a story about special relativity,who can explain it？

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.

 

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

 

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Read Janus's reply.

 

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ok, a new answer. I have got 3 different answers for the third scene.

 

ok, #43 is for DaveC426913.

 

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.)

 

Are you sure?

 

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.

 

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?

 

X=Y?

 

God, is that your explanation?

 

I have asked Janus. X=Y? Let's look forward to his answer.

 

I will read and analyze this passage carefully and give my answer.

 

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?

 

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.

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Below is a link, for your studying pleasure.
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

 

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

 

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.

 

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.

 

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.

 

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.

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?