Why is light traveling through the ether not just at speed c?

[RooftopDuvet]

I've just started reading the dustiest book on relativity i could find at the local library and like most others I've seen it includes the Michelson Morley experiment. I just need a bit of clarification on the formulae they use.

The equations t1=2l/(c^2 - v^2)^1/2 and t2=2lc/c^2 - v^2 seem logical enough but i don't understand why the first one applies to the velocity of light perpendicular to the ether.

The book I'm reading gets the equation from the situation of a boat of speed c crossing a flowing river of speed v. For the boat to travel perpendicular to the banks it has to drive slightly upstream so that its resultant velocity acts in the right direction - that being (c^2 - v^2)^1/2; if you're answering my question you've probably seen this.

If the boat were to instead just aim straight in the direction of the opposite river bank, the river would cause it to travel diagonally; though it's horizontal motion would be completely independant of its vertical motion. Isn't this the same situation as with the perpendicular path of the light through the ether? So why do we apply the equation t1 to the velocity of the light? Why don't we just used the speed c?

 

[Ich]

Because the light has to come back to the point it was sent off. It´s a two way experiment.

 

[RooftopDuvet]

My question was more... when the light reflects off the mirror it begins to travel perpendicular to the motion of the ether; these two motions are independant aren't they?

It's not as though the light is being bent at a slight angle more, so that it's resultant motion is in the perpendicular direction, like with the boat crossing the river. The light is reflected at 45 degrees and so its full velocity c acts vertically. If vertical and horizontal motion are independant then why do we use the equation t1 in this sense?

 

[clj4]

My question was more... when the light reflects off the mirror it begins to travel perpendicular to the motion of the ether; these two motions are independant aren't they?

It's not as though the light is being bent at a slight angle more, so that it's resultant motion is in the perpendicular direction, like with the boat crossing the river. The light is reflected at 45 degrees and so its full velocity c acts vertically. If vertical and horizontal motion are independant then why do we use the equation t1 in this sense?

Because the MM experiment deals with the motion of TWO items:

-the light fron (always moving at c)
-the interferometer mirrors :
:- moving at -v for the receding mirror
:- moving at +v for the "incoming" mirror
:- moving in a more complicated way for the mirror perpendicular to the motion

This is why you see all the c-v, c+v and sqrt(c^2-v^2) in the books.
It is unfortunate that the books do not explain this part.

 

[Ich]

My question was more... when the light reflects off the mirror it begins to travel perpendicular to the motion of the ether; these two motions are independant aren't they?

Of course not. Light should travel through the ether at overall speed c. When you add a small vertical speed component vy, you decrease necessarily the horizontal component vx (Pythagoras: c^2 = vx^2 + vy^2).