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Electric resonance in a moving frame

  1. Nov 17, 2009 #1
    The resonating frequency of a series LC circuit is given by:
    f=(c/2πn)[(dD/Aa)^1/2]
    This frequency measured in a moving frame is same and hence c is concluded to be constant
    but the distances and areas may change due to length contaction and the values of inductance and capacitance should not be the same.What say?
     
    Last edited: Nov 17, 2009
  2. jcsd
  3. Nov 17, 2009 #2
    I think it would help if you explained the notation you are using, e.g.

    Capacitance:

    [tex]C = \varepsilon A/D [/tex]

    where [itex]\varepsilon[/itex] is the permittivity of the gap between the capacitor plates, D is the separation of the plates and A is the area of the plates.

    Inductance of a simple coil:

    [tex]L = n^2 \mu a/d[/tex]

    where n is the number of turns in the coil, [itex]\mu[/itex] is the permeability of the core material, a is the cross sectional area of the coil and d is the length of the coil.

    Resonating frequency:

    [tex]f = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(n^2 \mu a/d)(\varepsilon A/D)}} = \sqrt\left(\frac{dD}{n^2 \mu \varepsilon A a}\right)} [/tex]

    which is similar to the equation you gave, but I am not sure where the speed of light c comes in, or why you have a factor of 1/2 unless you are talking about the angular frequency which has a factor of 1/(2itex]\pi[/itex]).

    I think you will find that the relativistic equation for the LC frequency is slightly different from the classical one and you probably have to allow for whether the capacitors and inductors are parallel or orthogonal to the motion. Either way, we can not have a sensible discussion, until we are clear on all the variables being used in you equation.
     
  4. Nov 17, 2009 #3

    Vanadium 50

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    No it's not. It's time dilated.

    That has nothing to do with the constancy of c.
     
  5. Nov 17, 2009 #4
    I think he meant the proper frequency as measured by an observer co-moving with the electrical circuit, will stay the same, which of course it will.
     
  6. Nov 18, 2009 #5
    It's actually the other way, (LC)^-1/2 is the resonating angular frequency.
    (μ ε)^-1/2 is c
    Yes.
    I found on an australian university website this explanation for the constancy of c,which does not seem to be a good one.
     
    Last edited: Nov 18, 2009
  7. Nov 18, 2009 #6
    OK substituting c for (μ ε)^-1/2, I get:

    [tex]f = \frac{c}{n}\sqrt\frac{dD}{ A a}} [/tex]

    or in your notation:

    f=(c/n)[(dD/Aa)^1/2]

    where as you have:

    f=(c/2πn)[(dD/Aa)^1/2]

    so we are getting close. Can you clear that up?
     
  8. Nov 18, 2009 #7
    You missed a line.Read the first line.
    They said that because the frequency measured in both the frames is the same, and all other values are the same, c should be constant.Not impressive; some or all of distances and areas change under length contraction.
    This raises an interesting point:We know the frequency is the same, but still it isn't due to contraction according to the formula
     
    Last edited: Nov 18, 2009
  9. Nov 18, 2009 #8

    Vanadium 50

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    Yes, but then nothing is length-contracted, so what's the point?
     
  10. Nov 18, 2009 #9
    Ah, OK... so

    [tex]f = \frac{c}{2\pi n}\sqrt\frac{dD}{ A a}} [/tex]

    or in your notation:

    f=(c/2(pi)n)[(dD/Aa)^1/2]

    where as you have:

    f=(c/2πn)[(dD/Aa)^1/2]

    I am just trying to get an equation we both agree on and then we maybe take a look at how it transforms under relativity.

    Good point.

    As vanadium points out, the resonating frequency will NOT be the same when measured by an observer with motion relative to the circuit.
     
  11. Nov 18, 2009 #10
    Assuming:

    [tex]f = \frac{c}{2\pi n}\sqrt\frac{dD}{ A a}} [/tex]

    and knowing the frequency in the moving frame (f') will time dilate such that:

    [tex] f ' = f \sqrt{1-v^2/c^2} = f/\gamma [/tex]

    then

    [tex]f ' = \frac{1}{\gamma} \frac{c }{2\pi n}\sqrt\frac{dD}{ A a}} [/tex]

    If the both the inductor of length d and the capacitor with gap D are orientated with d and D parallel to the motion ( and area a and A orthogonal to the motion) then:

    [tex]f ' = \frac{c }{2\pi n}\sqrt\frac{(d/\gamma)(D/\gamma)}{ A a}} [/tex]

    which is the same as the above equation.

    If the d and D are both orthogonal to the motion then:

    [tex]f ' = \frac{c}{2\pi n}\sqrt\frac{dD}{ (A/\gamma) (a/\gamma)}} [/tex]

    then there is a problem because [itex] f ' = f/\gamma [/itex] is no longer true.

    hmmm.. something else must be changing or the relativistic equation must be completely different to the classical one which is sometimes the case...
     
  12. Nov 18, 2009 #11
    I worked it that way too, frequency reduces by a factor of inverse of lorentz factor, but in any case, apllying length contraction doesn't let the equation hold true.
    BTW, is it even possible to measure the charge on the capacitor and voltage across it in a moving frame?If not, the whole argument fails
     
  13. Nov 18, 2009 #12
    As I understand it, charge is an invariant quantity in relativity.
     
  14. Nov 18, 2009 #13

    Dale

    Staff: Mentor

    This is probably correct. I know that the derivation of the formulas for the resonance frequency involve the energy in the electric and magnetic fields, but for a moving capacitor there can be a magnetic field and for a moving inductor there can be an electric field both depending on the orientation. This would necessarily change the resonance equation.
     
    Last edited: Nov 18, 2009
  15. Nov 18, 2009 #14
    I looked up what effect change in resistance of the circuit would have and apparently it has no effect on the resonant frequency if we treat the circuit as LCR resonator, so that line of thought would appear to be a dead end.

    Anyway resistance is proportional to Dp/A where D is the length of the resistor, p is the resistivity of the resistor material and A is the cross sectional area of the resistor. If we consider the effects of length contraction the resistance of the resistor is very different depending on whether it is orientated parallel or orthogonal to the motion. It turns out that if 4 resistors of equal resistance are arranged in a Wheatstone bridge configuration (see http://en.wikipedia.org/wiki/Wheatstone_bridge ) then the apparatus would make a very good absolute motion detector, a bit like the MM experiment. Since this is probably not the case then the other variable p (resistivity of the material) must change in a complex way to compensate for the changes in the dimensions of the resistors. Does anyone have any information on this?
     
  16. Nov 19, 2009 #15
    Let me try something. We have a dc source of potential V,a wire of length l carrying a current I with a drift velocity v in the unprimed frame.
    For simplicity, consider the whole wire with a uniform directional element joining the two terminals.
    Now the observer moves with a velovity v1 parallel to the wire.
    The electrochemical potential of the cell remains the same.
    Its length reduces to l/(gamma), and V is the potential between the extereme points of this reduced length.
    The initial drift velocity of the electrons is given by
    v=I/nqA
    A is the area of cross section
    n is the number of charge carriers per unit volume
    The quantity nq does not change in the primed frame.
    Here the area also remains A.
    But the drift velocity changes by the velocity addition formula by a factor to be multiplied to the original current to get the current in the primed frame.
    After this is done, divide the resistance by the same factor.
    What this analysis results is that the resistance cannot be simply computed by using length contraction.
    I think the resonating frequency in a moving frame can be found out similarly while not really caring for the inductances and capacitances in the primed frame(unsure).
     
    Last edited: Nov 19, 2009
  17. Nov 19, 2009 #16

    Dale

    Staff: Mentor

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