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

#### TonyYuan

##### Gravitational Fields and Gravitational Waves
Registered Senior Member
Long long ago, there was a story that four spaceships were sent out from the earth to fly in different directions. Their relative speed to the earth was 0.9c. According to the calculation of "special relativity", their time became slower. It's only 0.435 * the time of the earth. t' = t*sqrt*(1-v^2/C^2)

But one day the earth exploded, there was no residue left, so the four spacecraft lost their original reference. They could only make reference to each other, and found that there was a relative speed between them, so they were surprised to find that the time between them was different.

God, without the earth, the time on the spaceship will be different. With the earth, it will be the same again.

Who can explain it?
Thanks very much.

Long long ago, there was a story that four spaceships were sent out from the earth to fly in different directions. Their relative speed to the earth was 0.9c. According to the calculation of "special relativity", their time became slower. It's only 0.435 * the time of the earth. t' = t*sqrt*(1-v^2/C^2)

But one day the earth exploded, there was no residue left, so the four spacecraft lost their original reference. They could only make reference to each other, and found that there was a relative speed between them, so they were surprised to find that the time between them was different.

God, without the earth, the time on the spaceship will be different. With the earth, it will be the same again.

Who can explain it?
Thanks very much.
As long as the spaceships never meet up again, they will all disagree as to how much time has passed for each other. Every ship will claim that the other ships are aging slower. So if ship A and ship B are heading in opposite directions as they leave the Earth, then they will measure their relative speed as being 0.9945c, and each will say that the other ship's clock is ticking 0.105 the rate of his own. This is true even when Earth is in the scenario. In fact, every ship will also say that the Earth is the one aging slower than they are by a factor of 0.435 If the ships all stop, turn around and meet back up back where they started, they then will all say that they aged the same amount upon return, and less than a clock that had been left behind.
The existence or non-existence of the Earth does not come into play at all. Their speeds aren't being measured with respect to the Earth itself, but the inertial frame that the Earth is at rest with respect to. That inertial frame is still valid even if there were no earth at all.

I think the problem here is that you only have a very superficial grasp of Special Relativity. You gave the time dilation formula, but don't actually understand what it really means, how to properly apply it, or the fundamental reasoning behind it. For example, your claim that the ships would all agree on their respective times, as long as the Earth is present.
In truth, after 1 year by his clock, ship A will say that ship B has aged only some 38 days, while ship B will say that after 1 year by his clock, Ship A has only aged 38 days. As measured from the Earth, (or the inertial frame the Earth is at rest with respect to), both ships A and B's clocks read 1 year, at the same time and when the Earth clock reads 2.3 yrs. ( Again, there doesn't actually have to be an Earth there, nor does this clock have to be located at the point where the ships started together, it only has to be at rest with respect to this inertial frame.)
Thus while anyone at rest to the initial reference frame will say that the clocks on ships A and B always read the same, Anyone at rest with respect to either of the ships will say that the ship clocks don't read the same time. This is the relativity of simultaneity in play. And until you come to terms with it, Special Relativity will continue to elude you.

As long as the spaceships never meet up again, they will all disagree as to how much time has passed for each other. Every ship will claim that the other ships are aging slower. So if ship A and ship B are heading in opposite directions as they leave the Earth, then they will measure their relative speed as being 0.9945c, and each will say that the other ship's clock is ticking 0.105 the rate of his own. This is true even when Earth is in the scenario. In fact, every ship will also say that the Earth is the one aging slower than they are by a factor of 0.435 If the ships all stop, turn around and meet back up back where they started, they then will all say that they aged the same amount upon return, and less than a clock that had been left behind.

Let me draw the scene you said.
<--A----------------------------------Earth---------------------------------------B--->
0.9C............................................................................................................0.9C

My question:
Here the relative speed of A to B is 0.9945c. How is it calculated?

Thanks.

Let me draw the scene you said.
<--A----------------------------------Earth---------------------------------------B--->
0.9C............................................................................................................0.9C

My question:
Here the relative speed of A to B is 0.9945c. How is it calculated?

Thanks.
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

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 v( 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

So if the scene is like this:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C
my question：
According to your formula, is the relative speed between A and B also 0.9945c ?

So if the scene is like this:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C
my question：
According to your formula, is the relative speed between A and B also 0.9945c ?
Of course not. Your diagram clearly indicates that A and B are at rest with respect to each other.

This is not affected by the fact that Earth is receding from them at .9c.

BTW, this is consistent with the formula. If you plug 0 (zero) into the formula for their relative velocity, you will find the resullt to be .... 0.

Also BTW, your diagram in post #3 is slightly wrong. Here it is corrected.

Let me draw the scene you said.
<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C

Of course not. Your diagram clearly indicates that A and B are at rest with respect to each other.

This is not affected by the fact that Earth is receding from them at .9c.

BTW, this is consistent with the formula. If you plug 0 (zero) into the formula for their relative velocity, you will find the resullt to be .... 0.

Also BTW, your diagram in post #3 is slightly wrong. Here it is corrected.

Please notice the definition of u, v in the formula.
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 v( the relative velocity between ship B and the Earth as measured by either the Earth or ship B) are both 0.9c

The relative velocity formula starts with one end as the reference, which is A.
So the in the formulat given by Janus, w = (u+v)/(1+uv/c^2), u represents the positive velocity of Earth relative to A, and v is the velocity of B relative to Earth.
In general, A->B->C where v is velocity of B relative to A, u is C relative to B, and w is C relative to A.

So in your example, v is -.9, and u is .9. (.9 - .9) / (1 + uv/c^2) = 0/something = 0, which is indeed the relative velocity between A and B.

are both 0.9c
Velocity has a directional component.
If the velocities of both craft are +.9c, then they are moving in the same direction.

Your diagram shows them moving in opposite directions, so they must have opposite signs.

Also, if you plug them unto the formula, you will get the correct answers.

The relative velocity formula starts with one end as the reference, which is A.
So the in the formulat given by Janus, w = (u+v)/(1+uv/c^2), u represents the positive velocity of Earth relative to A, and v is the velocity of B relative to Earth.
In general, A->B->C where v is velocity of B relative to A, u is C relative to B, and w is C relative to A.

So in your example, v is -.9, and u is .9. (.9 - .9) / (1 + uv/c^2) = 0/something = 0, which is indeed the relative velocity between A and B.

<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C
Your conclusion is that the relative velocity between A and B is 0 ?

<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C
Your conclusion is that the relative velocity between A and B is 0 ?
Dude.

I corrected it by making one of them -.9c

Velocity has a directional component.
If the velocities of both craft are +.9c, then they are moving in the same direction.

Your diagram shows them moving in opposite directions, so they must have opposite signs.

Also, if you plug them unto the formula, you will get the correct answers.

First scene:
<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C

Second scene:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C

First scene:
<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C

Second scene:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C
What do you think?

Your conclusion is that the relative velocity between A and B is 0 ?
In the example where they go in the same direction, yes. Not the original scenario where both u and v are positive.
Remember, one of them is velocity of Earth relative to the stationary thing, which presumably is A.

What do you think?

Do you mean to substitute this formula? w = (u+v)/(1+uv/c^2)

Ok, i see.

Do you mean to substitute this formula? w = (u+v)/(1+uv/c^2)
Yeah. I see the problem. That's actually the formula for adding velocities.
It assumes both objects are moving in a positive direction at different speeds.

i.e Earth..........A>..........B>>

First scene:
<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C
.9945c

Second scene:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C
0

Third scene:
Earth---------------------------------------A--->
.|....................................................0.9C
.|
.|
.|
.|
\/ B 0.9C

3rd case is beyond me. It isn't a simple formula. There are whole papers written on the subject.

3rd case is beyond me. It isn't a simple formula. There are whole papers written on the subject.

I think such a simple scenario is confusing for physicists, enough to explain how difficult special relativity is to deal with complex problems. But this problem is so simple for classical physics.