# Quantum Vacuum

1. Sep 22, 2014

### fog37

Dear Forum,

Vacuum is classically interpreted as the absence of everything. But the quantum view of vacuum sees it as a bubbling, dynamic entity from which particles and antiparcles emerge for very short intervals of time.
Relativity teaches us that energy and matter can be converted into each other (is it correct to say that matter and energy are the same thing?)

So if vacuum has energy does it mean the vacuum is itself matter or does it mean that it "hides" matter? I am familiar with the Casimir effect (vacuum energy (with or without mirrors) is said to be infinite. This implies that the vacuum of space could be an enormous source of energy, the so called zero point energy)

From Maxwell equations, we know that the electromagnetic field is able to auto-propagate in the vacuum (it does not need a supporting medium likes sound does). Is it possible that the EM field interacts with this vacuum as it propagates or is this very unlikely?

Thanks,
fog37

2. Sep 22, 2014

### The_Duck

Yes, in quantum mechanics, the word "vacuum" has a very technical meaning. It refers to the lowest possible energy state of some system.

No. Some things that we do not call "matter," such as light, still have energy.

Not in the way that the word "matter" is usually used. Usually we use the word "matter" to refer to things made of electrons, quarks, and similar particles.

I'm unclear what this would mean.

This is an oversimplification.

No. You can't use vacuum energy to power anything. As I said above, we use the word "vacuum" to mean the lowest energy state of some system. So you can't drain this energy away and use it to do useful work, because the vacuum already has as little energy as it is possible for it to have.

There is a sense in which light "interacts with the vacuum" as it propagates. You can try to read about vacuum polarization.

3. Sep 22, 2014

### Staff: Mentor

To understand it you need to know the modern view of energy:
http://www.sjsu.edu/faculty/watkins/noetherth.htm
Noether's Theorem is more general. It says that if a transformation of the coordinate system satisfies certain condition, namely being continuous, then necessarily there exist a quantity that is conserved. The exact quantity may not be known but nevertheless it is known that it exists. There may be other transformations, such as the inversion between right and left-handed coordinate systems, which leave the Lagrangian function unchanged but for which there are not conserved quantities.

Basically energy is the quantity that is conserved if the laws governing a system are time invariant - which intuitively you expect to be the case generally. When applied to relativity this shows mass is a form of energy - but not that energy is a form of mass in the same way light is a form of energy but energy is not a form of light.

The view where the vacuum is virtual particles popping in and out of existence is a by-product of the perturbation methods used in Quantum Field Theory - it is now thought they don't really exist because they do not appear in methods that don't use that approach.

Thanks
Bill

4. Sep 24, 2014

### harrylin

Ah yes, that's interesting! :)

How do those other methods explain vacuum polarization?

Thanks,
Harald

5. Sep 24, 2014

### Staff: Mentor

Good question.

Don't know.

Pedagogically one way to derive QFT is as the limit of interacting blobs of stuff. To solve such things perturbation theory is used. But there is another way to proceed. Instead of taking the limit you leave the spacing small and solve numerically on a computer. Its called Lattice Field Theory:
http://en.wikipedia.org/wiki/Lattice_field_theory

Its known when you do that virtual particles do not appear.

Thanks
Bill

6. Sep 24, 2014

### Staff: Mentor

I think you mean the transformation has to be continuous and has to leave the Lagrangian invariant, correct?

7. Sep 24, 2014

### Staff: Mentor

Well that was a quote from the link - not mine.

But yes its something like that - although it's been a while since I delved into the detail. If I remember correctly it doesn't have to be invariant to the change - it can change by a four-divergence and still hold.

Thanks
Bill