Length Contraction "Ether wind"

unscientific

1. The problem statement, all variables and given/known data

Due to the ether wind, anything that moves along it is "contracted".

3. The attempt at a solution

Taking L₁ to be contracted length,

(L₁)²
= (L_1x)² + (L_1Y)²
= (L_10,X/γ)² + (L_10,Y)²


But what they wrote is the opposite..

Attached Thumbnails

 

TSny

The equation (3-3) in the book is an expression for the "normal" (not contracted) length in terms of the components of the contracted length. Your expression is for the contracted length in terms of the components of the normal length.

unscientific

Are both equivalent??


I tried to prove but the square roots got me..

unscientific

1. The problem statement, all variables and given/known data

Apparently rotating the setup at an angle doesn't change the time difference between t₁ and t₂..

3. The attempt at a solution



But the square roots below are clearly different..

Simon Bridge

square roots below what?

unscientific

I tried changing everything in the square roots to only 'sin' or 'cos' but can't seem to remove the square roots at all..

Simon Bridge

I'll have to see your working - is there an attachment missing?
OH I see ... there is a tinypic.com link that is not rendering for me.

unscientific

Oh sorry about that, can you view this attachment?

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unscientific

bumpp

Simon Bridge

Oh there you are - couldn't access the thread for a while there.
I'll have a proper look when I have a bit of time.
[edit]Is this the reworking of the MM experiment using SR?

TSny

unscientific,

Note that the time, [itex]t_1[/itex], to travel out and back along [itex]l_1[/itex] is not [itex]2l_1/c[/itex] because ##c## is not the speed of light relative to the apparatus. You're going to have to figure out the speed of the light relative to the apparatus for the light going out along arm [itex]l_1[/itex] and the (different) speed for the light coming back. Likewise for the other arm.

unscientific

I assume:
1) only the component parallel to ether wind is affected by it
2) The light is aimed at a smaller angle to compensate for this

so the speed towards is:
√[ (c sinθ)² + (c cosθ - v)² ]

and the speed back is:

√[ (c sinθ)² + (c cosθ + v)² ]

Not sure if that's right..

TSny

That would be ok if θ here is the angle that the light is "aimed at". But in that case, it's not the angle of inclination of the arm as shown in the diagram.

unscientific

I'm not sure what you mean..If c is indeed affected by ether wind won't θ be off-balanced?

TSny

Indeed. You'll have to fire the light beam in some direction [itex]\phi[/itex] so that the wind will blow it into the direction θ (along the arm). I was interpreting the direction "aimed at" as the direction fired ([itex]\phi[/itex]). Your expression for the light speed relative to the apparatus is ok if your θ is replaced by [itex]\phi[/itex].

You might try drawing a velocity addition triangle and using the law of cosines to express the light speed relative to the apparatus in terms of θ without worrying about the value of [itex]\phi[/itex].

harrylin

If I correctly understand what you say, then you are copying an error of Michelson! See:

1. http://en.wikisource.org/wiki/The_Re...niferous_Ether :
" If, however, the light had traveled in a direction at right angles to the earth's motion it would be entirely unaffected."

I think that this is also what you are saying now.

2. http://en.wikisource.org/wiki/Influe...ocity_of_Light :

"In deducing the formula for the quantity to be measured, the effect of the motion of the earth through the ether on the path of the ray at right angles to this motion was overlooked." [etc]

Does that help?

unscientific

Can't seem to get it to be independent of ∅... it turns out as (θ-∅)