Are there any infinite quantum fields?

**John J. Bannan** · 07-28-08, 07:50 PM

And if so, how do you know they are infinite?

**CptBork** · 07-28-08, 07:52 PM

How do you know you'd even be able to understand a detailed answer?

**John J. Bannan** · 07-28-08, 07:54 PM

Originally Posted by CptBork

How do you know you'd even be able to understand a detailed answer?

Well - I can certainly understand - "no, there is no such thing as an infinite quantum field" or "yes, there is such a thing as an infinite quantum field." Is your answer so complicated that you can put your answer in either of these two forms?

**CptBork** · 07-28-08, 08:03 PM

And if I said either yes or no, what would your next question be?

**Cannon** · 07-28-08, 08:04 PM

Is a paradox infinite? or just 1+1?

**John J. Bannan** · 07-29-08, 05:41 AM

Originally Posted by CptBork

And if I said either yes or no, what would your next question be?

If no, I wouldn't have a next question. If yes, I would ask you whether you are referring to size or degrees of freedom.

**~~Vkothii~~** · 07-29-08, 07:03 AM

Aren't the electromagnetic and inertial fields infinite in range?
(That means any interaction is not constrained by time or distance, btw)

**~~Reiku~~** · 07-29-08, 07:06 AM

yes.

**BenTheMan** · 07-29-08, 10:21 AM

What's an inertial field? Do you mean gravitational?

**~~Vkothii~~** · 07-29-08, 04:58 PM

Are gravity and inertia related?
Is there a way to model gravity that looks completely different to an inertial model? Or is the hair that's being split here about gravity being a force, but inertia isn't?

OK correction #56: There are two quantum fields with infinite range, the EM and Higgs fields.
Inertia and gravity are classical, there are classical and quantum versions of EM.

**~~Reiku~~** · 07-29-08, 07:25 PM

Originally Posted by Vkothii

Are gravity and inertia related?
Is there a way to model gravity that looks completely different to an inertial model? Or is the hair that's being split here about gravity being a force, but inertia isn't?

OK correction #56: There are two quantum fields with infinite range, the EM and Higgs fields.
Inertia and gravity are classical, there are classical and quantum versions of EM.

Yes, gravitational forces and inertia are related through what is called the Equivalence Principle.

**BenTheMan** · 07-30-08, 09:13 AM

Originally Posted by Vkothii

Are gravity and inertia related?
Is there a way to model gravity that looks completely different to an inertial model? Or is the hair that's being split here about gravity being a force, but inertia isn't?

OK correction #56: There are two quantum fields with infinite range, the EM and Higgs fields.
Inertia and gravity are classical, there are classical and quantum versions of EM.

I will admit to not really knowing what "inertia" is, beyond a textbook definition. So let us forget that bit.

I agree that the EM field has an infinite range. This is just saying that the electromagnetic force has an infinite range, and (thus) that the photon is massless. (These things are all intricately related!)

I'm not so sure about the Higgs field. I would have said the gravitational field has an infinite range. Why do you say the Higgs field?

**John J. Bannan** · 07-30-08, 11:55 AM

Originally Posted by BenTheMan

I will admit to not really knowing what "inertia" is, beyond a textbook definition. So let us forget that bit.

I agree that the EM field has an infinite range. This is just saying that the electromagnetic force has an infinite range, and (thus) that the photon is massless. (These things are all intricately related!)

I'm not so sure about the Higgs field. I would have said the gravitational field has an infinite range. Why do you say the Higgs field?

Why does an infinite EM field necessarily mean a photon is massless?

**AlphaNumeric** · 07-31-08, 09:08 AM

The photon is massless because of a gauge symmetry protection known as the Ward Identity. Similarly for the gluons.

**Guest254** · 07-31-08, 09:19 AM

Dons thicky cap: I thought the Ward identity was a statement about conservation of charge? Please be gentle - my memory is rusty and I barely understood the result first time round!

**BenTheMan** · 07-31-08, 09:24 AM

Originally Posted by John J. Bannan

Why does an infinite EM field necessarily mean a photon is massless?

AN's answer is correct. A mass term for the photon breaks gauge invariance. The argument is only a few lines long, so I'll make it explicit.

The lagrangian for EM is

$\mathcal{L} = \frac{-1}{4}F_{\mu\nu}F^{\mu\nu}$

where

$F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$

Gauge invariance is the statement that the action above L is invariant under the transformation

$A_{\mu} \rightarrow A_{\mu} - \partial_{\mu}\alpha$

where $\alpha$ is some function.

Then all you have to do, to check that $\mathcal{L}$ is gauge invariant is check

$\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} \rightarrow \partial_{\mu}(A_{\nu}-\partial_{\nu}\alpha) - \partial_{\nu}(A_{\mu}-\partial_{\mu}\alpha) = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}.$

Then it is straightforward to check that adding a mass term for the photon breaks gauge invariance. If you actually plan on doing the proof (three lines, max), the Proca lagrangian (which describes a massive photon) is given by

$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}m^2A_{\mu}A^{\mu}$ .

**BenTheMan** · 07-31-08, 09:28 AM

Originally Posted by Guest254

Dons thicky cap: I thought the Ward identity was a statement about conservation of charge? Please be gentle - my memory is rusty and I barely understood the result first time round!

Ahhh they're the same thing

I can't reconstruct the argument in my head (I SHOULD be able to), so I'll just look up the answer when I get into my office.

**Guest254** · 07-31-08, 09:28 AM

Originally Posted by BenTheMan

Gauge invariance is the statement that the action above L is invariant under the transformation

$A_{\mu} \rightarrow A_{\mu} - \partial_{\mu}\alpha$

where $\alpha$ is some function that obeys $\partial_{\mu} \partial_{\nu} \alpha = 0$ .

Dons thicky cap for the second time in the same thread: I'm having a bad day! I don't see why you need any restriction on $\alpha$ (other than being sufficiently smooth), because the partial derivatives commute. Are you making a specific choice of gauge? Why is that necessary?

Again, be gentle...

**BenTheMan** · 07-31-08, 09:31 AM

Originally Posted by Guest254

Dons thicky cap for the second time in the same thread: I'm having a bad day! I don't see why you need any restriction on $\alpha$ (other than being sufficiently smooth), because the partial derivatives commute. Are you making a specific choice of gauge? Why is that necessary?

Again, be gentle...

Ack you're correct.

This is what I get when I try to do QFT before coffee.

To all: In the interest of being correct, I am going back to edit my post. The incorrectness will be saved in this post for posterity!