The Fifth Scene Of Special Relativity

TonyYuan

TonyYuan

Gravitational Fields and Gravitational Waves
Registered Senior Member
The fifth scene: 2020.3.9
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.

------->velocity positive direction
Earth ……………………………….A…………………………………….B
……………………………………....u...…………………........…......…..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

Presume: u=0.2C, w=0.8C
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.
you can seee that v and u are in the different direction.But A sees B moving away at 0.8C, what happened?

Presume: 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.

You can see different u and different w, the direction of v is different, which is very interesting.
Deduction:
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.
....if w=0.2C, then x=0, v=0;
It is so interesting, the value of w will determine the direction of v.

???
We need to think: is this a Digital Game?
 
Last edited:
TonyYuan said:
The fifth scene: 2020.3.9
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.

------->velocity positive direction
Earth ………….A…………….B
. . .
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
Click to expand...
Well, that's considerably more clear than the first attempt at describing it, which never said what relation w represented, which is why you got so many different answers. BTW, x = -w above. Not sure why that needs to be a separate variable.
Presume: u=0.2C, w=0.8C
w = (u-v)/(1-uv/c^2) = (0.2C - v)/(1-0.2C*v/C^2) = 0.8C
Click to expand...
Two problems.
1) You got the formula wrong. It starts with u+v, not u-v.
2) You named your velocities the same names as the symbols used in the formula, tempting you to plug them into inappropriate places. And sure enough, you do this. What you're trying to do (solve for v) should work, but you get the signs wrong. All velocities are positive in this example.

To compute B relative to Earth: (0.2c + 0.8c) / (1 - (0.2c * 0.8c)/c^2) = 0.862c

You can check that substituting 0.862 for v and it works, except for you getting the formula wrong.

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.
Click to expand...
See? You got the wrong answer from getting the formula wrong.

you can seee that v and u are in the different direction.But A sees B moving away at 0.8C, what happened?
 
Last edited:
Halc said:
Well, that's considerably more clear than the first attempt at describing it, which never said what relation w represented, which is why you got so many different answers. BTW, x = -w above. Not sure why that needs to be a separate variable.

Two problems.
1) You got the formula wrong. It starts with u+v, not u-v. //u-v is right. See Janus's post. if u and v have the same direction, you know w=(u-v)/(1-uv/c^2)=0
2) You named your velocities the same names as the symbols used in the formula, tempting you to plug them into inappropriate places. And sure enough, you do this. What you're trying to do (solve for v) should work, but you get the signs wrong. All velocities are positive in this example.

To compute B relative to Earth: (0.2c + 0.8c) / (1 - (0.2c * 0.8c)/c^2) = 0.862c //0.8C is the value of w, it is the velocity of B relative to the A.

You can check that substituting 0.862 for v and it works, except for you getting the formula wrong.
See? You got the wrong answer from getting the formula wrong.
you can seee that v and u are in the different direction.But A sees B moving away at 0.8C, what happened?
Click to expand...
Halc, I have pointed out the mistake to you.
 
TonyYuan said:
Halc, I have pointed out the mistake to you.
...
//u-v is right. See Janus's post. if u and v have the same direction, you know w=(u-v)/(1-uv/c^2)=0
Click to expand...
Janus was commenting on an example when one of the velocities was negative, and thus subtracted a positive speed as a shorthand for adding a negative velocity. Your example here has all positive velocities, the way I did it, but doing it the way you are doing it (assigning 0.8 to w and solving for v) requires that u be a negative value since Earth relative to A is a negative velocity. Both should yield the same answer.

In a scenario where there are objects A,B,C, to compute the relative velocity of C relative to A given the other two velocities, use the formula I gave and let u = velocity of B relative to A, and v be the velocity of C relative to B.
That's it.

If you do that in the example with Janus that you reference, you wanted to compute C relative to A where Earth is B. Both A and C are positive velocities relative to Earth, so plugging in the first term (B relative to A) yields a negative number, which is why Janus subtracted a positive number instead.

Your answer doesn't make sense. You seem to conclude that B must be moving to the left, which is silly. That's a pretty good indication that you're not done it correctly.
 
In addition to what Halc and Janus (in a different thread) have already pointed out, another thing that makes TonyYuan's "scenes" difficult to parse is that there are two versions of the velocity composition equation, the regular one, and the inverse:

The velocity composition equation:
u = (v + u') / (1 + (vu'/c²))

The inverse velocity composition equation:
u' = (u - v) / (1 - (uv/c²))

Where:
v = the velocity of B relative to A, (and A is taken to be stationary)
u' = the velocity of C relative to B, (and B is taken to be stationary)
u = the velocity of C relative to A, (and A is taken to be stationary)

Notice that both equations share the same variables and their meanings, but that they are arranged differently. In other words, one cannot simply change the signs in one equation to get the other. When I asked TonyYuan why he changed the signs, he never explained that he was using the inverse equation. He just referred me to a post by Janus which used the regular equation, not the inverse!
 
TonyYuan said:
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

Presume: u=0.2C, w=0.8C
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.
you can seee that v and u are in the different direction.But A sees B moving away at 0.8C, what happened?
Click to expand...

Okay, so (u) is the velocity of A relative to E, (v) is the velocity of B relative to E, (w) is the velocity of B relative to A, and I am to presume: (u)=0.2c and (w)=0.8c then solve for (v).

The velocity composition equation:
u = (v + u') / (1 + (vu'/c²))

The inverse velocity composition equation:
u' = (u - v) / (1 - (uv/c²))

Where:
v = the velocity of A relative to E, (and E is taken to be stationary)
u' = the velocity of B relative to A, (and A is taken to be stationary)
u = the velocity of B relative to E, (and E is taken to be stationary)

So (u) is actually v:
v = 0.2c

And (v) is actually u:
solve for u

And (w) is actually u':
u' = 0.8c

That means I want to use this equation:
u = (v + u') / (1 + (vu'/c²))
u = (0.2 + 0.8) / (1 + (0.2*0.8))
u = 1.0 / (1 + .16)
u = 1.0 / 1.16
u = 0.862c
 
Neddy Bate said:
The velocity composition equation:
u = (v + u') / (1 + (vu'/c²))

The inverse velocity composition equation:
u' = (u - v) / (1 - (uv/c²))

Where:
v = the velocity of B relative to A, (and A is taken to be stationary)
u' = the velocity of C relative to B, (and B is taken to be stationary)
u = the velocity of C relative to A, (and A is taken to be stationary)
Click to expand...

I just realised there is also another inverse velocity composition equation:
v = (u - u') / (1 - (uu'/c²))
 
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.:)

I have replied on the previous thread. Thanks, Janus teach me much. He is a great scholar.
 
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