inherently quantum-mechanical about this issue. Anything with a sufficiently high mass-energy will follow a trajectory that differs significantly from a geodesic. "Sufficiently high" means really, really high, e.g., like one of the members of the Hulse-Taylor pulsar, which don't follow geodesics because they're radiating gravitational waves. (If you replace one of the members of that system with a body having half the mass, it would have a different trajectory, because it would radiate at a much lower rate.) A single photon has such a low mass-energy that there's no way you'd ever be able to detect its deviation from a geodesic.

Hmmm, so you're not using E=hv? Are you using eg. Eq 3.81, 3.86 of http://arxiv.org/abs/gr-qc/9704043 or Eq 11, 35 of http://arxiv.org/abs/0808.0997 ?

BTW, in Taylor-Hulse, are the pulsars not treated as point masses? I was under the impression that point masses radiating gravitational waves still follow geodesics of the full spacetime (background + gravity wave).

 

[atyy]

I am sure you were. The post was not directed directly at you. Most of the stuff I post is for the general benefit of less advanced members, basically because that is what I am too. No offence intended.

Oh I did think it was directed at me, but I wasn't offended Just trying to make it clear what I'm confused about.

 

[cragar]

Hi Cragar,

Lets look at the Newtonian equations for gravity which is adequate for the masses and velocities involved in dropping small objects on the surface of the Earth. The force of gravity is given by:

\frac{GMm}{r^2}

where M in this case is the mass of the Earth and m is the mass of the small object.

The acceleration of the small object towards the Earth is given by:

\frac{GM}{r^2}

Note that the mass of the small object in absent in this term, so the acceleration of the small object towards the Earth is independent of the mass of the small object and so the acceleration of all objects dropped towards the Earth is identical.

The acceleration of the Earth towards the small object is:

\frac{Gm}{r^2}

From the above it can be seen that the acceleration of the Earth towards the small object IS dependent on the mass of the small object. Now if we have a lead weight of mass m1, the combined acceleration (acceleration of the lead weight towards the Earth AND acceleration of the Earth towards the lead weight) is:

a_1 = \frac{GM}{r^2}+ \frac{Gm_1}{r^2} = \frac{G(M +m_1)}{r^2}

and the combined acceleration of a bag of feathers (of mass m2) and the Earth towards each other is:

a_2 = \frac{GM}{r^2}+ \frac{Gm_2}{r^2} = \frac{G(M +m_2)}{r^2}

Now a1 is slightly different from a2 and this means that if a lead weight and a bag of feathers were dropped one at a time, the time for the lead weight to fall would be slightly less than the time for the bag of feathers to fall.

Now if the lead weight and the bag of feathers are dropped at exactly the same time, the acceleration of the Earth towards the combined mass of the lead and feathers is:

\frac{G(m_1+m_2)}{r^2}

and the acceleration of the lead weight towards the Earth, combined with acceleration of the Earth towards both small masses is:

a_3 = \frac{GM}{r^2} + \frac{G(m_1+m_2)}{r^2} = \frac{G(M +m_1+m_2)}{r^2}

and the acceleration of the feathers towards the Earth, combined with acceleration of the Earth towards both small masses is:

a_4 = \frac{GM}{r^2} + \frac{G(m_1+m_2)}{r^2} = \frac{G(M +m_1+m_2)}{r^2}

Now a3 and a4 are identical, which means that if the lead and the feathers are dropped at exactly the same time, then they will hit the floor at exactly the same time.

Take home messages:
Two objects of different masses fall at different rates if dropped one at a time.
Two objects of different masses fall at the same rate if dropped at the same time.
The acceleration, rather than the force, is the important thing to consider when comparing falling rates.
The gravitational field of a object does not act on itself, where falling is concerned.

Thanks for your answer , this must have taken you quite some time .

 

[bcrowell]

If the rest mass of the photon were merely nonvanishing (but constant wrt frequency) then, the effects would be exactly the same as explained, the photons would describe trajectories that are independent of their respecive frequencies.

That's incorrect. If two photons have the same world-lines and the same nonzero rest mass, then their four-velocities are equal, so their momentum four-vectors p^i=mdx^i/ds are also equal. That means their energies p^0 are equal, so they have the same frequency. Only in the case of zero rest mass can you have equal four-velocity and unequal four-momentum; in this special case, the proper time s is constant, so you can't differentiate with respect to it, p^i=mdx^i/ds isn't a valid way of calculating momentum, and the above argument for f₁=f₂ fails to hold.

The above should be fairly obvious if you consider other particles with nonzero rest mass. For example, electrons with differing frequencies have different energies, so they can't possibly follow identical world-lines.

 

[bcrowell]

Hmmm, so you're not using E=hv? Are you using eg. Eq 3.81, 3.86 of http://arxiv.org/abs/gr-qc/9704043 or Eq 11, 35 of http://arxiv.org/abs/0808.0997 ?

I think the only necessary mathematical ingredient in the argument in #39 is that the rate of radiation of gravitational waves is a nonlinear function of the mass of the radiating object.

BTW, in Taylor-Hulse, are the pulsars not treated as point masses? I was under the impression that point masses radiating gravitational waves still follow geodesics of the full spacetime (background + gravity wave).

Okay, there are two different things we could ask:

(1) Given a certain background spacetime, can we toss in a variety of objects, give them the same initial conditions, and have them follow identical world-lines, which are geodesics of the background spacetime that would have existed if they hadn't been thrown in?

(2) Suppose we toss an object into a background spacetime, thereby modifying the metric. Does the object then follow a geodesic of the resulting modified metric?

The answer to #1 is clearly no in general, by the argument in #39. I think this is a totally standard fact you can find in any GR textbook: test particles only follow geodesics in the limiting case where the mass of the test particle is small.

For #2, I'm not sure this is a well-defined question. For comparison, think of E&M, which is also a classical field theory. It doesn't make sense to think of an electron as being influenced by its own electric field. Such an effect would have to vanish by symmetry, and it would also be formally infinite at zero range. In GR, we translate the point-charge into a point-mass. A point-mass has a metric surrounding it that is locally well approximated by a Schwarzschild metric, and therefore the mass itself lies at the singularity. The singularity isn't actually a point in the spacetime, so there's no way to say whether it follows a geodesic or not.

If you don't want a singularity, then you could, say, use a rigid body with some finite size. But then the different parts of the object are being acted on by nongravitational forces from the other parts. Therefore none of them follow geodesics.

 

[starthaus]

That's incorrect. If two photons have the same world-lines and the same nonzero rest mass, then their four-velocities are equal, so their momentum four-vectors p^i=mdx^i/ds are also equal. That means their energies p^0 are equal, so they have the same frequency.

What I said is that the "h" parameter does not depend on photon frequency, it only depends on photon rest mass (whatever that might be), so the equation of motion does not depend on frequency.

Only in the case of zero rest mass can you have equal four-velocity and unequal four-momentum; in this special case, the proper time s is constant, so you can't differentiate with respect to it, p^i=mdx^i/ds isn't a valid way of calculating momentum, and the above argument for f₁=f₂ fails to hold.

You are not contradicting what I said.

The above should be fairly obvious if you consider other particles with nonzero rest mass. For example, electrons with differing frequencies have different energies, so they can't possibly follow identical world-lines.

You are still not contradicting what I said. The equation I presented does not apply to electrons, applies to photons only. There is no dependency on energy, nor is there any dependency on frequency anywhere in it. Perhaps you should re-read what I said.

 

[atyy]

I think the only necessary mathematical ingredient in the argument in #39 is that the rate of radiation of gravitational waves is a nonlinear function of the mass of the radiating object.

Wouldn't one also need a formula relating the frequency of the electromagnetic wave to its mass or stress-energy-momentum?

Okay, there are two different things we could ask:

(1) Given a certain background spacetime, can we toss in a variety of objects, give them the same initial conditions, and have them follow identical world-lines, which are geodesics of the background spacetime that would have existed if they hadn't been thrown in?

(2) Suppose we toss an object into a background spacetime, thereby modifying the metric. Does the object then follow a geodesic of the resulting modified metric?

The answer to #1 is clearly no in general, by the argument in #39. I think this is a totally standard fact you can find in any GR textbook: test particles only follow geodesics in the limiting case where the mass of the test particle is small.

For #2, I'm not sure this is a well-defined question. For comparison, think of E&M, which is also a classical field theory. It doesn't make sense to think of an electron as being influenced by its own electric field. Such an effect would have to vanish by symmetry, and it would also be formally infinite at zero range. In GR, we translate the point-charge into a point-mass. A point-mass has a metric surrounding it that is locally well approximated by a Schwarzschild metric, and therefore the mass itself lies at the singularity. The singularity isn't actually a point in the spacetime, so there's no way to say whether it follows a geodesic or not.

If you don't want a singularity, then you could, say, use a rigid body with some finite size. But then the different parts of the object are being acted on by nongravitational forces from the other parts. Therefore none of them follow geodesics.

I was thinking of the comment at the end of section 5.3.6 of http://relativity.livingreviews.org/Articles/lrr-2004-6/ "It should be noted that Equation (550) is formally equivalent to the statement that the point particle moves on a geodesic in a spacetime with metric g+h". In 5.4.5, Eq (550) is rederived using a black hole, as you said.

 

[bcrowell]

You are not contradicting what I said.

Here's what you said in #43: "If the rest mass of the photon were merely nonvanishing (but constant wrt frequency) then, the effects would be exactly the same as explained, the photons would describe trajectories that are independent of their respecive frequencies."

Here's what I said in #47: "If two photons have the same world-lines and the same nonzero rest mass, then their four-velocities are equal, so their momentum four-vectors LaTeX Code: p^i=mdx^i/ds are also equal. That means their energies LaTeX Code: p^0 are equal, so they have the same frequency."

My statement contradicts yours. Your statement says that two photons with equal and nonvanishing rest masses could have different frequencies and the same trajectories. My statement says that if two photons have equal and nonvanishing rest masses, they must have the same frequency.

 

[bcrowell]

Wouldn't one also need a formula relating the frequency of the electromagnetic wave to its mass or stress-energy-momentum?

Sure.

 

[Frame Dragger]

Given the lack of Galactic "rainbows", why is this not taken as strong observational evidence that the photon's rest-mass = 0?

 

[bcrowell]

Given the lack of Galactic "rainbows", why is this not taken as strong observational evidence that the photon's rest-mass = 0?

The type of observation you're referring to does set an upper limit on the photon's rest mass. It's just that the best current upper limit comes from a different technique, and is many, many orders of magnitude better than the one set by that type of observation.

 

[starthaus]

My statement contradicts yours. Your statement says that two photons with equal and nonvanishing rest masses could have different frequencies and the same trajectories.

Correct. The equation of motion does not depend on frequency, it depends on the rest mass only.

My statement says that if two photons have equal and nonvanishing rest masses, they must have the same frequency.

...which is perfectly orthogonal to the first half of the statement.

 

[Frame Dragger]

The type of observation you're referring to does set an upper limit on the photon's rest mass. It's just that the best current upper limit comes from a different technique, and is many, many orders of magnitude better than the one set by that type of observation.

Well, that would certainly make the old rainbow point moot I guess!

 

[atyy]

Given the lack of Galactic "rainbows", why is this not taken as strong observational evidence that the photon's rest-mass = 0?

It is.

Einstein: E.E=p.p+m.m

de Broglie: E~f, p~1/L

f.f=(1/L).(1/L)+m.m

If m=0, then f.f=(1/L).(1/L), so f=(1/L), so fL=velocity of wave=constant independent of frequency.

If m!=0, then f=sqrt(k.k+m.m), which clearly doesn't simplify to fL=constant.

 

[starthaus]

Einstein: E.E=p.p+m.m

de Broglie: E~f, p~1/L

The moment you put down E~f, you cannot have anything but m=0.

fL=velocity of wave=constant independent of frequency.

fL=speed (a scalar), not velocity ( a vector). So , yes, speed is constant , velocity doesn't have to be. We cannot draw any conclusions about the velocity from the above derivation.

 

[starthaus]

Here's what I said in #47: "If two photons have the same world-lines and the same nonzero rest mass, then their four-velocities are equal, so their momentum four-vectors LaTeX Code: p^i=mdx^i/ds are also equal. That means their energies LaTeX Code: p^0 are equal, so they have the same frequency."
.

The invariance of the (E,p) combined with the equality of the rest masses and momenta results into the equality of E's.
BUT, since the photons in our discussion are assumed to have m=!0, the equality of their energies no longer implies the equality of frequencies.Indeed, for a "massive" photon E=!hf.

So, you have not refuted my original statement, that two "massive" photons do not necessarily have to have the same frequency in order to describe the same trajectory. This is due to the fact that their equation of motion does not depend on frequency.

 

[bcrowell]

My statement contradicts yours. Your statement says that two photons with equal and nonvanishing rest masses could have different frequencies and the same trajectories.

Correct. The equation of motion does not depend on frequency, it depends on the rest mass only.

Statement A, which you seem to have accepted as a paraphrase of your own statement, is this: "two photons with equal and nonvanishing rest masses could have different frequencies and the same trajectories."

Statement B is the one quoted above, directly from your post.

Statement C, from my #51, is that if "two photons with equal and nonvanishing rest masses" follow the same trajectory, then "they must have the same frequency."

Your implication in #55 is that statements A and B are logically equivalent. This is incorrect. They are not logically equivalent.

Statements A and C are in contradiction. Statement A is false. Statement C is true.