Speed Limit

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Stoney3K
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Joined: Sun Jun 28, 2009 9:24 pm

Post by Stoney3K »

chrismb wrote:[So I guess the next question is to understand why/whether you think those 'maximum speeds' (relative to lab-frame, of course) will be the same for two runs in the same accelerator for two beam currents of different intensity.
As far as I understand special relativity, they are not: If a particle beam with mass X travelling at energy Y has a 'maximum speed' of .99c, a beam twice as heavy will have a maximum speed of (.99 * .99)c. It will, however, never reach the speed of light, but as you pump more energy into it, the particles' mass will increase with the same factor.

It was long thought that the only reason photons could travel at the speed of light was because they have zero mass. But if photons were mass-less, they would also have zero impulse and solar sails could not work.
Because we can.

GW Johnson
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Post by GW Johnson »

I was taught that photons are massless, but of nonzero momentum. That's why they do travel at lightspeed, yet can have an impact force against solar sails or other objects. And they do seem to have such impact.
GW Johnson
McGregor, Texas

KitemanSA
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Location: OlyPen WA

Post by KitemanSA »

GW Johnson wrote:I was taught that photons are massless, but of nonzero momentum. That's why they do travel at lightspeed, yet can have an impact force against solar sails or other objects. And they do seem to have such impact.

hanelyp
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Post by hanelyp »

Stoney, I suggest you study up on relativity theory, and a bit of calculus.

gamma = 1/sqrt(1-V^2/C^2)
energy = gamma*M0*C^2
momentum = gamma*M0*V

at light speed gamma = 1/0 = infinity, which with a little calculus reduction with 0 mass gives a finite non zero energy and momentum.

Stoney3K
Posts: 154
Joined: Sun Jun 28, 2009 9:24 pm

Post by Stoney3K »

hanelyp wrote:Stoney, I suggest you study up on relativity theory, and a bit of calculus.

gamma = 1/sqrt(1-V^2/C^2)
energy = gamma*M0*C^2
momentum = gamma*M0*V

at light speed gamma = 1/0 = infinity, which with a little calculus reduction with 0 mass gives a finite non zero energy and momentum.
True, that was the formula I was looking for.

However, if you have a particle beam with a set intensity and travelling at a certain velocity, thus yielding an energy E, doubling that intensity (mass) would only have the same effect as trying to increase the beam's velocity, e.g. adding more energy.

The actual velocity will increase, but to the point that it asymptotically approaches c (due to the gamma correction term) but has no possible way of reaching it.

Whether or not is is possible to exceed c is currently unknown, as someone already pointed out, gamma (and hence the mass) would become imaginary in such a situation. This does not mean it's physically impossible, only that it is currently beyond our observation and therefore also beyond our model of (mathematical) comprehension.

If there was a manner to accelerate a particle beyond the speed of light, it must have some way of dealing with the discontinuity *at* c. What happens beyond that limit is not known, because we have no means of confirming that in an experimental way.

It is a good piece of food for thought if you consider that all of our scientific models, including quantum and particle physics, are based on observation and our own way of implementing observation. Relativity is based on our way of observing by taking light scattered from an object with our eyes, and hereby 'assuming' light is a universal constant with the same properties throughout the entire extent of known space. In quantum physics, Heisenberg's uncertainty principle can only be explained by the fact that we use light to measure and observe a particle.

Science in itself is only assessing the interaction of some subject to other objects or instruments, and pulling stuff apart down to the particle level and studying the interactions between particles. What would happen if we were to consider a particle or object in empty space by itself?
Because we can.

GW Johnson
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Post by GW Johnson »

Think about this: special relativity describes what inertially-moving object A looks like to an static observer in inertial reference frame B. That's almost exactly what Einstein's paper said, except his language was a whole lot more technical.

The operative word in my description is "looks like". Traveling objects do not really foreshorten or get heavier, but they sure look that way to an outsider. On board, it is the universe that "looks" weird. Who is correct? That is the traveling paradox.

The traveling paradox is resolved by who had to do the accelerating in order to come home. He is the one who actually travels into the stationary homebody's future.

The description of what something "looks like" to another observer presumes that the other observer can actually see the traveling object at all. That is the fundamental assumption.

In a mathematical solution to any set of equations, obtaining square root of negative one when the answer was presumed real, simply means one of your fundamental assumptions is incorrect.

And that's what I think the non-real results from the special relativity equations at V> = c truly mean. Your fundamental assumption (that you can actually see the traveling object) is wrong.

That says nothing at all about whether one could actually push an object beyond lightspeed, in my opinion.

It just means navigation at speeds above lightspeed is going to be hell until we can add some new physics theories to this picture.
GW Johnson
McGregor, Texas

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