# Length Contraction Ether wind

Length Contraction "Ether wind"

## Homework Statement

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

## 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..

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TSny
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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.

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..

Michelson Morley Experiment: Rotating setup

## Homework Statement

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

## The Attempt at a Solution

But the square roots below are clearly different..

Simon Bridge
Homework Helper

square roots below what?

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..

Simon Bridge
Homework Helper

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.

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?

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bumpp

Simon Bridge
Homework Helper

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?

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TSny
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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.

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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..

TSny
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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.

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?

TSny
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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 $\phi$.

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_Relative_Motion_of_the_Earth_and_the_Luminiferous_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/Influence_of_Motion_of_the_Medium_on_the_Velocity_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?

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-∅)

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 $\phi$.
I think this question works on the assumption that speed of light 'c' is unaffected by aether. Rather, they are trying to do length contraction here...

This brings me to my second post:

## Homework Statement

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

## The Attempt at a Solution

But the square roots below are clearly different..

TSny
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Gold Member

But all I'm saying is that if you let c' represent the speed of light along the arm $l_1$, then you can express c' in terms of c and θ using relative velocity arguments. See the attached diagram.

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I haven't yet had time to look carefully at your links. They look interesting! But all I'm saying is that if you let c' represent the speed of light along the arm $l_1$, then you can express c' in terms of c and θ using relative velocity arguments. See the attached diagram.
In this part they assume speed of light 'c' is the same. They are assuming only the distance contracts..

This is from A.P. French's Relativity text...chapter 3. I'm finding this text very very difficult, especially chapter 2 and 3 (I've skipped the entire chapter 2) I've already completed the homework problems from Halliday, Resnick and Walker and I find them to be much easier. This is demoralizing..

TSny
Homework Helper
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French's exercise is a little tedius to carry through, but it will work out.

What you want to show is that the difference in round trip time for the two arms will be the same no matter how you rotate the apparatus if you postulate length contraction in the ether wind. So, you still have the effect of the ether wind. Relative to the apparatus, the speed of light, c', will not be equal to the speed relative to the ether, c. Use the velocity triangle and the law of cosines as shown in the attachment to express c' in terms of c, v, and θ.

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TSny
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Perhaps a shorter way to derive c' in terms of c, v, and θ is to apply the pythagorean theorem directly to the right triangle abd in the attached figure.

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TSny
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In this part they assume speed of light 'c' is the same. They are assuming only the distance contracts..
No, French is still taking into account that the speed of light relative to the apparatus will be affected by the ether wind. Note how he uses equations (3-1) which are the same as (2-9) and (2-10). But he is now adding an additional effect: the Lorentz-FitzGerald contraction. With the contraction effect added, there will be no fringe shift as the apparatus is rotated relative to the direction of the ether wind.
I've already completed the homework problems from Halliday, Resnick and Walker and I find them to be much easier. This is demoralizing..
French's exercise is a step up from Halliday. It's a rather challenging exercise, so expect a little pain. You should feel good about completing the Halliday exercises and I'm sure you'll make it through French's exercise if you stick with it.

No, French is still taking into account that the speed of light relative to the apparatus will be affected by the ether wind. Note how he uses equations (3-1) which are the same as (2-9) and (2-10). But he is now adding an additional effect: the Lorentz-FitzGerald contraction. With the contraction effect added, there will be no fringe shift as the apparatus is rotated relative to the direction of the ether wind.

French's exercise is a step up from Halliday. It's a rather challenging exercise, so expect a little pain. You should feel good about completing the Halliday exercises and I'm sure you'll make it through French's exercise if you stick with it.
Ok, thank you for guiding me! I will give it another go!

I worked it out and got some crazy-*** square roots...