A Physics Quiz of a different type

[the man]

Wouldn't it be slightly slower than c because of the air and lens of the flash light.

[Ether]

Originally Posted by the man

Wouldn't it be slightly slower than c because of the air and lens of the flash light.

I thought Al was in outer space where there is no air.

[the man]

Still has a flash light with a lens, although I suppose he could remove it.

[the man]

Of course you still have the glass of the bulb...

[Ether]

Originally Posted by the man

Of course you still have the glass of the bulb...

The speed of the light is slower only while it is in the glass. Once it leaves passes through the glass it resumes its normal speed.

[the man]

That makes sense.

[EricH]

I'll take a stab at your problem, Ether.

System 1 (XY): The relative speed of the two particles is c/2-(-c/2) [speed of A - speed of B], which results in 2*c/2 or c, relative to each other, as viewed from a neutral position.
System 2 (X'Y'): In this case, the system is fixed onto Particle B in the above problem, which is assumed to be moving at -c/2 (from the reference point in System 1's point of view). The relative speed remains the same, as the motion did not change (if viewed from System 1's reference point), so observers on Particle B see Particle A leaving at speed c.

Which brings up the question: Do observers on Particle B really see Particle A, or does it just vanish as it moves?

There are some related thought experiments, such as the speed of a bullet fired going forwards versus the speed of a bullet fired going backwards (to an outside observer); in Al's particular case, the real question is does an observer in another reference frame witness light traveling at 2*c? (And if so, does that redefine the speed of light in that observing reference frame? )

[Ether]

Originally Posted by EricH

System 2 (X'Y'): In this case, the system is fixed onto Particle B in the above problem, which is assumed to be moving at -c/2 (from the reference point in System 1's point of view). The relative speed remains the same, as the motion did not change (if viewed from System 1's reference point), so observers on Particle B see Particle A leaving at speed c.

Surprisingly, the above answer is not correct. It is the heart of Einstein's theory of relativity and why it was so profound a discovery that changed the face of physics.

An observer sitting on Particle B (i.e., in reference frame X'Y') will observe Particle A's speed to be

(c/2+c/2)/(1+(c/2)(c/2)/c^2) = 0.8*c

[Aren Siekmeier]

Originally Posted by Ether

Surprisingly, the above answer is not correct. It is the heart of Einstein's theory of relativity and why it was so profound a discovery that changed the face of physics.

An observer sitting on Particle B (i.e., in reference frame X'Y') will observe Particle A's speed to be

(c/2)(c/2)/(1+(c/2)(c/2)/c^2) = 0.8*c

^ninja'd (sort of)

And yeah that. That's the velocity transform: u' = (u+v)/(1+uv/c^2), u is the particle's velocity, v is the velocity of the prime frame relative to the non-prime frame.

[Ether]

Originally Posted by compwiztobe

^ninja'd (sort of)

And yeah that. That's the velocity transform: u' = uv/(1+uv/c^2), u is the particle's velocity, v is the velocity of the prime frame relative to the non-prime frame.

There was a typo in my original post.

The relativistic addition of velocities is (u+v)/(1+u*v/c^2).

[Aren Siekmeier]

Heh, yeah, a little dimensional analysis would tell you that much.... I'm tired.