# What is relativistic mass and why it is not used much?

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It happens that the term relativistic mass is used, in particular in introductory text on special relativity. It should be noted that whether or not to use relativistic mass to a large extent is a matter of convention, convenience, and semantics as long as it is used properly and does not have any impact on the predictions of special relativity. This FAQ is intended to clarify the concept and explain why it has essentially fallen out of fashion in the scientific community.

The concept of relativistic mass
In special relativity, the expressions for the total energy and momentum of an object can be written as
$$E = m_0 \gamma c^2, \quad \vec p = m_0\gamma \vec v,$$
where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ and $m_0$ is the invariant mass (or rest mass) of the object. If we introduce the relativistic mass
$$m = m_0\gamma$$
these expressions turn into
$$E = mc^2, \quad \vec p = m\vec v,$$
which many find more familiar since the first is an iconic relation in special relativity and the latter coincides with the classical expression for momentum.

Shortcomings of relativistic mass
In classical mechanics there are two different concepts of mass, gravitational mass, which determines the force on an object due to gravity ($\vec F = m\vec g$), and inertial mass, which determines the resistance to acceleration of an object ($\vec a = \vec F/m$). Since special relativity cannot include Newtonian gravity, the concept of gravitational mass is moot. In addition, the resistance to acceleration depends on both the velocity of the object as well as the direction of the force and so relativistic mass cannot in general correspond to a generalisation of either. In the rest frame of an object, we regain the relation between force and acceleration of classical mechanics, but this is well described by the invariant mass and there is no need to introduce relativistic mass for this purpose.
A common misconception that can be attributed to the concept of relativistic mass is that an object changes its internal structure by gaining mass when it travels at relativistic speeds. The object’s internal structure is independent of its velocity and it will always appear to be the same in its rest frame. The source of this confusion is that relativistic mass depends on the frame in which the object is observed and the concept of mass is typically regarded as a property of an object. See also our FAQ on the mass energy equivalence.

Because of the above, relativistic mass is generally not used in scientific communication as quoting it would also require quoting the velocity of the object. Furthermore, as evidenced by $E = mc^2$, the relativistic mass is simply proportional (and thus equivalent) to the total energy of an object, which is a concept far less liable to misunderstandings as energy is frame dependent also in classical mechanics.

How kinematics are treated in special relativity
Instead of introducing the relativistic mass of an object, we simply consider the invariant mass $m_0$ to be the mass of an object. The total energy and momentum of an object is given by
$$E^2 = m_0^2 c^4 + p^2 c^2, \quad \vec p = m_0 \gamma \vec v.$$
The relativistic relation between the force acting on an object and its acceleration is given by
$$\vec F = \frac{d\vec p}{dt} = m_0 \frac{d(\gamma\vec v)}{dt},$$
which is equivalent to the definition in classical mechanics for small velocities. If the force is collinear with the velocity, this can be written as
$$F = m_0\gamma^3 a.$$
Note here that we would not recover $F = ma$ if we introduced relativistic mass. As $\gamma \to \infty$ when $v \to c$, the faster an object is travelling, the more force is needed to accelerate it. This, or equivalently the fact that the total energy goes to infinity as $v\to c$, is the more appropriate argument for why massive objects cannot be accelerated to the speed of light.

Associate professor in theoretical astroparticle physics. He did his thesis on phenomenological neutrino physics and is currently also working with different aspects of dark matter as well as physics beyond the Standard Model. Author of “Mathematical Methods for Physics and Engineering” (see Insight “The Birth of a Textbook”). A member at Physics Forums since 2014.

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58 replies
1. PAllen says:

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

where U is 4-velocity and A is 4-acceleration.

2. Orodruin says:
PAllen

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

where U is 4-velocity and A is 4-acceleration.

Indeed, this was previously an FAQ and the target audience is mainly people who are not familiar with 4-vectors. It was written due to the endless stream of people we get asking questions on the subject of changing mass in relativity.

It should be noted that if the system is not closed (e.g., an object absorbing external radiation), as in classical mechanics, you would obtain
$$frac{d(mU)}{dtau} = m A + V frac{dm}{dtau},$$
also analogous to the Newtonian case.

3. m4r35n357 says:
Orodruin

Orodruin submitted a new PF Insights post

4. Orodruin says:
m4r35n357

a is always the acceleration in a particular inertial frame for the purposes of this post. No knowledge about proper acceleration is assumed.

5. Dale says:
PAllen

Just a comment that if you use 4-force instead of 3-force, you do recover a Newtonian like equation:

F = dp/dτ = d(m[SUB]0[/SUB]U)/dτ = m[SUB]0[/SUB] dU/dτ = m[SUB]0[/SUB]A

where U is 4-velocity and A is 4-acceleration.

Yes. But usually by the time someone learns four-vectors they are not confused about relativistic mass vs. invariant mass.

In fact, I sometimes wonder if it isn't easier to just teach four-vectors than have such discussions at all.

6. Orodruin says:
harrylin

Well done Orodruin! :smile:

I think it should be pointed out that while I wrote the original FAQ, several PF science advisors contributed opinions and suggestions on how to refine it.

7. maline says:

Another way of expressing SR kinematics in  Newtonian terms  (using 3-vectors in some one frame) is by phrasing the second law  as F=dp/dt, and defining momentum by p=Ev/c^2, with E the total energy. The gamma factor can even be derived from this using the work/energy relation. Since these equations are simple (no square roots!), this has led me to speculate that such a definition of momentum  should be seen as more "causative" than mv*gamma, although of course they are both frame dependent. This would give E/c^2, the relativistic mass, a status that would justify its onetime popularity. ;-)

8. DrStupid says:
maline

and defining momentum by p=Ev/c^2, with E the total energy

That doesn't need to be defined. It can be derived from the classical definition of momentum.

9. maline says:

my point is that you can describe the dynamic laws of SR, in terms of three-vectors, without mentioning the gamma factor.

10. PAllen says:
maline

my point is that you can describe the dynamic laws of SR, in terms of three-vectors, without mentioning the gamma factor.

A peculiarity of this approach (which, of course, does not make it wrong) is that the simple non-radiating case of forced motion is not simple at all. You have:

F = dp/dt = E' v + E v'

with all terms important for the simple non-radiating impulsive force (I use c=1). Meanwhile, with 4-vectors, the non-radiating case becomes the very simple:

F = m A

which has to do with the fact that the 4-vector approach sees gamma as an intrinsic feature velocity, by virtue of proper time. It is metric in nature rather than part of dynamics. In the language of 4-vectors, if we talk about velocity or acceleration divorced from any particular mass or energy, the gamma factors are built in.

11. exponent137 says:

Can you write more about discreancy between special relativity and Newtonian gravity? Maybe, if a very fast comet is flying close to earth, its gravitational force is proportional to earth to m_0 and not to \gamma m_0$m_0$ and not to $\gamma m_0$?Is something else?

12. DrStupid says:
exponent137

$m_0$ and not to $gamma m_0$?
Is something else?

$left( {1 + beta ^2 } right) cdot gamma cdot m_0$

13. Orodruin says:

In general relativity gravitation is not coupled to mass only. Instead, the source of space-time curvature is energy, momentum, and stress. Only for weak gravitation and small velocities does it reduce to the Newtonian case where mass is the source of gravity. You simply cannot describe gravity in special relativity by changing the mass for the relativistic mass. There should be some threads here on that subject if you use the search function.

14. exponent137 says:

I searched these threads, and partly I understand better, but I do not find everything. As Dr. Stupid wrote above for the above example of Schwarchild geometry, factor $1+\beta^2$ means the main discrepancy between Newtonian gravity + SR vs. GR+"SR". Can you have still any concrete examples for this discrepancy. What is at big velocities and small gravity. (One example gave Dr. Stupid.)

15. Orodruin says:
exponent137

Can you have still any concrete examples for this discrepancy.

The perhaps most famous differences between the predictions of Newtonian gravity and GR are:
– The perihelion shift of Mercury's orbit
– The amount of gravitational deflection of light passing by the Sun
In the first case, both the velocity and gravitational fields are relatively small so the effect is very small – it requires very good precision to make the measurement. The second case was one of the classic tests of GR and involves light, so the speeds involved are large (large means comparable to light speed).

16. pervect says:

I have a suspicion that a significant number of the readers here on PF are rather rusty on their calculus.  (Some readers may not have had calculus at all – but trying to totally eliminate calculus from the exposition seems to me to require a totally different approach, if it's possible at all).As a consequence of this (presumed) rustiness,  simply mentioning that F = dp/dt may not entirely get through to such readers, who are still mentally attached to the idea of F = ma, both because of both familiarity and because it deosn't require them to try and remember their calculus.A little more hand-holding, reviewing the classical perspective and explaining that the origin of F=ma was in fact F=dp/dt, might be helpful.   As a consequence, it might be more clear when one explains that F=dp/dt no longer leads directly to F=ma in relativistic physics.  If nothing else, it may inspire some readers to  brush up on the necessary mathematical underpinings (this is probably rather optimistic).  In other cases it may at least get across the idea of what's missing and preventing a fuller understanding of the issues.I'm not sure of the details of how to actually  go about writing all of this, but I thought I'd mention the idea in case someone was motivated.

17. harrylin says:
pervect

[..] As a consequence of this (presumed) rustiness, simply mentioning that F = dp/dt may not entirely get through to such readers, who are still mentally attached to the idea of F = ma, both because of both familiarity and because it deosn't require them to try and remember their calculus.

A little more hand-holding, reviewing the classical perspective and explaining that the origin of F=ma was in fact F=dp/dt, might be helpful. As a consequence, it might be more clear when one explains that F=dp/dt no longer leads directly to F=ma in relativistic physics. If nothing else, it may inspire some readers to brush up on the necessary mathematical underpinings (this is probably rather optimistic). In other cases it may at least get across the idea of what's missing and preventing a fuller understanding of the issues.

I'm not sure of the details of how to actually go about writing all of this, but I thought I'd mention the idea in case someone was motivated.

Some textbooks (such Alonso&Finn which I used as a student) present the SR derivations of acceleration from F=dp/dt; that's a useful stating point.

18. vanhees71 says:

Sure, you have (in the usual (1+3)-dimensional formulation) for, e.g., a particle in an electromagnetic field (neglecting radiation corrections)
$$frac{mathrm{d} vec{p}}{mathrm{d} t}=q left (vec{E}+frac{vec{v}}{c} times vec{B} right ), quad vec{p}=m frac{vec{v}}{sqrt{1-vec{v}^2/c^2}}, quad vec{v}=frac{mathrm{d} vec{x}}{mathrm{d} t}.$$
Of course, I've used the invariant mass $m=text{const}$ here. I never ever use something else as mass than this one and only invariant mass, but the (1+3)-formalism in a given inertial frame is often useful. For details on SR dynamics, also in manifestly covariant form, see my (unfinished) SRT FAQ: