a story about special relativity,who can explain it？

What, are we doing your homework now?

Your description of the diagram is very ambiguous, but since you're asking about speed of B relative to Earth, I take the 0.8c figure to be B relative to A (v), in which case B relative to Earth (w) is 0.862c. Surely you could have plugged in those numbers into the formula yourself. Or not, since you clearly botched the job in post 91 and also in 98. Maybe you need to review your algebra skills.

Tony, I want to hear your voice. What does Newton say? What does he say in scene 1?
Your're the one seemingly pretending to live in a different universe, denying empirical observation. Are you just trolling or do you have a different purpose in this deliberate misrepresentation of physics?

 

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This formula is given by Janus, and I am putting forward an experimental model around this formula. I think it's hard for Janus to give a reasonable explanation for this fifth model.
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.

u,v,w are defined in the formula gave by Janus.
u is A to earth
v is B to earth
w is B to A

 

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Of course it has "no energy" today. That is because the use of SR settled the problem. It is one of its classic applications. So you can't just ignore it if you think SR is wrong - you need an alternative way to explain it.

 

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Tony, in this formula:
w = (u + v) / (1 + (uv/c²))
v is the velocity of a spaceship relative to earth
u is the velocity of a projectile launched from the spaceship, relative to the spaceship
w = the velocity of the projectile relative to earth

Note that I do not get a negative value for w in this solution:
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

So, if:
v=+0.200c and u=+0.714c then w=+0.800c
Or if:
v=+0.714c and u=+0.200c then w=+0.800c

Or, if:
v=+0.200c and u=+0.800c then w=+0.862c
(As Halc said)

The only thing you have to do is be careful about which variable is which, and use + and - signs to represent left-to-right and right-to-left, respectively. You don't have to draw diagrams as you have been doing. Just specify the variables.

 

Given "u" the velocity of A relative to the earth, "w" the velocity of spaceship B relative to A, find "v" the velocity of B relative to the earth.

please see the post #40 by Janus.
......
u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

v = 0.714c It means that the velocity direction of B is opposite to that of A. But now I'm going to show that A, B have the same velocity direction to the earth.

 

Your answer is error.

 

Thank you. I'd have to rethink my answer if you agreed with it.

I notice you don't give your voice. What does Newton say about scenario 1, the simplest answer?

 

Newton said u+v

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That's right, which means it should be rather trivial to accelerate something to greater than light speed. Funny that they give things far more energy than should be needed to do this, and yet they never go faster than c.
Newton's view was falsified 150 years ago. Heck, even Newton knew his views didn't make correct predictions in boundary cases.
Get with the program. You're currently numbering yourself among all the other denial nuts out there. You've got plenty of company, but none of it good.
I notice you get your 'science' from their web sites. Your paper is full of parroted nonsense from them.

If you want to show a contradiction in a theory, know the theory first. Until then, if you get seemingly wrong numbers, it's you making the mistake (probably deliberately), not a mistake in the theory. People far smarter than any of us have tried to take this stuff apart, and have failed.

 

Velocity of A with respect to Earth (motion to the right seen as positive) 0.2c
Velocity of B with rspect to Earth 0.8c

Velocity of Earth as measured by B -0.8c ( by B's reckoning, it is the Earth moving to the left.)
Velocity of A as measured by B, - 0.714c (-.8c+0.2c)/(1- 0.2c(-0.8c)/c^2)
A is moving to left as measured by B. Difference in speed between A and Earth ( the closing speed) 0.8c- 0.714c = 0.086c

Velocity of Earth with respect to A as measured by A -0.2c ( earth moving to the left)
Velocity of B with respect to A as measured by A 0.714 c
Difference in speed between Earth and B as measured by A, 1c

A and B agree as to the speed they are moving apart
Earth and A agree as to the speed they are moving apart
Earth and B agree as to the speed they are moving apart.
Everyone agrees that everyone is moving apart from everyone else.

The speed the Earth says A and B are moving apart will not agree with the answer either A or B says
The speed A says the Earth and B are moving apart will not agree with the answer either the Earth or B gives.
The speed B says the Earth and A are moving apart will not agree with the answer either the Earth or A gives.

When working with velocity vectors, it is important to keep everything straight.
When you changed w to 0.1c relative to the Earth, that means means B's rightward velocity with respect to the Earth is less than A's.
Even in Newtonian physics, that means that According to B, the Earth has a leftward (minus) velocity and A has a rightward (positive velocity) with respect to itself.
If A is in between the Earth and B then everybody agrees that Earth and A are moving apart and A and B are moving together ( until A passes B and they begin to move apart.

There is nothing strange about this.

Even going back to the original scenario. Lets assume you only know that relative to B A has a velocity of -0.714c, and the Earth a velocity of - 0.8c relative to B, what would be the velocity of between A and the Earth?
If you use ( (-0.714c)+(-0.8c)/(1+(-.8)(-0.714)/c^2), you get -0.9636c, which is the wrong magnitude. The reason is that you are trying to find the difference between the two given velocities, not the sum.
So instead you have to use either:
(-0.714c- (-(-0.8c)/(1- (-0.8c(-0.714c) = 0.2c
or
(-0.8c- (-(-0.714c)/(1- (-0.714c(-0.8c) = -0.2c
The first answer is A's velocity relative to the Earth as measured by the Earth (to the right, thus positive)
The second answer is the Earth velocity relative to A as measure by A ( to the left, thus negative.

This is no math game, it is just properly applying the math to the scenario you are considering

"Difficult to understand" does not mean "impossible to understand", especially since there are a great deal of people in the world who do understand it.
Being difficult to understand is not a disqualifying characteristic. The universe is under no obligation to be easy to understand.
Is Relativity counter-intuitive? Yes. But is also completely logical and very accurately describes how our universe behaves.
One the main reasons people have difficulty with Relativity is that it involves a complete rethink of the very nature of time and space compared to the ideas that preceded it, and many people have trouble discarding the old way of thinking about things and accepting the new. They keep trying to make Relativity to be consistent with the old model of time and space, and when find that the two models are incompatible, they jump to assume that there is something "wrong" with Relativity rather than abandon their present way of thinking about things.

There is nothing in your scenario #5 that shows any flaw in SR. The fact that you not comfortable with the answers SR gives doesn't make it flawed. The universe is not concerned with what you are or are not comfortable with. If only it had consulted with you before it adopted its operating principles.

 

Yep that's the thing...smarter people then any pretender that wants to attempt to invalidate SR on a web site. It ain't gonna happen.
He's also not the first person who has come here claiming SR is invalid....I recall another called chinglu.

 

You still have yet to support that assertion.

 

Janus, your answer is error.
w is the velocity of B relative to the A ===> w=0.8C
Given "u" the velocity of A relative to the earth, "w" the velocity of spaceship B relative to A, you should find "v" the velocity of B relative to the earth.

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.

u,v,w are defined in the formula gave by Janus.
u is A to earth
v is B to earth
w is B to A

 

You still haven't been able to face the questions I asked.
v = 0.714c It means that the velocity direction of B is opposite to that of A. But now I'm going to show that A, B have the same velocity direction to the earth.

 

Question; If the earth's gravity captures light, how is it we receive sunlight? Does the sun not have gravity?

I can see the effect in a black hole which may swallow spacetime geometrics , but as long as spacetime is unbroken light will follow its geodesics.

 

It's not capture, it's affected. It's like light being bent by a star. It is like "light will follow its geodesics."

Can you understand i posted Fifth scene:
Earth.....................................A---->u=0.2C............................B---->w=0.8C

 

So light is affected by gravity, what's new? Did Einstein not prove this?

 

I'll send you Einstein's history. He also realized the mistake.
Einstein’s VSL attempt in 1911
There’s a Wikipedia article on the Variable Speed of Light, which is often abbreviated to VSL. If you take a look at an old version dating from 2014, you can see a section entitled Einstein’s VSL attempt in 1911. This says Einstein first mentioned a variable speed of light in 1907 and reconsidered the idea more thoroughly in 1911. However it then goes on to say Einstein abandoned the idea in 1912 because it only predicted half the deflection of light by the Sun. However it isn’t true. Einstein didn’t abandon the idea. That’s why you can find him saying the same thing year after year:

1912: “On the other hand I am of the view that the principle of the constancy of the velocity of light can be maintained only insofar as one restricts oneself to spatio-temporal regions of constant gravitational potential”.

1913: “I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis”.

The last quote is the English translation of what Einstein said in German in 1916: “die Ausbreitungsge-schwindigkeit des Lichtes mit dem Orte variiert”. That translates to “the propagation speed of light with the place varies”. Einstein never did abandon his variable speed of light. The people who tell you that grew up before the Einstein digital papers were online. The general relativity they were taught wasn’t the same as Einstein’s.

The speed of a light wave depends on the strength of the gravitational potential along its path.

Einstein’s postulate only lasted two years
Yes, Einstein said the speed of light is constant in 1905 when he was doing special relativity, but by 1907 he was broadening his horizons and looking into what would become general relativity. That’s when he wrote a paper on the relativity principle and the conclusions drawn from it. He used Φ (phi) to denote gravitational potential, and he said this: “These equations too have the same form as the corresponding equations of the nonaccelerated or gravitation-free space; however, c is here replaced by the value c[1 + γξ/c²] = c[1 + Φ/c²]. From this it follows that those light rays that do not propagate along the ξ-axis are bent by the gravitational field”. Only two years after his special relativity postulate, there’s Einstein talking about a speed of light that varies with gravitational potential. This wasn’t some one-off. He said the same thing in 1911. That’s when he wrote a paper on the influence of gravity on the propagation of light. He said this: “If c₀ denotes the velocity of light at the coordinate origin, then the velocity of light c at a point with a gravitation potential Φ will be given by the relation c = c₀(1 + Φ/c²). The principle of the constancy of the velocity of light does not hold in this theory in the formulation in which it is normally used as the basis of the ordinary theory of relativity”. He said the principle of the constancy of the velocity of light does not hold. And it’s clear from the context that the word velocity is as per “high velocity bullet”. It’s the common usage as opposed to the vector quantity. Einstein was talking about the speed of light, which is why he was referring to c.

 

That figure is wrong according to the setup you describe.

The value of .714c is positive and so is A's velocity relative to Earth, so don't know why you conclude they're going in opposite directions. You've applied the formula incorrectly, so you're getting these wrong answers. You got -.714 in the post prior, by again applying the formula incorrectly.
Really, it isn't hard. u is 0.2. v is 0.8, per your description. Try plugging in those values.

 

Question1: Are there any straight lines in the universe?

Question2: Does velocity have anything to do with gravitational curvature of spacetime?