What Determines the Speed of Light in QED?

[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

 

[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.

 

[techwonder]

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

 

[quantumdude]

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.

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.

How is this determined?

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.

What is the speed of light really

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

 

[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?

 

[NateTG]

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?

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.

 

[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

 

[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?

 

[quantumdude]

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?

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'.

 

[quantumdude]

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?

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.

Would this lead to a negative mass for some virtual photons?

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.

 

[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 ΔE and Δt?

 

[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 Hawking '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.

The Bob (2004 ©)

 

[Dina-Moe Hum]

Hey people.

I may not be able to help much (as I am only 15) but here is what I think.

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 Hawking '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.

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

The Bob (2004 ©)

: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

 

[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

The Bob (2004 ©)

 

[quantumdude]

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

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.

Is this "off-shell" idea strictly a QED idea? I've never heard of it in classical mechanics.

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.

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 ΔE and Δt?

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 Δt.

 

[turin]

... neither the violation of the dispersion relation nor the violation of the conservation of energy is observable in the time Δt.

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 a lot when discussing virtual particles. Is that all QFT boils down to: perturbation theory?

 

[quantumdude]

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?"

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.

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?

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.

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.)

That's OK.

I have had a little introduction to perturbation theory, and I have noticed that it gets brought up a lot when discussing virtual particles. Is that all QFT boils down to: perturbation theory?

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.

 

[turin]

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.

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.

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.

I do not see the connection between
lifetime<&Delta;t
and
undetectable.

That's what the calculation of processes boils down to, but the theory itself is a lot more profound than that.

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

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.

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?

 

[quantumdude]

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.

I do not see the connection between
lifetime<Δt
and
undetectable.

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).

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

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.

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?

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".

 

[techwonder]

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

 

[The Bob]

So do we now know the speed of light because that was the question and I cannot keep up with QEDs and the triangles because I have not studied it. Is the speed of light 29,980,000m/s in a vacuum or not because that is what I use to calculate Time Dilation.

Cheers

The Bob (2004 ©)

 

[selfAdjoint]

That value is an approximation, but a good one. c is a constant of nature and probably doesn't come out even in human units like kilometers per second or miles per hour (why would it?). Physicists continually try to improve their measurements of its value. Your value should work fine for working out dilations.

 

[techwonder]

c=299 792 458 m/s.

I suggest going to http://physics.nist.gov/cuu/Constants/ for precise constants

I believe that the meter is defined as a number of periods of some well defined light source (Cesium atoms?) so c will not change one bit.

 

[The Bob]

That value is an approximation, but a good one. c is a constant of nature and probably doesn't come out even in human units like kilometers per second or miles per hour (why would it?). Physicists continually try to improve their measurements of its value. Your value should work fine for working out dilations.

c=299 792 458 m/s.

I suggest going to http://physics.nist.gov/cuu/Constants/ for precise constants

I believe that the meter is defined as a number of periods of some well defined light source (Cesium atoms?) so c will not change one bit.

Cheers you two

I understand that we can never measure the speed of light to a very high degree of accuracy and that c is the best way to describe it (as light is a natural object (<---- wrong word I know) and therefore has no human limits on it). It was just that the original question was what is the speed of light so I was seeing if we had put a number on it.

(Anyone want to know about Time Dilation just ask [That is to say you want lessons on stuff you already know from a 15 year old person. Never mind])

The Bob (2004 ©)

 

[turin]

The uncertainty principle states that the product of the energy and the lifetime of the virtual photon must satisfy the inequality ΔEΔt>(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).

I think this is what I was looking for (for the time being). Thanks Tom. BTW, I'm assuming you meant:

The uncertainty principle states that the product of the energy and the lifetime of a detectable particle must satisfy the inequality ΔEΔt>(hbar).

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

[π(x,t),φ(y,t)]=-i(hbar)δ(x-y)

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

I don't think I am quite at a level where I can really appreciate this.

 

[vanesch]

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?

Worse, they can have IMAGINARY masses !

(m^2 = negative :-)

But, as someone pointed out, virtual photons are calculational devices in perturbation expansions...

cheers,
Patrick.