Are photons affected differently by gravity?

[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!

[bcrowell]

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.

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.

I guess the common explanation is that light just follows a straight path through universe, thereby being "affected" by bent space-time/gravity..

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

[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?

[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?
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.

Or would blue light have a greater gravitational redshift/energy loss?

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.

[Altabeh]

Thank you, bcrowell!

Would then blue light be more bent than red light? Or would blue light have a greater gravitational redshift/energy loss?

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

[cragar]

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

[bcrowell]

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

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

[utesfan100]

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?

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.

[Altabeh]

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

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

[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:
Once upon a time a student lay in a garden under an apple tree
reflecting on the difference between Einstein's and Newton's views
about gravity. He was startled by the fall of an apple nearby. As he
looked at the apple, he noticed ants beginning to run along its
surface. His curiosity aroused, he thought to investigate the
principles of navigation followed by an ant. With his magnifying
glass, he noticed one track carefully, and, taking his knife, made a
cut in the apple skin one mm above the track and another cut one mm
below it. He peeled off the resulting little highway of skin and laid
it out on the face of his book. The track ran as straight as a laser
beam along this highway. No more economical path could the ant have
found to cover the ten cm from start to end of that strip of skin. Any
zigs and zags or even any smooth bend in the path on its way along the
apple peel from starting point to end point would have increased its
length.

"What a beautiful geodesic," the student commented.

His eye fell on two ants starting off from a common point P in
slightly different directions. Their routes happened to carry them
through the region of the dimple at the top of the apple, one on each
side of it. Each ant conscientiously pursured his geodesic. Each went
as straight on his strip of appleskin as he possibly could. Yet
because of the curvature of the dimple itself, the two tracks not only
crossed but emerged in very different directions.

"What happier illustration of Einstein's geometric theory of gravity
could one possibly ask?"

murmured the student.

"The ants move as if they were attracted by the apple stem. One might
have believed in a Newtonian force at a distance along his track. This
is surely Einstein's concept that all physics takes place by 'local
action'. What a difference from Newton's 'action at a distance' view
of physics! Now I understand better what this book means"

@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.

[atyy]

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

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.

[bcrowell]

Yes, for a fixed number of photons, because the energy would be greater.
How? Classically, the electromagnetic field should curve spacetime as a solution of Einstein-Maxwell equations. But does frequency enter the classical energy-momentum tensor?
No, frequency doesn't. But for a fixed number of photons, the energy is proportional to the frequency.

[Frame Dragger]

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

[bcrowell]

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

Isn't this all pretty standard textbook stuff?

[atyy]

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

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

[bcrowell]

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

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.

[Frame Dragger]

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.

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?

[bcrowell]

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?

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.

[Frame Dragger]

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.

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...

[yuiop]

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.

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.

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 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.

[Frame Dragger]

Galactic Rainbows... would be amazing! Darned laws of physics! :cry:

[atyy]

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.

So you would say that light rays of different frequencies don't travel differently (say in gravitational lensing) because the classical state does not have a definite number of photons?

[Frame Dragger]

So you would say that light rays of different frequencies don't travel differently (say in gravitational lensing) because the classical state does not have a definite number of photons?

I can't tell if you're asking a question from lack of knowledge... because the feeling I keep getting here is, "It's A TRAP!!" :wink:

[cragar]

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.


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.

[Frame Dragger]

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.

You're asking why lead and feathers fall at the same rate in a vacuum, essentially?

There was an example in another thread of dropping a particle from a given hight through the earth... an its rebound (oscillation) and so forth. I'm going to see if I can find that, and link it here, because it contains a better answer to your question than I'm likely to give.

The simple answer is that Earth always exerts a "steady" um... "pull" or pseudoforce, etc.. lets say "pull" of 9.8m/s^2. I could drop a planet from freefall, and it would fall at the same rate. A feather, or lead, or a melon, etc... all are following a geodesic determined by Earth's mass. A feather DOES have a gravitational field, and as you say it would exert some minute "force".

To use a bigger object, think of these nutty 2012 people, ok? They think some massive body is going to pass by Earth, and disrupt our orbit around Sol. In that scenario, BOTH bodies are falling towards each other (neglecting the momentum of the killer body), hence the idiotic concern that earth could be perturbed from its orbit (more than it is over time naturally). Of course, if something really massive DID make a pass, especially something far more massive than Earth, and CLOSE, we'd move. You'd notice it. :rofl:

Of course, finally... think of the moon. It's gravity effects Earth, and we sure as hell effect the moon (quakes, orbit, etc). If we took the two bodies at rest, and released them, they would fall towards each other. If the moon posessed a greater mass, the Earth would fall faster towards the moon. HOWEVER... the moon would fall no faster than 9.8m/s^2. The net "attraction" would appear to a human observer, to be a combination of the two. That's all orbit really is... falling past each other constantly.

So... the simplest answer is: Lead causes the earth to fall faster towards IT, but not visa versa. Each mass exerts constant gravity based on conditions, and that field appears to extend over an infinite distance, and always be additive. Gravity is odd. :tongue:

Note: I am ignoring a lot here for the sake of simplicity, but I fet I should give SOME answer.

[cragar]

thanks for your answer .

[atyy]

I can't tell if you're asking a question from lack of knowledge... because the feeling I keep getting here is, "It's A TRAP!!" :wink:

No, definitely not, but I appreciate the compliment o:) I've long known about this heuristic, but I don't know how it can be made solid. The formalisms I know are:

1) classical GR + geodesic equation
2) classical GR + equation of state
3) QFT on curved spacetime
4) GR + electrodynamics as effective quantum field theory

(1) is a ray limit of (2), (2) and (3) are respectively classical and semiclassical limits of (4). In (1) and (3) the field does not modify spacetime, so presumably we should use either (2) or (4).

For (2), maybe a solution of the Einstein-Maxwell equations might work?
http://arxiv.org/abs/gr-qc/9704043
http://arxiv.org/abs/0808.0997

[yuiop]

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.

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.

[Frame Dragger]

No, definitely not, but I appreciate the compliment o:) I've long known about this heuristic, but I don't know how it can be made solid. The formalisms I know are:

1) classical GR + geodesic equation
2) classical GR + equation of state
3) QFT on curved spacetime
4) GR + electrodynamics as effective quantum field theory

(1) is a ray limit of (2), (2) and (3) are respectively classical and semiclassical limits of (4). In (1) and (3) the field does not modify spacetime, so presumably we should use either (2) or (4).

For (2), maybe a solution of the Einstein-Maxwell equations might work?
http://arxiv.org/abs/gr-qc/9704043
http://arxiv.org/abs/0808.0997

Hmmm... for your question re: #2... it SEEMS like it should work, but I really lack the expertise. As much as it pains me, I believe this is an area with which Conway is intimately familiar, even if we disagree on his theoretical outcroppings of it (specifically a solution for #4).

[bcrowell]

So you would say that light rays of different frequencies don't travel differently (say in gravitational lensing) because the classical state does not have a definite number of photons?

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?

[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?! :wink:

[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 lazer 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 lazer 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 lazer 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 lazer 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 travelling 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. :smile:

@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! :smile:

[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 :smile: 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! :smile:

[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.

[atyy]

Sure.

So I presume the mass-frequency relation wouldn't be E=hv, since that's quantum mechanical. 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 ?

[starthaus]

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.

I never implied that, quite the opposie. Please read post #59.

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

Please read post #59 again. I pay a lot of attention to what you write, could you please pay more attention to what I write?

[orson.octaviu]

(General Relativity should be used to determine the actual amount of bending, which is twice that you get using only Newtonian physics, but still not enough to produce orbit in weak gravity fields like that from the sun.)

[Altabeh]

Wow... what an incredible load of crap.

Do you think using this kind of squalid language can actually reinforce the probablity of the failure of the idea given by orson.octaviu? I've been following the thread in an on and off manner (due to studying heavily these days) but the statement is neither incorrect (in case you meant this) nor off-topic in the real sense of it! The poster should have a reason for bringing it to our attention so no need to offend him!

AB

[starthaus]

(General Relativity should be used to determine the actual amount of bending, which is twice that you get using only Newtonian physics, but still not enough to produce orbit in weak gravity fields like that from the sun.)

Should we take this statement as a criticism against the absence of symbolic solutions? Because very precise numeric solutions certainly exist (http://www.black-holes.org/numrel2.html).

[Frame Dragger]

Do you think using this kind of squalid language can actually reinforce the probablity of the failure of the idea given by orson.octaviu? I've been following the thread in an on and off manner (due to studying heavily these days) but the statement is neither incorrect (in case you meant this) nor off-topic in the real sense of it! The poster should have a reason for bringing it to our attention so no need to offend him!

AB

Ahhh... you're right. I'm going to delete my post which was insulting, and accept that I've been a little heated by Conway.

@orson.octaviu: I apologize for both my language, and the tone and content of my response.

EDIT: Thank you Altabeh. I do mean that.

[George Jones]

Should we take this statement as a criticism against the absence of symbolic solutions? Because very precise numeric solutions certainly exist (http://www.black-holes.org/numrel2.html).

I am not sure what you mean. Here is my take on orson.octaviu means.
(General Relativity should be used to determine the actual amount of bending, which is twice that you get using only Newtonian physics,

The angular deflection in a weak Schwarzschild field is given by

\frac{4GM}{c^2 b},

where b is the impact parameter. An argument that uses a combination of Newtonian mechanics, the equivalence principle, and special relativity gives a deflection angle of

\frac{2GM}{c^2 b}.

The two results differ by a factor of two, just as orson.octaviu said.
but still not enough to produce orbit in weak gravity fields like that from the sun.)

If the mass of a spherically symmetric object lies within

r = \frac{3GM}{c^2,}

then light orbits at this value of r (photon sphere).

The surface of a spherically symmetric extended object like the Sun lies outside this value of r, so light cannot orbit an object like the Sun. Again, what orson.octaviu wrote is correct.

[starthaus]

I am not sure what you mean. Here is my take on orson.octaviu means.


The angular deflection in a weak Schwarzschild field is given by

\frac{4GM}{c^2 b},

where b is the impact parameter. An argument that uses a combination of Newtonian mechanics, the equivalence principle, and special relativity gives a deflection angle of

\frac{2GM}{c^2 b}.

The two results differ by a factor of two, just as orson.octaviu said.

No argument about this.


If the mass of a spherically symmetric object lies within

r = \frac{3GM}{c^2,}

then light orbits at this value of r (photon sphere).

The surface of a spherically symmetric extended object like the Sun lies outside this value of r, so light cannot orbit an object like the Sun. Again, what orson.octaviu wrote is correct.

I interpreted as him saying that GR does not produce symbolic solutions to the problem of planets orbiting the Sun. Rather than second guessing, let's ask him what he meant.

[George Jones]

I interpreted as him saying that GR does not produce symbolic solutions to the problem of planets orbiting the Sun. Rather than second guessing, let's ask him what he meant.

I think that orson.octaviu's entire post, deflection and orbits, was about the photons in the title of this thread.

[starthaus]

I think that orson.octaviu's entire post, deflection and orbits, was about the photons in the title of this thread.

Then it is all correct.

[yuiop]

Guys, guys ... this is no way to welcome a new member to PF!
Even if what orson.octaviu said was completely wrong or outrageous I would of thought it would be normal to go out of your way to be courteous to someone making their very first post on this forum and give them the benefit of the doubt. As George points out, what he said is actually accurate in the context of this thread, which makes the reception he got even less justified. For what it is worth, I believe the apology of Dragger is sincere and if orson decides to come back (and I wouldn't blame him if he didn't) then I would would like to extend a welcome to him.

[Frame Dragger]

Guys, guys ... this is no way to welcome a new member to PF!
Even if what orson.octaviu said was completely wrong or outrageous I would of thought it would be normal to go out of your way to be courteous to someone making their very first post on this forum and give them the benefit of the doubt. As George points out, what he said is actually accurate in the context of this thread, which makes the reception he got even less justified. For what it is worth, I believe the apology of Dragger is sincere and if orson decides to come back (and I wouldn't blame him if he didn't) then I would would like to extend a welcome to him.

Wow, I feel like a truly collossal ***. You're right as well...
I'm sorry if I set a bad tone, drove him away, etc. Thank you for recognizing my sincerity... I really don't feel proud of my reaction. Hell, I don't even know how old he is!... I could have just shot down a kid who was RIGHT. :frown:

For Douglas Adams fans... (Aries Rising Record Group Holdings!!!!)

@orson: If you read this, I am not representative of the PF community, but Kev is. This place really is worth it, even given my response.

[bcrowell]

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.

I never implied that, quite the opposie. Please read post #59.
I did read post #59. I'm glad that you've clarified now that you don't think A and B are logically equivalent. So now we come back to the point that A and C contradict one another, A being false and C being true.

Please read post #59 again. I pay a lot of attention to what you write, could you please pay more attention to what I write?
I have read your posts very carefully. They contain incorrect statements, which I've pointed out.

[starthaus]

I did read post #59. I'm glad that you've clarified now that you don't think A and B are logically equivalent. So now we come back to the point that A and C contradict one another, A being false and C being true.

That's your opinion, I am of the opinion that the reverse is true, A is true and C is false as per post #59. When you get in the realm of "massive" photons you need to know that completely different rules apply (different Maxwell equations, different Lagrangian, etc.) . Good luck.

[Frame Dragger]

This seems like a time to request the opinion of an advisor whom you mutually respect, instead of "agreeing to disagree" in physics, doesn't it? Doc Al, or ZapperZ, etc...

[bcrowell]

This seems like a time to request the opinion of an advisor whom you mutually respect, instead of "agreeing to disagree" in physics, doesn't it? Doc Al, or ZapperZ, etc...

That's a reasonable suggestion, although I'm actually going to stop following this thread.

[Frame Dragger]

That's a reasonable suggestion, although I'm actually going to stop following this thread.

Thank you. I understand that advice is just that, and you're free to make a choice, especially if this thread is pissing you off or boring you, or seems pointless now. That said, if it's reasonable...?

[George Jones]

Okay, I think I know what is going on. Part of the problem (and only part) is that folks in this thread are using (I think) two different correct definitions of energy: 1) energy as the conserved quantity due to the t-independence of the Lagrangian for a massive particle moving in Schwarzschild spacetime; 2) energy of a massive particle as measured by a local observer.

A detailed exposition by me would require considerable time for the organization of my thoughts, and for the texing of mathematics in an explanatory post, and I am very hard-pressed for time right now. Still, since there seems to be some interest, I might ...

[Frame Dragger]

Okay, I think I know what is going on. Part of the problem (and only part) is that folks in this thread are using (I think) two different correct definitions of energy: 1) energy as the conserved quantity due to the t-independence of the Lagrangian for a massive particle moving in Schwarzschild spacetime; 2) energy of a massive particle as measured by a local observer.

A detailed exposition by me would require considerable time for the organization of my thoughts, and for the texing of mathematics in an explanatory post, and I am very hard-pressed for time right now. Still, since there seems to be some interest, I might ...

I for one, would be very interested, although that amount of work would be wasted on me alone. Either way, thank you for arriving!

[dx]

Okay, I think I know what is going on. Part of the problem (and only part) is that folks in this thread are using (I think) two different correct definitions of energy: 1) energy as the conserved quantity due to the t-independence of the Lagrangian for a massive particle moving in Schwarzschild spacetime; 2) energy of a massive particle as measured by a local observer.



I don't think the definition (1) is appropriate in this case, since that is not the definition of energy which corresponds to the relations E = hf and P = hσ. Since the argument is about the frequency of a particle, we have to use the definition of energy and momentum which correspond to the phase gradient of the particle's de Broglie wave.

The dispresion relation for a particle of mass m is \omega^2 = k^2 + ({m^2}/{\hbar^2}) . At any given point, the 4-velocity v (the group velocity of a wavepacket with the above dispersion relation) uniquely determines the phase gradient through the relation d\theta = -g \cdot v where g is the metric tensor. So if two particles have 4-velocities v = v', then their frequencies must be equal:

f_1 = (d \theta \cdot \partial_t)h^{-1} = ((-g \cdot v) \cdot \partial_t)h^{-1} = ((-g \cdot v') \cdot \partial_t)h^{-1} = (d \theta' \cdot \partial_t)h^{-1} = f_2

i.e. bcrowell's statement C is correct.

[starthaus]

I don't think the definition (1) is appropriate in this case,.

You need to use a special field of QFT that deals with massive spin 1 bosons. The standard Maxwell equations , Lagrangians, etc, no longer apply as we know it.

[George Jones]

I don't think the definition (1) is appropriate in this case, since that is not the definition of energy which corresponds to the relations E = hf and P = hσ. Since the argument is about the frequency of a particle, we have to use the definition of energy and momentum which correspond to the phase gradient of the particle's de Broglie wave.

The dispresion relation for a particle of mass m is \omega^2 = k^2 + ({m^2}/{\hbar^2}) . At any given point, the 4-velocity v (the group velocity of a wavepacket with the above dispersion relation) uniquely determines the phase gradient through the relation d\theta = -g \cdot v where g is the metric tensor. So if two particles have 4-velocities v = v', then their frequencies must be equal:

f_1 = (d \theta \cdot \partial_t)h^{-1} = ((-g \cdot v) \cdot \partial_t)h^{-1} = ((-g \cdot v') \cdot \partial_t)h^{-1} = (d \theta' \cdot \partial_t)h^{-1} = f_2

i.e. bcrowell's statement C is correct.

Yes, I agree. I was going to make a more detailed, less sophisticated (than your nice post) post that related the two definitions of energy.
You need to use a special field of QFT that deals with massive spin 1 bosons. The standard Maxwell equations , Lagrangians, etc, no longer apply as we know it.

A plane-wave for a massive spin-1 field satisfies the dispersion relation that dx wrote down. See equation (6.61) and below of the book Field Quantization by Greiner. \hbar \omega is an eigenvalue of the Hamiltonian of of a massive spin-1 quantum field, and, consequently, the energy E and \hbar \omega of a massive "photon" are related by E = \hbar \omega. See equation (26) on page 164 of Greiner.

[yuiop]

If photons have non zero rest mass, can they in principle be slowed to local velocities of less than c in a vacuum?

[starthaus]

Yes, I agree. I was going to make a more detailed, less sophisticated (than your nice post) post that related the two definitions of energy.


A plane-wave for a massive spin-1 field satisfies the dispersion relation that dx wrote down. See equation (6.61) and below of the book Field Quantization by Greiner. \hbar \omega is an eigenvalue of the Hamiltonian of of a massive spin-1 quantum field, and, consequently, the energy E and \hbar \omega of a massive "photon" are related by E = \hbar \omega. See equation (26) on page 164 of Greiner.

I don't have access to the above book, can you do me the favor of posting the Maxwell equations the author uses and the associated Lagrangian ? Thank you.

[yuiop]

..... So if two particles have 4-velocities v = v', then their frequencies must be equal:

f_1 = (d \theta \cdot \partial_t)h^{-1} = ((-g \cdot v) \cdot \partial_t)h^{-1} = ((-g \cdot v') \cdot \partial_t)h^{-1} = (d \theta' \cdot \partial_t)h^{-1} = f_2


Here is a much less sophisticated argument.

The de Broglie relationship for a matter wave is:

f = \frac{mc^2}{h\sqrt{1-v^2/c^2}}

where m is the rest mass. This can be solved to give:

\frac{v}{c} = \sqrt{1-\left(\frac{mc^2}{hf}\right)^2 }

If it is assumed that rest mass is proportional to frequency for photons with non zero mass, then two photons with the same frequency must have the same characteristic velocity (and vice versa) as dx claims.

If the further (probably reasonable) assumption is made that all photons with any frequency have the same rest mass, then the characteristic velocity of photons is frequency dependent. This suggests that photons with non zero rest mass would be deflected to different extents by a gravitational lens and a certain amount of "rainbowing" would present (but probably difficult to detect) as suggested by bcrowell earlier

[George Jones]

I don't have access to the above book, can you do me the favor of posting the Maxwell equations the author uses and the associated Lagrangian ? Thank you.

Look at

http://books.google.com/books?id=VvBAvf0wSrIC&printsec=frontcover&dq=greiner+field&cd=1#v=onepage&q=&f=false

starting on page 152.

[starthaus]

Here is a much less sophisticated argument.

The de Broglie relationship for a matter wave is:

f = \frac{mc^2}{h \sqrt{1-v^2/c^2}}



The above is equivalent with the dispersion equation 6.61 on page 154 of the reference (thanks, George Jones and dx, the Proca formalism was exactly what I was talking about all along).

Either way, one obtains:

m=\frac{hf\sqrt{1-v^2/c^2}}{c^2}

or

m=\sqrt{\omega^2-k^2} (from 6.61, page 154)

This is particularly disturbing since it says that the rest mass of the "massive" photon is not only a function of its frequency (f) (we kind of expected that) but also of its speed (v) (!). Eq (6.61) says the same exact thing since k is momentum.
We need to abandon this line of thinking and look at the equations of motion as derived from the Euler-Lagrange equations. You can see that the Lagrangian for the massless photon (corresponding to the Maxwell equations, page 149, eq.1) is quite different from the Maxwell-Proca Lagrangian (by the presence of the term in "m"). Earlier in this thread, before the discussion veered into speculations relative to "massive" photons, I posted (http://www.physicsforums.com/showpost.php?p=2655053&postcount=38) the equation of motion for the massless photon as derived from the Maxwell Lagrangian. I would like to challenge someone else, to write down the equations of motion as derived from the Maxwell-Proca Lagrangian. Only then, we can answer if the "massive" photons are "rainbowed" or not in a gravitational field. Unfortunately, since they do not exist, we cannot test the predictions of the Proca theory on this particular case.

[George Jones]

m=\sqrt{\omega^2-k^2} (from 6.61, page 154)

This is particularly disturbing since it says that the rest mass of the "massive" photon is not only a function of its frequency (f) (we kind of expected that) but also of its speed (v) (!). Eq (6.61) says the same exact thing since k is momentum.

This in not the right interpretation. Supposed I watch a rock with rest mass m zip by me. Then, (with c = 1)

m = \sqrt{E^2 - p^2}.

The rest mass of the rock is constant, so this says that if E changes, then p changes in such a way that m remains constant. E and p are not independent.

Similarly, the dispersion relation for a "photon" with rest mass m is

m = \sqrt{\omega^2 - k^2}.

The rest mass of the "photon" is constant, so this says that if \omega changes, then k changes in such a way that m remains constant. \omega and k are not independent.

[starthaus]

This in not the right interpretation. Supposed I watch a rock with rest mass m zip by me. Then, (with c = 1)

m = \sqrt{E^2 - p^2}.

The rest mass of the rock is constant, so this says that if E changes, then p changes in such a way that m remains constant. E and p are not independent.

Similarly, the dispersion relation for a "photon" with rest mass m is

m = \sqrt{\omega^2 - k^2}.

The rest mass of the "photon" is constant, so this says that if \omega changes, then k changes in such a way that m remains constant. \omega and k are not independent.

Yes, I am quite aware that this is the conventional interpretation. Nevertheless, in the Proca formalism, we are way outside of the conventional interpretation. Photons have "mass" and they still travel at c (or not). So, my challenge stands, please use the Proca-Maxwell lagrangian in order to derive the equation of motion.

[yuiop]

...

m = \sqrt{\omega^2 - k^2}.

The rest mass of the "photon" is constant, so this says that if \omega changes, then k changes in such a way that m remains constant. \omega and k are not independent.

a bit like frequency f and wavelength \lambda are not independent in the c=f\lambda equation in a vacuum. They change in such a way that c is always constant.

[George Jones]

We need to abandon this line of thinking and look at the equations of motion as derived from the Euler-Lagrange equations. You can see that the Lagrangian for the massless photon (corresponding to the Maxwell equations, page 149, eq.1) is quite different from the Maxwell-Proca Lagrangian (by the presence of the term in "m"). Earlier in this thread, before the discussion veered into speculations relative to "massive" photons, I posted (http://www.physicsforums.com/showpost.php?p=2655053&postcount=38) the equation of motion for the massless photon as derived from the Maxwell Lagrangian.

No, the Maxwell Lagragian gives the equation of motion (i.e., the field equation) for the electromagnetic field, not equation of motion for the zero rest mass particles in Schwarzschild spacetime.
I would like to challenge someone else, to write down the equations of motion as derived from the Maxwell-Proca Lagrangian. Only then, we can answer if the "massive" photons are "rainbowed" or not in a gravitational field. Unfortunately, since they do not exist, we cannot test the predictions of the Proca theory on this particular case.

The Maxwell-Proca Lagrangian gives the equation of motion (i.e., the field equation) for a massive spin-1 field, not the equation of motion for the zero rest mass particles in Schwarzschild spacetime.
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

These equations of motion for massive and zero rest mass particles are derived from Lagrangians constructed from the Schwarzschild metric, not from the Maxwell and Maxwell-Proca Lagrangians. The first equation is appropriate for massive spin-1 quanta (massive "photons"), while the second equation is appropriate for massless spin-1 quanta (massless photons).

The dispersion relation for massive photons written in the form

k = \sqrt{\omega^2 - m^2}

shows that spatial momentum (as measured in a particular frame) depends on frequency, and thus h depends on frequency.

[starthaus]

These equations of motion for massive and zero rest mass particles are derived from Lagrangians constructed from the Schwarzschild metric, not from the Maxwell and Maxwell-Proca Lagrangians.

Correct, so we can narrow the challenge to a simple question: find out the expression for "h" in the Proca formalism. Does "h" in post #38 depend on the speed of "massive" photons or not?

[George Jones]

Correct, so we can narrow the challenge to a simple question: find out the expression for "h" in the Proca formalism.

This can be done fairly easily, but I am quite drained right now. If no one does it sooner, I will post tomorrow.
Does "h" depend on the speed of "massive" photons or not?

Yes, it depends on spatial velocity (with respect to a particular frame).