What is the Einstein Velocity Addition Formula and How Does It Work?

[freydawg56]

http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/einvel.html#c1

ok, so I'm looking at this, and i see that
where c = speed of light
if ship A is moving at .7c to the right ===> <=== and ship B moving .7c to the left

and they are both on the same line and about to collide, at what velocity (in C) will they hit each other.

the reference point is the collision point of impact not some distant observer.

Theory i read appears that C + X = C but this seems to violate common sense.
I can even imagine that you can't possible surpass the speed of light, but can someone make this a little more obvious.

Thanks, and this is not a HW question, I'm in EE. I'm just asking for the fun of it.

 

[BAnders1]

Theory i read appears that C + X = C

What is X exactly? By the way, common sense will lead you to dark alleys and beat you to death in physics.

 

[freydawg56]

X is anything positive , so basically C+ 2C = C because i thought the speed of light is the max possible speed.

 

[HallsofIvy]

The problem is that "C+ V" is shorthand for adding the velocities according to Einstein's formula for addition of velocities. If two objects move directily toward one another, with speed u and v relative to some third frame of reference, then the speed of each, in the other's frame of reference ("U+ V") is
\frac{u+ v}{1+ \frac{uv}{c^2}}.


In particular, if one is a light beam, with speed u= c, then its speed ("C+ V"), relative to the other, is
\frac{c+ v}{1+ \frac{cv}{c^2}}= \frac{c+v}{1+ \frac{v}{c}}= \frac{c(c+v)}{c+ v}= c

 

[edpell]

And we should say Lorentz velocities. He did the addition first. Physics is a team sport (sort of).

 

[TMFKAN64]

ok, so I'm looking at this, and i see that
where c = speed of light
if ship A is moving at .7c to the right ===> <=== and ship B moving .7c to the left

and they are both on the same line and about to collide, at what velocity (in C) will they hit each other.

the reference point is the collision point of impact not some distant observer.

In the reference frame where the collision point is stationary, clearly ship A and ship B are approaching each other with a relative velocity 1.4c. This doesn't contradict anything in special relativity though, since there is no single object being measured at a speed greater than c. (And of course if you are in one of the ships, the velocity of the other ship can be determined using Einstein's velocity addition formula, and it will be less than c, as expected.)

 

[TcheQ]

You could (should for simplicity?) say one is moving -.7c and one .7c with respect to the collision point. You would then calculate their adjusted mass+momentum (and resulting change from collision) as the function F=2*f(0.7c), and not F=f(.7c+.7c) as you suggest

 

[E=mc^84]

Just use the relativistic addition of velocities formula and plug in the relative velocities of each reference frame...

 

[Naty1]

Halls of Ivy..glad you posted that formula...I was just thinking earlier today using that would make these repeated discussions on relative speeds easier...and if anybody searches they'll actually be able to do their own computation...

 

[Naty1]

Why isn't the formula and explanation posted in Frequently asked Questions??

I can't find the input page for doing it, but if someone can identify it and it's not alread there, I'll draft a submission...