Suppose one were to an accelerate a bullet to 99.9999999999999999999% C. In essence, to have accelerated it with an enormous energy so that it's relativistic mass is tremendous. Now, suppose one finally directs that bullet to a bulls-eye target on the Earth. If relativistic mass is frame-dependent, which mass would count for what happens when that bullet strikes the target? For ultimately, there can be one result, no? Accodring to the Relativity Calculator at http://www.stardestroyer.net/Empire/Science/Relativity.html A 1kg object accelerated to 299999999.99999999 metres per second, would have a relativistic mass of 3.355E+7, a time dilation factor of 0 (actually .000000000001 or something but it rounds off), a Newtonian Kinetic Energy of 4.500E+16j, a relativistic KE of 3.020E+24j, a Newtonian Momentum of 3.000E+8 (kg·m/s), and a relativistic momentum of 1.007E+16 ( kg·m/s).
Well, relativistic momentum would be conserved in the collision between the bullet and the target, so if you want to you can say that relativistic mass is what counts, but the issue is somewhat semantic. The relativistic momentum of the bullet is: p = gmv where g is the relativistic Lorentz factor, m is the rest mass and v is the velocity. You can read this expression in one of two ways. The first way is: p = g(mv), in which the Newtonian momentum mv is multiplied by the Lorentz factor. In that case, rest mass m "counts", but to do the calculation properly you need to factor in relativistic effects. Alternatively, you can read the equation as: p = (gm)v = Mv, in which the rest mass is multiplied by the Lorentz factor to give the relativistic mass, and that multiplies the velocity. (The definition of relativistic mass is M = gm.) Either way, the calculation you need to perform is the same.
James R.: Is not the relativistic momentum itself frame-dependent? And would not that result then in prior to the impact, an observer riding the bullet and an observer on the target disagreeing on the momentum of the bullet?
To back up James R using alternative phraseology: If you're on earth you might calculate the relativistic mass of the bullet. If you're riding the bullet you might calculate the relativistic mass of the earth. Neither viewpoint is ideal. It's like one "frame" is looking at the event from one end, and the other is looking at the event from the other end. What you need to try and do, is mentally step out of the frames and look at the event as a whole. You've got a mass m moving at velocity v with respect to another mass M, and afterwards you've got a masses m1, m2, m3, etc moving at velocity v1, v2, v3 etc in relation to each other. KABOOM!
So basically: It does not matter whether one considers the bullet or the Earth moving towards one another, in either case the resulting explosion will be the same? Why then is relativistic mass and real mass different? Or is relativistic mass then simply a matter of momentum without any consideration besides that?
Prince_James: Yes, relativistic momentum (in fact, even non-relativistic momentum) is frame dependent. It doesn't matter. As long as you do your calculations in one frame of reference and don't incorrectly mix frames, you'll get an answer valid for that particular frame. You can then transform that answer for any other frame, as required. So basically: It depends what you mean by "the same". By "real mass", I assume you mean rest mass. They are different because they are defined differently. Rest mass is a relativistic invariant scalar. Relativistic mass is not. You're probably getting a hint of why physicists don't like relativistic mass very much. The problem is that the concept of relativistic mass actually mixes up information about mass and velocity, whereas when you use rest mass this doesn't happen.
James R.: How might one transform the answer to another frame in such a case as above? The resulting release of energy will be equal in either case. Understood. Gotcha.
How much energy does it take to accellerate a bullet to this velocity? How much energy does the universe have in total? If the bullet has all the energy then where oh where is the target?:bugeye:
Quantum Quack: It wouldn't take that much energy on a universal scale. A lot of energy, yes, but one star could easily provide it given enough time.
Prince James and James R: there's been a lot of debate about rest mass and relativistic mass. The subject gets pretty deep, so I won't dive into it here.
I was just responding with the obviously misguided notion that it takes an infinite amount of energy to accellerate an object of mass to 'c' therefore when doing so the object has an infinite amount of mass at 'c'. And that it would take an infinite amount of time to do so. Considering that we currently think that the universe is "finite" in both mass and time this would indicate that to accellerate an object to 'c' would suck up all the available resources , mass and time that the universe has and still not achieve 'c' for our object because the disparity between notions of "finite" and "infinite". The target in your gedanken becomes a part of the resources necessary to accellerate the object so therefore the target as such is the "bullet" Certainly I have seen this discussed before by others here at sci forums and maybe it is utter bullsh*t.....but thought it humorous to post anyway...hope you don't mind...
Not at all, although I think it is rather well established. E = MC2 predicts states that energy and matter are the same. Energy is purey released from matter via annihilation and matter is purely attained from energy by accelerating an object at mass towards C.
Hi QQ, To accelerate a 1kg mass to 99.9999999999999999999% c takes about: 2 x 10^37 joules, or the energy output from the Sun for 1700 years, or the energy output of Quasar 3C 273 in one fortieth of a second, or One ten-thousandth the energy released in a supernova the energy equivalent of the mass of all Earth's oceans. I'm not certain of those numbers, they were pretty quick calculations. Hi James, Sounds like a good exercise. I'm not so hot at transforming momentum and kinetic energy in SR, so I'll give it a bash. Will post results later. In the meantime, try analysing a non-relativistic collision in different reference frames: A 5000kg truck is moving at 10m/s along the road. A 1000kg car is moving at 20m/s in the opposite direction. They collide head on at those speeds, sticking together after the collision. Find the energy released in the collision. Repeat in the car's pre-collision rest frame. Repeat in the truck's pre-collision rest frame. Repeat in the car-truck post-collision rest frame. The results should be the same in each reference frame. Have you done this sort of problem before, or would you like some help?
