If an object's mass, when measured in its frame of reference (even if that frame is moving), is invariant, then why does it require more and more energy to accelerate it? I can understand this when measured by an observer in a reference frame in relative motion to it, they would measure the mass as increased but to the moving object, there would be no change. I'm running myself in circles a little bit here but maybe someone can help?
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asked Oct 29, 2018 at 0:39
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3 Answers
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One needs to carefully distinguish between coordinate acceleration (which is frame dependent) and proper acceleration (which is frame invariant).
Coordinate acceleration is the 2nd derivative with respect to time of the object's position in some (inertial) coordinate system.
Proper acceleration is essentially the acceleration measured by an ideal accelerometer attached to the object - all observers agree on what the accelerometer reads at some event.
In principle, an object can have constant proper acceleration of say, $1\,g$. However, it is not possible in principle for an object to have constant coordinate acceleration since that implies the object eventually reaches and then exceeds $c$.
Thus, an object with constant proper acceleration is observed to have arbitrarily small coordinate acceleration as the object's speed gets arbitrarily close to $c$.
As an aside, the notion of a relativistic mass that is speed dependent is generally considered outdated. It's certainly not necessary.
answered Oct 29, 2018 at 1:39
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1$\begingroup$ It's not only unnecessary to use, but, if used at all, also insufficient alone. One would also need the longitudinal mass, as well as masses for all other angles between the accelerating force and the body's velocity. So much hassle for so little gain. $\endgroup$– Joker_vDCommented Oct 29, 2018 at 9:11
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(This answer elaborates on the footnote in Alfred Centauri's answer.)
This is partly a matter of definition. Sometimes the same word "mass" is used with a different meaning, and I think this difference in language may be contributing to the confusion. There are two different quantities associated with the word "mass":
A quantity that physicists usually call "mass", denoted $m$, that is an intrinsic property of the object and does not depend on how it is moving. (I haven't actually conducted a survey of professional physicists, but in my experience, this is how they usually use the word.)
A synonym for the object's energy $E$, but expressed in mass-like units as $E/c^2$. This is sometimes called the object's "relativistic mass", denoted $m_R$, and it does depend on how the object is moving (because the object's energy does).
When the object is not moving, these two different quantities are equal to each other: $m_R=m$. When the "relativistic mass" language is used, the quantity $m$ (which physicists usually just call "mass") is called "rest mass."
Here's the same answer again, with equations to help clarify things:
The energy $E$, momentum $p$, speed $v$, and mass $m$ of an object are related to each other according to these equations: $$ E^2 - (pc)^2 = (mc^2)^2 \hskip2cm v = \frac{pc^2}{E} $$ where $c$ is the speed of light. The $m$ in the first equation is what physicists usually mean when they use the word "mass". It is an intrinsic property of the object and does not depend on the object's speed. The object's energy $E$ and momentum $p$ do depend on the speed, and they do so in such a way that the combination $E^2-(pc)^2$ does not depend on the speed. That's why this particular combination is interesting, and that's why the $m$ on the right-hand side of the equation deserves a special name: mass.
To relate this to the "relativistic mass" $m_R$ (which, again, is not used by the majority of physicists in my experience), re-arrange the second equation shown above to get $$ p = \frac{E}{c^2}v. $$ If we use $m_R$ as an abbreviation for $E/c^2$, then this becomes $$ p = m_R v, $$ which looks superficially like the more familiar low-speed approximation $p=mv$. This resemblance might be comforting to some, but it is also misleading, because the energy $E$ (and therefore $m_R$) is a function of $v$. The momentum $p$ is not really proportional to the velocity $v$ (except approximately when $v\ll c$).
answered Oct 29, 2018 at 1:58
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Power plant producing electric power for relativistic trams says: Trams have large longitudinal relativistic masses, that's why it takes a lot of energy to accelerate a tram.
This is based on Doppler shift of EM-waves and the fact that said waves have momentum:
A tram moving away from the power plant says: I am not receiving much energy from those overhead AC-lines, the frequency is low and so is voltage. But the momentum of the energy is pushing me forwards with a small force.
A tram moving towards the power plant says: I am receiving lot of energy from those overhead AC-lines, the frequency is extra high and so is voltage. But the momentum of the energy is pushing me backwards with a large force.
(Energy flows in the overhead line and tram moves either to the upstream direction, "towards the plant" or to the downstream direction, "away from the plant")
