In 1851, Fizeau performed an experiment to study how sending light through a mov
ID: 3278053 • Letter: I
Question
In 1851, Fizeau performed an experiment to study how sending light through a moving fluid affects the speed at which the light propagates. Based on the logic of Galilean velocity addition, Fizeau expected that if light travels at a speed c' 0, the light should instead travel at the modified speed upsilon_tot = upsilon + c'. Fizeau's careful measurements demonstrated that light does not move at this predicted total speed, and this problem remained unsolved for half a century. Einstein pointed to the Fizeau experiment as one of his inspirations for special relativity, and he finally solved the mystery when he showed that his relativity velocity-addition formula correctly accounts for Fizeau's results. In this problem, you will follow in Einstein's footsteps by explicitly comparing Fizeau's predicted formula upsilon_tot, F (upsilon) = upsilon+c' with Einstein's corresponding formula upsilon_tot, E (upsilon) = (upsilon + c')/(1 + upsilon c'/c^2), where both formulas are regarded as functions of the fluid speed upsilon for constant c'. We'll see that we can calculate the precise discrepancy between Fizeau's formula and Einstein's formula even though we cannot in practice make the fluid speed upsilon anywhere close to the speed c'. (a) For the case of water, light travels at a speed c' = (3/4) c, meaning 75% of the usual speed at which light would travel through the vacuum of space. If we were to plot upsilon_tot, F (upsilon) as a function of upsilon, what would be the numerical slope of our graph? (b) Continuing to assume that our fluid is water, calculate the precise slope of Einstein's function upsilon_tot, E (upsilon) for values of upsilon near 0 by explicitly evaluating the derivative partial differential upsilon_tot, E (upsilon)/partial differential upsilon at upsilon = 0, and compare your result with part (a). How significant is the discrepancy between Fizeau's slope and Einstein's slope?Explanation / Answer
a) for fizeau's formula
Vtot(v) = v + c'
c' = 3c/4
Vtot(v) = v + 3c/4
d(Vtot(v))/dv = 1 [ slope is 1]
b) For einsteins' formula
Vtot(v) = (v + c')/(1 + vc'/c^2)
c' = 3c/4
so, Vtot(v) = (v + 3c/4)/(1 + 3vc/4c^2) = (4vc + 3c^2)/(4c + 3v)
d(Vtot(v))/dv = [(4c + 3v)[4c] - (4vc + 3c^2)*3]/(4c + 3v)^2 = [7c^2]/(4c + 3v)^2
as v approaches 0
d(Vtot(v))/dv = 7c^2/16c^2 = 0.4375
there is a significant difference in the slope values as can be seen in the calculations
also, Fizeau's alope is constant , always, where as einsteins slope depends on velocityu of fluid