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I In the Casimir effect, can you slide plates unopposed by force?

  1. Dec 6, 2017 #1
    How would sliding the plates parallel to each other in order to separate them (they are prevented from contacting to avoid friction) require the same amount of energy as pulling them apart? You're not pushing against the force (the net force at the edges pulling it back is balanced by opposite force on opposite edges). Im sure it has to be the same amount of work in both cases to conserve energy, I just dont know how it works.
     
  2. jcsd
  3. Dec 6, 2017 #2

    mfb

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    Staff: Mentor

    As soon as the plates are shifted with respect to each other there is a force component against the direction you slide them. Both edges contribute to that.
     
  4. Dec 7, 2017 #3

    Demystifier

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    In textbooks you will only find a calculation of the force for infinite plates. But real plates, which can be separated by sliding, are not infinite. For real plates there is a component of Casimir force that is not normal to the plates. This component is most pronounced near the edges of the plates. The "opposite" forces on opposite edges are forces on different plates, so they don't cancel out. (To see this, try to draw a picture with forces, being careful about which force acts on which plate!) Therefore, you are pushing against the force.

    In fact, the Casimir force is a red herring here. The same apparent "problem" exists also for classical force between oppositely charged plates.
     
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