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How does the Casimir force work?

  1. May 14, 2003 #1


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    How do you imagine the Casimir force working? By what mechanism does it arise, as you picture it?

    If you had two square plates each 1.6 meters (5 feet) on a side and you could position them side by side almost touching but with say 1/60 of a millimeter gap, by what force would they attract each other?

    I ask because there seems to be some misunderstanding at PF or at least a difference of opinion as to the mechanism causing the force. Someone was telling me that vacuum energy exerts a positive pressure :wink: and that is what pushes the plates together (!) because the vacuum energy is thinner between the plates than on the outside.

    Probably the negative pressure associated with the vacuum energy density is irrelevant here---though suspected to play a vital role in cosmology. Instead what matters is the fourth power dependence of the energy density between the plates. The fact that the energy density is so much less when they are closer together! That's my take---what do you think and how do you explain the force?

    I'll calculate the force in natural units. If the separation distance is L and the area is A, then the formula for the force is simply

    (pi2/240) A/L4

    the constant in front is about 1/24, since pi-square is about ten.
    it is just some fairly harmless numerical constant
    the main thing you have to do is divide the area of one of the plates by the fourth power of the separation.

    In natural units (hbar=c=G=1) the side of the square is E35
    and the area is E70

    The separation of 1/60 of a millimeter is E30
    so the fourth power is E120

    So the force is E70 divided by E120, or E-50 (and dont forget the numerical constant)

    the natural unit of force (plays a central role in the Einstein eqn of GR and in the friedmann equations, discussed in other PF threads) in metric equivalent is 12E43 newtons.

    So when you say the attraction Casimir force is (1/24)E-50
    then it works out in metric terms to (1/24) x 12E-7 newtons
    or 1/20 of a micronewton. But the metric equivalent is a side issue. the answer might just as well be left in natural units as (1/24)E-50.

    E-50 is sort of like a micronewton, so what difference does it make whether you say (1/24)E-50 or (1/20) micronewton.
    Damgo told me a while back in another thread that he didnt
    find it prohibitively difficult to adapt to natural units---those used
    in his QFT text---and I have been trying this out and in fact it
    is not difficult at all. The calculations are all much easier too.

    As a comparison test, try calculating the same Casimir force in
    metric units. It will be a bit messy but you should get the same answer of 1/20 micronewton.
    Last edited: May 14, 2003
  2. jcsd
  3. May 14, 2003 #2
    You can find the announcement of the first measure in here:

    http://www.aip.org/enews/physnews/1996/split/pnu300-3.htm [Broken]
    Last edited by a moderator: May 1, 2017
  4. May 14, 2003 #3


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    Thanks rutwig,
    and even there it says:
    [[...However, when the vacuum is bounded by a pair of conducting surfaces, the only electromagnetic waves that can exist are those with wavelengths shorter than the distance between the surfaces. The exclusion of the longer wavelengths results in a tiny force between the conductors....]]

    Does that not seem a mistake to you?

    I thought that only those waves can exist whose wavelengths fit into the separation an integer number of times.
    Like 1/3 L, or 1/5 L

    Simply to be shorter than L does not seem right to me. does it to you?
    Last edited by a moderator: May 1, 2017
  5. May 15, 2003 #4
    It's not a mistake. You just have to add up the momentums of all the wavelengths of virtual particles which are smaller than the distance between the plates, and subtract that from the background particles.
  6. May 15, 2003 #5
    There are several way to understand this.

    1. the most straightforward way is to just calculate energy-momentum tensor of the field. Generically E-M tensor is the product of fields, and as you probably know, they are product of creation and anihilation operators. When you define anihilation and creation operator you expand fields in the eigen function basis (Fourier decomposition in the simple flat plate case). Existance of two parallel conducting plate will impose a different boundary condition than asymptotically infinite flat space. After all those neat tricks like normal ordering you will get a term which is formally diverge. But, you can make sense this term by identify with corresponding calculation for no plate case and subtract that value. Guess what you will get a finite term with negative sign - it is pulling two plate into each other.

    2. Indeed what happen above is that vacuum of the system changes when you put different boundary condition. Remember vacuum in QFT is defined as a state that is anihilated by anihilation operator. And anihilation operator is defined through expansion of field operator in eigen function basis. (For those of you wondering why we define vacuum this way, it is really a statement saying vacuum is the state with no particle. It is easy to understand a statement if you use usual Fock basis.) So, it is obvious that having different boundary condition changes a vaccum. Casimir energy is the difference between vacuum of new system and simple flat space. (i.e. we want simple Minkowski space to have a zero vacuum energy.)

    3. So, now you can guess you will have different Casimir energy if you change boundary condition yet again. Indeed, another famous exact result is for spherical conductor. It turns out the Casimir force there is repulsive rather than attractive as in parallel plate case. The sign really depends on geometry even though the dependence on length scale is always 1/L^4 (for force).

  7. May 15, 2003 #6


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    Aiiieeee! now we have three inconsistent explanations!
    You say exclude the contribution from particles with wavelengths
    SMALLER than the gap

    The 1996 news item offered by Rutwig says exclude the contribution from wavelengths LARGER than the gap (i.e. too big to go in)

    And the Usenet Physics FAQ at John Baez site says exclude the contribution from wavelengths which are not integral divisors of the gap-----ie. not of the form L/3, L/4, L/5, L/6 and so on.
    this is the idea I came in with: that only these virtual particles
    (... L/3, L/4, L/5, L/6...) could contribute to the sum
    I thought that because I had read the FAQ about it
    But then Rutwig gave a link which contradicted that and
    said to exclude the wavelengths that are simply too long,
    which is quite a different matter,
    and then SchwarzschildR you say to subtract out the contribution
    from those that are too short!

    My hunch is that of the three explanations the Phil Gibbs Physics FAQ is
    most reliable so I will go get the link and post it here


    Exerpt:"...Casimir realised that between two plates, only those virtual photons whose wavelengths fit a whole number of times into the gap should be counted when calculating the vacuum energy, The energy density decreases as the plates are moved closer which implies there is a small force drawing them together..." This is dated 1997 and revised 2002
    Last edited: May 15, 2003
  8. May 15, 2003 #7
    To make sense:

    You got one explination for outside, and one for inside, and the overall condition of inside.

    Inside the plate, the wavelengths must be integral values as per quantum physics, and these must be smaller than L. To get the net outside force, you simply take the normal vacuum E-M tensor and exclude the ones inside the plates, i.e. subtracting a negative more or less.

    In other words, exlcluding the inside ones gives you the outside, excluding the outside values gives you the inside values, and the wavelengths that can fit in L must follow L/n where n is the set of {positive} integers.
  9. May 16, 2003 #8
    So looks like the classical equivalent is:
    F = (πhcA)/(480L4
    from marcus's link
  10. May 16, 2003 #9


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    Yes honorable Foxan
    the formula I was using is verbatim the same
    but using hbar instead of h.

    (and I was setting hbar=c=1 because I use those units)

    (πhcA)/(480L4) equals

    (π2πhbarcA)/(480L4) equals

    (π2hbarc A)/(240L4) equals

    (π2A)/(240L4) which is about

    (1/24) A/L4

    god save the empire and the greek BC astronomers :smile:
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