A Physics Quiz of a different type

[Al Skierkiewicz]

This brings up an interesting situation. If you were traveling at the speed of light with a flashlight in your hand... When you turned it on would no light be emitted? If you are at half the speed of light and turn it on, does the light leave at half the speed of light?

[Ether]

See answers in embedded in red below:

Originally Posted by Al Skierkiewicz

This brings up an interesting situation. If you were traveling at the speed of light with a flashlight in your hand... When you turned it on would no light be emitted? Light would be emitted at speed c, in your reference frame. If you are at half the speed of light and turn it on, does the light leave at half the speed of light? No. The light leaves at speed c, in your reference frame.

The answer to your question hinges on the same principle as the answer to the question posted here one hour previously.

[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]

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.

[nssheepster]

Okay, are you ready?
If you go at the speed of light, assuming you can in the first place, you will either dissolve at a sub-atomic level instantaneously, or merge into light.

Brings a new meaning to "don't go into the light...", doesn't it?

[Ether]

Originally Posted by nssheepster

Okay, are you ready?
If you go at the speed of light, assuming you can in the first place, you will either dissolve at a sub-atomic level instantaneously, or merge into light.

Brings a new meaning to "don't go into the light...", doesn't it?

Sitting in your chair looking at your computer monitor, you are traveling at the speed of light (with respect to the photons coming off your monitor).

Have you dissolved yet ?