# Classical aberration of starlight

mmwave
I have been trying to determine the change in angle required for a telescope due to the aberration of starlight when it is filled with water. The empty telescope is easily done with the law of sines.

The starlight reaches Earth at an arbitrary angle of theta from the vertical with a speed of c. The horizontal is the relative speed of Earth & star V. The hypoteneuse of the triangle is c' the Galilean relativity speed of V+c. The angle of the hypotenuse from the vertical is given by Theta_prime - Theta = V/c * cosine Theta.

Now fill the telescope with water & calculate the new angle theta_prime. I can't find any way to solve this! The only tools I have are law of sines and the law of cosines.

When theta equals zero I can see Theta_prime = Vn/c where n is the index of refraction of the water. I can extrapolate that the answer I want is
Theta_prime - Theta = Vn/c * cos Theta but I can do the analytic geometry to prove it.

Suggestions? (The special relativity answer is much easier to derive but I really want to know how to solve the classical case.)

mmwave
While it's somewhat satisfying that no one else knows the answer but not as satisfying as learning the analytic geometry trick to make this doable.

Gold Member
Unless you have fish in your ancestry, I can't for the life of me figure out why you'd want to have water in your telescope.

mmwave
Danger said:
Unless you have fish in your ancestry, I can't for the life of me figure out why you'd want to have water in your telescope.

The difference between classical & relativistic predictions for aberration in a water filled telescope is a 'proof' of the correctness of special relativity. I'm trying to work out the classical prediction and it seems much harder than the relativistic one!

(My aquatic ancestors are dolphins thank you very much!
)

mmwave.

Gold Member
Hmm... it's a new one on me. Never heard of it (along with millions of other things). I have no idea what any of those math things mean. Out of curiosity, though... when you mentioned an 'empty' tube as opposed to a water-filled one, did you mean air-filled or actually evacuated? I mean, are you comparing the wave-propogation in water to that in air or in vacuum?

mmwave
Danger said:
Hmm... it's a new one on me. Never heard of it (along with millions of other things). I have no idea what any of those math things mean. Out of curiosity, though... when you mentioned an 'empty' tube as opposed to a water-filled one, did you mean air-filled or actually evacuated? I mean, are you comparing the wave-propogation in water to that in air or in vacuum?

The change in aberration, if any, depends on the relative speed of light in free space and in the tube. When Sir Airy did the experiment, it was water filled versus air filled but considering the difference in index of refraction between air and vacuum, it doesn't really matter.

mmwave.

mmwave
It may be bad etiquette to keep answering my own post but the answer finally came to me when I ignored the book's approach to the problem. For the record:

The time of flight is L/u where L is the length of the telescope, u is the speed of light in whatever medium u = c/n. The horizontal distance moved by the scope is vt. Sin theta equals vt/L = vL/u / L = v/u = vn/c. It's so simple...