Michelson-Morley Experiment

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Re: Michelson-Morley Experiment
« Reply #60 on: July 26, 2019, 03:16:54 AM »
Observer O never see light travel at angle, they just see it moving down  straight with velocity less than c.
No they don't. They will always see it travelling at c.
Again, if you want to convert between the reference frames, you need to use the relativistic equations. You can't just ignore them like you are doing.

Stop using classical equations for a relativistic phenomenon.

The problem is if observer O and P measuring the same speed of light like what  postulated by special relativity, then how we can determining the correct value for angle alpha. If for observer P see light travel at  hypotenusa with velocity c, and observer O see light travel down also with velocity c, then that means cos alpha = c/c. So alpha always  = 90 degree, whatever angle alpha in reality is.
No it doesn't, as they are in 2 different reference frames.
You can calculate the angle for each of them.
The one in the reference frame seeing light travel along the hypotenuse at a speed of c will not see the light travelling straight down at a speed of c, nor will they see the vertical component of the velocity of the light being c.
They will be able to calculate the angle just fine. It will not necessarily be 90 degrees.

Basically the same argument applies for the moving observer.
They are not seeing the light travel along the hypotenuse at a speed of c.

Stop pretending that the 2 frames are the same or that you can use classical equations for this relativistic phenomenon.

Again, what you are doing is no better than the following:
Quote
Observer A sees observer B moving towards them at a speed of v, and shining a torch towards them. Observer B sees light travelling away from them at a velocity of c, so that must mean that observer A sees light travelling at a velocity of v+c.
It is pure garbage which completely ignores the relativistic effects.

If you stop ignoring the relativistic effects and start treating them properly as 2 separate reference frames, there will be no issue.

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sokarul

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Re: Michelson-Morley Experiment
« Reply #61 on: July 26, 2019, 03:30:31 AM »
As the link you quoted says, c is constant.

So tan alpha = c/c = 1
cos alpha also  = c/c = 1
and sin alpha also = c/c = 1

then this will destroying trigonometric law.
Impossible we can determined what angle alpha is.
Itís literally in the link you provided. You are thinking classical. You need to be thinking relativistic.
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fjr66

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Re: Michelson-Morley Experiment
« Reply #62 on: July 26, 2019, 05:18:46 AM »
Observer O never see light travel at angle, they just see it moving down  straight with velocity less than c.
No they don't. They will always see it travelling at c.
Again, if you want to convert between the reference frames, you need to use the relativistic equations. You can't just ignore them like you are doing.

Stop using classical equations for a relativistic phenomenon.

I give some easier example.
If beam of light travel down parallel with y axis in O frame. Then someone that at rest in O just see this light at rest respect to X axis, there is no motion along x direction. There is no velocity component of this light parallel to X axis. But what if beam of light travel at tilted path that make an angle with y axis. Then this beam will have velocity component that parallel with x axis and y axis, lets say it v_x and v_y that is less than c. So it is possible for some obserber to move with velocity = v_x and see this beam of light moving down with velocity = v_y. 

I think it is not hard to understand.

Again, what you are doing is no better than the following:
Quote
Observer A sees observer B moving towards them at a speed of v, and shining a torch towards them. Observer B sees light travelling away from them at a velocity of c, so that must mean that observer A sees light travelling at a velocity of v+c.
It is pure garbage which completely ignores the relativistic effects.

If you stop ignoring the relativistic effects and start treating them properly as 2 separate reference frames, there will be no issue.

It is for two observer that moving parallel to each other with beam of light also in parallel direction. What I discussed was how about a beam that make an angle with observer.

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fjr66

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Re: Michelson-Morley Experiment
« Reply #63 on: July 26, 2019, 05:27:02 AM »
No it doesn't, as they are in 2 different reference frames.
You can calculate the angle for each of them.
The one in the reference frame seeing light travel along the hypotenuse at a speed of c will not see the light travelling straight down at a speed of c, nor will they see the vertical component of the velocity of the light being c.
They will be able to calculate the angle just fine. It will not necessarily be 90 degrees.
The problem is how to determined velocity component in term of velocity. If you using trigonometric law in consistent way, then we get v_x = v cos alpha and v_y = v sin alpha.  But for light this will make v_x and v_y less than c (the value that postulated by special relativity).  So there is observer capable to move with velocity = v_x (because it is less than c) and seeing beam of light just move in vertical direction with v = v_y less than c.

Is there any relativistic trigonometric, so v_x and v_y also v?

Re: Michelson-Morley Experiment
« Reply #64 on: July 26, 2019, 05:29:37 AM »
I give some easier example.
If beam of light travel down parallel with y axis in O frame. Then someone that at rest in O just see this light at rest respect to X axis, there is no motion along x direction. There is no velocity component of this light parallel to X axis. But what if beam of light travel at tilted path that make an angle with y axis. Then this beam will have velocity component that parallel with x axis and y axis, lets say it v_x and v_y that is less than c. So it is possible for some obserber to move with velocity = v_x and see this beam of light moving down with velocity = v_y. 

I think it is not hard to understand.
No, it isn't hard to understand. You are wrong as you are using classical equations to describe a relativistic phenomenon.

Why don't you start with the simple 1D case. Once you grasp that, and realise you need to use relativistic formulas, you can move on to the 2D case.

The observer does not see the beam of light moving with a velocity of v_y as they are moving and their motion needs to be accounted for.
You are correct that the horizontal component of the velocity of the light will be 0 for this observer, but not the vertical velocity.

The proper formula would be this:
v_y*sqrt(1-v_x^2/c^2)/(1-v_x^2/c^2).

Multiplying the top and bottom by c^2 we get:
c*v_y*sqrt(c^2-v_x^2)/(c^2-v_x^2).

Noting that c^2=v_x^2+v_y^2, we get c^2-v_x^2=v_y^2.
Subbing that in we get:
c*v_y*sqrt(v_y^2)/v_y^2=c*v_y^2/v_y^2=c.

That means for the observer moving along, they will see the light travel vertically at a speed of c, not v_y.

There is no violation here.

Repeatedly ignoring this and continuing to use classical equations will not mean you are correct, it will mean you are wilfully ignorant.

If you have a rational object, then provide it, but stop appealing to classical equations that do not apply.

The problem is how to determined velocity component in term of velocity.
No it isn't.

If you try doing that in either reference there is no problem.

The problem is you pretending the 2 reference frames are the same and using classical mechanics.

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fjr66

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Re: Michelson-Morley Experiment
« Reply #65 on: July 26, 2019, 12:43:51 PM »
Why don't you start with the simple 1D case. Once you grasp that, and realise you need to use relativistic formulas, you can move on to the 2D case.
There is no 1D analogy for this. 2D is minimum requirement for velocity composition.

The observer does not see the beam of light moving with a velocity of v_y as they are moving and their motion needs to be accounted for.
You are correct that the horizontal component of the velocity of the light will be 0 for this observer, but not the vertical velocity.
Then you agree with me. If possible for light to have 0 velocity along x direction then it is also possible to make it have a velocity a little larger than that. For example if beam of light make 45 degree angle with y axis. Then its velocity along x axis was c * 1/2 sqrt(2). And it is possible for object with a mass to travel at that velocity parallel x axis And for him, he will measure the speed of light less than c along y axis.   

Impossible for light to travel parallel y axis and its velocity along x axis also c.

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sokarul

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Re: Michelson-Morley Experiment
« Reply #66 on: July 26, 2019, 12:45:40 PM »
The link you posted says light travels at c no matter what. Why do you ignore this?
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fjr66

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Re: Michelson-Morley Experiment
« Reply #67 on: July 26, 2019, 02:30:03 PM »
The link you posted says light travels at c no matter what. Why do you ignore this?
If light travel parallel y axis, then it velocity along x axis is zero. It is very obvious.

Re: Michelson-Morley Experiment
« Reply #68 on: July 26, 2019, 02:48:53 PM »
There is no 1D analogy for this. 2D is minimum requirement for velocity composition.
I didn't say a 1D analogy. I said the 1D case.
i.e. a system where all the velocities are in a straight line.

This is what most people are initially taught when taught about relativity and changing reference frames.

If you have an observer A, who is "stationary", watching an observer B move at a velocity of v to the right, and have a beam of light moving at a velocity of c to the left, does observer B see the light move at a velocity of c, or c+v?

One of these is the result of a classical formula.
The other is the result of a relativistic formula.

Then you agree with me.
No, I don't.
I have repeatedly explained why you are wrong and you just repeatedly ignore it.
Why?

You are approaching the problem from a purely classical point of view, ignoring the required relativistic corrections to correctly determine the speed in a given reference frame.
As such, your results are wrong.

And no, you can't just grab velocities from different reference frames and pretend they are in the same reference frame either.

The moving observer will observe the light moving with a speed of c, just like I demonstrated.

Again, why do you ignore this and repeatedly assert the same refuted lies?

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fjr66

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Re: Michelson-Morley Experiment
« Reply #69 on: July 26, 2019, 03:01:07 PM »
The link you posted says light travels at c no matter what. Why do you ignore this?
If light travel parallel y axis, then it velocity along x axis is zero. It is very obvious.

Look at this picture


In primed coordinate, u_x' was stated in term of einstein velocity addition. At the numerator there is u_x + v. This is because v and u_x in opposite direction. If u_x and v in the same direction, then we substract it. So if u_x equal with v,
numerator become zero, and this means u_x' will zero whatever happen in denominator. So that means the beam of light have zero velocity component along x direction. Someone move with v = u_x only see light move downward along y axis with velocity given by u_y'.

But,
The value of u_y' = u_y / gamma (1 + u_x v / c^2)
= c sin theta / gamma (1 - c^2 cos^2 theta /c^2)
= c sin theta/ gamma (sin^2 theta)
= c /gamma (sin theta)

Because the value of gamma >= 1 then this is make u_y' always less than c.

So observer moving in x direction with v = u_x just see the light moving downward parallel y axis with speed less than c. Thus violating special relativity postulate.
« Last Edit: July 26, 2019, 03:05:43 PM by fjr66 »

Re: Michelson-Morley Experiment
« Reply #70 on: July 26, 2019, 03:44:17 PM »
Look at this picture
How about you try to do so, and make sure you understand it.

More importantly, look at the line above the one you circled, i.e. this one:


Notice how u'y is not the same as uy
Notice how they are 2 different values?

That means your approach is wrong.
It means that the velocity of light for the moving observer who sees it move purely in the y direction IS NOT GOING TO BE uy. IT WILL BE DIFFERENT!

Again, this is not a difficult concept to grasp.

I already provided the math above to show what this is.
Doing it for light, and where the observer is moving with the same x velocity as light, will result in the y component of the velocity being equal to c.

The value of u_y' = u_y / gamma (1 + u_x v / c^2)
= c sin theta / gamma (1 - c^2 cos^2 theta /c^2)
There is no need to bring trig functions into this at all.

All it does it cause greater confusion.

Because the value of gamma >= 1 then this is make u_y' always less than c.
Ignoring one part of the denominator wont magically make it vanish.
Sin(x) will always be less than or equal to 1. Does that mean u'y will always be greater than c? NO!

Do the math properly, making sure you finish rather than stopping prematurely.
That means expanding gamma (i.e. putting in what gamma actually is) and going until you have it expressed in terms of uy, ux and c.

Again, I have already done that for you.
The final result was:
u'y=c

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sokarul

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Re: Michelson-Morley Experiment
« Reply #71 on: July 26, 2019, 05:10:39 PM »
The link you posted says light travels at c no matter what. Why do you ignore this?
If light travel parallel y axis, then it velocity along x axis is zero. It is very obvious.
Yes vectors exist. Thanks for pointing out something we already know.
Sokarul

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