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Thermal expansion (and the Michelson Morley experiment)

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  • #1
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Homework Statement



(B) http://imgur.com/VL4aG
although I showed the rest of the problem for context if needed

Homework Equations


ΔL=αLΔT


The Attempt at a Solution



I'm a little confused by the wording of the problem. When we find the shift in ΔN, are we actually finding Δ(ΔN)?
Anyway, I started off by defining the original ΔN=2Lβ^2/λ as Ni, the initial ΔN value.
Then, I just replaced L with L+ΔL, which is the original arm length plus the change due to temperature. I called this new equation Nf

Nf=2(L+ΔL)β^2/λ=2(L+αLΔT)β^2/λ

Nf=2L(1+αLΔT)β^2/λ = 2Lβ^2/λ + 2Lβ^2ΔT/λ

Now to find the difference between the two, I subtract Ni from Nf

ΔN = Nf-Ni = 2Lβ^2/λ + 2Lβ^2ΔT/λ -2Lβ^2/λ = 2Lβ^2ΔT/λ

However, my answer for ΔN has an extra factor of β^2 in the numerator where the given expression in (B) does not. What did I miss? Am I not reading the question properly?
 

Answers and Replies

  • #2
TSny
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...I started off by defining the original ΔN=2Lβ^2/λ as Ni, the initial ΔN value.
Then, I just replaced L with L+ΔL, which is the original arm length plus the change due to temperature.
Note that the formula ΔN=2Lβ2/λ was derived under the assumption that both arms of the apparatus have the same length. So, if you replace L by L + ΔL in that formula, you are just getting ΔN for an apparatus in which both arms have undergone the same thermal expansion ΔL.

You want to consider the effect of increasing the length of just one arm. This will cause a fringe shift even without an “ether wind”. So, assuming no ether wind (i.e., β = 0), how much extra time would it take for light to travel down and back along an arm that has been lengthened by ΔL? How many fringe shifts does this correspond to? Compare this to what you are supposed to show.

If you were to include the effect of the ether wind in this calculation, it would add corrections of order β2. Because of the square, this is a "second-order" effect. Since β is very small, the second-order correction may be neglected compared to the “zeroth-order” effect that you get by neglecting the ether wind.
 
  • #3
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Ah I see, I completely forgot that only one arm was being affected. I understand what you're saying, but I'm having trouble expressing it mathematically. How do I know which arm to lengthen? The same change in L will affect the parallel and perpendicular arms differently (different denominators, 1-B^2 vs sqrt(1-B^2), but the problem says there is only one value for the change in N.
 
  • #4
TSny
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Ah I see, I completely forgot that only one arm was being affected. I understand what you're saying, but I'm having trouble expressing it mathematically. How do I know which arm to lengthen? The same change in L will affect the parallel and perpendicular arms differently (different denominators, 1-B^2 vs sqrt(1-B^2), but the problem says there is only one value for the change in N.
It won't matter which arm expands. You can let β=0 since the effect you are trying to calculate is dominated by the difference in length of the arms rather than the effect of the motion of the earth. As I mentioned before, the effect of the motion of the earth for this calculation only adds a very small "second-order" correction. So, you can get the desired result by just letting β=0. ΔN will be caused by the fact that the light will take a longer time to travel down and back the longer arm compared to the other arm. All you need to do is calculate the time difference and convert that into a number of fringes. [EDIT: Or, if you don't want to bother with calculating the time difference, just consider the total difference in distance traveled by the light in the two arms. Then consider how many wavelengths of light this difference in distance represents. Each wavelength of difference in distance corresponds to one fringe shift in the interference pattern.]
 
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