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A Casimir effect and dimensional analysis

  1. Oct 22, 2016 #1
    How can you use dimensional analysis to estimate the force between two plates, as a function of the area of the plates and their separation?
     
  2. jcsd
  3. Oct 22, 2016 #2
    This is my hunch:

    Estimating the mass of the plates to be ##0.1\ \text{kg}## and the time taken to move ##1\ \text{m}## be ##100\ \text{years} \sim 10^{9}\ \text{s}##, the force is

    ##F=\frac{MA}{T^{2}L}=10^{-8}\frac{A}{L}##

    in SI units.

    What are your thoughts?
     
  4. Oct 22, 2016 #3

    Vanadium 50

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    First, is this homework?

    Second, dimensional analysis does not mean "make up some numbers and see what happens". It means to ask yourself questions like "if I double the distance between the plates, what happens to the force?". What you wrote down, for example, says the force is independent of distance between the plates. Does that seem reasonable to you?
     
  5. Oct 22, 2016 #4
    No, this is not homework.

    Well, the force is independent of the separation of the plates in my formula. In fact, I have ##F\sim\frac{A}{L}##, where ##L## is the separation between the plates.
     
  6. Oct 22, 2016 #5

    Vanadium 50

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    Which is why your formula is wrong.
     
  7. Oct 22, 2016 #6
    Oh no, I meant 'dependent'. The formula has ##L##, the separation between the plates.
     
  8. Oct 22, 2016 #7

    Vanadium 50

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    But the force between the plates does not go as 1/L. You're writing nonsense.
     
  9. Oct 22, 2016 #8
    I'm using the following dimensional analysis:

    ##\displaystyle{[F]=\frac{kg\ m}{s^{2}}}##,

    so that the force ##F## should scale as the length of some quantitiy in the problem. Now there are two parameters ##A## and ##L## in the problem. ##A## is the area between the plates and ##L## is the separation between the plates, so I found it reasonable to say that ##\displaystyle{F \sim \frac{A}{L}}##.

    Where's my mistake?
     
  10. Oct 22, 2016 #9
    The problem is that there may be dimensionful constants, such as ##\hbar## and ##c## involved. You realise that your analysis neglects the presence of the ##\mathrm{kg\, s}^{-2}## terms in the dimensions of the force - you can't just extract the length dimension and make a claim based on that alone (because other physical quantities, apart from the constants, can also be 'coupled' to the length dimension).
     
  11. Oct 22, 2016 #10
    But the Casimir force cannot only depend on ##\hbar## and ##c##. I tried dimensional analysis using

    ##F=\hbar^{\alpha}c^{\beta}##

    and I get three equations for the two variables ##\alpha## and ##\beta## by using dimensional analysis. The equations are inconsistent with each other.

    Do you suppose that Newton's gravitational constant ##G## may also be involved?
     
  12. Oct 22, 2016 #11
    Of course it doesn't; it also depends on the separation of the plates and the area of the plates.
    As a first guess (which turns out to be correct), we wouldn't naturally include it. At the end of the day, dimensional analysis is not an exact science - and we can only guess based on the supposed physics involved in the situation. In this case, the Casimir effect is thought to be due to the zero-point energy / (electromagnetic) vacuum fluctuations, and so it seems natural to include ##\hbar## and ##c##.
     
  13. Oct 22, 2016 #12
    Ah! I see!

    But dimensional analysis also suggests that

    ##\displaystyle{F = \frac{kg m}{s^{2}}}##.

    Are you saying that I should encode ##\displaystyle{\frac{kg}{s^{2}}}## into factors of ##\hbar## and ##c##?
     
  14. Oct 22, 2016 #13
    No, there is no reason to separate out that particular part of the dimension of the force. (Besides if you tried that, you'll find that it's impossible) Just use the dimensional analysis relation you did earlier but including all four quantities this time: ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}##. We intuitively expect the force to scale linearly with the surface area of the plate, so we can further take ##\gamma = 1## to simplify the situation.
     
  15. Oct 22, 2016 #14
    But isn't this choice of equation ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}## completely arbitrary?

    After all, F depends on mass, length and time. ##\hbar## and ##c## might often appear in some physical quantities due to the combinations ##Js## and ##ms^{-2}## which are commonly found in physical quantities in quantum-field-theoretic systems.

    So, to me, ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}## looks like an arbitrary choice of equation to solve.
     
    Last edited: Oct 22, 2016
  16. Oct 22, 2016 #15
    Well it is arbitrary to some extent - but not completely arbitrary. Its just like an ansatz, an educated guess that we make based on what we know (or think we know) about the underlying physics of the phenomenon. There's certainly no guarantee that dimensional analysis will end up giving us right results especially if we chose the wrong starting quantities to consider.
     
  17. Oct 22, 2016 #16
    But then, are you by any chance using the fact that there are three units - mass, length, time - in force and so we need three equations for the separate units of mass, length and time. Now, this is a quantum-field-theoretic effect so that ##\hbar## and ##c## are natural choices to consider and then ##L## is something that we just intuitively expect the force to depend on. Is that it?
     
  18. Oct 22, 2016 #17
    I wouldn't think this is a particularly important consideration - after all, certain quantities carry several dimensions, so we don't necessarily need three equations.
    Mainly this I would say. As with all guesswork formulations, if somehow we cannot get a reasonable solution for the powers, then its back to the drawing board to re-examine what other quantities we might have missed out and then include them in and repeat the analysis.
     
  19. Oct 22, 2016 #18
    But in this case,

    ##[F] = [\hbar]^{\alpha} [c]^{\beta} [L]^{\gamma}##

    ##kg ms^{-2} = (Js)^{\alpha} (ms^{-1})^{\beta} (m)^{\gamma}##

    ##kg ms^{-2} = (kgm^{2}s^{-1})^{\alpha} (ms^{-1})^{\beta} (m)^{\gamma}##

    so that the equation decouples into three separate equations for ##kg##, ##m## and ##s##.

    This shows that we need three parameters (in this case, they are ##\hbar##, ##c## and ##L##) since we need three powers to solve for from the three equations.

    This is why I say that we need three parameters for this problem (in this case, they are ##\hbar##, ##c## and ##L##).
     
  20. Oct 22, 2016 #19
    What are your thoughts?
     
  21. Oct 22, 2016 #20
    I still don't think it is always necessary for there to be three quantities involved. After all, another way to write the force is in terms of mass x acceleration - that's two quantities. It's also possible that four or more quantities are required. Basically, there is no unique way to decompose one quantity in terms of other quantities, so it still depends on the choices that we make based on the physics involved.
     
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