a story about special relativity,who can explain it？

[TonyYuan]

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.

Please read the conditions carefully.

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. v and u are in the different direction.

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 another example:
Earth.....................................A---->u=0.2C............................B---->w=0.1C
if u=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 and u are in the same direction.

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

 

[TonyYuan]

Question1: Are there any straight lines in the universe?
maybe
Question2: Does velocity have anything to do with gravitational curvature of spacetime?

maybe

 

[TonyYuan]

------->velocity positive direction

Earth ……………………………….A…………………………………….B
……………………………………u=0.2C………………………………..v
u is the velocity of A relative to the Earth
v is the velocity of B relative to the Earth
x is the velocity of Earth relative to B
w is the velocity of B relative to the A

w = (u+x)/(1+u*x/C^2)
if u = 0.2C, put it into the formula. w = (0.2+x)/(1+0.2x) ==>5w = (1+5x)/(5+x)===>x=(25w-1)/(5-5w)
so if x>0, then (25w-1>0 and 5-5w>0) or (25w-1<0 and 5-5w<0), simplify this inequality (w>0.4 and w<1) or (w<0.4 and w>1, it is not correct). So we get:
if 0.4<w<1, then x>0.
if w<0.4 or w>1(it is not correct), then x<0.
Finally, we can get:
....if 0.4C<w<1C, then x>0, v<0; this mean u and v have the contrary velocity direction.
....if -1C<w<0.4C, then x<0, v>0; this mean u and v have the same velocity direction.

It is so interesting, the value of w will determine the direction of v.
If we think about the speed it can't be faster than the speed of light. So -1<(25w-1)/(5-5w)<1, simplify this inequality, get -0.2<w<0.2. Finally, we can get:
…. only in this case -0.2C<w<0.2C can it be established, under this conditions v>0.

Conclusion:
When u=0.2C, must v>0, must -0.2C<w<0.2C. Otherwise, it will no longer conform to special relativity. It is so interesting.
Further, you can prove that:
….if u>0, then must v>0 and must –u<w<u.
….if u<0, then must v<0 and must u<w<-u. ..........
It's interesting that the velocity of u determines the direction of v and the velocity range of w. ......

Digital game.

 

[TonyYuan]

Corrections:

------->velocity positive direction

Earth ……………………………….A…………………………………….B
……………………………………u=0.2C………………………………..v
u is the velocity of A relative to the Earth
v is the velocity of B relative to the Earth
x is the velocity of Earth relative to B
w is the velocity of B relative to the A

w = (u+x)/(1+u*x/C^2)
if u = 0.2C, put it into the formula. w = (0.2+x)/(1+0.2x) ==>5w = (1+5x)/(5+x)===>x=(25w-1)/(5-5w)
so if x>0, then (25w-1>0 and 5-5w>0) or (25w-1<0 and 5-5w<0), simplify this inequality (w>0.4 and w<1) or (w<0.4 and w>1, it is not correct). So we get:
if 0.4<w<1, then x>0.
if w<0.4 or w>1(it is not correct), then x<0.
Finally, we can get:
....if 0.4C<w<1C, then x>0, v<0; this mean u and v have the contrary velocity direction.
....if -1C<w<0.4C, then x<0, v>0; this mean u and v have the same velocity direction.
It is so interesting, the value of w will determine the direction of v.

Conclusion:
You can see the w determines the the direction of v.
......

Digital game.

 

[TonyYuan]

Corrections:

------->velocity positive direction

Earth ……………………………….A…………………………………….B
……………………………………u=0.2C………………………………..v
u is the velocity of A relative to the Earth
v is the velocity of B relative to the Earth
x is the velocity of Earth relative to B
w is the velocity of B relative to the A

w = (u+x)/(1+u*x/C^2)
if u = 0.2C, put it into the formula. w = (0.2+x)/(1+0.2x) ==>5w = (1+5x)/(5+x)===>x=(25w-1)/(5-5w)
so if x>0, then (25w-1>0 and 5-5w>0) or (25w-1<0 and 5-5w<0), simplify this inequality (w>0.4 and w<1) or (w<0.4 and w>1, it is not correct). So we get:
if 0.4<w<1, then x>0.
if w<0.4 or w>1(it is not correct), then x<0.
Finally, we can get:
....if 0.4C<w<1C, then x>0, v<0; this mean u and v have the contrary velocity direction.
....if -1C<w<0.4C, then x<0, v>0; this mean u and v have the same velocity direction.
It is so interesting, the value of w will determine the direction of v.

Conclusion:
You can see the w determines the the direction of v.
......

Digital game.

Corrections:

------->velocity positive direction

Earth ……………………………….A…………………………………….B
……………………………………u=0.2C………………………………..v
u is the velocity of A relative to the Earth
v is the velocity of B relative to the Earth
x is the velocity of Earth relative to B
w is the velocity of B relative to the A

w = (u+x)/(1+u*x/C^2)
if u = 0.2C, put it into the formula. w = (0.2+x)/(1+0.2x) ==>w = (1+5x)/(5+x)===>x=(5w-1)/(5-w)
so if x>0, then (5w-1>0 and 5-w>0) or (5w-1<0 and 5-w<0), simplify this inequality (w>0.2 and w<5) or (w<0.2 and w>5, it is not correct). So we get:
if 0.2<w<1, then x>0.
if w<0.2 or w>1(it is not correct), then x<0.
Finally, we can get:
....if 0.2C<w<1C, then x>0, v<0; this mean u and v have the contrary velocity direction.
....if -1C<w<0.2C, then x<0, v>0; this mean u and v have the same velocity direction.
It is so interesting, the value of w will determine the direction of v.

Conclusion:
You can see the w determines the the direction of v.

Before read this post ,you should read post #121 first.

 

[Janus58]

Please read the conditions carefully.

w is the velocity of B relative to the A ===> w=0.8C

As measured by which reference frame? A's rest frame? or the Earth's rest frame?

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?

Again, as measured from which frame?

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. v and u are in the different direction.

The form of the Equation you are using is not consistent with the question you are asking.
if w is the relative velocity of B to A ( as measured by A) and u is the relative velocity of A to the Earth as measured by the Earth then v, the relative velocity of B to the Earth as measured by the Earth is found by
v = (u+w)/(1+uw/c^2) = 0.862c
if you rearrange this to solve for w (for what reason I have no idea, as the the equation above gives you the answer directly.)
Then you must use
w=(v-u)/(1-uv/c^21)= 0.8c
You reversed the order of u and v in the top part of that equation which resulted in you getting the wrong answer for v. [/quote]

But A sees B moving away at 0.8C, what happened?
[/quote]
What happened is that you screwed up in applying the math to the problem.

Corrections:

------->velocity positive direction

Earth ……………………………….A…………………………………….B
……………………………………u=0.2C………………………………..v
u is the velocity of A relative to the Earth
v is the velocity of B relative to the Earth
x is the velocity of Earth relative to B
w is the velocity of B relative to the A

w = (u+x)/(1+u*x/C^2)

Again, wrong.
Given your definitions of u,v,x,and w
The equation should read.
v = (u+w)/(1+uw/c^2)
If you reassign the meanings of the variables, you also have to properly reflect that reassignment in the equation.
You keep screwing it up when it comes to this.
The reason you keep getting answers that seem bizarre is that you are really bad at knowing how to properly arrange the equations so that they accurately represent the scenario. The problem isn't with SR, but with your inability to apply it correctly.

 

[phyti]

In the 1st scene, if A and B are moving in opposite directions away from earth at .9c,
each measures the velocity of the other using

v=(v1-v2)/(1-v1v2) with v2 the one measuring

With v1=-.9, and v2=.9,

A gets (.9+.9)/ (1+.81) = 1.80/1.81=.994 for B,

B gets (-.9-.9)/ (1+.81) = -1.80/1.81=-.994 for A.

The results are reciprocal as required by SR.

 

[(Q)]

The assumption that the speed of light is constant is wrong. That is to say, the theoretical basis of special relativity is not tenable.

Since you have yet to support this crackpot claim and your errors keep piling up due to the fact you have no idea what you're doing, it's probably safe to conclude you're just another anti-relativity crank.

 

[Janus58]

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.

For light path to be significantly bent, it has to pass very near the star and and not "far away". The equation is:
Angle of deflection = 4GM/rc^2
where M is the mass of the object and r is the distance the light passes from its center. The angle is in radians.
With our own Sun a light ray just skimming it surface is deflected by just 1.7 arc-seconds (~ 1/2100 of a degree)
A ray just skimming the surface of the Earth would be deflected by just 0.0006 arc-seconds ( less than 1/6,000,000 of a degree).
So while the gravity of a mass can deflect light, this effect is just too insignificant to account for the effect you are trying to posit.
It like trying to say that it was the collision with a bumble bee during its flight which caused a plane to be 10 min late arriving at the airport. While that collision would have had a very small brief effect on the plane's speed, It would have come nowhere close to producing enough of an effect to make the plane 10 min late.
It's not enough to say "A can effect B, therefore A can cause C", unless you can show that the effect that A has on B is significant enough to produce result C.

Not only that, but you can't just jump from " gravity bends light paths" to " Earth's gravity drags light with it" because one does lead to the other.

An example of using facts to jump to the wrong conclusion:

There is something called the "Moon illusion". It is that the Moon appears to be much larger when near the horizon than when high in the sky.
There have been people that try to explain it like this:
1. Light is refracted by mediums such as water and air.
2. Lens can magnify images by refracting light, and it is the curvature of the lens that produces this effect.
3. When we look at the Moon when it it near the horizon, we are not only looking a it through more air, but also through more of a curve of the atmosphere.
4. Thus the Moon looks larger because this refraction through a curved medium produces a magnifying effect.

This is wrong on a couple of accounts:
The Index of refraction for air is pretty small, thus an "air lens" capable of producing noticeable magnification of the moon's image would have to be significantly more curved than the atmosphere is.
While light coming at us from an object near the horizon is bent on it way to us, it is all in one direction, which just makes objects look slightly higher in the sky than they really are. As this effect gets weaker the higher above the horizon, the lower part of the Moon tends to be effected more than the upper part. The visual effect is that the Moon near the horizon looks a little "squashed".
So while being near the horizon does produce a slight visual distortion to the Moon's image, it is not a magnification effect.
So while points 1,2 and 3 aren't wrong per se, the conclusion they arrive at is.

 

[Write4U]

Gravitational lensing is caused by a massive body between a distant object and ourselves. It can create the appearance of two or more objects where there is really only one. The light from the object gets bent round the massive body in between.

The massive body, such as a galaxy or black hole, creates a very strong gravitational field in space. The exact nature of the effect depends on:

• relative distance and position between observer, lens and lightsource
• size of the lens
• mass inside the lens

Weak Lensing

Distortions.
Weak lensing does not produce multiple images of the same object. Instead it produces a highly deformed or stretched image from an object far beyond the lens. In 1986 this was discovered in cluster Abell 370. Later it was understood that this was a greatly deformed image of a galaxy far beyond the cluster.

The object can appear larger or smaller, as is shown in the table. Weak lenses allow us to observe very distant galaxies, which we would be unable to observe without such lens. By bending the light, the amount of light (magnitude) from the source increases. This way a very distant and faint galaxy can become visible, while we normally would not be able to observe it.

https://simple.wikipedia.org/wiki/Gravitational_lensing

Just throwing it into the mix....

 

[TonyYuan]

As measured by which reference frame? A's rest frame? or the Earth's rest frame? //my ans: A
The form of the Equation you are using is not consistent with the question you are asking.
if w is the relative velocity of B to A ( as measured by A) and u is the relative velocity of A to the Earth as measured by the Earth then v, the relative velocity of B to the Earth as measured by the Earth is found by
v = (u+w)/(1+uw/c^2) = 0.862c
if you rearrange this to solve for w (for what reason I have no idea, as the the equation above gives you the answer directly.)
Then you must use
w=(v-u)/(1-uv/c^21)= 0.8c
You reversed the order of u and v in the top part of that equation which resulted in you getting the wrong answer for v.

x = -v
Your scene is: x = (u+w)/(1+uw/c^2) = 0.862c ==> w=(x-u)/(1-ux/c^2)
E---------------------->A-------------------->B-------------------->E
.........u=0.2C...........................w=0.8C....................x=0.86C;
The combination of these three data (0.2, 0.8, 0.86) according to the formula must be right.

My scene is :w=(u+x)/(1+ux/c^21)=0.8C
B---------------------->E--------------------->A--------------------->B
................x=0.714C...................u=0.2C.....................w=0.8C
The combination of these three data (0.714, 0.2, 0.8) according to the formula must be right.
That's why we're different: (x-u)/(1-ux/c^2) != (u+x)/(1+ux/c^21).
Why does this happen? The reason for this difference is that my scene has changed the position of the earth.
Janus's : Earth......A...........B
mine: B..........Earth........A
So we only know the relative velocity between A,B and we don't know the position of the earth, we can't calculate the relative velocity to the earth. Even they're very close.

 

[TonyYuan]

For light path to be significantly bent, it has to pass very near the star and and not "far away". The equation is:
Angle of deflection = 4GM/rc^2
where M is the mass of the object and r is the distance the light passes from its center. The angle is in radians.
With our own Sun a light ray just skimming it surface is deflected by just 1.7 arc-seconds (~ 1/2100 of a degree)
A ray just skimming the surface of the Earth would be deflected by just 0.0006 arc-seconds ( less than 1/6,000,000 of a degree).
So while the gravity of a mass can deflect light, this effect is just too insignificant to account for the effect you are trying to posit.
It like trying to say that it was the collision with a bumble bee during its flight which caused a plane to be 10 min late arriving at the airport. While that collision would have had a very small brief effect on the plane's speed, It would have come nowhere close to producing enough of an effect to make the plane 10 min late.
It's not enough to say "A can effect B, therefore A can cause C", unless you can show that the effect that A has on B is significant enough to produce result C.

Not only that, but you can't just jump from " gravity bends light paths" to " Earth's gravity drags light with it" because one does lead to the other.

Light deflected by just 0.0006 arc-seconds =6*10^-6.
The earth's revolution degree: 360/365*24*60*60 = 1.1415*10^- 5.
They're only twice as close. It can't be ignored!

Now let's analyze Eddington observation.
There are two reasons that can lead to the bending of the object's moving path, one is the effect of the force, resulting in the speed in the vertical direction, the other is the refraction caused by the inhomogeneous medium or field, resulting in the bending of the path. Then the bending of light is probably caused by the latter. The sun is surrounded by a circle of gravitational field, the stronger the gravitational field is. When the light passes through the gravitational field of the sun, the light bends like refraction due to the inhomogeneity of the gravitational field. The bending of this path is different from the bending caused by the direct action of gravity. So it is not suitable to calculate the bending angle of light with the classical theory of flat throw.
https://photos.app.goo.gl/1kPbdLVXdxm2qqVD6
In this paper, we have analyzed that no matter sun moves in the same or opposite direction or at a certain angle, the speed of light relative to sun will not change. In the calculation of the theory of flat throw, because there is an acceleration process under the pull of gravity at the beginning, before the acceleration caused by gravity becomes negative, the light will be more far away from the sun, less affected by gravity, so the deflection angle of light will be smaller. This is the reason why the deflection angle of light calculated by the classical theory of flat throw is too small. There is nothing wrong with Newtonian mechanics, but it was not used correctly that time.
Your data supports the idea that light is trapped by gravity.
The rotation of the earth does not affect the strength of the gravitational field, so we should consider the rotation of the earth around the sun.

I give up looking for breakthroughs in the formula of special relativity. Those are mathematics, there will be no flaws. We focus on the bending of the light.

 

[TonyYuan]

Light deflected by just 0.0006 arc-seconds =1.667*10^-7 degree.
The earth's revolution degree: 360/365*24*60*60 = 1.1415*10^- 5 degree/s.
If the time of light passing through the earth is 0.01 seconds, about 3000 km,the angle of rotation of the earth is 1.1415*10^- 7 degree. It can't be ignored!

Now let's analyze Eddington observation.
There are two reasons that can lead to the bending of the object's moving path, one is the effect of the force, resulting in the speed in the vertical direction, the other is the refraction caused by the inhomogeneous medium or field, resulting in the bending of the path. Then the bending of light is probably caused by the latter. The sun is surrounded by a circle of gravitational field, the stronger the gravitational field is. When the light passes through the gravitational field of the sun, the light bends like refraction due to the inhomogeneity of the gravitational field. The bending of this path is different from the bending caused by the direct action of gravity. So it is not suitable to calculate the bending angle of light with the classical theory of flat throw.
https://photos.app.goo.gl/1kPbdLVXdxm2qqVD6
In this paper, we have analyzed that no matter sun moves in the same or opposite direction or at a certain angle, the speed of light relative to sun will not change. In the calculation of the theory of flat throw, because there is an acceleration process under the pull of gravity at the beginning, before the acceleration caused by gravity becomes negative, the light will be more far away from the sun, less affected by gravity, so the deflection angle of light will be smaller. This is the reason why the deflection angle of light calculated by the classical theory of flat throw is too small. There is nothing wrong with Newtonian mechanics, but it was not used correctly that time.
Your data supports the idea that light is trapped by gravity.
The rotation of the earth does not affect the strength of the gravitational field, so we should consider the rotation of the earth around the sun.

I give up looking for breakthroughs in the formula of special relativity. Those are mathematics, there will be no flaws. We focus on the bending of the light.

Light deflected by just 0.0006 arc-seconds =1.667*10^-7 degree.
The earth's revolution degree: 360/365*24*60*60 = 1.1415*10^- 5 degree/s.
If the time of light passing through the earth is 0.01 seconds, about 3000 km,the angle of rotation of the earth is 1.1415*10^- 7 degree. It can't be ignored!

 

[TonyYuan]

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

A race between Newton and Einstein, who will win?
https://photos.google.com/photo/AF1QipPiogT4OFvz8gAvi-LHHCWNFKLEQUk6IGdGcgEw

 

[Write4U]

Not only that, but you can't just jump from " gravity bends light paths" to " Earth's gravity drags light with it" because one does lead to the other.

The image of a massive object dragging something in a fluid medium presented a strange visual.

Does a wave a boat makes in the water drag any water molecules with it or do the watermolecules stay basically stay in the same place and the boat and the wave roll over the water molecules?

Can the boat push any watermolecules? Let alone pull them?

 

[phyti]

Tony#123;

Define Earth as the ref. frame, speed of A relative to E is a=.2,
speed of B relative to E as b=?,
speed of B relative to A as u=.8.
Solve for b.
One of the restrictions in SR is, you can't add velocity measurements from different frames, since measurements are frame dependent.

Simple addition by E would make b= 1.0, which is illegal.
A form of reverse engineering can done in math.
Using the standard (b-a)/(1-ab), the question is,
what speed b, would result in A's value of u?

1. u = (b-a)/(1-ab)

which yields

2. b = (a+u)/(1+ua) = (.2+.8)/(1+.16) = .86

The revised equation for b, is derived in a more complicated manner in SR texts, and labeled as the 'composition of velocities' from the 1905 paper by Einstein onward. In reality it is just algebra.

Also, no speed determines direction, it is inserted as an independent variable.

 

[TonyYuan]

We are waiting for Janus's answer. He hasn't given an answer yet.

 

[TonyYuan]

I will analyze it according to the specific data given by Janus. See if it's a rigorous calculation and a rigorous logical deduction.The scenario I designed is not complicated, so I believe Janus can give the answer directly.
I also give the calculation of the bending of light. The gravitational field of the earth really affects the bending of light, and this influence is decisive and cannot be ignored.
It is hoped that our discussion will affect the future of physics.So I'll put all my abilities into this discussion. I will be a good match for Janus. Also, for me, he is the best opponent.

 

[(Q)]

I will analyze it according to the specific data given by Janus. See if it's a rigorous calculation and a rigorous logical deduction.The scenario I designed is not complicated, so I believe Janus can give the answer directly.
I also give the calculation of the bending of light. The gravitational field of the earth really affects the bending of light, and this influence is decisive and cannot be ignored.
It is hoped that our discussion will affect the future of physics.So I'll put all my abilities into this discussion. I will be a good match for Janus. Also, for me, he is the best opponent.

Dunning-Kruger.

 

[Write4U]

Dunning-Kruger.

Don't discourage him. IMO, he is making a serious effort and appears to have some knowledge of the subject.
I don't (can't) follow any of the mathematics , but find the narratives interesting.

As to "changing the future of physics" might be a tad optimistic, but we can dream....