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Casimir Effect

  1. Jul 14, 2003 #1
    This is a simple (and possibly not very bright) question:

    We know that F=A/d^4 for the Casimir Force (~10^-7 N for two cm^2 plates).

    Does the gravitational attraction between the masses of the plates not figure?

    I once thought that this was not an EM effect as such or at least doesn't depend on EM waves passing 'through' between the plates. Is that right?

    Are various other possible forces factored out, or simply just don't apply?

    I should know the answer, but for now, I'm stumped.

    Thanks, Rudi
     
  2. jcsd
  3. Jul 14, 2003 #2
    The Casimir Force is dependent on neither graviational nor electrostatic attraction. From what I vaguely remember, it is a quantum effect that can be thought of as being due to a difference in virtual photon pressure.

    eNtRopY
     
  4. Jul 14, 2003 #3
    Interesting question, Rudi.

    Casimir predicted (according to QED) a reduced vacuum energy density between two closely spaced plates (due to truncating of virtual photon modes)which results in an attractive force. That force increases (by the inverse 4th power) as the distance d, between the plates decreases. You formula is not exactly correct. The correct formula is:

    F = [pi]hcA/(480 d^4)

    h=Planck's; A = area of plates; c= speed of light; d = distance between plates.

    To answer your question. Usually any measurement is done with flat neutral plates to null electrostatic force; (the distance must be very small to get a Casimir force that is measureable.) However, I've never seen any calculation done to accomodate adjustments due to gravitational attraction, which I assume would be too small to matter. How...ever, just to make sure, try using small plate mass (say, somewhere between 10 to 50 grams) and 1 cm^2 to calculate grav.F using newtonian eqn. Depending on the plate thickness, R (distance between the center of plate thickness) could be about maybe .5 to 6 mm. (F=Gmm/R^2)

    Then compare that grav. F value with the Casimir Force (using the formula above) OR: set the 2 eqns. equal to one another to see at what plate seperation the Casimir force would be comparable to mutual gravitational attraction, (see below.) It may be interesting.

    Creator

    Gmm/R^2 = [pi]hcA/480 d^4
    solving for d and assuming d<<R:
    d = 4th root[[pi](R^2)hcA/480 Gm^2]

    However then you will need to determine if, at that separation, the force will be great enough to be measureable.
     
    Last edited: Jul 14, 2003
  5. Jul 15, 2003 #4
    To Creator:

    Many Thanks: I didn't realize that the simple equation, which I have seen cited, wasn't correct. Your equation makes a lot more sense, explaining the EM phenomenon relationship.

    In particular, thanks for the gravity measure suggestion. (If I don't find it in the literature; we have Igor and Spook, down in the basement, who'll gladly do the experiment for an extra ration of gruel or swill.*) If anything significant is found, I'll let you know.
    As you suggest, it may be considered insignificant on paper.

    Of course, Casimir effect has gained new prominence lately: Applications in Microtechnology.

    *( Sorry, weak attempt at humor.)

    Again Many Thanks, Rudi.
     
    Last edited by a moderator: Jul 15, 2003
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