Re: A Physics Quiz of a different type
 

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.

Re: A Physics Quiz of a different type
 

That makes sense.

Re: A Physics Quiz of a different type
 

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? :ahh:)

Re: A Physics Quiz of a different type
 

Special relativity all comes from Einstein's two postulates:
1. The laws of physics are the same in all inertial reference frames (keeping this around from Galilean relativity).
2. Light travels at the same speed in all reference frames.

Because all of the math and conclusions about various relativistic phenomena are drawn from these, this is the two sentence answer sought by the OP.

msimon's original Wikipedia quote basically sums up the theoretical implications of superluminal speeds and why it is generally dismissed as impossible. For more blood and gore, read on.

To Al, because light travels at the same speed (in vacuum) in all reference frames, no matter how fast you are going, light will always emanate from a source you are carrying in all directions at the speed of light. To a stationary observer you are passing, the same is true. This is related to the fact that light has no medium (there is no ether), so an source's motion cannot be measured relative to anything meaningful. This was observed in the Michelson-Morley experiment and was a driving force behind Einstein's insight.

To Ether, a Lorentz transform will spit out the answer. Let's establish two events in space time (unprimed coordinates are in frame XY, primed coordinates in frame X'Y'):
Event 1 with both particles at the origin, x1 = x1' = ct1 = ct1' = 0.
Event 2 with particle A after time T in frame XY, x2 = c/2*T, ct2 = cT, x2' = ?, ct2' = ?
Frame X'Y' is moving with velocity c/2 to the left so B = -0.5 (B = v/c), y = 1/sqrt(1-B^2) = sqrt(4/3) (Beta and gamma are usually the letters used here but B and y will suffice).
The Lorentz transform (which one can derive from the mathematical and physical properties of Einstein's postulates) is:
x' = y*x - yB * ct
ct' = -yB * x + y * ct
So plug in above and x2' = y*cT* (1/2-B) = ycT, ct2' = y*cT * (-B/2 + 1) = 5/4*ycT.
Particle A's velocity in frame X'Y', then, is (x2'-x1')/(t2'-t1') = 4/5 c.
NOT c as one would expect from Galilean relativity. One could also apply the canned velocity transform (based on this) and get the same result.

The increasing mass concept is something I feel is not quite accurate. It exists to reconcile conceptually whats going on with Newtonian physics. Really, the momentum that is conserved is with respect to a proper velocity (because no one can agree on a common clock anymore), and this screws up all our nice F = ma tools (but F is still dp/dt !). As you go faster and faster, space-time is distorted such that the extra acceleration is accounting for more and more energy. Calculating the Energy of a relativistic particle (again from Lorentz transforms) gives K= (y-1)mc^2, where m is always just the rest mass, and the quickly diverging behavior is due to y as opposed to m getting bigger. E = mc^2 is just a statement about the rest mass of particles, since when v = 0, y = 1, so K = 0. The total energy is E = ymc^2.

Well that was fun.

Re: A Physics Quiz of a different type
 

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

Re: A Physics Quiz of a different type
 

^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.

Re: A Physics Quiz of a different type
 

There was a typo in my original post.

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

Re: A Physics Quiz of a different type
 

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

Re: A Physics Quiz of a different type
 

OK, so because E=ymC^2 (y being gamma), and because gamma approaches infinity as v approaches C, we end up where you need an infinite amount of energy to accelerate just past C.

As for Al's' thing with the flashlights, an observer standing still would see a solid beam of light, just hanging there in the air, like an infinite, glowing sausage.*



*Not really, but it's cool to think about that, no?

Re: A Physics Quiz of a different type
 

Just equal to c, for any object with non-zero rest mass.

Re: A Physics Quiz of a different type
 

Well, wouldn't you be able to reach c, but not exceed it?
That means you CAN reach c. Going past c is what requires infinite energy, reaching c needs just a smidgen less than infinity.

And yes, we're assuming the object rest mass is nonzero

Re: A Physics Quiz of a different type
 

As v -> c, y -> inf, so we cannot even get to c because the "infinite energy" is when you reach c.

At v=c, B=v/c=1, so y=1/sqrt(1-B^2) is already undefined (or infinity, if that's how you want to define dividing by zero).

Re: A Physics Quiz of a different type
 

Notes embedded in red below

I see Aren beat me to it.

Re: A Physics Quiz of a different type
 

Its an asymptote.

Re: A Physics Quiz of a different type
 

Specifically, a vertical asymptote.