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In non-relativistic physics, momentum defined as $p = mv $, so as object velocity increased, then the momentum increased. Or if mass of object increased/decreased, then the momentum too increased or decreased (as in rocket). And this concept very useful when discussing elastec collision between two object to determine how the state of both object after collision.

But in relativistic theory, momentum defined as $$ p = \frac{ m v}{\sqrt{1 - \frac{v^2}{c^2 }} } .$$ In this formulation, $m$ keeps constant, because there is no relativistics mass. So classical approximation for this variable is $$ \frac{ m v}{\sqrt{1 - \frac{v^2}{c^2 }} } \approx m v$$ or $$v_\text{rel} = \frac{ v}{\sqrt{1 - \frac{v^2}{c^2 }} } \approx v ,$$ because $m$ not changed as Einstein say.

So, what is physical meaning of this $v_\text{rel} $. Is it imply that $\textbf{ velocity }$ of object changed when its $\textbf{velocity}$ approach the velocity of light?

Edit: And what constitute $\gamma v $ for the planet orbiting the sun, if the mass stay constant?

Qmechanic
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Mohammad Fajar
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1 Answers1

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Momentum puts a numeric value on inertia in the sense that it represents how much impulse is needed to stop an object (or conversely was needed to get it up to speed from rest in the first place).

For a constant force, impulse is the applied force times the duration of application (i.e. how hard and long you have to push), and because of Newton's second law we see that a net impulse of $J = F \Delta t$ applied to an object of mass $m$ can accelerate it from rest to speed $$v = a\, \Delta t = \frac{F}{m} \Delta t \;. $$ Of course such acceleration is a gradual process, shortly after the initial application of the force the object has a small speed, and the next short period of application adds a little more speed and so on.

The only thing that is different about the relativisitic case is the way in which velocity adds. Adding velocity $u$ to an aboject already going at $v$ doesn't result in a total of $V_\text{classical} = u + v$ as appears to be the case at low relative speeds, but in $$ V_\text{relativisitc} = \frac{u + v}{1 + \frac{uv}{c^2}} .$$ If you apply the correct (relativistic) velocity composition rule to the accelerating object you end with it's velocity asymptotically approaching $c$ rather than growing without bound, but the impulse needed to get it to those speeds continues to be $Ft$ and does increase without bound, so $$ p = \frac{mv}{\sqrt{1 - (v/c)^2}} $$ as you said.

To re-iterate the fact I started with momentum measures the impulse needed to stop an object, and that interpretation is completely consistent in either classical or relativistic physics.

Aside: we measure the relativistic momentum of particles in accelerators all the time (as in it's a work-a-day process) using their bending radius in magnetic fields. Einstein's theory passes this test with flying colors.

dmckee --- ex-moderator kitten
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