Does relativistic mass phenomena only appear while accelerating or even when the object is travelling at constant velocity (say 90% speed of light)?

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4Can you explain what you mean by "relative mass phenomena"? – probably_someone Jun 14 '19 at 13:08
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15Briefly, it only depends on the velocity, not how you reach that velocity. Modern treatments of relativity avoid the concept of relativistic mass because it can be confusing and misleading. See https://physics.stackexchange.com/q/133376/ – PM 2Ring Jun 14 '19 at 13:09
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6To add to @PM2Ring's comment, relativistic mass is literally just another name for the total energy of a particle. – probably_someone Jun 14 '19 at 15:42
2 Answers
The formula for relativistic mass is $$ M_r = \frac{M_0}{\sqrt{(1 - v^2 / c^2)}}. $$ where $$ \begin{align} M_r &= \text{relativistic mass}\\ M_0 &= \text{rest mass}\\ v &= \text{velocity}\\ c &= \text{speed of light} \end{align} $$ So the relativistic mass is affected only by the speed.
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8True, but friends don't let friends use relativistic mass. ;) BTW, you can use the MathJax dialect of $\LaTeX$ here, eg $m_r=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ – PM 2Ring Jun 14 '19 at 14:31
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I tried to google that real quick, but didn't that it was MathJax, thanks. I will edit it now. Edit: someone beat me too it. – Dr_Bunsen Jun 14 '19 at 14:33
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Agreed that it's an outdated concept, but it probably doesn't deserve to be downvoted since this directly answers the question in the OP. – jacob1729 Jun 14 '19 at 15:57
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1@jacob1729 but it does promulgate the completely outdated notion which is something we ought to discourage, no? – Kyle Kanos Jun 14 '19 at 17:33
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@KyleKanos perhaps. I think this comment thread does that well enough, whilst I tend to assume very downvoted answers are factually wrong. – jacob1729 Jun 14 '19 at 17:58
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2@jacob1729 I agree with Kyle that we should discourage the use of relativistic mass, but I wouldn't downvote an answer that uses it in response to a question like this that directly asks about relativistic mass (assuming the answer is technically correct). However, I do expect such answers to at least mention that relativistic mass is a deprecated concept. – PM 2Ring Jun 17 '19 at 10:14
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I am not that literate in special relativity, and I didn't know it was an outdated concept. Any recommended literature to update my knowledge? – Dr_Bunsen Jun 17 '19 at 13:43
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1@Dr_Bunsen basically any SR textbook made after the 1950s. Surely any modern treatment (say last 20 years). – Kyle Kanos Jun 18 '19 at 11:42
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Hmm, I picked up my old textbooks, and I could find relativistic, and this exact equation. But reading on the internet, I found that only rest mass is still used, and the gamma factor is added to the momentum formula. Time to start writing in my textbook ;) – Dr_Bunsen Jun 18 '19 at 14:46
There is no relativistic mass effect, it came into existence because of inability of an object to gain unlimited speed on applying constant force for unlimited time.
If a constant force $F$ is applied on an object then it causes change in state of motion, momentum. This is good for very short impulse in time but for further duration of time, the present momentum as inertia of an object opposes further change in momentum. Let above force applied for some distance for given time and gain possible maximum speed having momentum $p_0$, then equation of motion is, $$\frac{dp}{dt}+\alpha p=F$$ $$\implies \frac{dp}{dn}=p_0\left(1-\frac{p}{p_0}\right)\tag1$$where $p_0$ is maximum momentum an object can have or say it is momentum gain at first step of impulse where no inertia due to momentum for given force, $dn$ is number of steps or interval of time for force is applied and given by,$$dn=\frac{t}{t_0}=\frac{dn}{dP}\frac{dP}{dE}dE=\frac{dE}{E_0}$$where $t_0$ is time to produce momentum $p_0$ from force $F$, $P$ is rate of energy conversion, $\frac{1}{E_0}$ or $\frac{dn}{dP}$ is minimum amount of energy transferred to produce momentum $p_0$, its thermal equivalent is $\frac{1}{kT}$ and electric equivalent is $\frac{1}{eV}$.
Solving above equation (1) gives,$$p=p_0\left(1-e^{-n}\right)=p_0\left(1-e^{\frac{E}{E_0}}\right)\tag2$$
This equation shows that even apply force for very long time, decline in change in momentum is not due to other losses but entropy which inhibit further energy conversion into work, that is gain in speed.
Now if there is no applied force in (1) and an object have momentum $p_0$ from initial condition, then solution is,$$p=p_0e^{-n}=p_0e^{-\frac{E}{E_0}}\tag3$$
From (1), as $F$ is source term and constant so it forces response to constant after some time and the term containing rate of change in momentum to be zero, that is known as equilibirium condition also, thus $p_0=t_0F$.
The value of maximum momentum from force $F$ can be express into form of,
$p=\int_0^{\infty}\bar np_ndn\tag4$
where $\bar n$ is average number of impulses at $n^{th}$ impulse as there is influence of previous impulses at that point and also there is loss of momentum between two impulses, $p_n=tF=nt_0F=np_0$ is sum at $n^{th}$ impulse. To have expression of $(4)$, we rearrange $(2)$ to have,
$p_0=\int_1^{\infty}p\frac{e^n}{e^n-1}dn\tag*{}$
and multiply this with loss coefficient of $(3)$ and generalize it as $(4)$, so
$\displaystyle p=\int_1^{\infty}\frac{n}{e^n-1}p_ndn=\int_1^{\infty}\frac{n^2F}{e^n-1}dn\tag5$
where, $\bar n=nf(n)=n\frac{1}{e^n-1}$, and $F=p_0$ for $t_0$ having unit value.
Using $(5)$ for energy, replace $F$ by power $P$ instead. Unlike Planck's law of radiation, $(5)$ has variable average value with variable probability function.
Implication of this is that, the value of integral is finite for infinite variable n which is time here. So value of integral solely depends upon value of force, more the force more the speed gained by any object. There is upper limit on speed no matter how long force is applied it can't increase speed infinitely.
Actually in most cases, force is related to square of speed and that is evident in fluid dynamics, terminal velocity or drag. Circular motion clearly shows that acceleration is square of speed at unit distance and opposite to lever function, speed decreases as we go further from centre.
Point is clear that there is relation of density of an object and its ability to gain energy or kinetic energy shows that how mass carry energy in form of speed. But on basis of that to claim that there is relativistic mass because it was thought that on applying force for infinite time not producing infinte speed or infinite gain in energy of an object, there would be mass increase and only significantly at relativistic speed is absurd.
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4The concept of relativistic mass has absolutely nothing to do with entropy. – J. Murray Dec 16 '22 at 16:10
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Absolutely, this appear because energy is not converting into speed or kinetic energy. Now they started to say that mass is invariant and energy increases. Only energy imparted by an object is kinetic energy even in intensity of light. – Neil Libertine Dec 16 '22 at 16:25
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@NeilLibertine this seems related to Lorent's transformation effect on boosting when view from stationary observer. The rocket that is doing the accelerating does not feel a thing, and accelerates normally in its calculation. It is the ground observer that sees odd things like the rocket unable to maintain acceleration even when using continuous thrust to do so. – James Dec 16 '22 at 20:55
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@James My point is that there is no perpetual motion and nothing gets its momentum instantaneously, not even light if it is particle. Lorentz boost was invented for many reasons one of it is compensate for entropy and keeping galilean and newtonian inertia. – Neil Libertine Dec 17 '22 at 02:40