## Relativity question from a physics No0b.

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FrankManic
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### Relativity question from a physics No0b.

Okay.... So.

We have two objects. by executive fiat we accelerate each mass instantly to .6C, with each mass moving directly away from each other. So... It seems to me that the two masses are traveling at 1.2C relative to each other. Which seems really wonky.

Basically, I'm falling down at the relativity part. As far as we know nothing can travel faster than C, the speed of light. But... if we've got two things moving away from each other at more than .5 C you could, at least I think you could, shoot a photon from object A towards object B, and as long as object B didn't slow down the photon would never reach it...

So I'm asking you to beat down my thought experiment with your sweet, sweet physics loving. Please try to keep it comprehensible to someone who barely struggled through Algebra II.

Woofsie
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### Re: Relativity question from a physics No0b.

As far as I know, velocities stop adding that way when you get close to c. So their velocity relative to each other would be more like .9c (or something).

FrankManic
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### Re: Relativity question from a physics No0b.

Woofsie wrote:As far as I know, velocities stop adding that way when you get close to c. So their velocity relative to each other would be more like .9c (or something).

Aggghh!

Sometimes, I think my life would be much, much easier if I could really understand relativity. Usually after I've been drinking heavily.

Seraph
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### Re: Relativity question from a physics No0b.

The forumula you use to add velocities is:
$u+v \over 1+(u/c)*(v/c)$
where u and v are the velocities of your two objects.

So for the u=v=0.6 you would get something close to 0.88c.

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### Re: Relativity question from a physics No0b.

What Seraph said.

That is actually the true way to add any velocities together - the simple, linear version is an approximation.

If you could measure with infinite precision, you would find that the relative velocity difference between you moving at 10 m/s and a ball moving away at -10 m/s wouldn't be 20 m/s, but would have a small relativistic correction.

In reality, this correction is really, really small until things start going a significant percentage of c, so you don't need to worry.

ThinkerEmeritus
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### Re: Relativity question from a physics No0b.

When dealing with special relativity, it is unbelievably important to keep track of who is measuring what velocity. When you set two balls off in opposite directions, each with a speed 0.6c, you mean that you measure their speeds to be 0.6c in opposite directions. Then their relative velocity as measured by you is 1.2c. The relative velocity means that the distance between the two balls as measured by you is increasing at a rate of 1.2c (furlongs per fortnight or whatever units you prefer). Now if there is a measuring instrument on one of the balls, it will measure the velocity of the other ball as given by the formula Seraph quoted, so to the measuring instrument the distance between the balls is increasing at some rate less than c.

Just for completeness, the measuring instrument finds that you are moving away at 0.6c, just as either of you would expect.

The annoying innovation of special relativity is that there are three velocities at which the balls are moving apart, one as measured by you and two more as measured from each of the balls. The latter two are +0.88c and -0.88c, differing only by direction in this case because the velocities of the balls as measured by you differ in direction but not in magnitude.
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gmalivuk
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### Re: Relativity question from a physics No0b.

In addition to any number of threads that already exist about relativity, there's also the stickied one at the top of the science forum for common questions, which specifically mentions relativity. Subsequent discussion of this can go there.
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