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Old Dec 2nd 2015, 03:33 AM   #11
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Originally Posted by topsquark
What does "proper mass of the photon" mean?
My dear Dan! You've managed to surprise me!! Surely you don't mean that you don't mean that you don't know what the term proper mass is? If that's the case then it's merely a synonym for the term proper mass. I use the term proper mass because the term "rest mass" has no meaning for particles which don't have a rest frame. It's what you'd call the "mass of the photon," i.e. m[sub]g[/sub] where g = the greek letter gamma

The Proca Lagrangian is defined as (this is from Jackson's EM text, 3rd Ed.)
. Just click on the Attached Thumbnail.

Originally Posted by topsquark
If we like we can solve the Proca equation with a massive particle, but it won't act like a photon because it isn't a photon. One of the immediate consequences is that the massive "photon" can't travel at c.
That's incorrect. You have the wrong idea of what a photon is. A photon is the quanta of the EM field. Currently it is hypothesized that it's proper mass is zero. However, if experiments show that it's proper mass is not actually zero then that doesn't mean that it's not a photon. It just means that the hypothesis that the proper mass of the photon is zero is wrong.

Please download Classical Electrodynamics - Third Edition by John David Jackson (1999) at: http://bookos-z1.org/book/655173/890650
and turn to the section 1.2 starting on page 5 entitled Inverse Square Law or Mass of Photon. In particular see page 7 which reads

Originally Posted by John David Jackson
The surface measurements of the Earth's magnetic field give slightly the best value (...), namely,

m[sub]g[/sub] < 4x10[sup]-51[/sup]
See also the mass of a photon? by Don Koks (very smart and very kind man)
http://math.ucr.edu/home/baez/physic...oton_mass.html
Attached Thumbnails
Proca Equations and nonzero mass of photon-proca_lagrangian.jpg  
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Old Dec 2nd 2015, 11:35 AM   #12
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We're crossing lines between theory and experiment. My comments refer to the current theory of EM that says the photon has no rest mass. Your comments (if I'm correct) are based on what we know experimentally. Obviously they are not always quite the same. I was confused about the term "proper mass" as relating to the photon because the photon has no mass.

A photon is the quanta of the EM field. Currently it is hypothesized that it's proper mass is zero. However, if experiments show that it's proper mass is not actually zero then that doesn't mean that it's not a photon. It just means that the hypothesis that the proper mass of the photon is zero is wrong.
This is quite true. However my comment was directed more to the fact that the Proca equation for a particle with a mass has to be solved differently from the massless Proca equation. If the photon is massless then it has two degrees of freedom...If the photon has a mass then it has three degrees of freedom. My point is that you can't simply take a "massive" photon and reduce it to a massless photon by taking the limit as the mass goes to 0. There is a singularity involved.

See also the mass of a photon? by Don Koks (very smart and very kind man)
http://math.ucr.edu/home/baez/physic...oton_mass.html
An interesting read. Thanks for the link.

-Dan
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Old Dec 2nd 2015, 12:23 PM   #13
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I think that you're really confused about all of this. So all I'm going to say is the following
Originally Posted by topsquark
This is quite true. However my comment was directed more to the fact that the Proca equation for a particle with a mass has to be solved differently from the massless Proca equation. If the photon is massless then it has two degrees of freedom...If the photon has a mass then it has three degrees of freedom. My point is that you can't simply take a "massive" photon and reduce it to a massless photon by taking the limit as the mass goes to 0. There is a singularity involved.
The Proca Lagrangian is the Lagrangian of the electromagnetic field for which the proper mass of the photon is non-zero. I have no idea what you mean by the photon having three degrees of freedom if the proper mass is not zero.

Your comment the Proca equation for a particle with a mass has to be solved differently from the massless Proca equation. is beyond me. Please clarify. Perhaps you don't know this but if you take the Proca Lagrangian and set the proper mass of the photon to zero then you end up with the Lagrangian for the EM field.

What is this "singularity" that you're talking about??

I suggest that you read Jackson's text on the Lagrangian for the EM field (just to be sure that we all know this, the correct name is actually the Lagrangian "density") in addition to the Proca Lagrangian density.

Originally Posted by topquark
My point is that you can't simply take a "massive" photon and reduce it to a massless photon by taking the limit as the mass goes to 0.
Ummmm .... whoever said anything about taking a limit anyway? All I've discussed here are two situations: zero proper mass and non-zero proper mass.

The following is exactly and entirely what I've been saying from the start. Let m = proper mass of photon. Then:

If m is 0 then the Lagrangian density is the Lagrangian density for the EM field.

If m is not 0 then the Lagrangian density is the Proca Lagrangian density for the EM field

I have no idea where you got all that other stuff from.
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Old Dec 2nd 2015, 02:08 PM   #14
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Why is it that I can never find my Weinberg when I'm looking for it?

Okay, without the text I can't really back up my statements too well, but here is the thing in a nutshell.

The EM field is represented by the Proca equation with m = 0. We can solve this equation but, as you well know, not completely. Normally we would choose the Coulomb gauge (one condition), and perhaps the axial gauge (a second condition.) These means out of the four degrees of freedom as specified by the Proca equation we have 4 - 2 = 2 degrees of freedom left. We usually talk about them as "helicities" and choose them to be -1 and +1. This is not to be confused as spin but their properties pretty much follow as if they do have spin = 1...they just don't have a spin 0 component. Examples of these are photons (obviously), gluons, etc. (Gravitons are -2 and +2 helicity.)

The Proca equation with a mass is somewhat similar, but here we no longer have the ability to choose the axial gauge, meaning we have 4 - 1 = 3 degrees of freedom. We call these the "spin" and such particles have spin -1, 0, +1 components. Examples here are the vector bosons, of which the most famous are the W's and Z from weak theory and, of course, any photons in which you are considering to have a mass.

What's happening here (and here's where I need Wienberg) is that the symmetry group of massless Proca and massive Proca fields are fundamentally different and this is where the singularity shows up. We can talk about a massive photon but I don't know of any treatment that discusses how we can add a small mass to the photon and still preserve the symmetry group, which means to my mind that we can't call a massive photon a photon.

I am aware that, specifically in Condensed Matter Physics, that we can have effectively massive photons: The propagation of photons in a material will induce, as viewed holistically, a mass term for the photon. But this not saying that photons have an actual mass.

Weinberg derives the relativistic wave equations from SR symmetries. ("The Quantum Theory of Fields", Vol 1, ISBN: 0521670535) I like his approach as it really gives a logical treatment of where the equations come from. I can give you more details of the symmetry groups once I find the book. (Again!)

-Dan
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Last edited by topsquark; Dec 2nd 2015 at 02:10 PM.
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Old Dec 3rd 2015, 03:44 AM   #15
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I've been watching arguments flying way over my head in this thread,
but I note something that I hadn't fully appreciated before
photons are treated fundamentally differently from other "particles"

In a simplistic manner I took the eponymous
E=mc^2
and the energy of a photon:
e=h.c/lambda

put them together and you get:
m=h/(c.lambda)

Planks Constant is tiny and the speed of light is huge, thus the mass of a photon (even for very high frequencies) is minuscule, but not zero...

However I notice that this seems to be avoided by talking about the momentum of the photon:
E=pc

Does this imply that E=mc^2 does not apply to photons?
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Old Dec 3rd 2015, 10:12 AM   #16
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Originally Posted by MBW View Post
I've been watching arguments flying way over my head in this thread,
but I note something that I hadn't fully appreciated before
photons are treated fundamentally differently from other "particles"

In a simplistic manner I took the eponymous
E=mc^2
and the energy of a photon:
e=h.c/lambda

put them together and you get:
m=h/(c.lambda)

Planks Constant is tiny and the speed of light is huge, thus the mass of a photon (even for very high frequencies) is minuscule, but not zero...

However I notice that this seems to be avoided by talking about the momentum of the photon:
E=pc

Does this imply that E=mc^2 does not apply to photons?
The energy-momentum equation in SR is E^2 = (mc^2)^2 + (pc)^2, so if m = 0 then E = pc. If it isn't moving then p = 0 and this gives E = mc^2.

-Dan
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Old Dec 3rd 2015, 01:20 PM   #17
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Sorry that this a a bit off to the side of the main thrust of the thread,
however would I be correct in surmising that since they have no (rest) mass, photons do not generate a gravitational attraction?
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Old Dec 3rd 2015, 02:48 PM   #18
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Originally Posted by MBW View Post
Sorry that this a a bit off to the side of the main thrust of the thread,
however would I be correct in surmising that since they have no (rest) mass, photons do not generate a gravitational attraction?
Photons carry energy, thus they participate in the gravitational force. See, for instance, gravitational lensing.

To be brief, photons (or any massless particles really) follow a geodesic which is the equivalent to a "straight line" in Euclidean geometry. A gravitational field curves space-time and the photons follow that curvature.

-Dan
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Old Dec 4th 2015, 03:42 PM   #19
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Originally Posted by MBW View Post
Sorry that this a a bit off to the side of the main thrust of the thread,
however would I be correct in surmising that since they have no (rest) mass, photons do not generate a gravitational attraction?
That is a common misconception. Alan Guth did a video for my website on this. See: http://www.newenglandphysics.org/com...s/DSC_0003.MOV
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Old Dec 4th 2015, 04:09 PM   #20
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Originally Posted by topsquark
The EM field is represented by the Proca equation with m = 0.
Actually there is a name for the Proca Lagrangian with m = 0. It's called the Lagrangian for the EM field. When the Proca Lagrangian is plugged into Lagrange's equation its called the Proca equation

Originally Posted by topsquark
We can solve this equation but, as you well know, not completely.
That is quite incorrect. I've done this and it works fine. Have you tried it yourself? Just take the Lagrangian density for the electromagnetic field (which is identical to the Proca equation with m = 0) and plug it into Lagrange's equation and Maxwell's equation is the result.

You appear to keep thinking of this as a problem in quantum mechanics when you keep referring to spin. This is a problem in classical electrodynamics.

Originally Posted by topsquark
Normally we would choose the Coulomb gauge ..
What does this have to do with anything that I said? I.e. I started by saying
Find the Proga Lagrangian. In it there is a term for the proper mass of the photon. Using the Proca-Lagrangian you can arrive at a set of EM equations which are different than the ones you're already familiar with.
You then asked me What does "proper mass of the photon" mean? Then you said
Originally Posted by topsquark
If we like we can solve the Proca equation with a massive particle, but it won't act like a photon because it isn't a photon. One of the immediate consequences is that the massive "photon" can't travel at c.
This is when you started to get screwy on me. The Proca Lagrangian only applies to the electromagnetic field and nothing else. And it only applies to the EM field if its found that the proper mass of the photon is not zero. It can't be used for the W's and Z of the weak nuclear force because its not defined for them.

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