# Vacuum fluctuation

Sterj
[SOLVED] vacuum fluctuation

1. If their is build energy in vacuum (from Heisenberg's uncertainty principle), for example:

build: 10 J ==> here for: d(t)=h/(4pi)/10J
There must be built a particle and an anti particle
==> particle has 5 J and antiparticle 5 J
Is that right?

Is the particle at same place like the antiparticle (while life time) or how do they disappear?

Casimir effect:
Why is there a force? (I mean, in the middle there are high energy waves and outside there are low energy waves, but why is there a force)?

And if there are high energy waves in the middle of the plates and outside low energy waves, from where take the high energy wave the energy?

Related Quantum Physics News on Phys.org
mathman
Casimir effect takes place because between the plates only waves that have wavelengths that divide evenly into the spacing can exist, while outside all wavelengths are possible. As a result, there are more photons and therefore more pressure outside than in between.

Sterj
If a wave has a too large wave length, what's with it, it can't disappear (it doesn't go between the plates, where else?)?

mathman
The Casimir effect is a result of virtual photons. The ones that can't fit just don't come into being in the first place. Look up Casimir effect on Google. You will get a much more complete explanation.

Sterj
I looked about 3 hours and coudn't find an answer. (What happens with photons that don't fit between the plates?)

Staff Emeritus
Gold Member
Dearly Missed
Sterj said:
I looked about 3 hours and coudn't find an answer. (What happens with photons that don't fit between the plates?)
Look at what mathman said. It is precisely the Casimir effect that virtual photons that won't fit between the plates won't manifest between the plates.

Sterj
@selfAdjoint, I know that. My question is, what happens with photons that don't fit between the plates (where do they go if they collide with the plates?)

Sterj said:
(where do they go if they collide with the plates?)
Sterj
I’m just starting to understand the Casimir effect myself, so let me try for you and I’ll look to, SA or mathman, to correct me if I say something dumb.

The effect is based on “virtual” particles and although there is no such thing as an “anti-photon” in the real with the virtual there can be. When the ‘appear’ they form and quickly destroy each other for a net zero as if thy had never been there. So the idea is when they form near a large flat plate they can hit said plate, but the photon and the ‘anti’photon again with a net zero effect as in no heat added or light reflected. Or if reflected they of course quickly destroy each other to a net zero. BUT they do move the plate just a bit. So why isn’t the plate being pushed around the room – well their hitting the other side as well.

Now with TWO plates the idea is to get them so close together that the pairs cannot be created in there at all. So now the “pressure” is on either side of the two of them and not on the inside thus the force drawing the two plate to each other.

Now assumed in this is some awfully intense and rigorous testing and measuring. Like being sure it is testing in a complete and near true vacuum, or at least accounting for it.

I assume you could, even should, get the same effect from bringing two plates so close together that the larger molecule components of air get forced out of in-between the plates. With those same molecules banging away on the outside a similar force pulling the plates together would be expected.
It’d be interesting to see how this was eliminated or accounted for.
Any ideas SA or Mathmen?
Can the Photon pairs even come out in a non-vacuum?
I.E. even air might stop them from forming.

Hope this helps Sterj
RB

Staff Emeritus
Gold Member
Dearly Missed
RandallB said:
Now with TWO plates the idea is to get them so close together that the pairs cannot be created in there at all. So now the “pressure” is on either side of the two of them and not on the inside thus the force drawing the two plate to each other.
Pretty good explanation up to there. You don't get rid of all the virtual particles between the plates, just those with wavelengths longer than the spacing betweeen the plates. So there's fewer v.p.'s between than outside, and therefore greater pressure from v.p.'s outside than between.

Pretty good explanation up to there. You don't get rid of all the virtual particles between the plates, just those with wavelengths longer than the spacing betweeen the plates. So there's fewer v.p.'s between than outside, and therefore greater pressure from v.p.'s outside than between.
Makes good sense,
Do you know if during the testing getting a good vacuum was an issue, or did they just account for any air pressure air issues somehow?
RB

Sterj
Can we speak of virtual particle in this case (RandballB's text). I always thought, that the casimir effect happens only cause of the Zero point oscillation of the electro magnetic field. The energy of one of those oscillation is:
E=1/2 h(bar)w and this is not the energy of a virtual photon, so if an electro magnetic field is in it's ground state there aren't really photons, are there?

Sterj said:
Can we speak of virtual particle in this case (RandballB's text). I always thought, that the casimir effect happens only cause of the Zero point oscillation of the electro magnetic field. The energy of one of those oscillation is:
E=1/2 h(bar)w and this is not the energy of a virtual photon, so if an electro magnetic field is in it's ground state there aren't really photons, are there?
I really think some things need to be cleared out here.

The Casimir effect happens because of the fluctuations in energy density of the vaccuum. Don't think that a virtual photon has one definite energy value. This is impossible since the virtual particles need to be described in terms of an energy density. This implies that if you were to measure the energy of a virtual photon, you'd get one value but if you measure the energy of the same virtual photon some time later, you'll get another energy value (you can't actually do this, but it is just to illustrate my point). This is a mere consequence of the HUP.

In the case of the vaccuum the energy density is : $$2 \int \frac{d^3 k}{(2 \pi)^3} \frac{\hbar \omega_k}{2}$$

This expresses the energy gensity of the vaccuum fluctuations of the EM-field. In other words, if the lowest energy level is omega, then an EM-field contributes the above energy density. One can apply this formula in the case of two plates by keeping in mind the boundary conditions of the EM-field at the plates. This is how you get the formula for energy density in the case of the Casimir-effect.

The energies of the virtual photons really can be any value denoted by this integral. Also, the very reason for the appearance of these virtual particles is the energy fluctuation, so if the vaccuum had one fixed energy value that is constant in time and place, you would not acquire them virtual particles.

Virtual particles are off-mass shell, meaning they do not respect then Einstein energy relation. This implies that they do not respect energy conservation. A particle of mass m can emit a virtual particle of mass > m. This happens with the W-bosons in weak interactions. Energy is conserved between the very first and very last step of some interaction but not in between, during the actual interaction

marlon

Sterj
Thanks,

But take the em field between two plates. Now let's chose one oscillation of this field between the plates. This harmonic oscillation is in its ground state and has a well defined wavelength, thus a well defined omege. The energy of this one harmonic osscilation is 0.5h(bar)w.

And now if we take all the oscillations between the plates the total energy is the energy of all harmonic oscillators.

That's what my book says.

What can we understand of the ground state of the h.oscill. in an em field? (now without fluctuations). I mean, if there are no photons in a em field the field oscillates. Now, this zero point oscillaten is free of photons, but what are this oscillations? (I mean in an em field it can't give something else than photons).

Sterj
"One of the fundamental results of quantum theory is that at n=0 a harmonic oscillator has a non--zero energy. In quantum field theory, this result implies that an electromagnetic field exists even if there are no photons in the field."

from: http://webphysics.davidson.edu/Projects/AnAntonelli/node38.html

He also says, that the zero point energy of the em field don't "carry" photons.

But the vibrations of the EM-field ARE the photons. There are no photons IN the EM-field. You misinterpreted this text.

In this case, if one says there are no photons, then it means that n= 0. But these are the REAL photons. The lowest energystate of an harmonic oscillator is measured with respect to (or relative to) the non-zero expectation value of the vaccuum. This is NOT the vaccuum energy, it is the energy of an harmonic oscillator PLACED in the vaccuum. I gave you the integral that calculates the energy of the vaccuum.

marlon

Last edited:
The energy density of the vaccuum is given by (EM-field) :

sum over n of$$\int \frac{dk_y dk_z}{(2 \pi)^2} \sqrt{(\frac{\pi n}{d})^2 + k_y ^2 + k_z^2}$$

plates are perpendicular to x axis

marlon

Sterj
Mhh,
"But the vibrations of the EM-field ARE the photons. There are no photons IN the EM-field. You misinterpreted this text."

Ok, you mean that there aren't any real photons in a harmonic oscillator placed in vacuum. And the virtual photons are the vibrations. Sounds strange - I always thought that photons are wave packets and not oscillators.

Sterj said:
Mhh,
"But the vibrations of the EM-field ARE the photons. There are no photons IN the EM-field. You misinterpreted this text."

Ok, you mean that there aren't any real photons in a harmonic oscillator placed in vacuum. And the virtual photons are the vibrations. Sounds strange - I always thought that photons are wave packets and not oscillators.
the problem is that you are "mixing" QFT with QM. In QM, going from one energy level to another is done by absorbing or emitting a photon. Thus, a photon is a particle that represents a quantum of energy.

However, in QFT, every particle is the excitation of a field. The EM-field can be seen as a mattress of a gazzillion harmonic oscillators. If you disturb the matress, it will start to vibrate and the energy coming from this vibration really mimicks a particle of that energy and with certain mass m. In the case of the EM-field, the associated particle has mass m=0 and is the photon. The "dirturbance" can be executed by placing two electrons on the mattress

regards
marlon

Sterj
AH, thanks. That was really my problem.

Sterj

just to be complete. You know that a third plate is brought in when we prove the Casimir effect in order to screen off our plates from the 'environement' It is proven then that the energy density is has a different value in the regions separated by the plates. That is what causes the observable Casimir effect and proves the notion of virtual particles due to energyFLUCTUATIONS

marlon

Staff Emeritus
Gold Member
Dearly Missed
The Casimir plates idea is essentially static. Is there any proposal for a dynamic version, in which some changing phenomenon which produces a (possibly moving) region in which the vacuum energy density is less than in its surroundings?

For example consider the tuned bunches of electrons in a LINAC; does the region between two bunches have lower vacuum energy than its surroundings for the same basic reason the Casimir region is?

For example consider the tuned bunches of electrons in a LINAC; does the region between two bunches have lower vacuum energy than its surroundings for the same basic reason the Casimir region is?
In all honesty, i really don't know what you are talking about.
I suppose you can see them electrons as the 'perturbations' (you know the sources and sinks in QFT expressed by the J's in the path integral formalism) of the vaccuumfield ? But then , ???

marlon

Sterj
I found a .pdf-file and there is written:
Define the vacuum state as the state with no photons in any mode. Thus the vacuum energy is:
and then the pic.

Why can he drop the term 1/2?

#### Attachments

• 44 KB Views: 509
Sterj
Is that because there is always a particle and its antiparticle (and thus in the case of photons, there are two photons because the photon is its own antiparticle).

i don't get the summation over lambda ?

Are these the polarizations ?

marlon