Wednesday, August 10, 2011

Solutions to Special Relativity Problems (1)

We now go to the solution of the earlier special relativity problems. To recap:

1. In the Michelson-Morley experiment, the length L of each arm of the interferometer was 11 meters. Sodium light of wavelength 5.9 x 10^-7 m (590 nm) was used. The experiment would have revealed any fringe shift > 0.005 fringe.

What upper limit does this place on the Earth's velocity through the supposed Ether?


Given L = 11m and lambda = 5.9 x 10^-7 m

limit for lowest resolution: 2 delta d = 0.005 fringe


d = (0.005 fringe)/ 2 = 0.0025 fringe

and since:

delta d = Lv^2/ (lambda c^2)

v = {(delta d)(lambda) c^2)/ L}^½

v = {(0.0025)(5.9 x 10^-7 m)(3 x 10^8 m/s)/11 m }^½ = 3.5 km/s

2. Using Fig.1, say the time of travel to the right is: t(r) = L/(c - v) and the time of travel to the left is t(L) = L/(c + v).

a) Find the "total time of travel" by adding both left and right contributions.


We are going along (parallel) or opposed (anti-parallel) to, the "ether wind" direction from the diagram and this is horizontal so designate it by direction x, e.g.

t(total x)= L/ (c + v) + L/(c - v)

The common denominator is:

(c - v)(c + v) = c^2 -cv + cv + v^2 = c^2 - v^2

Then t(total x) = 2Lc/ (c^2 - v^2)

t(total x) = 2L/c (1 - v^2/c^2)^-1

b) find the time consumed for "a half-trip".


half trip time = (t(total x))/2 = ½ [2Lc/ (c^2 - v^2)]

(t(total x))/2 = Lc/(c^2 - v^2)

But recall that the interferometer has parallel and perpendicular components so that total time registering both is:

T(total x-y) = t1 + t2 where:

t1 = 2Lc/ (c^2 - v^2)

t2= 2L/ (c^2 - v^2)^½

So: t1 + t2 = 2Lc/ (c^2 - v^2) + 2L/ (c^2 - v^2)^½

= 2L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

so T/2 = L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

c) Find the time consumed for a round trip.

round trip is 2(t1 + t2) = 2(2L{c + (c^2 - v^2)^½)/ (c^2 - v^2))

= 4L{c + (c^2 - v^2)^½)/ (c^2 - v^2))

d) Add the two "half trips" and what do you obtain?

We get:

2 x T/2 = 2L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

e) Why does this not agree with the value obtained for (a)?

Doesn't agree because each summing (or halving) as computed above takes into account differing directional components of velocity with respect to the interferometer.

No comments: