# What's the speed of light?

1. Jun 18, 2004

### techwonder

In QED the speed of light is not fixed to be c - at least according to QED by Feynman. He says that the average is c, but light can move at other speeds. How is this determined? What is the speed of light really

2. Jun 18, 2004

### Redfox

It is a supposed limit of speed for any electromagnetic interaction in vacuum, including light and radiowaves. It is generally accepted that C is the limit, around 330.000 km/s. Light can travel at lesser speeds in matter due to physical barrier that matter creates, it refers to the speed of the general stream of light passing that matter.

3. Jun 18, 2004

### techwonder

I'm talking only about speed of light in vacuum. QED does not fix this speed. BTW, c=299 792 458 m/s

4. Jun 18, 2004

### Tom Mattson

Staff Emeritus
The speed of real photons is fixed to be c. Virtual photons, on the other hand, have nonzero mass, and so do not travel at c.

It is determined by a QED perturbation expansion, in which virtual photons arise. I wouldn't ascribe too much reality to them, because nature doesn't know anything about "perturbation theory", so QED does not definitively predict their existence. They are convenient mathematical constructs.

It is c, the speed of real photons in vacuum.

5. Jun 18, 2004

### turin

Tom,
If what techwonder has said in the first post is true, then there would have to be virtual photons that travel faster than c, correct? Would this lead to a negative mass for some virtual photons?

6. Jun 18, 2004

### NateTG

QM - or at least some versions of it allow for virtual particles with negative energy. This is, for example, a part of Hawking's description of Hawking radiation where particles with negative energy get sucked into the black hole, and the remaining particle becomes real.

7. Jun 18, 2004

### reilly

There are cases in which QED perturbation theory works quite nicely -- electron-electron or electron-proton scattering, for example. The one-photon exchange potential is the Coulomb potential, and, in fact both quantum and classical treatment of scattering from a Coulomb potential give identical cross-sections -- the famous Rutherford cross-section.

The virtual photon exchanged in electron-electron scattering has negative mass. To show this takes a certain amount of facility with relativistic kinematics -- the computation is best done in the center-of-mass system in which the total mometum is zero -- the particles are moving toward each other with equal and opposite momentum. If the electrons have momentum of magnitude p, and the scattering angle is theta, then the so called momentum transfer, which is the mass of the virtual photon is

-2*p**2 (1-cos[theta])

We are talking a simple Feynman diagram, the one photon exchange diagram which is explained in any text on QED.

I apologize if my post is somewhat cryptic -- the topic is a bit technical.

Regards,
Reilly Atkinson

8. Jun 18, 2004

### billy_boy_999

i thought that the average of c comes from summing all the probability amplitudes of the infinitude paths for the photons...so the photons would almost always average speed c but the probability waves travel at all speeds...wrong?

9. Jun 19, 2004

### Tom Mattson

Staff Emeritus
It sounds like you're talking about the difference between group velocity and phase velocity. In which case, you'd be correct: Different frequencies of light travel at different speeds, but the pulse as a whole moves at 'c'.

10. Jun 19, 2004

### Tom Mattson

Staff Emeritus
Yes. But I want to emphasize that virtual photons are mathematical concoctions that are unobservable, even in principle. We can't know if they even exist.

I was a bit sloppy with my statement before when I said they have "nonzero mass". I meant that the square of their 4-momentum q2 is not zero. Virtual photons are said to be "off their mass shell". For off-shell particles, the relation:

q2=E2-p2=m2

does not hold.

With that: Yes, q2 for a virtual photon can be either positive or negative.

11. Jun 20, 2004

### turin

Tom,
Do virtual particles have direct, real effects (that cannot be explained merely with "real" particles)?

Is this "off-shell" idea strictly a QED idea? I've never heard of it in classical mechanics. If there is an occasion which violates the dispersion relation, then I probably have a misunderstanding that runs rather deeply. Is this related to the often-mentioned uncertainty relation between &Delta;E and &Delta;t?

12. Jun 21, 2004

### The Bob

Hey people.

Every particle must have direction and velocity. If not they are either in a surrounding area of 0°K (or -273.16°C) and are not moving or they do not exist. It is as simple as that. Actually if you think about it even particles in 'absolute zero' areas must move in relation to the Universe. Everything moves relative to it no matter what inert frame of reflection it is in. Therefore I conclude that every particle (real or superfithal) must have direction and velocity relative to something else. If you haven't already read Stephen Hawkings 'A Brief History Through Time' Chapter 1 'Our Picture Of The Universe'. That shows a clear picture that everything is relative (even if we do live in an infinite universe).

Hope this helps or makes any sense at all because I think I may have tried to bite off more than I can chew.

If this has nothing to do with this Topic I apologise. If it was not much help (as I am only 15) then never mind but that is what I think.

Hope I can one day be as intelligent as all of you.

Last edited: Jun 22, 2004
13. Jun 21, 2004

### Dina-Moe Hum

:surprise:
Hi The Bob, I think that culture and intelligence are two different things, even if, maybe, they are influenced by each other. And you are a pretty smart kid to be 15. Keep on studying, you're on the right way

14. Jun 21, 2004

### The Bob

Hey Dina-Moe Hum,

Thanks for the compliments but what, may I ask, does culture and intelligence have to do with virtual photons? I was trying to say that virtual photons must have a direction and velocity in relation to everything else, just like anti-matter, which is said to exist but cannot bee seen (like gravity).

Thanks a lot, though, for the reply becuase I was expecting a lot of abuse and insults about my post.

Cheers

Last edited: Jun 21, 2004
15. Jun 21, 2004

### Tom Mattson

Staff Emeritus
It's not quite that simple. The thing about QED is that it can't be solved analytically, so we have to do a perturbation expansion. The virtual photons are the result of that mathematical trick.

But yes, QED makes accurate predictions that no other theory makes.

It's a "QFT" idea, not limited to QED. In other words, there are "off shell" particles in Electroweak theory (which includes QED) and QCD as well.

Yes, inasmuch as neither violation is observable. In other words, neither the violation of the dispersion relation nor the violation of the conservation of energy is observable in the time &Delta;t.

16. Jun 21, 2004

### turin

I'm a little bit confused about this last part here. If it is not observable, then how can we say that it happens? I suppose you mean "not directly observable?" But no phenomenon/equation, AFAIK, is directly observable, only the effects. Are you saying that not even the effects are observable? Do they average out to elude detection?

Sorry for being so stubborn. I've never had any QFT, only two semesters of QM. (I want to take some QFT soon, but not at my present university.) I have had a little introduction to perturbation theory, and I have noticed that it gets brought up alot when discussing virtual particles. Is that all QFT boils down to: perturbation theory?

17. Jun 21, 2004

### Tom Mattson

Staff Emeritus
I mean that no virtual photon can register a signal in a detector, like a real photon can. In other words, virtual quanta are not asymptotically free (they can't escape to r=infinity) like real quanta can.

Not quite. If they exist, they are emitted, and then reabsorbed, in a time Δt=(h-bar)/ΔE, where ΔE is the energy of the virtual photon. This means that, due to quantum uncertainty, they can't be detected, even in principle.

As Halliday and Resnick put it, the quantum world allows you to borrow an amount of energy ΔE, provided you pay it back in a time Δt. That's what is going on here.

That's OK.

That's what the calculation of processes boils down to, but the theory itself is a lot more profound than that. The formal properties of canonical quantization and Lorentz invariance, which combine to make the most accurate theory ever devised, make a strong statement about the correctness of both quantum theory and SR. Other formal properties, such as the Spin-Statistics Theorem, the Pauli Exclusion Principle, and the CPT theorem, can be derived from QFT as well.

18. Jun 21, 2004

### turin

I'm not very experimentally savvy, so please try to bare with me. As I have understood the theory behind QM, the position of a particle in a HO potential, for instance, will collapse upon measurement to a particular point in space. However, being in a HO potential, the particle is bound (the probability of the particle being an infinite distance away is zero, and the probability of finding the particle beyond some finite distance eventually becomes negligible). So, what I understand from this is that the position of a bound particle, and therefore dare I say the particle itself, can be detected (or else what does the position measurement mean?). Even though the particle is prevented from r -> infinity by the HO potential, it can still be detected.

Now I ask, is this just a bull****, mickey-mouse description of a particle in a HO potential? I can't think of an experiment that I have never done to verify that a particle in a HO potential can, in fact, be detected, but I don't see anywhere in the theory, at least, that forbids it.

I do not see the connection between
and
undetectable.

I don't understand how the theory can say anything that the calculations don't already say.

I've heard of canonical quantization before, but I never have quite understood it. Should I have learned about it after 2 semesters of QM?

19. Jun 21, 2004

### Tom Mattson

Staff Emeritus
Turin, your description of the HO problem is fine. But observation of virtual photons is not like observation of bound (real) particles in a potential. In fact, virtual photons are not bound states at all! But in QED virtual photon exchange, every virtual photon that is emitted is doomed to be absorbed a short time later, and this time is in accordance with the uncertainty principle.

The connection is that the virtual particle does not live long enough to be detected. The uncertainty principle states that the product of the energy and the lifetime of the virtual photon must satisfy the inequality &Delta;E&Delta;t>(hbar). Virtual photons violate this inequality, and so no experiment, no matter how clever, can directly reveal their existence (by "directly" I mean a signal in a detector).

Sure you do, you've already seen it with regular QM. For instance, you can formally derive the continuity equation from the Schrodinger equation without calculating a spectrum from a given potential and boundary conditions.

You have definitely seen a simple analog: [x,p]=i(hbar).

Canonical quantization is an extension of this to field variables and conjugate momenta. In noncovariant form, it looks like:

[&pi;(x,t),&phi;(y,t)]=-i(hbar)&delta;(x-y)

which is the so-called "equal time commutator".

20. Jun 21, 2004

### techwonder

That was a much more profound answer than I expected! Thanks everybody!