Some help on Michelson-Morley experiment

[mathlete]

Not exactly, but related.

The first one seems to be a basic math reduction problem I can't figure out. I'm trying to get the time distance between to objects that travel in perpendicular directions with something affecting their speed (the ether). I have:
$$\Delta t = (\frac{2cd}{c^2-v^2}) - (\frac{2d}{\sqrt{c^2-v^2}})$$
(If anyone knows this experiment and knows what I'm trying to find, please tell me if this is wrong... I don't know if that's the cause of my problems, but I don't really think so. If not just ignore this part)

Now I'm supposed to reduce this to (when v<<c):
$$\Delta t \approx (\frac{d}{c})(\frac{v}{c})^2$$
But I can't seem to do it. Apparently I have to use the fact that:
$$(1+z)^n \approx 1+nz$$

The second issue I'm having is in the experiment itself. They say that "Hencethat the total change in delay time between the two paths (of light) observed as the interferometer rotates should be twice the difference calculated using the expression c (the one I'm supposed to find above)... show that this result implies that the motion of the ether at the surface of Earth is less than one sixth the speed of Earth in its orbit"

Now I have no problem doing this, I just need to plug in numbers.. but I don't know what to plug in exactly. I can solve for v (speed of ether). I have the wavelength of light they use (589 nm), and from this can find the period to use for the change in time. I know c=speed of light. But what is L supposed to be, cause it's not given in the problem. Is it the wavelength of light? When I use it I get an answer that's way too large.

Any help is appreciated

 

[mathlete]

Anyone have an idea?

 

[jtbell]

$$\Delta t = (\frac{2cd}{c^2-v^2}) - (\frac{2d}{\sqrt{c^2-v^2}})$$

First note that

$$(c^2 - v^2) = c^2 \left(1 - \frac{v^2}{c^2}\right)$$

Next, note that

$$\frac {1}{1-\frac{v^2}{c^2}} = \left(1-\frac{v^2}{c^2}\right)^{-1}$$

and

$$\frac {1}{\sqrt{1-\frac{v^2}{c^2}}} = \left(1-\frac{v^2}{c^2}\right)^{-1/2}$$

Since v << c under the conditions of the M-M experiment, this allows you to use the binomial approximation

$$(1+z)^n \approx 1+nz$$

You can use - instead of + in this formula.

But what is L supposed to be

In the equation you give, you have a d, not an L. I assume that's that what you're really asking about. It's the round-trip path length of the light through one of the two paths through the interferometer. We only count the sections that are different for the two paths, so it's twice the length of one of the "arms" (from the beamsplitter to a stationary mirror).

The lengths of the two "arms" have to be nearly equal in order to get a decent interference pattern, so it doesn't matter in practice which length you use.

 

[mathlete]

First note that

$$(c^2 - v^2) = c^2 \left(1 - \frac{v^2}{c^2}\right)$$

Next, note that

$$\frac {1}{1-\frac{v^2}{c^2}} = \left(1-\frac{v^2}{c^2}\right)^{-1}$$

and

$$\frac {1}{\sqrt{1-\frac{v^2}{c^2}}} = \left(1-\frac{v^2}{c^2}\right)^{-1/2}$$

Since v << c under the conditions of the M-M experiment, this allows you to use the binomial approximation

$$(1+z)^n \approx 1+nz$$

You can use - instead of + in this formula.

Ah, ok. I'll work on it some more, I'm probably missing a simple step or something

In the equation you give, you have a d, not an L. I assume that's that what you're really asking about. It's the round-trip path length of the light through one of the two paths through the interferometer. We only count the sections that are different for the two paths, so it's twice the length of one of the "arms" (from the beamsplitter to a stationary mirror).

The lengths of the two "arms" have to be nearly equal in order to get a decent interference pattern, so it doesn't matter in practice which length you use.

Sorry, I forgot to mention in the original question that L=2d. That's what I thought too, that it's the length of the arm I should be using, but I'm not given that in the problem so I can't find a numerical answer for velocity of the ether