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Energy from Casimir effect

  1. Aug 30, 2007 #1
    Hi,

    Could somebody explain the flaw in the following free-energy situation? (Apologies for bad ASCII art - # represents empty space)

    (1) I set up two reflecting plates a short distance apart, and allow them to gain velocity towards each other, due to the Casimir effect.

    #############--------------><-------------######## PLATE 1


    #############------------------------------######## PLATE 2


    (2) Once the plates are close, I extract the small amount of kinetic energy and stop the motion of the plates (the mechanism for this is not important here).

    #############--------------><-------------######## PLATE 1
    #############------------------------------######## PLATE 2


    (3) I then split one of the plates and slide the sections sideways away from the other plate. The Casimir effect now no longer occurs between the two plates.

    -------------->#########################<------------- PLATE 1
    #############------------------------------######## PLATE 2


    (4) I move the other place back to its orginal position - this does not require any energy as the Casimir effect no longer applies.

    -------------->#########################<------------- PLATE 1


    #############------------------------------######## PLATE 2


    (5) I slide the two parts of the split plates back into position. I'm now back at stage 1.

    #############--------------><-------------######## PLATE 1


    #############------------------------------######## PLATE 2


    I presume the flaw is at stage (3) and this must somehow require energy, but I do not initially see how at this stage.

    Can anyone else help explain this?
     
  2. jcsd
  3. Aug 30, 2007 #2
    #3 will require a force. Remember that forces are just derivatives of energy -- the energy of the system has gone up, so you will have to exert a force to overcome it.
     
  4. Aug 30, 2007 #3
    Thanks genneth, but I'm not sure that explains it.

    #3 does require a force but this force could be tiny, as the force does not have to overcome the pressure differential caused by the Casimir effect, as it is applied at right-angles to the Casimir force.

    I guess the real question is what (if any) is the relationship between the size of the Casimir force and the size of the required right-angled plate-moving force?
     
  5. Aug 30, 2007 #4
    The force will be exactly correct. Just calculate the relevant energy densities.
     
  6. Aug 30, 2007 #5
    I can't calculate the require motive force without knowing what physical phenomenon would cause a resistive force to be present against the sideways motion? In the absence of a resistive force, the required motive force is infinitessimal.

    Can you tell me what physical phenomenon would cause a resistive force to the motion suggested in #3?
     
  7. Aug 30, 2007 #6
    It's similar to parallel capacitor plates. If you charge up a capacitor, and "slide" the plates apart, you'll experience a force. The mechanism is different, but formally the the mathematics is the same.
     
  8. Aug 31, 2007 #7
    Hi,

    The capacitor plate analogy is interesting but I do see some differences. For instance, consider the case where the capacitor plates have been moved halfway across:

    ###########################<-------------- POSITIVE PLATE 1
    ####################||||||||||||||||||||||||| NEUTRAL FRICTIONLESS SPACERS
    #############------------------------------#### NEGATIVE PLATE 2

    Plate 1 has a +ve charge and Plate 2 has a -ve charge. If I stopped applying a motive force to plate 1 at this point, the attractive electrostatic force between plates 1 and 2 would cause plate 1 to slide back to its starting position. Thus there is a clear resistive force to the shearing motion that must be overcome.


    In the Casimir case, if I stop moving the two neutral plates at the same point:
    ###########################<-------------- NEUTRAL PLATE 1
    ####################||||||||||||||||||||||||| NEUTRAL FRICTIONLESS SPACERS
    #############------------------------------#### NEUTRAL PLATE 2

    I cannot see what mechanism would cause a force to apply sideways against plate 1. The Casimir effect would still tend to press plates 1 and 2 together, but with less strength than before.

    So I don't think the two situations are directly analogous.
     
  9. Sep 3, 2007 #8
    So does anyone know what physical phenomenon could cause a resistive force to arise when shearing two plates that have been brought closer together by the Casimir effect?

    In the absence of a reason for a resistive force, I would have to conclude that useful energy can be extracted from the vacuum energy! (Or that I do not fully understand the Casimir effect).
     
  10. Sep 4, 2007 #9
    Do you accept that force is the derivative of energy with displacement? Do you accept that when you slide the plates, the region between the plates (which have a lower energy density) reduces? Do you then accept that the energy of the entire system has been raised? Then there will exist a force when you slide that plate, one that precisely balances the apparent energy gain.
     
  11. Sep 4, 2007 #10
    *Do you accept that force is the derivative of energy with displacement?
    Yes - but it is also the derivative of momentum with respect to time, and many other derivatives (as are all physical measures).

    *Do you accept that when you slide the plates, the region between the plates (which have a lower energy density) reduces?
    Certainly.

    *Do you then accept that the energy of the entire system has been raised?
    Not necessarily - it could be that the energy of the vacuum has been lowered to balance out the gain in the energy of the plates.

    *Then there will exist a force when you slide that plate, one that precisely balances the apparent energy gain.
    I agree this is entirely possible but do not understand the mechanism that would cause it to occur. You seem to be saying, the energy must balance so there must be a force so we don't need to understand any deeper. But the explanation for the Casimir effects I have read would not cause a lateral force to occur between the two plates. So what effect causes this force to arise? A mystic 'balancing the energy' force is not enough explanation for me!

    A question for you - do you agree that if I set up the two examples of charged and neutral plates as in my previous post and had both environments starting from rest, that the charged plates would spontaneously move laterally but the neutral plates would not?
     
  12. Sep 5, 2007 #11
    The mechanistic reason -- going from down to up -- is due to edge effects.
     
  13. Sep 6, 2007 #12
    Hippodag: the "reason" is because you are re-enabling the Casimir effect. This is no different from cocking a spring.

    What causes the force? I think what you mean by "causes" is "what QFT exchanges cause a balancing force that resist the plate being pushed back in?" Not "find various ways to predict that this force will exist" but "how does it manifest?"
     
  14. Sep 7, 2007 #13
    Hi JDługosz,

    You are absolutely correct, I am interested in knowing the QFT exchanges that would cause the balancing force.

    A bit of a googling search shows some interesting papers, this one:

    http://www.cmth.ph.ic.ac.uk/photonics/Newphotonics/pdf/pendry97a.pdf

    talks about 'quantum friction' - about how shearing two close plates even in a perfect vacuum causes frictional effects due to QFT exchanges of virtual particles. The basic cause here is doppler effects - moving plates shift the virtual particles and hence cause an effect on the other plates. At first glance this seems an ideal reason, but a couple of reasons stop it applying here - (1) it applies to dialectric plates, but for the Casimir effect I have conductors, and (2) it is proportional to the velocity of movement - so I could just move the plates verrrrry slowly, and hence have a minimal force to work against.


    This one:

    http://www.citebase.org/fulltext?format=application/pdf&identifier=oai:arXiv.org:quant-ph/0606235

    has nice pictures showing the energy density contours between two plates undergoing Casimir interactions. This shows the fact that the energy density is lower even outside the region directly between the plates (this may be obvious, but I hadn't thought of it).

    Again, at first glance again this looks like a good reason for a resistive force, as the lower energy density past the edge of the moving plate could cause a resistive lateral force to apply. However there is an equal and opposing low density region at the other end of the plate, cancelling out this effect.

    So I don't understand the physics that would cause a resistive force to arise. Whilst 'edge effect' sounds like a perfectly plausable reason for this force, I don't understand what it really means in terms of quantum field theory. Note - this may not actually be a very easy question - it's OK to admit that you don't understand as well! (In fact it's a good thing to admit limits to ones knowledge).

    For example, if the two plates were very close (even by Casimir scales) then clearly moving the plates vertically apart would involve directly reinstating the low energy region of space, and require energy, very close to the 'cocking a spring' analogy. But, moving the plates laterally does not directly create a low energy region of space, so what is the precise mechanism that causes it to require energy?

    So does anyone here understand well enough to explain in straightforward terms so that someone who understands college level physics QED would understand? (i.e. like Feynman might have)?
     
  15. Sep 10, 2007 #14
    OK I've figured out one reason for a resistive force by considering quantisation at different angles.

    I've done some maths on this effect and (according to my calcs) it accounts for 80% of the Casimir energy gain. However this still leaves a net gain in energy by linking then separating the plates. Of course I may have made mistakes in the maths or failed to take something else into account...

    I can post some details of my calculations if anyone is interested - let me know.
     
  16. Sep 20, 2007 #15
    Calculation of the Casimir effect

    Letting you know that I am interested (but don't invest special effort because of this.
    The interaction energy between the two finite plates should clearly be more complicated than between two infinite planes, would be interesting to see the details.
     
  17. Sep 22, 2007 #16
    Hi Slaviks,

    The basic idea is to consider quantisation at every different angle in turn.

    #################--A---------------------------####
    ##########################################
    #############--B----C----D-----------------########

    If the distance between the plates (AC) is a, then the Casimir force per unit area is (-hbar * c * pi^2) / (240*a^4), or X/a^4 where X is a constant. So the only two variables that affect the size of the Casimir force is the distance between the plates and the area between the plates.

    If I consider quantisation in the AC direction (perpendicular to the plates) then the Casimir force acts entirely to push the plates together. If I consider quantisation in the AB direction, the force has a lateral component that will cause the plates to slide together. Finally, if I consider quantisation in the AD direction then there is a lateral component of the force in the opposite direction, causing the plates to slide apart.

    The key concept I realised here is that the area between the plates in the AD direction is less than the area between the plates in the AB direction, so the two lateral Casimir forces do not net-out to zero; there will be a net lateral Casimir force causing the plates to slide together.

    Calculating the size of this force involves integrating over every angle between the two plates to calculate (i) the distance between the two plates at this angle, (ii) the effective area between the plates at this angle, and (iii) the lateral component of the resulting Casimir force at this angle.

    It's also a bit tricky as the plates are in 3-d space, so you have integrate over angles in the plane shown above (AB, AC, AD etc.) but also over other solid angles, i.e starting at point A and ending at points behind and in front of the plane of the screen.

    I have had a go at generating and solving this integral - not totally easy as the trigonometry and calculus involved are a bit fiddly. Anway, I got an answer that seemed sensible, and showed a lateral force that accounted for most of the Casimir energy gain in moving the plates together. I am not confident that I got the calculations correct, as I had to do numerical integration across two dimensions and my numerical methods mathematical skills are not the greatest.

    Of course there are several assumptions implicit in this approach, the main one being that quantisation effects between parallel conductors work in the same fashion at different angle - does anyone have enough QM knowledge to know if this would be true?

    -Hd
     
  18. Sep 24, 2007 #17
    Tricky Casimir

    Hi!

    Surely, there should be a kind of Casimir attraction also in the
    lateral direction.

    However, you seem to make an assumption with
    makes you estimate a somewhat uncontrolled approximation:



    The quoted Casimir force per area is derived for infinite plates, totally disregarding
    the boundary effects. Only then it is strictly normal to the surface and uniform in magnitude. However, there still could be a way to check the accuracy of your assumption: you analysis of two attracting plates perfectly applies to a classical capacitor. There could me elaborate treatments of boundary effects in capacitors in text books (it might be exactly solvable by the means of holomorphic transformations), you can check against them. Or even do the full classical math themselves: two opposite plates attarct just as (an infinite) set of infinitizemal point-charges, so do an exact 3D integration.

    Well, in principle I do. The path to exact answer is to first solve the
    boundary value problem for the eigenmodes of the 3D electromagnetic field
    around two laterally shifted perfectly conducting stripes. Then you assign \hbar \omega to each mode an carefully (an array of math tricks is available here) subtract an infinite sum from an infinite integral. The result is the interaction energy, whence you derive the force. Alas, this a priory kills your proposal for energy extraction: the calculated force will be conservative by derivation.

    BTW, I look for volunteers to help me solving
    exactly a toy model of Casimir/van der Waals crossover:

    https://www.physicsforums.com/showthread.php?t=183407

    I believe it's an interesting problem with a publishable outcome.
     
  19. Sep 24, 2007 #18
    I believe those two edge effects add, not subtract. They are not just mirror images of each other, they are mirror images but with the role of each plate swapped.

    To simplify, consider the plates as composed of many individual cubes. There is a Casimir force between each cube and every other cube. No matter how you move the cubes around, whether in unison or independently, the total energy you extract or provide is the sum of all the energy values extracted or provided to each possible pair of cubes.

    So if you move the cubes around but eventually bring them back to the same configuration that you started in, then the net energy provided is zero.
     
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