# a story about special relativity,who can explain it？

Discussion in 'Physics & Math' started by TonyYuan, Mar 17, 2020.

1. ### HalcRegistered Senior Member

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166
Thank you for confirming every bit of stupidity of which I accused you in the prior post.

That's right, but motion of planets is not necessary for them to attract each other. I put a rock on a tower. Gravity has the rock exerting force on the tower, but it isn't moving (let's assume no spin to eliminate all kinetic energy). Neither Earth nor the rock emits gravitational waves in this scenario, but the gravity force is still there, and only the tower keeps the rock from falling.

Yes, energy for Earth's gravitational waves comes from orbital motion, and it is one of the four things altering Earth's orbital radius. It is the least of those four things.

Last edited: Mar 26, 2020 at 1:26 AM

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3. ### TonyYuanRegistered Member

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160
Have you ever seen a UFO video provided by the United States, the black dots fly away vertically from the surface of the earth or the sun, why are they flying away vertically instead of taking a parabola like launching a rocket on our earth.
The effect of gravity in this direction is minimal. Crazy guess ...
https://photos.app.goo.gl/PwuKEnYsbQQvh2eVA
https://www.bilibili.com/video/av41125616/
https://www.bilibili.com/video/BV1pt411g7B2/?spm_id_from=333.788.videocard.1

Last edited: Mar 26, 2020 at 1:56 AM

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5. ### TonyYuanRegistered Member

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160
The motion of the planet in the Y direction will cause a gravitational field fluctuation, so will the movement in the X direction not produce a gravitational field fluctuation? Definitely also possible.
You are helping me improve
Tony's quackery: Doppler Effect Of Gravitational Field.
F = G * M * m / r ^ 2 * f(v)
= (G * M * m / r ^ 2) * ( v_gravitational field - v ) / c
= (G * M * m / r ^ 2) * x / c

Last edited: Mar 26, 2020 at 2:15 AM

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7. ### TonyYuanRegistered Member

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160
Why UFO flying away vertically instead of taking a parabola like launching a rocket on our earth?

If the spacecraft can be accelerated to 3 * 10 ^ 4 meters / second in 1 second, then it can be accelerated to 6 * 10 ^ 4 m/s in the next second. After 10,000 seconds, it can reach the speed of light in about 2.7 hours. But according to F = (G * M * m / r ^ 2) * x / c, the smaller the effective speed x, the smaller the gravitational force F will be. (x = v_gravitational field - v)

F = (G * M * m / (r + 0.5 * a * t ^ 2) ^ 2) * (c-a*t) / c, since the diameter of the sun is 1.392 * 10 ^ 9 meters, ignore 0.5 * a * t ^ 2, this simplifies to get:
F = G * M * m / (r ^ 2)-G * M * m / (r ^ 2) * a * t / c. Suppose K = G * M * m / (r ^ 2),
So F = K-K * at / c, a = F / m
F0 = K-K * a0 * t0 / c
F1 = K-K * (a0 + delt_a) * (t0 + delt_t) / c
= K- (K * a0 * t0 / c + K * a0 + delt_t / c + K * delt_a * t0 / c + K * delt_a * delt_t / c)
delt_F = F0-F1 = K * a0 * delt_t / c + K * delt_a * t0 / c + K * delt_a * delt_t / c
≈K * a0 * delt_t / c + K * delt_a * t0 / c, suppose X = K / c, then we get:
det_F = F0-F1 = X * a0 * delt_t + X * t0 * delt_a.

When t = 0, a0 = 0, det_a = 0
When t = 1, a0≈X, delt_a≈X, delt_F = 2X.
When t = 2, a0≈2X, delt_a≈2X, delt_F = 4X.
When t = 3, a0≈4X, delt_a≈4X, delt_F = 8X.
...
It can be seen that the speed of force reduction is X * 2 ^ t, when X * 2 ^ t = K / c * 2 ^ t = G * M * m / r ^ 2 = K, we can get:
K / c * 2 ^ t = K, c = 2 ^ t, t = log2 (c) Then t = 28.16 seconds. In other words, from the surface of the sun, after 28.16 seconds of acceleration, the gravitational force of the sun on the spacecraft becomes 0, and the spacecraft reaches the speed of light c.

It's crazy, we can fly at the speed of light c.

Last edited: Mar 26, 2020 at 5:10 AM
8. ### exchemistValued Senior Member

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9,084
Something seems finally to have gone FUNG! in his brain.........

9. ### TonyYuanRegistered Member

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160
This is obviously impossible. Because there is an error in the derivation.

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It's very difficult to reach the speed of light.

10. ### TonyYuanRegistered Member

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160
I really don't see how these formulas violate these laws you said. Can you speak more clearly?
f (v) = ( v_gf - v ) / c , v_gf and v all are velocity , not speed.
F = G * M * m / r ^ 2 * f(v) = (G * M * m / r ^ 2) * ( v_gf - v ) / c.
Let the magnitude of their velocity vector difference be x, then get F = G * M * m / r ^ 2 * f(v) = (G * M * m / r ^ 2) * x / c .

11. ### TonyYuanRegistered Member

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160
Doppler Effect Of Gravitational Field. It will unify classical physics and relativity.
https://photos.app.goo.gl/vTBpR1KnAo293hnS6
1.
In a uniform gravitational field, the speed of light does not change.
2.At different gravitational field strengths, the speed of light is different.
(1)The stronger the gravitational field, the slower the light speed, and the weaker the gravitational field, the faster the light speed.
(2)Light passing through an uneven gravitational field will cause refraction bending due to different speeds.
3.The Doppler effect exists in the gravitational field.
(1)If away from each other, the strength of the gravitational field will be smaller than that at the relatively stationary state, the larger the relative speed , the weaker the gravitational field;
(2)If close to each other, the strength of the gravitational field will be greater than that at the relatively stationary state, the greater the relative speed, the stronger the gravitational field.
4.There is a uniform gravitational field on the surface of the earth, and the speed of light measured in different directions is constant.
5.To calculate the motion of celestial bodies with universal ravitation, the Doppler effect of the gravitational field needs to be considered. The greater the relative speed, the more pronounced.
https://photos.app.goo.gl/8sGswt8dhJRsxHkJ8

The Doppler effect is classical physics. This theory can be applied to mechanical waves, electromagnetic waves, and gravitational waves as well. Although gravitational waves are not mentioned in classical physics, the model I created for gravitational waves is a model of classical physics. The knowledge you need to understand it does not go beyond the scope of classical physics.

The Doppler effect of the gravitational field I described is clear. The gravitational formula needs to include a velocity (not speed) parameter. F = G * M * m / r ^ 2 * f (x), x is the speed of light minus the component of v in the c direction.
When x = 0, it is the gravitational strength when it is relatively stationary, F = G * M * m / r ^ 2, so f (0) = 1.
When x = c, the gravitational wave cannot reach the object, so F = 0, so f © = 0.
F = G * M * m / r ^ 2 * f(x) = (G * M * m / r ^ 2) * x / c .

12. ### paddoboyValued Senior Member

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23,610
Tony, we have a saying in Australia for those attempting to do what you are pretending to do....You're pushing shit uphill....or if you like, You're pissing into the wind.

Let me add as I said before...when an amateur like me, is able to pick you up on a couple of points that a high school student should know, then some serious doubts exist as to your capabilities and knowledge.

13. ### TonyYuanRegistered Member

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160
I don't think we are here to discuss who has better mastered the knowledge in books. We are exploring science. Just as you are learning Newtonian mechanics, you are generally scolding Newton for his mistakes.
Look at your answer to that race game between Newton and Einstein, and you will believe that you are right. Don't you find it funny?

You have been emphasizing that classical physics is not as accurate as relativity in calculating celestial motion and bending of light. Then allow us to make corrections to classical physics, we introduce the Doppler effect of the gravitational field, why is this not possible?
We are not going to make the ridiculous assumption that "the speed of light is constant." , this is as funny as God said.

The special theory of relativity is the stupidest theory I've ever seen. Make one hypothesis and use another hypothesis to round that lie.

paddoboy, I hope you can read carefully what Einstein said after completing General Relativity. He himself thinks that "the speed of light is constant" is funny, and he will be surprised by the fanaticism of his supporters.

Last edited: Mar 27, 2020 at 6:18 AM
14. ### TonyYuanRegistered Member

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160
Race game between Newton and Einstein.
https://photos.app.goo.gl/YtQNDmrjQ48xkhhx6
Let's look at the answer below, it is the correct answer given on the basis of a correct understanding of the special relativity. Don't you find it funny?

If the vectors shown represent component vectors in Earth frame, then both get to the line at the same time, but only in Earth frame.
If, on the other hand, both Newton and Einstein are going straight to the side, but they each have an identical gun that shoots a bullet at the line at 0.2c, then Newton's bullet gets there first in Earth frame. Again, the answer is different in different frames.

If their bullets were fired at you, do you know who shot you? paddoboy, please tell us.

Last edited: Mar 27, 2020 at 6:31 AM
15. ### (Q)Encephaloid MartiniValued Senior Member

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20,068
And, the anti-relativity crank exposes himself. What will he say next...?

He'll say something only Dunning and Kruger can explain.

16. ### TonyYuanRegistered Member

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160
If their bullets were fired at you, do you know who shot you? paddoboy, please tell us.
Or you (Q) can tell us. Maybe you can ask your master Janus before giving an answer.

17. ### TonyYuanRegistered Member

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160
Many people may think that the gravity between the planet and the sun depends only on their mass and distance. GR tells us that the huge mass of the sun will cause a huge space-time curvature, which will advance forward as the sun moves, and the distant space will gradually return to flat. So for a certain space position that the sun passes, his space-time will go through a process as shown in the figure.
https://photos.app.goo.gl/QK2czMKm8wZ8di2c8
It looks very much like a wave, which is as fast as light. GR also tells us that the planets only move along geodesics in curved space. But you should think that the space-time curvature will also occur around the planet, and the curvature of the planet and the curvature caused by the sun will be superimposed on each other, as shown in the figure.
https://photos.app.goo.gl/bBaCgHcamdvSxaoh7

We know that the planet is orbiting the sun, so will this superimposed curved shape be related to the movement of the sun and the planet?
1. Assuming the sun and the earth are relatively still, we can get a curved shape.
2. Assuming there have a relative velocity between the sun and the earth, we can get a curved shape again.
Do you think their shapes will be the same? Obviously different! Since they are different, the phenomenon of gravity caused by time and space bending will not be the same! Therefore, when calculating the gravitational force between two objects, we must take their respective velocities into account.
So general relativity also supports the Doppler effect of gravitational fields. In other words, adding the Doppler effect to universal gravity is equivalent to general relativity.
F = (G*M*m / r^2) * f(x) = (G*M*m / r^2) * (x / c) . x is the speed of light c minus the component of v in the c direction.
https://photos.app.goo.gl/zDqx1UGW8LkRs7g4A

Last edited: Mar 27, 2020 at 3:03 PM
18. ### TonyYuanRegistered Member

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160
F = (G*M*m / r^2) * f(x) = (G*M*m / r^2) * (x / c)
If the Doppler effect of the gravitational field is taken into account to calculate the precession of Mercury, will it be consistent with the observed results?
I need your help, come on.

Last edited: Mar 28, 2020 at 12:00 AM
19. ### HalcRegistered Senior Member

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166
We tried. You ignore the help.
Mercury either spins itself to shrapnel or exits the solar system before it can do this. All the planets would do this, except they prob

Run a simulation and see for yourself.

Side note: It doesn't allow anything to accelerate to the speed of light. You already demonstrated the inability to perform relativistic addition.

Last edited: Mar 28, 2020 at 12:16 AM
20. ### TonyYuanRegistered Member

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160
Halc, thanks very much, can you share your calculations with me? I'm interested to try the calculations myself.

Note: When calculating, you can't get x directly from c - the speed of Mercury (4.789*10^4 m/s).

Last edited: Mar 28, 2020 at 12:11 AM
21. ### HalcRegistered Senior Member

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166
I didn't compute any specific numbers. I just assumed a random planet that had some spin and had an eccentric orbit.
You seem inconsistent in a definition of what x is.
"x is the speed of light c minus the component of v in the c direction"
c isn't a direction. I assumed from prior discussion that x was ... 'minus the recession speed from the object from which the gravitational force is being computed', or something like that. So x is less than 1 for two objects moving apart, and greater than 1 for approaching objects.

The orbital speed of Mercury ranges from about 39 to 59 km/sec, so there is no one speed for it.

22. ### TonyYuanRegistered Member

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160
"x is the speed of light c minus the component of v in the c direction"
This is the velocity of the gravitational field, the speed is c, and the direction is the direction of the gravitational field, as shown in the figure.
https://photos.app.goo.gl/WL29FA6ZTf46odiT7

23. ### TonyYuanRegistered Member

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160
The Doppler effect of the gravitational field can well explain the deviation of Mercury precession.
https://photos.app.goo.gl/e7QhPSgN6FrQs9tL7

The calculation of the precession of Mercury's perihelion is actually based on Newton's law. The result is that precession per century = 5,557.62". 90% of this is caused by the precession of the coordinate system. The perturbations caused by the perturbations of Venus, Earth, and Jupiter are actually 5,600.73", which is subtracted to 43.11" per century.

If the Doppler effect of the gravitational field was taken into account in the calculation, there would be no 43.11 "deviation.