2012-08-12 10:31:01 8 Comments

I learned recently that when an object moves with a velocity comparable to the velocity of light the (relativistic) mass changes. How does this alteration take place?

### Related Questions

#### Sponsored Content

#### 1 Answered Questions

### [SOLVED] Relativistic mass effect

**2019-06-14 13:02:50****Tushar Sharma****102**View**0**Score**1**Answer- Tags: special-relativity mass reference-frames acceleration velocity

#### 2 Answered Questions

### [SOLVED] Is this relativistic mass?

**2019-04-08 20:32:12****Brain Stroke Patient****183**View**4**Score**2**Answer- Tags: special-relativity mass inertial-frames mass-energy

#### 2 Answered Questions

### [SOLVED] Relativistic mass

**2019-03-04 22:54:23****Ballanzor****144**View**0**Score**2**Answer- Tags: special-relativity mass

#### 13 Answered Questions

### [SOLVED] Why does the (relativistic) mass of an object increase when its speed approaches that of light?

**2010-12-06 14:19:43****Kit****127088**View**27**Score**13**Answer- Tags: special-relativity speed-of-light mass velocity

#### 2 Answered Questions

### [SOLVED] Are relativistic momentum and relativistic mass conserved in special relativity?

**2018-06-12 19:15:04****Timothy****494**View**2**Score**2**Answer- Tags: special-relativity mass momentum conservation-laws collision

#### 2 Answered Questions

### [SOLVED] Can relativistic mass be treated as rest mass?

**2016-07-04 21:10:13****Will****180**View**0**Score**2**Answer- Tags: special-relativity

#### 2 Answered Questions

### [SOLVED] If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light?

**2014-10-11 09:38:05****rahulgarg12342****624**View**1**Score**2**Answer- Tags: special-relativity mass speed-of-light reference-frames mass-energy

## 9 comments

## @Riad 2018-04-03 20:10:11

This is a simple way to understand mass. We know Light (and EM in general) does not have a rest mass. But a standing wave or a trapped beam of light between two mirrors (as in lasers), does have a rest mass. So we conclude that rest mass is nothing more than a trapped/arrested momentum, or energy if you like since the two are derivable from each other. The trapping can be done either by walls like mirrors or a cavity wall, or can be done by trapping in a circular motion (without the need for walls). Thus we say here that mass and relativistic mass must be equivalent to an un-trapped momentum, whereas a rest mass is equivalent to a trapped/arrested momentum. Many examples confirm this picture. First you have the annihilation and creation experiments where rest mass becomes pure energy flux and visa versa. Then you have the case where the mass of a proton being much larger than the total rest masses of the internal constituents- as the excess rest mass comes from the huge momentum of constituents moving at relativistic speeds. By the same logic, the electron itself must be no more than a trapped momentum. This of course goes well with our knowledge of the electron.. as having a real mechanical spin (as per Einstein-de-Hass experiment), and also having an internal clock- the Zitterbewegung that is also related to the spin. Then you have the case of a double trapping.. where you have rest masses with high speeds trapped in a larger structure- as in the case of the nucleus. This again makes the mass of the nucleus larger than the sum of the rest masses of the internal components. But such difference starts becoming less and less as the structure becomes large. A hot matter for example, has a larger rest mass than a cold one, but the difference can't be measured- being very small. We also note that here we agree with Einstein formula E=m c^2, since in a standing wave you have double the kinetic energy.. that is E=2* .5 m v^2=m v^2=mc^2 as the speed in our case is that of light.

## @anna v 2012-08-12 11:52:14

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

$F=ma$ , 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:

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".

## @Qmechanic 2012-08-12 11:59:17

I updated the question slightly.

## @Abhimanyu Pallavi Sudhir 2013-09-01 04:19:10

Why the downvote?

## @Ben Crowell 2013-09-01 16:34:22

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.## @anna v 2013-09-01 16:38:13

@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.

## @Ben Crowell 2013-09-01 16:40:22

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.

## @anna v 2013-09-01 16:44:16

@BenCrowell The beauty of a successful/validated theory is that it hangs together consistently describing the data and predicting new behaviors. It would be invalidated if one datum was not fitted/predicted. It is trite, but a theory can never be proven correct, it is validated by data; it can be falsified by one wrong datum, i.e. behavior of nature.

## @Dims 2014-12-20 02:31:52

Additionally to Alfred Centauri's answer I can say, that mass

$m_{rel}=\gamma m$

is AUTOMATICALLY implies directional inertia, since it is not constant.

Any non-constant mass causes a propulsion force.

From definition of force

$F=\frac{dp}{dt}=\frac{dm_{rel}v}{dt}=v\frac{dm_{rel}}{dt}+m_{rel}\frac{dv}{dt}$

we have

$F_{prop}=v\frac{dm_{rel}}{dt}$

It is zero for constant mass, but not zero for non-constant.

This is essential part of mechanics, and is used in rocket engineering.

## @anotherguy 2014-12-19 23:37:59

Whether or not it's gaining mass depends on how one defines mass. The "relativistic mass" is simply the total energy, with a factor of c squared thrown in.

I recall reading that Einstein himself was against the idea of such a concept, and is quoted as saying that the only mass one should consider is the "rest mass". I would have to agree.

Relativistic mass is basically something that is only used when scientists are explaining things to non-scientists. The reason you can't reach the speed of light is not because an object is inherently changing, it's because of the relationship between relative speed and energy.

Bear in mind also, that, if we use relativistic mass, we are no longer justified in saying light is massless. Light has relativistic mass, because it has energy.

## @user12262 2013-09-09 19:04:36

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.)## @Art Brown 2013-09-03 02:33:57

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:

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 $T=W$.

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 $\frac{1}{2} mv_f^2$.

## @Ben Crowell 2013-09-03 15:54:13

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.

## @dmckee 2013-09-11 14:32:52

@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"?## @Ben Crowell 2013-09-11 15:50:13

@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.

## @dmckee 2013-09-11 16:22:30

@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

usefulthing 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.## @Ben Crowell 2013-09-11 22:39:26

@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.

## @dmckee 2013-09-12 02:48:24

@Ben Then you need to have the same level of equipment to go looking for a proof in SR---that was what was bothering me: you seemed to be asking for a proof with fewer tools then you used in classical mechanics. You need to have the equivalent of universal gravitation so that you can show that the work appears in the context of conservative fields. Presumably you use E&M for that rather than gravitation because it has the Lorentz symmetry built in.

## @Ben Crowell 2013-09-22 00:35:34

@dmckee: Re the relativistic validity of W=Fd, see physics.stackexchange.com/questions/13695/… . My answer to my own question gives an argument that I find satisfying.

## @Ben Crowell 2013-09-01 16:09:02

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

normof 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

## @Constantine 2013-09-01 00:19:53

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.

## @Ben Crowell 2013-09-01 14:51:14

The OP knows that it follows from the postulates of SR, but wants to know how. This doesn't address the question.

## @Alfred Centauri 2012-08-12 12:23:50

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 massor less precisely, therest 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 $\vec p$:

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

## @Niel de Beaudrap 2012-08-12 13:31:57

+1 Good answer. Note that other quantities, such as temperature,

canbe 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 dealinge.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.## @Alfred Centauri 2012-08-12 21:45:08

@NieldeBeaudrap, good comment and I'm considering editing my answer to address it.

## @Ron Maimon 2012-08-13 04:31:26

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.

## @Alfred Centauri 2012-08-13 11:01:09

@RonMaimon, I didn't write "$\gamma m$ is a directional mass", I wrote "it

leadsto the notion of directionally dependentinertia".## @Ben Crowell 2013-09-01 16:36:52

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

whythere 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.## @Dims 2014-12-20 02:25:28

You are generally right, except saying that "directionally dependent inertia" is unnatural. Since we have changing mass, we should take propulsion into account: any changing mass makes a propulsion.