The Mass of a Photon

Suppose I begin with a Langrangian:

\(L= \frac{1}{2} \partial_{\mu} \phi^{\dagger} \partial^{\mu} \phi - V(\phi)= \frac{1}{2}[\dot{\phi}^2 - (\partial_x \phi)^2...] - V(\phi) = \frac{1}{2}(\partial \phi)^2 - V(\phi)\)

We shall now assume that \(\phi\) is a complex field so it is equivalent to \(\phi=\phi_R+\phi_I\) with a conjugate of \(\phi^{\dagger}=\phi_R-\phi_I\). Using standard U(1) symmetry, we can perform the normal transformations (i.e)

\(\partial \phi'=e^{i\theta} \partial \phi\)

and its conjugate phi dagger

\(\partial \phi'^{\dagger}=e^{-i\theta}\partial \phi^{\dagger}\)

Then one comes to learn through a more vigorous mathematical framework, but nothing too difficult that a photon has no mass term. In a more simplified version, if we assume a wave \(k\) for a single field, and that \(k\) is momentum, then if momentum goes to zero then it means that the quanta has zero energy of a very long wavelength. (This part troubles me... but not as much as the conclusions will soon).

Physics explains to us, that if a particle at rest has no energy, then it corresponds to a massless particle. Very interesting way to imagine this... but is it a correct way to view it? If we were not talking about massless particles then it seems almost certain a particle at rest would contain an energy, so to say a particle at rest with no energy corresponds to a massless particle confuses me. Equally, if this was a massless particle, then how could it be at rest with no energy? How does this fit properly and correctly into the standard models framework for massless particles? Draw a straight line, and set that line to the ground state of \(\phi\) so that \(\phi= 0\) along that line. If there are waves that variate and cross that line, the gradient part of the Langrangian costs large amounts of energy. When it costs energy, this is supposed to represent particles with mass. That's rather straight forward, but massless particles also cost energy as well surely? Mass enters the equations via a mass term natural given as \(\frac{M^2}{2}\phi^2\). One can think of \(\sqrt{M^2\phi^2}\) being an energy oscillation.

The latter part does not bother me as much as the speculations drawn on the massless particle side of things, especially when we are to be formulating the idea in our heads that particles at rest have no energies, but in contradiction very large wavelengths. Can anyone explain this ambiguity to me?

 

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Is this reiku??

 

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Why would this be a Japanese Palm Healing...? This is mathematics.

 

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No, science says a particle with no rest mass can never be at rest, in any inertial frame.

\(\phi\) is not the photon field, so mass terms relating to \(\phi\) have nothing to do with the photon.

Previously you mentioned "Then one comes to learn through a more vigorous mathematical framework, but nothing too difficult that a photon has no mass term.". I suggest you go read that bit of the 'more rigorous mathematical framework' because its the bit you seem to have completely missed in your understanding.

 

You're fiesty for the morning. Leonard Susskind would disagree with you...

46 mins in he says ''a particle at rest with no energy corresponds to a massless particle,'' and repeats this with different terminology.

I will send you a link in a PM since I cannot post any yet... or is that restricted as well... we will see ... if not I will show you were the lecture is.

 

Yeah restricted as well. It is lecture 7 of ''new revolutions in particle physics'' at youtube. Follow it up, skip to the 46th minute marker and he will confirm what I said in the OP.

 

A particle with no mass is only able to move at one speed, the speed of light. Hypothetically if you can set \(\mathbf{p}=0\) in \(E^{2} = m^{2}+\mathbf{p}^{2}\) and you find E=0 then m=0 but this is not a configuration possible in the real world, as you cannot set p=0 if m=0.

Do you have any response to the rest of my post? I, like Prom, wonder if you understand this because you don't seem to. The Lagrangian you have given is for a scalar field, the photon is a vector field. Yes, you give a U(1) transformation for the scalar field but its global, it commutes with the derivative. That means you don't have a photon field from it, as the photon field is a local U(1), it doesn't commute trivially with the derivative. And like I said, nowhere have you talked about the Lagrangian relevant to the photon, the mass term you give is for the scalar field.

I might be 'fiesty' in the morning but I'd rather be right and feisty than wrong and deceptive.

 

I'll answer the rest of your post... there was not much to it. None of this is difficult... but until then, would you please follow the link and hear what prof. Susskind says so that the general essence of your post against me has been tackled. You will see he says exactly what I say, so I cannot be held accountable for errors.

 

It would have been nice for you to have thought this through a little. I came here expressing confusion, not to hype up mistakes or be misleading, as you have harshly addressed. If you had thought the post through a little, you would have came to the conclusion either I have been given wrong information, or that you are jumping the gun a little in your answer. I assume that most recent post of yours addressed the issue in a different light - so I assume you have watched the video...? This was in fact your reason to state that I have not understood the rigorous mathematical side of it, but that has practically just smacked you in the face. It surely isn't a matter of that at all, if Sussking confirms what I said before.

I know it's not a photon field. It's a gradient potential.

Please don't ask me to. Having to write out Gauge Invariances and whatnot would just kill me at this hour of the morning. Unlike you, we humans need to get a good couple of coffee's down us before we start debating such things. If you can, just trust me when I say ''I understand it.'' If you don't I'll write it all out later. I simply don't have the time right now.

 

And so you don't reply. Nice one.

Say did anyone else bother looking at the link? It's a bizarre way to look at massless radiation. Surely the standard model cannot be right in such assumptions?

 

It's called having a job dipshit. Sorry I didn't wait around for your reply like you have for mine, I have other things to do with my time.

I'll reply this evening when I get home, I only have 5 minutes left of my lunch break and you're not worth it.

 

Dipshit? You're the one who found a large enough time to write that drivel you did this morning... you telling me you couldn't have found a little bit more time?

Never mind, I'll wait.

 

Though, when you do arrive, please find time to look at the link - I'll refresh your memory - it is called ''lecture 7, new revolutions in particle physics'' on youtube. It is at the 46th minute marker. I'll be enthusiastic for your return.

 

I'm curious.

When a lot of physicists get together in one place, is there this much hostility?
I mean you get the impression from here and other physics sites that there would be lots of name-calling followed by frequent calls to "step outside and settle this".....

I guess I just can't see why anyone can get this worked up over what the mass of a photon is (or isn't).

Arthur

 

While I have every sympathy with the view that we have seen this poser (or his clone) here before, I am slightly troubled by the following

So special relatively postulates (reasonably enough) that there is no coordinate transformation that will bring a particle travelling at light-speed to rest in its own reference frame. And the only known such particles are massless and are called photons. Fair enough

But does the converse automatically follow? Namely, a particle with no mass must travel at light speed? This is not obvious to me.

 

I know this is a little off topic:

There are points that pose that "Genius" can occasionally suffer from side effects that others don't suffer.

For instance the Alexander the Great complex where a person feels destined to do something "Great", You also have the "I'm a smarter Genius than you" complex which most of the time they end up looking a complete arse over (Elitism), There is then the potential difficulties in being socially adaptive with those of IQ's below or above your own (Some blame it upon them spending so much time swatting up on their given choice of Elitism), There are then those that actually are busy doing a job in the field that get a little fed up with those that seemingly have way too much time to argue postulations that have already been solved.

Hostility is just pretty much down to incompatible archetypes interacting with each other. You'll find that a Scientific Consensus is often built by people that get on with each other, you'll then get a few complaining that the Consensus is wrong about one or other point and suggest that scientists of the consensus "Just don't like to rock the boat". They tend to neglect mentioning the reason they weren't apart of the consensus is because they would a poison person. (A person that is argumentative, rude, obnoxious and purposely winds people up, who harm the overall objective of a group through their hostility)

As for the topic:
If a photon had mass it wouldn't be able to transverse fibre optics at light speed as the fibre can be seen as a resistance force. So you could try using Fibre Optics as an apparatus to understand a photons nature, considering that most of our communications networks are being or have been converted to it.

 

That was my first thought... but:

So, it's Vkothii, then.

 

Simon says, you are wrong.

 

Gluons are massless and known to exist. Gravitons are theoretically predicted to be massless too.

 

That wasn't required. The expression you gave was simply not the U(1) associated to the photon field. It would have been sufficient to say \(\phi \to e^{i\theta(x)}\phi\), ie make the x dependence manifest. The fact you went on to say \(\partial \phi \to e^{i\theta}\partial \phi\) means that you could not have been talking about the photon U(1) because only the global U(1) commutes with the partial derivative. If you were working with the relevant gauge symmetries you should have said \(D\phi \to e^{i\theta}D\phi\) because its the covariant (with respect to the associated vector field) derivative which commutes with the photon U(1) gauge transformation, not the partial derivative.

I understand what Susskind was getting at, I don't think you do. I explained how you could interpret it that way but when you call into play other physical considerations, like the fact you can't have m=p=0 and still have something there (a mass-less, energy-less, momentum-less object?) or that special relativity has massless objects move along one and only one kind of trajectories, null ones. An object at rest is not moving along a null trajectory, its moving time-like.

Now if you're up to speed on these things and you think I'm not then how about you use your understanding to engage in discussion? After all if I'm wrong it shouldn't be difficult to show so if you understand this stuff.

No, its called 'work'. My employers aren't too bothered if I turn up 5 minutes late but I'd rather turn up on time than have spent more time pointing out mistakes you made. After all, I've got time in the evenings (ie like now) to do that.

I can see what the issue might be, after all doesn't the name 'rest mass' imply its define in a rest frame? A better name is 'invariant mass', in that its a quantity which is Lorentz invariant. Using the definition \(-m^{2} = p_{a}p^{a}\) (using the -+++ signature) it follows that 'm' is Lorentz invariant. It also follows that the 4-vector p is null if m=0 because only null vectors satisfy \(v_{a}v^{a} = 0\) (tautologically) thus all massless particles have null 4-momentum and vice versa.

And as a vague aside I'm aware I probably should have dialled down the language somewhat in my previous post but I spent the morning wrestling with the Mac OS file structure's minor differences compared to that of Ubuntu and wasn't filled with patience.