I recently learned that relativistic mass isn't actually real or popular and that mass is actually invariant and not dependent on velocity. My entire sense of why we couldn't travel faster than light was because I thought that the faster we got the more massive we became. Now I'm lost and trying to understand why we can't travel faster than light.
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1$\begingroup$ Possible duplicate of If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light? $\endgroup$– AccidentalFourierTransformCommented Jan 7, 2018 at 21:25
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$\begingroup$ It's just semantics. Relativistic mass is passe, but not wrong. Different definitions change the way the equations look, but not any of the predictions. $\endgroup$– Ben51Commented Jan 7, 2018 at 21:33
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$\begingroup$ @Ben Perhpas not exactly wrong, but not exactly right, either. If you apply a force in the direction of motion, the object's mass will appear to be the relativistic mass. If you apply a force at right angles to the direction of motion the object's mass will appear to be the rest mass. Which is it? The best solution, perhaps, is to abandon the concept of relativistic mass. The theory works fine without it. $\endgroup$– garypCommented Jan 7, 2018 at 23:21
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$\begingroup$ I don't disagree. Just pointing out that people who used to do it that way weren't doing physics wrong. $\endgroup$– Ben51Commented Jan 7, 2018 at 23:59
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1 Answer
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Well:
$E = \gamma m_0 c^2,$
regardless. $m_0$ is the rest mass and $\gamma$ is the Lorentz factor. So, in the early days, people said let's have a relativistic mass:
$$ m = \gamma m_0, $$
so as to preserve the legendary:
$$ E= mc^2$$
at all velocities (an other formulae), not just at $v=0$. That's less attractive than working with invariants of the Lorentz transformation, such as:
$$ E^2-(cp)^2 = (m_0c^2)^2 $$.
At this point, it's safe to drop the $0$-subscript and consider mass to always be rest mass.
In summary, the difference is where you put the $\gamma$: in the so-called relativistic mass, or explicitly in the formula for total energy. Either way, total energy diverges at $v=c$.
