I would like some clarification on the behavior of relativistic mass as it approaches the speed of light. I am a high school student with an interest in physics so chances are I will have a hard time understanding any formal math in special relativity but I appreciate it anyway.
In special relativity, an object at any non-zero velocity (within the universal speed limit) experiences a length contraction. I would like to know how a mass behaves when an object approaches high speeds, as relativistic mass increases and the length decreases (contracts). Thanks!
Best Answer
This isn't actually correct. The object does not experience length contraction since the object is at rest with respect to itself.
It is correct to say that, in an inertial reference frame (IRF) in which the object is uniformly moving, the observed length, in the direction of the motion, will be contracted from the length in the IRF in which the object is at rest.
But the object does not experience length contraction since uniform motion is relative. There are an infinity of relatively moving IRFs in which the object is in relative motion and each one observes a different length contraction.
Likewise, a mass is at rest with respect to itself. In an IRF in which the mass is uniformly moving, the total energy of the mass is given by
$$E = \sqrt{(pc)^2 + (mc^2)^2} = \gamma mc^2$$
where
$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
In the reference frame in which the momentum is zero (particle is at rest), this simplifies to
$$E = mc^2$$
That's really all there is to it. The invariant mass $m$ is the same for all observers.
There is a notion of relativistic mass which is just $\gamma m$ but some question if this notion is useful. From the Wikipedia article "Mass in special relativity":
At any rate, as the relative speed $v$ approaches $c$, the relativistic mass $\gamma m$ goes to infinity since, as you can see from the expression for $\gamma$, the denominator approaches zero.