a story about special relativity,who can explain it？

[Janus58]

Fourth scene:
A------------------L=9LS---------------->Earth<------------------L=9LS-----------------------B
v=0.9C............................................................................................................v=0.9C

With Earth as the reference, A's time goes slower, and with Earth as the reference, B's time goes slower. So when they meet, the time is same.
but
With A as the reference, B's time goes slower, and with B as the reference, A's time goes slower. When they meet, please tell me whose time is slower?

First off, I will assume that you assume that all three clocks read the same at the start of this scenario, as determined by the Inertial frame which the Earth is at rest.
But, as I already mentioned in the Earlier post, due the Relativity of simultaneity, this is not the case according to A. When A's clock reads zero, both the Earth clock and the clock on B will already read some time after zero.
For example, according to the Earth, it takes 10 sec for A and B to reach it. during that time A and B's clocks will tick off 4.36 sec and each reads 4.36 secs when they reach the Earth.
A agrees as to the respective times on each clock when they meet. Thus according to A, its clock starts at 0 and ticks off 4.36 sec until it meets up with the Earth.
This means a couple of things:
Since A also measures it velocity with respect to the Earth as being 0.9c, the distance between itself and Earth when A's clock reads 0 is only 3.924 ls (this is a consequence of length contraction)
The Earth clock tick's 0.436 the rate of A's clock and accumulates ~ 1.9 sec from the time A's clock reads 0 and A meets up with the Earth.
What this means is that, according to A, The Earth clock already reads 8.2 sec when A's clock reads 0. (Relativity of Simultaneity).
Also, according to A, B is traveling towards it at 0.9945c, and due to time dilation accumulates only ~0.468 sec as A's clock advances 4.36 sec. B's clock then already read a bit over 3.9 sec when A's clock read 0.
B was also only 4.336 ls away from A when A's clock read 0. (only 0.4112 ls further away than the Earth was at that same moment).

The upshot is that while everyone will agree as to what all the clocks read when they are all together, they won't agree upon the respective readings for any moment before that.

Whenever you work out such a scenario, you can't just consider time dilation alone, you have to take everything into account, including length contraction and the relativity of simultaneity.

 

[TonyYuan]

You use the Relativistic velocity equation.
w = (u+v)/(1+uv/c^2)
Here u ( the relative velocity between ship A and the Earth as measure by either the Earth or ship A) and u( the relative velocity between ship B and the Earth as measured by either the Earth or ship B) are both 0.9c

Thus w( the relative velocity between ship A and B as measured by either ship) is
(0.9c+0.9c)/(1+ 0.9c(0.9c)/c^2) = ~0.9945c

Fifth scene:
Earth.....................................A---->u=0.2C............................B---->w=0.8C
What is the velocity v of B relative to the earth?

w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.8C
0.2-v=(1-0.2v)*0.8
0.84v = -0.6
v = -0.714C, the velocity v of B relative to the earth is -0.714C, velocity direction is opposite to A.

But A sees B moving away at 0.8C, what happened?
You can see different u and different w, v can get different directions, which is very interesting. For example:
if v=0.2C,w=0.1C
w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.1C
0.2-v=(1-0.2v)*0.1
0.98v = 0.1
v = 0.102C,v direction is same to A.

Is this a math game?

 

[exchemist]

Fifth scene:
Earth.....................................A---->u=0.2C............................B---->w=0.8C
What is the velocity v of B relative to the earth?

w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.8C
0.2-v=(1-0.2v)*0.8
0.84v = -0.6
v = -0.714C, the velocity v of B relative to the earth is -0.714C, velocity direction is opposite to A.

But A sees B moving away at 0.8C, what happened?
You can see different u and different w, v can get different directions, which is very interesting. For example:
if v=0.2C,w=0.1C
w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.1C
0.2-v=(1-0.2v)*0.1
0.98v = 0.1
v = 0.102C,v direction is same to A.

Is this a math game?

Tony you run the risk of exhausting people's patience with all these scenarios. What are you trying to achieve with them?

 

[TonyYuan]

Tony you run the risk of exhausting people's patience with all these scenarios. What are you trying to achieve with them?

I know some students who participate in the Olympic physics competition. They are very confused about special relativity. They are all very top students. They are difficult to understand. I think there must be something wrong. As a teacher, I want to find out whether it is a mathematical game. If it is, let it be classified as a game. If not, let's explore.

I've been thinking about how to find out the flaws today, and I think scenario 5 should be able to illustrate the problem.

 

[Neddy Bate]

Tony,

In post #62, why did you change the plus signs to minus signs in the equation quoted by Janus58?

 

[TonyYuan]

Tony,

In post #62, why did you change the plus signs to minus signs in the equation quoted by Janus58?

Janus, he's been thinking about direction.
u is A to Earth.
v is Earth to B
w is B to A.

In post #40,you can see:
u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

 

[exchemist]

I know some students who participate in the Olympic physics competition. They are very confused about special relativity. They are all very top students. They are difficult to understand. I think there must be something wrong. As a teacher, I want to find out whether it is a mathematical game. If it is, let it be classified as a game. If not, let's explore.

It is unclear what you mean by a "mathematical game". SR obviously provides a self-consistent system of calculation. Any system of mechanics must do that. That does not make it a "game", obviously. It is a model of physical reality that can be tested experimentally. As it has been.

 

[Neddy Bate]

Tony,

In post #40, you leave the plus signs in the equation, and then put the minus sign on only the velocity which goes from right to left. Why not do that same thing in #62?

 

[TonyYuan]

Tony,

In post #40, you leave the plus signs in the equation, and then put the minus sign on only the velocity which goes from right to left. Why not do that same thing in #62?

you can still use w = (u+v)/(1+uv/c^2) ,the conclusion is consistent. It depends on whether you think of them(u,v,w) as speed or velocity .

Second scene:
Earth---------------------------------------A--->
-----------------------------------------------B--->
.....................................................0.9C
Please answer the relative speed between A and B .
w = (u+v)/(1+uv/c^2) = 0C

Fifth scene is very like second scene.The different is known u, w, solve v.

 

[TonyYuan]

It is unclear what you mean by a "mathematical game". SR obviously provides a self-consistent system of calculation. Any system of mechanics must do that. That does not make it a "game", obviously. It is a model of physical reality that can be tested experimentally. As it has been.

I looked at the history of special relativity and Einstein's uncertainty about the constant speed of light. I don't think the theory is too reliable.
The greatest contribution of special relativity is the mass energy equation, but this equation can be easily deduced. We also interpret the Morley experiment. LIGO and star bending light phenomenon give us great enlightenment, and put forward the view that the earth's gravity holds the light.

 

[(Q)]

the earth's gravity holds the light.

Wtf does that mean?

 

[TonyYuan]

Wtf does that mean?

The light far away from the star can be bent by the gravitational field of the star, which can prove the influence of gravity on the light, while the light so close to the earth will be more affected by the gravity of the earth.
Just like the speed of sound in an airplane, no matter which direction it is measured, it is the same. This is a similar analogy.
LIGO found that gravitational wave is also a good proof.

 

[(Q)]

the light so close to the earth will be more affected by the gravity of the earth.

How is light more affected by the gravity of the earth?

 

[TonyYuan]

Tony,

In post #40, you leave the plus signs in the equation, and then put the minus sign on only the velocity which goes from right to left. Why not do that same thing in #62?

Do you have any other questions? I like to discuss. Your opinions can always bring me ideas.

 

[TonyYuan]

How is light more affected by the gravity of the earth?

Because gravity is inversely proportional to the square of distance. The light near the star is 10 times the diameter of the star, while the light on the earth is 1 times the radius of the earth. They differ by 20 * 20 = 400 times.
sorry, more than 400 times is right. star is bigger than earth.

 

[(Q)]

Because gravity is inversely proportional to the square of distance. The light near the star is 10 times the diameter of the star, while the light on the earth is 1 times the radius of the earth. They differ by 20 * 20 = 400 times.

So, what is your point?

 

[TonyYuan]

So, what is your point?

The light on the earth is pulled by the earth's gravity, and the measured speed of light is the same everywhere on the earth. It's like measuring the speed of sound in an airplane, it's all constant.

I have designed an instrument for measuring the speed of light with an accuracy of 0.01 m/s or higher. Of course, this is all theoretical data. Finally, it depends on the test results.

 

[(Q)]

The light on the earth is pulled by the earth's gravity, and the measured speed of light is the same everywhere on the earth. It's like measuring the speed of sound in an airplane, it's all constant.

You're still not making a point, what does that have to do with your claim, "the earth's gravity holds the light."?

 

[Neddy Bate]

Fifth scene:
Earth.....................................A---->u=0.2C............................B---->w=0.8C
What is the velocity v of B relative to the earth?

w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.8C
0.2-v=(1-0.2v)*0.8
0.84v = -0.6
v = -0.714C, the velocity v of B relative to the earth is -0.714C, velocity direction is opposite to A.

The answer to the fifth scene should be:

w = (u + v) / (1 + (uv/c²))
w = 0.8c
u = 0.2c

0.8 = (0.2 + v) / (1 + (0.2v))
0.8(1 + (0.2v)) = 0.2 + v
0.8 + (0.8*0.2v) = 0.2 + v
0.8 + 0.16v = 0.2 + v
0.16v - v = 0.2 - 0.8
0.16v - 1v = -0.6
(0.16 - 1)v = -0.6
-0.84v = -0.6
v = -0.6 / -0.84
v = 0.714c

 

[paddoboy]

The light far away from the star can be bent by the gravitational field of the star, which can prove the influence of gravity on the light, while the light so close to the earth will be more affected by the gravity of the earth.

I'm an amateur as far as SR/GR is concerned, but I do know that the correct interpretation is that light/photons simply follow geodesics in spacetime.

Just like the speed of sound in an airplane, no matter which direction it is measured, it is the same. This is a similar analogy.

I fail to see any connection. Are you perhaps thining of Doppler effect and cosmological and gravitational redshift? That's a different ball game as far as I am aware.

LIGO found that gravitational wave is also a good proof.

Gravitational radiation are simply ripples in spacetime caused by catastrophic collisions and/or asymmetric S/nova explosions.

Do you have any other questions? I like to discuss. Your opinions can always bring me ideas.

SR was brought to our attention in 1905 and since then hundreds of scientists have looked and studied it from all angles. I'm pretty sure if there were any inherent problems, we would know about them.