# Why does the (relativistic) mass change & why?

I studied that when an object moves with a velocity comparable to the velocity of light the (relativistic) mass changes...but I am really eager to know how does this alteration take place....If anyone could really answer my question I would be graceful towards him/her.

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I'm not comfortable enough on the topic of Special Relativity to give a full answer, but what you're referring to is not (necessarily) what is actually going on. When introducing the concept of an upper limit on velocity, many like to use the interpretation that the 'mass' of the object in motion is changing. Once you've studied the topic in a bit more depth, you see rather that it is better to talk about the relationship between energy and momentum, which does not require any mysterious changing masses. –  Daniel Blay Aug 12 '12 at 11:50
duplicated by physics.stackexchange.com/q/71772 –  Ben Crowell Sep 1 at 14:22
This paper gives an approach I hadn't seen before: Sonego and Pin, "Deriving relativistic momentum and energy," arxiv.org/abs/physics/0402024 –  Ben Crowell Sep 6 at 23:25

when an object moves with a velocity comparable to the velocity of light the (relativistic) mass changes [...]

This premise appears mistaken.

When and while some specific object which is identified by some specific (intrinsic, proper, invariant) mass $m$ moves with some specific constant speed $v$ (relative to a suitable system of participants who are capable of evaluating this speed of this object, in comparison to the speed of light in vacuum $c_0$)

then the so-called "relativistic mass" of this object, in this trial, $m / \sqrt{ 1 - (v/c_0)^2 }$, does not change, but remains constant as well.

Instead, different trials may be considered, in which the same specific object of specific (intrinsic, proper, invariant) mass $m$ moves with different speeds such that its "relativistic mass" consequently differs from trial to trial.

(The history of the notion "relativistic mass" and its limited utility in comparison to the notion of "(intrinsic, proper, invariant) mass" has already been addressed in other answers.)

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Here is a (still rather long) sketch of Einstein's original development of the relativistic kinetic energy, from his celebrated 1905 paper "On the Electrodynamics of Moving Bodies" linked to in Ben Crowell's answer. This answer's approach is also motivated by a paper linked by dmckee in chat.

Having noted the inconsistency of Maxwell's electrodynamics with standard Newtonian mechanics, Einstein offers up his principles for a new dynamics:

1. the dynamical laws should be the same in two coordinate systems in uniform relative motion (so there's no preferred system of "absolute rest")
2. the constancy of the speed of light $c$.

Note that the first is actually already satisfied by Newtonian dynamics; it's the second principle that's revolutionary.

From these principles he develops the Lorentz transformation, from which in turn flow (among many other things):

• time dilation: a moving clock appears to run slow. From the point of view of a reference frame in which a clock, moving with velocity $v$, marks a time interval $\Delta \tau$, the reference frame's clock system records a longer interval $\Delta t$: $$\Delta t = \gamma \Delta \tau \quad \text{ where } \gamma = \frac{1}{\sqrt{1-\left( \frac{v}{c} \right)^2}}$$ (I have to note here that Einstein actually uses the symbol $\beta$ for what we call $\gamma$; since $\beta$ now means something completely different, this change in notation causes me no end of confusion. I wonder when the change occurred?)

• the velocity addition formula, which for co-linear velocities $v$ and $w$ gives a resultant $V$: $$V = \Phi(v,w) = \frac{v + w}{1 +\frac{vw}{c^2}}$$ If $w << v$, we can write (in the limit $w \rightarrow 0$):
$$V = v + dv = v + \phi(v) w \quad \text{ where } \phi(v) = \left. \frac{\partial \Phi}{\partial w} \right|_{w=0} = \frac{1}{\gamma ^2}$$

Note that Newtonian dynamics is recovered by setting $\phi=1$, which amounts to $\gamma=1$.

Einstein next demonstrates that Maxwell's equations already satisfy both his principles, and that the electric field component $E$ in the direction of motion of a moving frame is the same in both moving and stationary frames.

With all that as preparation, it's time for the main event: consider a charge $q$ accelerated from rest by a uniform electric field $E$ across a potential energy difference $W=qEl$. By conservation of energy, the final kinetic energy $T$ of the charge will be $T=W$.

1. At any point in its motion, where the particle's instantaneous velocity is $v$ in the lab frame, one can establish a co-moving reference frame in which Newton's law applies (instantaneously) for the change in velocity $dw$ (in that frame): $$dw = \frac{q}{m} E d \tau$$
2. Transforming this expression to the lab frame using the above results, one finds: $$\frac{dv}{\phi(v)} = \frac{q}{m} E \frac{dt}{\gamma} \quad \text{ or } \quad dv = \frac{1}{\gamma^3} \frac{qE}{m} dt$$

Since the rate of energy change (the power) is: $$\frac{dT}{dt} = -\frac{dW}{dt} = \frac{d}{dt} (qEx) = qEv$$ we find, for the kinetic energy of the accelerated charge: $$T = \int_0^{t_f} qEv dt = m \int_0^{v_f} \gamma^3 v dv = mc^2 (\gamma - 1)$$ where the $\gamma$ in the result is evaluated at the final velocity $v_f$.

Note that in the Newtonian limit $\gamma=1$, the integral evaluates to the familiar $\frac{1}{2} mv_f^2$.

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Nice answer, +1. However, I've never been satisfied with the logical justification of the step in Einstein's original derivation where he assumes that the work-kinetic energy theorem holds without modification in relativity. –  Ben Crowell Sep 3 at 15:54
@Ben You said something similar to me the other day too, and while I commend the search for deeper meaning when-even and where-ever, this one is puzzling me a little bit. Do you know of a deeper reason for introducing the work-energy theorem in Newtonian mechanics that "well, it turns out to be useful"? –  dmckee Sep 11 at 14:32
@dmckee: In Newtonian mechanics the work-energy theorem is a theorem, which can be proved from Newton's laws. That's completely different logically from assuming that the it remains valid without any change in form when we generalize to SR. –  Ben Crowell Sep 11 at 15:50
@Ben it is trivial to show that $\Delta W = \frac{1}{2} m \Delta (v^2)$, but that isn't the magic. The magic is that this is a useful thing to know. That the energy feeds into a conservation rule. Without going the Noterian route, I find it hard to justify that the energy should matter without just going forward (show that it comes up in conservative fields and other places so that we can solve problems). Maybe that's just a sign of ignorance on my part, but that is why it doesn't bother me to start computing things with the theorem and seeing if it turns out to be useful in relativity. –  dmckee Sep 11 at 16:22
@dmckee: In the broadest context, conservation of energy is indeed "magic" in the sense that we just play with it, patch it up as needed, add new forms of energy, etc., as needed in order to make it a valid law. But in a more restricted context, e.g., Newtonian gravity, conservation of energy is a theorem that follows from Newton's laws, and the work-kinetic energy theorem is a part of the machinery of that theorem. The full dynamics of SR was found purely deductively before any expts were available to confirm any aspect of it. If some portion of that deduction is fallacious, that's a flaw. –  Ben Crowell Sep 11 at 22:39

Let's start by assuming the postulates of special relativity given in Einstein 1905a. One of these is that $c$ is the same in all frames of reference. There are really two things we would like to do: (1) prove that the usual formulas from Newtonian mechanics no longer give a usable description of dynamics, and (2) find out how to modify those formulas.

Task #1 is pretty straightforward. For example, suppose we have an elastic, one-dimensional collision between objects $M$ and $m$, with $M \gg m$, in a frame of reference where $m$ is initially at rest and $M$ has initial velocity $v$. If we assume the Newtonian expressions for momentum and kinetic energy, then the result of such a collision is that $m$'s final velocity is $v'=2v$. In the case where $v=c/2$, this would cause $m$ to fly off at $v'=c$. But this contradicts Einstein's second postulate, because if it's possible for material objects to move at $c$, then it's possible for observers to move at $c$, but then in such an observer's frame of reference, a ray of light could be moving at zero speed.

We can also see qualitatively from this argument that inertia must increase at speeds comparable to $c$. For consistency with the postulates of relativity, the actual result of this collision must be $v'<c$. The mass $m$ is acting as though it has more than the expected resistance to the change in its state of motion. There are two equivalent ways of stating this: (a) we can say that $m$ increases with speed, or (b) we can modify the equations for energy and momentum while considering $m$ to be a constant. It doesn't fundamentally matter whether we choose a or b; it just amounts to reshuffling a certain correction factor in certain equations. Up until about 1950, a was more popular, but these days all physicists use b.

So now we have task #2, which is to quantitatively fix up the dynamical formulas in Newtonian mechanics so that they are relativistically correct. There are a lot of different ways to do this. The route Einstein originally took was to demonstrate equivalence of mass and energy (Einstein 1905b). The paper is pretty readable, but if you really want to continue with this approach and develop a full treatment of momentum, in my opinion it gets a little cumbersome. A more modern approach, demonstrated in Einstein 1935, is to think in terms of four-vectors. This approach allows for a pretty compact derivation, at the expense of some abstraction.

The kinematical consequences of the postulates in Einstein 1905a are summarized by the Lorentz transformation, which converts the time and space coordinates of an event $(t,x,y,z)$ into coordinates $(t',x',y',z')$ in another frame that is in motion relative to the first at a velocity $v$. It's not my purpose to rederive the Lorentz transformation here, so I'll just appeal to its properties as needed. This makes it natural to start talking about vectors $\textbf{r}$ and $\textbf{r}'$ in four dimensions. These are called four-vectors. We really have to throw away the old notion of a three-vector, because a three-vector like $(x,y,z)$ doesn't have any well-defined transformation properties; we can't tell what it would look like in another frame without knowing $t$.

Just as Newtonian mechanics has uniform rules for operating on displacement vectors, force vectors, momentum vectors, etc., we expect that the Lorentz transformation will be applicable to all the corresponding objects in relativity. You can take this as a postulate if you like.

The fundamental laws of physics are conservation laws, such as conservation of momentum. The above considerations tell us that in order to generalize conservation of momentum to relativity, we're going to have to make a four-vector out of the Newtonian three-momentum. If the law is reexpressed in terms of a four-vector, then the equation will automatically be valid regardless of what frame we're in, since both sides of the equation will transform identically.

The Lorentz transformation of a zero vector is always zero. This means that the momentum four-vector of a material object can't equal zero in the object's rest frame, since then it would be zero in all other frames as well. So for an object of mass $m$, let its momentum four-vector in its rest frame be $(f(m),0,0,0)$, where $f$ is some function that we need to determine, and $f$ can depend only on $m$ since there is no other property of the object that can be dynamically relevant here. Since conservation laws are additive, $f$ has to be $f(m)=km$ for some universal constant $k$. In sensible relativistic units where $c=1$, $k$ is unitless. Since we want $\textbf{p}=m\textbf{v}$ to hold for four-vectors so as to recover the appropriate Newtonian limit for massive bodies, and since $v_t=1$ in that limit, we need $k=1$.

Transforming this momentum four-vector into some other frame, we find that its timelike component is no longer $m$. It equals $m$ plus an expression whose low-velocity limit is the kinetic energy. We interpret this expression as the relativistic kinetic energy. We no longer have separate conservation of mass, only conservation of mass-plus-energy or "mass-energy," $E$.

The Lorentz transformation always preserves the norm of a vector $\textbf{r}$, defined by $r_t^2-r_x^2-r_y^2-r_z^2$. For a body of mass $m$, the norm of the momentum four-vector will always be $m^2$, regardless of what frame we're in. The result is

$$m^2=E^2-p^2 \qquad ,$$

which is valid for both massive and massless particles. In the $m \ne 0$ case, one can then prove that $p=m\gamma v$. The mass $m$ is constant, which is the modern convention. In school textbooks that are still stuck in the 1940's, $m\gamma$ is referred to as the relativistic mass, $m$ as the rest mass.

Einstein, "On the electrodynamics of moving bodies," 1905; English translation at http://fourmilab.ch/etexts/einstein/specrel/www/

Einstein, "Does the inertia of a body depend upon its energy-content?," 1905; English translation at http://fourmilab.ch/etexts/einstein/E_mc2/www/

Einstein, "Elementary derivation of the equivalence of mass and energy," Bull. Amer. Math. Soc. 41 (1935), 223-230, http://www.ams.org/journals/bull/1935-41-04/S0002-9904-1935-06046-X/home.html

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Mass in physics is a mathematical construct, and mass of an object approaching $\infty$ as the speed of an object approaches $c$ is a mathematical consequence of the postulates of Special Relativity.

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The OP knows that it follows from the postulates of SR, but wants to know how. This doesn't address the question. –  Ben Crowell Sep 1 at 14:51

In relativistic mechanics, there is a conserved quantity, relativistic momentum:

$\vec p = \gamma m \vec v$

$\gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

where m is the invariant mass or less precisely, the rest mass.

Now, one interpretation is to identify $\gamma m$ as the relativistic mass, a speed dependent mass. But this is actually unnatural as it leads to the notion of directionally dependent inertia; objects having more inertia along the direction of motion.

In fact, it is more natural to identify $\gamma \vec v$ as the spatial components of a four-vector, the four-velocity $\mathbf{U}$.

Then, the four-momentum is just $m\mathbf{U}$ with spatial components $\vec p$:

$m\mathbf U = (\gamma m c, \gamma m \vec v)$

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+1 Good answer. Note that other quantities, such as temperature, can be anisotropic (directionally dependent); or at least working physicists in such disciplines as near-Earth space physics find it useful to talk of such things when dealing e.g. with the motions of ions in plasma bounded by the Earth's magnetic field. Nevertheless, the fact that it is simpler to speak of the four-velocity than an anisotropic mass is motivation enough to abandon the notion of relativistic mass. –  Niel de Beaudrap Aug 12 '12 at 13:31
@NieldeBeaudrap, good comment and I'm considering editing my answer to address it. –  Alfred Centauri Aug 12 '12 at 21:45
Not a good answer--- the relativistic mass is independent of direction, it's only when you take it out of the derivative, and interpret m as the ratio of F to a, rather than as the ratio of p to m, that the direction dependent business starts. So you shouldn't say that $\gamma m$ is a directional mass, or a transverse mass, because the generalization is of $p=mv$ not $F=ma$. Otherwise fine. –  Ron Maimon Aug 13 '12 at 4:31
@RonMaimon, I didn't write "$\gamma m$ is a directional mass", I wrote "it leads to the notion of directionally dependent inertia". –  Alfred Centauri Aug 13 '12 at 11:01
This is helpful because it points out to the OP that his/her question was phrased in obsolete terminology. However, it doesn't fundamentally address the question. The question is asking for why there is a factor of $\gamma$ involved, regardless of whether or not we group the factors of $p=(m\gamma)v$ and refer to $m\gamma$ as mass. –  Ben Crowell Sep 1 at 16:36

In Neutonian physics the mass of a particle of matter does not change . It is defined by

*F=m*a* , where F is the force necessary to apply to this specific mass m in order to accelerate it by an acceleration a.

When velocities approach the velocity of light, experiments have told us that the higher the velocity of the particle the more force must be applied for the same acceleration a.

The theory of special relativity addresses this behavior , and it has been validated again and again by experiments. From the link:

To an observer who is not accelerating, it appears as though the object's inertia is increasing, so as to produce a smaller acceleration in response to the same force. This behavior is in fact observed in particle accelerators, where each charged particle is accelerated by the electromagnetic force.

One can find the formula of the mass change in the above link.

Now there is no other answer to "why", then "because that is the way nature behaves".

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I updated the question slightly. –  Qmechanic Aug 12 '12 at 11:59
Why the downvote? –  DIMension10 Sep 1 at 4:19
Now there is no other answer to "why", then "because that is the way nature behaves". Not true. The relativistic behavior of inertia and momentum follows logically from the postulates of special relativity. –  Ben Crowell Sep 1 at 16:34
@BenCrowell special relativity and its postulates was fitted to the fact that experimental data showed such a behavior. It is not the data that follows special relativity, special relativity is the mathematical description of what we have observed/measured in nature, a shorthand for all that data. –  anna v Sep 1 at 16:38
Certainly Einstein's 1905 postulates (which are kinematical, not dynamical) were motivated by experiment. That doesn't mean that every fact in the theory of SR can only be proved by direct reference to experiment. –  Ben Crowell Sep 1 at 16:40
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