## Length Contraction "Ether wind"

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
= (L1x)2 + (L1Y)2
= (L10,X/γ)2 + (L10,Y)2

But what they wrote is the opposite..
Attached Thumbnails

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Homework Help 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.

 Quote by 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.
Are both equivalent??

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

## Length Contraction "Ether wind"

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..

 Recognitions: Homework Help square roots below what?

 Quote by Simon Bridge square roots below what?
I tried changing everything in the square roots to only 'sin' or 'cos' but can't seem to remove the square roots at all..

 Recognitions: Homework Help 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.

 Quote by 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.
Oh sorry about that, can you view this attachment?
Attached Thumbnails

 bumpp
 Recognitions: Homework Help 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. Is this the reworking of the MM experiment using SR?
 Recognitions: Gold Member Homework Help unscientific, Note that the time, $t_1$, to travel out and back along $l_1$ is not $2l_1/c$ 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 $l_1$ and the (different) speed for the light coming back. Likewise for the other arm.

 Quote by TSny unscientific, Note that the time, $t_1$, to travel out and back along $l_1$ is not $2l_1/c$ 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 $l_1$ and the (different) speed for the light coming back. Likewise for the other arm.
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..

Recognitions:
Gold Member
Homework Help
 Quote by unscientific so the speed towards is: √[ (c sinθ)2 + (c cosθ - v)2 ]
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.

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

Recognitions:
Gold Member
Homework Help
 Quote by unscientific I'm not sure what you mean..If c is indeed affected by ether wind won't θ be off-balanced?
Indeed. You'll have to fire the light beam in some direction $\phi$ 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 ($\phi$). Your expression for the light speed relative to the apparatus is ok if your θ is replaced by $\phi$.

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 $\phi$.

 Quote by unscientific I assume: 1) only the component parallel to ether wind is affected by it[..]
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?

 Quote by TSny Indeed. You'll have to fire the light beam in some direction $\phi$ 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 ($\phi$). Your expression for the light speed relative to the apparatus is ok if your θ is replaced by $\phi$. 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 $\phi$.
Can't seem to get it to be independent of ∅... it turns out as (θ-∅)

 Tags length contraction, michelson morley