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### Messages - fjr66

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1
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« 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.

2
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« 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.

3
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« 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.

4
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« 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?

5
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« 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.

6
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 07:19:13 PM »
Each observer sees the light with a different frequency.

I never talk about frequency or Doppler Shift. I talk about the correct way to determining angle alpha.

7
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 07:13:56 PM »
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.

8
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 06:16:02 PM »
Only if the light is travelling at an angle in the initial reference frame, in which case it is just looking at the y component of the light's velocity, which should be less than c.
Do you understand that?

Then if there is any observer that co moving with one velocity component, they will see light travel with velocity less than c. Like what I said before. For outside observer that stand in the road, they see the ball travel at angle with velocity v and have velocity component v_x and v_y which its value less than v. But what is for observer in the car that moving with velocity v = v_x? They never see the ball have velocity component v_x, they just see it moving vertically with velocity less than v that is v_y.

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##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 05:58:58 PM »
As you can see what you posted, c is always c. You can not have anything greater than c.

If you are getting a greater than c value for anything you are doing something wrong.
I'm not said its value greater than  c. But when light have speed less than c along specific direction, then it is also violating special relativity postulate.
That never happens either.

Keep in mind that is for a vacuum.  When light enters a medium it will slow down.

I never talk about light travel at vaccum. I talk about velocity projection. Is that have the same magnitudo with the velocity itself?

10
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 05:56:46 PM »
Look at this picture for inertial observer O:

Is this the picture for the inertial observer O?
If so, they see the light moving at that angle, with a speed of c.
They do not just see the vertical component.

If this is actually for an outside observing that is also observing O moving, then they would see the light moving down with a speed of c, at that angle.
In order to determine what the observer O is seeing you need to use the proper relativistic formula to determine what they would be seeing.
Using the non-relativistic formula will produce incorrect results.

This is not a difficult concept to understand.
Why do you repeatedly use a non-relativistic formula when it is firmly established that relativistic effects will play a significant role?

Observer O never see light travel at angle, they just see it moving down  straight with velocity less than c. It is observer P that at rest with the source that see light moving with angle alpha with velocity c. It is like when you throw a ball inside a moving car, then you never see horizontal component of velocity of the ball. You just see it moving straight vertically. But what if for observer outside that standing in the road, they will see the ball moving at angle relative to vertical.

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.

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##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 10:05:21 AM »
As you can see what you posted, c is always c. You can not have anything greater than c.

If you are getting a greater than c value for anything you are doing something wrong.
I'm not said its value greater than  c. But when light have speed less than c along specific direction, then it is also violating special relativity postulate.

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##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 09:24:02 AM »
Its best not to use the classical explanation, but rather the correct relativistic explanation.

Even in relativistic explanation, they still use u_y = c sin theta  which is less than c. 13
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 09:05:09 AM »
Im sure the angle formula is correct.

Im referring to the speed of light.
The simplified formula is:
V+c=c not v+c>c.

The speed of light is the same for any reference frame.

We talk about angle alpha. If you said alpha = arctan (v_y / v_x) then other trigonometric rules can be applied too.

So v_y = c * sin alpha < c (violating special relativity postulat).

But if you said v_y = c and v_x also = c, then arctan (c / c) = 45 degree. So the beam of light will arrive at angle 45 degree whatever initial angle is. And experimental data showing that this is not the case.

14
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 02:27:47 AM »
Its best not to use the classical explanation, but rather the correct relativistic explanation.
Can you give the correct explanation for angle alpha?

15
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 02:26:42 AM »

That is because in this reference frame light is not travelling along the hypotenuse. So the sqrt(v^2+c^2) does not correspond to any actual speed.
So we do not violate special relativity.
Look at this picture for inertial observer O: 16
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 25, 2019, 12:41:39 AM »
(v^2 )/(v^2 + c^2) + (c^2)/(v^2 + c^2) = 1. But this means r = sqrt(v^2+c^2 ) > c, violating postulat of special relativity.
No, it doesn't.
That is because in this reference frame light is not travelling along the hypotenuse. So the sqrt(v^2+c^2) does not correspond to any actual speed.
So we do not violate special relativity.

If we used r = c, so velocity of light still c in hypotenuse, we get
That you are now using a different reference frame and thus the velocity will be different.
Light will now be travelling down the hypotenuse with a speed of c. This will give it a horizontal component of v and a vertical component of sqrt(c^2-v^2).

So no violation of trig laws and no violation of relativity.

Stop just adding the velocities while ignoring relativity.

What you are doing is no better than claiming if you are moving towards the source of the light it has to be going faster than c. It is ignoring relativity.

There is no violation here.

Special relativity never put any concern to velocity direction (and its projection to any plane), it is just about magnitude. Thats why it creates so many paradox that impossible to solved.
No, the directionality is important as well. It is just when you have anything other than a straight line the math gets much harder so people like to focus on a straight line.

I provided what you need to do to deal with the directionality.

Directly below the section you quoted it provides the relativistic explanation and shows how the speed of light is kept constant.

If light travel at hypotenusa AC with velocity c. Then there is observer in horizontal line moving parallel with BC line, with velocity c sin alpha. For this observer light only have a vertical component parallel AB line  with velocity  c cos alpha which is less than c (also violating special relativity postulate).

17
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 24, 2019, 07:59:44 PM »
It needs to move relative to Earth to obtain the results of abberation, But stellar abberation contain wrong assumption. Look at above picture. Starlight come in direction AB with velocity c. And earth move in direction CB with velocity v. So for observer on earth starlight coming with velocity \sqrt{v^2 + c^2} that is larger than c.

You may said that we cannot use pythagorean rule in special relativity, maybe something like Einstein velocity addition is needed. But how can we determining angle  alpha (the angle at which the telescope must be tilted) if not using pythagorean rule.

If we make velocity v increased, then  angle alpha increased too. But this is also make velocity along AC larger than c, which is violating special relativity postulat.

So either special relativity is incorrect or stellar abberation is wrong.

And maybe in special relativity we must postulating  not just speed of light that constant for all observer but its direction too. How speed stay constant when direction changing, whereas direction determined by pythagorean rule.

Look at the picture below, it is from wikipedia article about stellar abberation.
https://en.wikipedia.org/wiki/Aberration_(astronomy) If earth move with velocity v relative to the star, then he will measure speed of light = u_x + v. But what if he at rest relative to the star, then he just measure velocity of light = u_x = c sin theta < c.

Special relativity never put any concern to velocity direction (and its projection to any plane), it is just about magnitude. Thats why it creates so many paradox that impossible to solved.

18
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 24, 2019, 07:14:41 PM »
Direction is determined from the component velocities, as is speed, but in different ways.
The speed is done by Pythagoras, but the direction is done by trig.
We can't separated trigonometric rule from pythagoras law. They mutually consistent.

Even if light travel along hypotenusa with velocity c, we get it projection to vertical line AB that is c cos alpha < c, which is violating special relativity postulat. So observer that at rest in line AB will measure speed of light less than c.

19
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 24, 2019, 07:00:40 PM »
The angle alpha is the inverse tan of the velocity of Earth divided by the speed of light.
You still used trigonometric law to determining angle alpha, if you consistent you will end up with the problem.
Remember in trigonometric we know that sin^2 alpha + cos^2 alpha = 1, where sin alpha = v/r, and cos alpha = c/r.
But where we get the value of r? If r = sqrt(v^2+c^2 ) then that law is valid, that is
(v^2 )/(v^2 + c^2) + (c^2)/(v^2 + c^2) = 1. But this means r = sqrt(v^2+c^2 ) > c, violating postulat of special relativity.

If we used r = c, so velocity of light still c in hypotenuse, we get
sin^2 alpha + cos^2 alpha = v^2/c^2 + c^2/c^2 > 1, which is violating trigonometric laws.

So either special relativity is incorrect  or trigonometric law can not be used in this case.

But the problem is how we determining angle alpha if not using trigonometric law?

20
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 24, 2019, 05:08:57 AM »
It needs to move relative to Earth to obtain the results of abberation, But stellar abberation contain wrong assumption. Look at above picture. Starlight come in direction AB with velocity c. And earth move in direction CB with velocity v. So for observer on earth starlight coming with velocity \sqrt{v^2 + c^2} that is larger than c.

You may said that we cannot use pythagorean rule in special relativity, maybe something like Einstein velocity addition is needed. But how can we determining angle  alpha (the angle at which the telescope must be tilted) if not using pythagorean rule.

If we make velocity v increased, then  angle alpha increased too. But this is also make velocity along AC larger than c, which is violating special relativity postulat.

So either special relativity is incorrect or stellar abberation is wrong.

And maybe in special relativity we must postulating  not just speed of light that constant for all observer but its direction too. How speed stay constant when direction changing, whereas direction determined by pythagorean rule.

For observer moving parallel with the source, it is safely to assume velocity of light is the same for both of them. But how about observer moving in perpendicular direction with the source. Is speed of light still constant for him.

21
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 24, 2019, 04:54:53 AM »
It needs to move relative to Earth to obtain the results of abberation, But stellar abberation contain wrong assumption. Look at above picture. Starlight come in direction AB with velocity c. And earth move in direction CB with velocity v. So for observer on earth starlight coming with velocity \sqrt{v^2 + c^2} that is larger than c.

You may said that we cannot use pythagorean rule in special relativity, maybe something like Einstein velocity addition is needed. But how can we determining angle  alpha (the angle at which the telescope must be tilted) if not using pythagorean rule.

If we make velocity v increased, then  angle alpha increased too. But this is also make velocity along AC larger than c, which is violating special relativity postulat.

So either special relativity is incorrect or stellar abberation is wrong.

And maybe in special relativity we must postulating  not just speed of light that constant for all observer but its direction too. How speed stay constant when direction changing, whereas direction determined by pythagorean rule.

22
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 22, 2019, 07:17:10 AM »
If there is no medium, then it is not a wave. Wave equation must differentiable at every point. But if there is no medium, then there is discontinuity at the tip. It is impossible element of wave to bent without medium.

Electromagnetic waves don't need medium.

Increase of electrostatic field at one point generates magnetic field in the next point,
decrease makes that magnetic field decrease and then increaee in negative direction, and so on.

That magnetic field does the same to the third point, making electric field increase,
then decrease and then increase in opposite direction, then decrease again, and so on.

The new electric field induces the magnetic oscillation in the fourth point, magnetic induces electric in fifth,
and the chain continues without a medium.

Electric to magnetic to electric to magnetic and so forth.
Hence the name: Electromagnetic Wave.

If there is medium, then it interferes with the electric-magnetic-electric chain generation and changes the speed of the wave.
Maxwell formulate all his theory based on ether. 23
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 22, 2019, 07:01:12 AM »
No, they don't contradict relativity at all.
All they contradict is the ballistic model of light.
Relativity only has light being constant in an inertial reference frame.
A rotating reference frame is not an inertial reference frame.
So light have a velocity not equal c in rotating frame (also constant permitivity and permeability). This means we must reformulate all Maxwell equation respect to rotating frame. But the earth itself rotating frame, and we must reformulate all Maxwell theory respect to earth, such an irony.

How about to make it simple. Let just say light travel at aether with velocity c, regardless it in rotating frame or not, then all the problem was solved.

24
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 22, 2019, 12:25:51 AM »
I don't see it. The conclusion of MM contradicts the explanation according to the then prevailing theory that earth moves through space and its medium. That is said elsewhere in the article. It suggests that the earth is at rest.
Or that there is no 'medium'(Aether).
If there is no medium, then it is not a wave. Wave equation must differentiable at every point. But if there is no medium, then there is discontinuity at the tip. It is impossible element of wave bent without medium.

Why is that? Why can't light be both a particle and a wave?

It is just description, not intrinsic meaning of light. Sometimes one description match with observation, other times it is not. Like how to describe light as particle when seeing diffraction phenomena. Even in quantum mechanics we can see particle as a wave because it is can be diffracted. But wave in electromagnetic theory must spread to infinity, not localised wave as in quantum mechanics.

If light have no medium to propagate, then impossible we can explain diffraction phenomena. Diffraction in classical description occur because wavefront disturb medium in front of it, and becomes the source of new wave. If there is no medium how can it be possible?

EM wave profuced by acceleration of charged particle. But how diffraftion in EM wave  can be described when  there is no charged particle in diffraction slit. There must medium in it to produce a new wave.

25
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 22, 2019, 12:10:24 AM »
I don't see it. The conclusion of MM contradicts the explanation according to the then prevailing theory that earth moves through space and its medium. That is said elsewhere in the article. It suggests that the earth is at rest.
Or that there is no 'medium'(Aether).
If there is no medium, then it is not a wave. Wave equation must differentiable at every point. But if there is no medium, then there is discontinuity at the tip. It is impossible element of wave bent without medium.

Why is that? Why can't light be both a particle and a wave?

It is just description, not intrinsic meaning of light. Sometimes one description match with observation, other times it is not. Like how to describe light as particle when seeing diffraction phenomena. Even in quantum mechanics we can see particle as a wave because it is can be diffracted. But wave in electromagnetic theory must spread to infinity, not localised wave as in quantum mechanics.

26
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 21, 2019, 11:54:10 PM »
Michelson Morley have a conclusion that the earth have no relative motion with ether which is medium for light wave to propagate. That means the earth is center of universe. Because if light have no medium to propagate, then wave equation have a discontinuity at its boundary.
No it doesn't.
Michelson and Morley came to the conclusion that no motion was detected between the earth and the ether.
And that is far from the same conclusion.
So light wave must have a medium, and this medium at rest respect to earth. That means the earth is center of universe.

27
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 21, 2019, 11:51:50 PM »
I don't see it. The conclusion of MM contradicts the explanation according to the then prevailing theory that earth moves through space and its medium. That is said elsewhere in the article. It suggests that the earth is at rest.
Or that there is no 'medium'(Aether).
If there is no medium, then it is not a wave. Wave equation must differentiable at every point. But if there is no medium, then there is discontinuity at the tip. It is impossible element of wave to bent without medium.

28
##### Flat Earth Debate / Re: How cloud rotate with the same phase as earth surface
« on: July 21, 2019, 11:06:49 PM »
Then there is much higher probability that the cloud going to the west than to the east as the earth rotated.
Why?
What are you basing this on?
Because the main factor: earth rotation to the east.

29
##### Flat Earth Debate / Re: How to solve twin paradox in special relativity
« on: July 21, 2019, 10:51:59 PM »
Even for the original twin paradox, they don't use time dilation formula one way. But in that case, they end up with different age.
Yes, because of the asymmetry.
The observer on Earth only sees the observer in the distant ship ageing rapidly for the return journey which appears to go very quickly.
For the observer on the ship, they see the observer on Earth ageing rapidly for the entire journey which takes half the apparent time for them.
So time dilation does not have physical meaning, it is just mathematical consequence of relativity. Each observer can say other time was slower than him.

30
##### Flat Earth Debate / Re: Michelson-Morley Experiment
« on: July 21, 2019, 10:46:36 PM »
Michelson Morley have a conclusion that the earth have no relative motion with ether which is medium for light wave to propagate. That means the earth is center of universe. Because if light have no medium to propagate, then wave equation have a discontinuity at its boundary.

For example if light travel at ether with velocity c, and and earth move have relative motion with  ether with velocity v then light  have velocity c+v relative to earth. But according to Maxwell theory, speed of light itself depend on two contant that is vacuum permitivity and vacuum permbeability. And these contant is the same for all inertial observer. It is means light itself have a constant velocity for all inertial observer. So either Maxwell theory is incorrect, or we must change our definition of time and space (that depend on Galilean relativity) and also the definition of wave equation itself (that in classical formulation should have a medium to propagate).

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