what would happen? Would time stop? Didn’t Einstein develop several theories on this question?
Your mass would become infinite, your volume would become zero, and
Yes, yes it would.
Sort of. His theory of special relativity gives us the answers to it.
–John
Well, this is all a matter of reference frames. If you were going to speed of light, you would not notice any changes. Your watch would tick like normal, and your rulers would all look the same. However, to an outside observer who is not traveling relative to you, you would appear two-dimensional (flattened perpendicular to your direction of travel) and your time would appear to be frozen.
I don’t believe that mass becomes infinite. If that were the case, then anything that was travelling the speed of light would encompass the entire known universe. Likewise, if your mass is infinite, your volume could not possibly be zero.
Time does not stop. Rather, time slows down for you, relative to the rest of the universe as you approach the speed of light. When you hit c, you can travel anywhere in the universe instantaneously…but time continues on for everyone else. So while it may take you no time to get to alpha centauri at c, it would still be 10 years difference between the time you left and the time you got back and everyone else ages.
So, with this in mind, going back to my first point, I think that rather than your mass becoming infinite, what happens is you have the ability, if you so desire, of being anywhere and everywhere in the universe at the same time (relative to yourself), thus making your mass potentially infinite. To do this, though, would take an infinite amount of time to accomplish for those watching you back on earth.
I could be wrong though, it’s been known to happen once or twice. I think I’m right. Go ahead rocket man, prove me wrong. 
The mass does, in a sense, become infinitely greater than the rest mass. Anything with rest mass travelling the speed of light WOULD encompass the entire universe (in two dimensions at least). Of course it would take infinite energy to achieve this. It wouldn’t even take an infinite amount of time, providing you had an infinite amount of power at your disposal (if you only had a finite amount of power, it would take an infinite amount of time).
These are all just the results of taking the limit of velocity approaching c- all the absurd results just go to show that it’s impossible to achieve.
Note that photons have zero rest mass, thus avoiding the infinite mass problem at lightspeed.
Arjuna34
Electrons have rest mass, though…?
It’s meaningless for something with rest mass to travel exactly at the speed of light, and it’s likewise meaningless for something without rest mass (like light itself) to travel at any other speed (yes, the speed of light in, say, glass, is slower, but that’s because photons are being absorbed and re-emitted by the atoms. In between the atoms, they’re still travelling at c). If you take something with rest mass, like an electron or starship, and accelerate it to some velocity close to (but not quite) c, then from the point of view of someone on the starship, everything on board the ship would look normal, but the rest of the Universe would be squished shorter in the direction of motion, and all the other clocks in the Universe would appear to be running slowly. From the point of view of someone who stayed back on Earth, by contrast, the starship would appear to be squished short, and the starship’s clocks would appear to be slow.
As for saying that the mass increases, that’s just a fudge to make the equations for momentum look the same. It makes more sense to say that the mass stays the same, but re-define relativistic momentum. The total energy, however, does increase without bound as velocity approaches c.
Now you’ve gone and confused me AGAIN.
If your total energy increases without bound, doesn’t your actual mass increase from E=mc[sup]2[/sup] ? After all, you’re packing a lot of K.E. into your spaceship.
If you do this, can’t you increase the mass of an object to such an extent that it’s shwartzchild radius expands to encompass it? Can you, or can you not, make a black hole by accelerating an object arbitrarily close to c? What is the difference between giving an object a REALLY big amount of K.E. and piling a whole lot of matter up in a small space?
You are not changing the amount of matter…there are still the same number of atoms involved as when you started out. Whether or not you are changing the mass is basically a matter of definition, i.e., if you decide you still want the equation for momentum to be p = mv (v being velocity) then you have to say that the mas m increases from the rest mass, but you can also define momentum as p = gammamv where gamma = 1/sqrt(1-v^2/c^2). Then the m in this equation is still the rest mass. Of course, in either case, you recover the equation p = mv with m the rest mass to an excellent approximation for any sort of speeds we normally deal with in our (current) everyday lives.
My belief would be that the Shwarzchild radius would depend on the rest mass, but I could be wrong on that…Once you start dealing with gravity, you are into general relativity and that’s out of my league.
You need to use the rest mass for Schwartzchild radius. Otherwise, you could turn anything into a black hole just by looking at it the wrong way. Say you have an electron in a particle accelerator cranked up to almost the speed of light: From the electron’s point of view, the entire Earth is moving at that speed, and thus would have an enormous kinetic energy. Nevertheless, the electron will not think that the Earth is a black hole, and be trapped: It can, provided that it doesn’t run into anything like the air, escape the planet.
I think you are asking how then, can they travel at the speed of light. The answer is they don’t. It’s just that they are very light and it takes a small amount of energy to accelerate them to speeds approaching the speed of light. They still cannot travel at light speed.
Damn. That’s my Doomsday Device shafted then.
If the good doctor will permit me to expound a bit, are you referring to electrical current? The way you press a switch and the signal travels through the wires at light speed?
The actual electrons don’t move that fast. They move pretty slowly, as a matter of fact. But one bumps into another and another and another, and the push travels at the speed of light. For example, Let’s say you have a million pool balls lined up perfectly in a row, and you shoot your cue ball at the one on the end. even though the balls in the row will not move any more than they normally would, the energy from the cue ball would be transmitted at an (apparently) absurdly high speed through all of them to the millionth ball, which rolls off almost immediately.
Electrical current works the same way. The electrons aren’t moving at c, their energy is.
The good doctor was surely referring to electrons in particle accelerators. A table-top aparatus can get electrons up to 0.6c no problem, and the 27km LEP (Large Electron-Positron collider) at CERN (which runs at 100 GeV, i.e., the electrons and positrons are accelerated through 100 billion volts) gets the electrons and positrons up to 0.999999999987c.
-P
Actually, the energy isn’t moving at c either (although it can be close in some instances) unless it gets radiated as photons (e.g. via an antenna).
Arjuna34
I understood what he said, and I wasn’t arguing with or contradicting him. I was just indicating that I had a different interpretation of the question that was asked. I guess I could have made it a little more clear.
Sorry!
Ah, yes. I incorrectly took the antecedent of “you” to be DrMatrix in the quote:
Sorry 'bout that!
-P
How do you calculate the relative speed between two such particles in a collider? Does one see the other approaching at 0.999999999999999999999999999999c or something like that?
Yes. All you need to do is add the two velocities relativistically. What you are really doing is putting yourself in a frame with one particle at rest (call this frame A) and the other particle moving at you extra fast. The only two things you know are this other particle’s velocity in the lab frame (0.999999999987c) and the velocity of the lab frame itself relative to your frame (frame A) (also 0.999999999987c). Thus, you see the particle moving at the relativistic sum of these two velocities. That is:
v' = answer
v1 = v2 = v = 0.999999999987c
v' = (v1 + v2)/(1 + v1*v2)
= 2*v/(1 + v^2)
It becomes useful to write the velocities as 1 minus a small number and to work just with that small number. Otherwise when you plug it into a calculator (well, mine at least) you just get v’=1 because of rounding. So:
v = 1 - a ==> a = 1.3*10^-11
v' = 1 - a' (must find a' in terms of a...)
A little algebra, and:
a' = a^2/(a^2-2*a+2) = 8.5*10^-23
So, one particle sees the other particle coming at it at (1-a’) = 0.999999999999999999999915c.
-P
You’ll notice that I’ve left off some c’s here and there. That’s just habit, sorry. (c often gets set to 1 to make the math look neater.) I hope it is still clear.
-P