Are photons affected differently by gravity?

[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.

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=Vv...er&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 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 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).