relativistic and not relativistic motions


by bernhard.rothenstein
Tags: motions, relativistic
bernhard.rothenstein
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#1
Aug6-06, 10:15 PM
P: 997
We say that the uniformly accelerated motion x=gtt/2 is not a relativistic motion because after a sufficiently long time of motion v=gt can exceed c. we say that x=cc/g(coshgt'/c-1) is a relativistic motion because the velocity of the motion it describes never becomes c. Do you know other such "relativistic motions?"
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Meir Achuz
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#2
Aug7-06, 08:40 AM
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Just integrate dp/dt=f(t) for any function f(t), and you will have a
"relativistic motion". If you want x(t), just find v=dx/dt from
v=p/\sqrt{p^2+m^2}, and integrate.
pervect
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#3
Aug7-06, 03:02 PM
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Alternately, given any function v(t) < c, one can compute the acceleration required to cause the specified motion. The only thing "special" about special relativistic motion is that |v(t)| < 1. One can also show that the rate of change of momentum with respect to time becomes infinite as v->c, i.e.

[tex]
\frac{dp}{dt} = \frac{dp}{dv} \frac{dv}{dt} = \frac{m}{{\left( 1 - \frac{v^2}{c^2} \right)} ^ \frac{3}{2}} \frac{dv} {dt}
[/tex]

Thus as v->c, dp/dt becomes infinite.

One does not really need dynamics to see this, the fact is that if one adds together any number of velocities less than 'c' using the SR velocity addition formula, one gets a resultant velocity less than 'c'.

The process of accelerating is just a process of "adding to" one's original velocity. One must use the SR form of the velocity additon law.

Delta-v = a * delta t

is true only in the objects rest frame, the SR velocity additon formula converts the delta-v in the objects rest frame into the delta-v in the coordinate frame.


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