a story about special relativity,who can explain it？

DaveC426913 said:
Yes.
Click to expand...
OK, I'm waiting for your answer "Yes".
Third scene:
Earth---------------------------------------A--->
.|....................................................0.1C
.|
.|
.|
.|
\/ B 0.1C
Please answer the relative speed between A and B .

You can use classical physics to decompose velocity. Can you use classical physics to synthesize velocity?
What is the relative speed of AB?

You are denying classical physics, but you are using it to prove your theory.

 
You are denying classical physics, but you are using it to prove your theory.
 
Janus58 said:
0.0c.
In the first case we are finding the velocity between A and B as measured by A or B, so the directions of the velocities must be taken as measured by which of the two we are making the measurement from.
In the case of A we get:
The Earth is moving to the right relative to A and B is moving to the right relative to the. If we assign motion to the right as positive then both u and v are positive.
In your second scene The Earth, as measured from A is moving to to the Left while B is still moving to the right relative to the Earth, for A, the scene looks like.
<--Earth---------------------------A
0.9c
-------------------------------------B-->
0.9c

u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

Third scene We will look at it from the rest frame for A where you get:
0.9c
<--Earth-----------------A
|
|
|
\/ B

Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.392c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.9c)^2+ (0.392c)^2) = 0.9817c
Click to expand...
ok, a new answer. I have got 3 different answers for the third scene.
 
TonyYuan said:
If the weakest gravitational wave could rise the fluctuation of the LIGO interference fringe of light, how should we neglect the gravity’s influences on light?

If light is held by gravity, then measuring the speed of light anywhere on the earth will be a constant.
Click to expand...

This kind of idea has already been considered and rejected. It is essentially the "Aether Drag" model.
The problem with it is that it is inconsistent with astronomical observations. There is an effect known as stellar aberration. It is caused by the Earth's orbital motion with respect to the stars and causes a slight apparent shift in the visual position of the stars. The amount of shift is tied to both the speed of light and the Earth's orbital motion. We know both numbers by independent means. The shift we actual measure matches perfectly with what we would predict it to be by those known values.
However, if the Earth's gravity "dragged" light with it, this would have the effect of changing the angle of shift we see from those stars. In other words, the actual angle we measure would not match the prediction. Since we do measure the predicted angle, there can be no aether( or gravitational) drag effect (Or at least not one strong enough to produce the effect you are suggesting.)
 
Janus58 said:
0.0c.
In the first case we are finding the velocity between A and B as measured by A or B, so the directions of the velocities must be taken as measured by which of the two we are making the measurement from.
In the case of A we get:
The Earth is moving to the right relative to A and B is moving to the right relative to the. If we assign motion to the right as positive then both u and v are positive.
In your second scene The Earth, as measured from A is moving to to the Left while B is still moving to the right relative to the Earth, for A, the scene looks like.
<--Earth---------------------------A
0.9c
-------------------------------------B-->
0.9c

u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

Third scene We will look at it from the rest frame for A where you get:
0.9c
<--Earth-----------------A
|
|
|
\/ B

Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.392c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.9c)^2+ (0.392c)^2) = 0.9817c
Click to expand...
Are you sure?
 
DaveC426913 said:
No. From straight math.
Figure out the hypotenuse length of a right triangle with sides .9
That will get you the classical receding velocity.
Then apply the formula.
Click to expand...
That can't be correct since the hypotenuse is one number (and a number greater than c at that), and the addition formula requires two numbers.

Anyway, best answers are as always from Janus, who seems to never get it wrong.
Be appreciated if any error in my post 23 was pointed out.
 
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Fourth scene:
Earth---------------------------------------A--->
.|....................................................0.9C
.|
.|
.|
.|
\/ B 0.8C
Please answer the relative speed between A and B .

Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.0.3488c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.9c)^2+ (0.3488c)^2) =X

Since the Earth-A pair share a 0.8c to the left velocity, this pair is time dilated by a factor of 0.6 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.54c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.8c)^2+ (0.54c)^2) =Y

X=Y?

 
Janus58 said:
0.0c.
In the first case we are finding the velocity between A and B as measured by A or B, so the directions of the velocities must be taken as measured by which of the two we are making the measurement from.
In the case of A we get:
The Earth is moving to the right relative to A and B is moving to the right relative to the. If we assign motion to the right as positive then both u and v are positive.
In your second scene The Earth, as measured from A is moving to to the Left while B is still moving to the right relative to the Earth, for A, the scene looks like.
<--Earth---------------------------A
0.9c
-------------------------------------B-->
0.9c

u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

Third scene We will look at it from the rest frame for A where you get:
0.9c
<--Earth-----------------A
|
|
|
\/ B

Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.392c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.9c)^2+ (0.392c)^2) = 0.9817c
Click to expand...

X=Y?
 
Halc said:
That can't be correct since the hypotenuse is one number (and a number greater than c at that), and the addition formula requires two numbers.
Click to expand...
God, is that your explanation?
 
Janus58 said:
This kind of idea has already been considered and rejected. It is essentially the "Aether Drag" model.
The problem with it is that it is inconsistent with astronomical observations. There is an effect known as stellar aberration. It is caused by the Earth's orbital motion with respect to the stars and causes a slight apparent shift in the visual position of the stars. The amount of shift is tied to both the speed of light and the Earth's orbital motion. We know both numbers by independent means. The shift we actual measure matches perfectly with what we would predict it to be by those known values.
However, if the Earth's gravity "dragged" light with it, this would have the effect of changing the angle of shift we see from those stars. In other words, the actual angle we measure would not match the prediction. Since we do measure the predicted angle, there can be no aether( or gravitational) drag effect (Or at least not one strong enough to produce the effect you are suggesting.)
Click to expand...
I will read and analyze this passage carefully and give my answer.
 
TonyYuan said:
I have asked Janus. X=Y? Let's look forward to his answer.
Click to expand...
I think if you just keep on coming up with more and more hoops for people to jump through, they will at some point have had enough. So I do hope this will be your last demand for another worked example.

Are you able to see the principle behind how to work this out, yet?
 
TonyYuan said:
Fourth scene:
Earth---------------------------------------A--->
.|....................................................0.9C
.|
.|
.|
.|
\/ B 0.8C
Click to expand...

Einstein derived the general equation for velocity composition in his seminal 1905 paper, On the Electrodynamics of Moving Bodies. Einstein was a genius, by the way.

dMojfpO.png


Below is a link, for your studying pleasure.
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
 
TonyYuan said:
ok, a new answer. I have got 3 different answers for the third scene.
Click to expand...
Go here: https://en.wikipedia.org/wiki/Velocity-addition_formula
scroll down the box "Transformation of Velocity (plane polar components)*
Inside you find the equation
u = sqrt(u'^2+v^2 +2 vu' cos(theta) - (vu' sin(theta)/c)^2)/(1-v/c^2 u' cos(theta)
Which deals with velocity additions in non-parallel velocities scenarios.
Theta is the angle between the two velocities (in this case, as measured by the Earth.). And as measured by the Earth it is 90 degrees. As the cos of 90 degrees is 0, we can simplify in this particular scenario to
u = sqrt(u'^2+v^2 - u'v/c^2)
where v is the relative velocity between A and the Earth, and u' is the vertical velocity between Earth and B as measured by the Earth.

Now let's go back to my solution:

It basically was ( using the same velocity conventions)
u = sqrt( v^2 + (sqrt(1-v^2/c^2)u')^2)
Which again reduces to

u = sqrt(u'^2+v^2 - u'v/c^2)

* Neddy Bate provided this same basic equation is his post
 
Neddy Bate said:
Einstein derived the general equation for velocity composition in his seminal 1905 paper, On the Electrodynamics of Moving Bodies. Einstein was a genius, by the way.


Below is a link, for your studying pleasure.
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
Click to expand...
I firmly believe in Newtonian classical physics, because it has been verified by numerous experiments, it is very consistent with the laws of things in nature, it can explain almost all physical phenomena. I believe it also includes quantum mechanics.
 
TonyYuan said:
Fourth scene:
sqrt ((0.9c)^2+ (0.3488c)^2) =X
...
sqrt ((0.8c)^2+ (0.54c)^2) =Y

X=Y?
Click to expand...
They do yield the same number, and you'd expect that, since it would be contradictory for B to be receding from A at one speed but A is receding from B at a different speed.
 
I expected the two data X,Y to be different, and the results were the same, which surprised me. This should be mathematically proven.

It's more like a math game:
u^2+[v*sqrt(1-u^2/c^2)]^2 = u^2+v^2 - u^2*v^2/c^2.

so X=Y it must be right.
 
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TonyYuan said:
I firmly believe in Newtonian classical physics, because it has been verified by numerous experiments, it is very consistent with the laws of things in nature, it can explain almost all physical phenomena. I believe it also includes quantum mechanics.
Click to expand...
Then you know nothing about chemistry. Very little in chemistry makes sense without quantum mechanics and the results it gives are not all what you get from classical mechanics.
 
Halc said:
The web sites make it look really complicated, but my intuitions generate the following:
Rotate the picture 45 degrees. If Earth is moving at .9c diagonally, it must have component speed of .6364c, right?
Similar, B has the same component speed in the final direction of motion.
Add two of those using the above formula and you get .9059c.

What's so complicated about that? Maybe I made some naive mistake.
Click to expand...
Error: I didn't account for the dilation from the vertical movement. All I did was add the velocities in the Earth frame rather than in the frame of the hypothetical midpoint between A and B. My answer here in post 23 is wrong.

TonyYuan said:
I expected the two data X,Y to be different, and the results were the same, which surprised me. This should be mathematically proven.
Click to expand...
Hmm, are you of the opinion then that Newton would have been OK with you having a speed of 5 km/hr relative to me and that not necessitating that I have a speed of 5 km/hr relative to you?
 
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