relmass

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. Currently teaching courses in mathematical methods for physicists, special relativity, and quantum field theory. A member at Physics Forums since 2014.
5 replies
  1. maline
    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. ;-)

  2. exponent137
    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?

  3. exponent137
    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.)

  4. pervect
    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.

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