My understanding is that the "relativistic mass" of an object means any of the following three quantities (which are all identical):

  • The "mass-energy", as defined by the formula $m = E / c^2$.
  • The inertial mass: the ratio between the force applied on an object, and the amount of acceleration it undergoes as a result.
  • The gravitational mass: the amount of tendency for an object to gravitate towards other objects.

(But the rest mass of an object is something different, and my understanding is that when physicists say "mass" nowadays, they almost always mean the rest mass.)

Now, as an object's speed increases, the object gains kinetic energy. So, it seems like it must gain inertial mass and gravitational mass, too. As the object's speed increases, the object requires more force in order to produce the same acceleration, and it also exerts a greater gravitational force on other objects. Right?

But speed is relative. For every possible speed, there is a frame of reference (an inertial frame of reference, in fact) in which I am traveling at that speed.

Now, I weigh about 100 kg. All of the above seems to mean that there is also a frame of reference in which I weigh 300 kg, and a frame of reference in which I weigh 10,000 kg, and so forth.

So that seems to imply that there must be a frame of reference in which gravity is pinning me down hard and I struggle to move, as well as a frame of reference in which I am instantly crushed flat by the force of gravity, and even a frame of reference in which I collapse into a black hole.

That doesn't happen, of course. I am comfortably sitting in a chair and typing on my laptop, in all frames of reference.

Where's the hole in my understanding? Do the inertial mass and the gravitational mass of an object depend on the frame of reference, like it seems like they would? If they do, is there a simple explanation of how this dependency is consistent with the observation that I can stand up and walk around regardless of frame of reference?

  • 1
    $\begingroup$ As you say, in some frame of reference your relativistic mass would make you a black hole. But if you're a black hole in any frame, you're a black hole in every frame. And presumably you aren't a black hole, since information can escape from you. ;) This is one reason we don't like relativistic mass. $\endgroup$
    – PM 2Ring
    Commented Nov 12, 2018 at 4:00
  • $\begingroup$ "But the rest mass of an object is something different, and my understanding is that when physicists say "mass" nowadays, they almost always mean the rest mass." That's right. Relativistic mass isn't a serious idea anymore, which has been pointed out in the previous comment. Your reasoning has shown why. $\endgroup$ Commented Nov 12, 2018 at 4:04
  • $\begingroup$ So it seems that relativistic mass does depend on the frame of reference. But are inertial mass and gravitational mass equal to relativistic mass? $\endgroup$ Commented Nov 12, 2018 at 4:07
  • 1
    $\begingroup$ The notion of relativistic mass is excessive. The relation between mass, energy, and momentum is $E^2=m^2c^4+p^2c^2$. Or in natural units simply $E^2=m^2+p^2$. So there is just one "mass". $\endgroup$
    – Kosm
    Commented Nov 12, 2018 at 4:18
  • $\begingroup$ @Kosm Given that, what do Newton's second law and Newton's law of gravitation look like? Does the law of gravitation depend on p as well as m? $\endgroup$ Commented Nov 12, 2018 at 5:46

2 Answers 2


Newton's second law is the definition of force \begin{equation} {\bf F}=\dot{\bf p},~~~\dot{\bf p}\equiv \frac{d\bf p}{dt}. \end{equation} At small velocities you can approximate ${\bf p}=m{\bf v}$, so the above eqn. gives ${\bf F}=m\dot{\bf v}=m{\bf a}$.

Relativistic formula for ${\bf p}$ says \begin{equation} {\bf p}=\gamma m{\bf v},~~~{\rm where}~\gamma\equiv\frac{1}{\sqrt{1-v^2/c^2}}. \end{equation} Taking time derivative of $\bf p$, and plugging it into the Newton's second law above gives you its relativistic version.

As for the gravitational law, you will need General Relativity. In GR gravitational law is Einstein Field Equations. When the gravitational fields are weak and the velocities are small, EFE reproduce Newton's gravitational law (Newtonian limit of EFE).

In the end, you don't need to introduce "relativistic" and "rest" masses. Just "mass". Also gravitational and inertial masses are the same (up to experimental uncertainties).


$\let\g=\gamma \def\mg{m_{\mathrm g}} \def\bF{\mathbf F} \def\ba={\mathbf a} \def\bp={\mathbf p} \def\br{\bar r} \def\bt{\bar t} \let\dss=\displaystyle$

The inertial mass: the ratio between the force applied on an object, and the amount of acceleration it undergoes as a result.

If you define inertial mass this way you will find - following the line sketched by @Kosm - that inertial mass is different if force is parallel or orthogonal to instantaneous velocity. More precisely, in transversal case inertial mass will be $m\g$, in parallel case will be $m\g^3$. A well known result, which led some to speak of a "longitudinal mass" and a "transversal mass".

As to gravitational mass (more exactly, passive gravitational mass) I looked at the issue years ago. I'm not in a position of exclude it had already been completely solved before, but I liked to deal with it myself, as an exercise. I have some notes on it, but in italian. So I'll try to give here a very short account of my solution.

The idea is to put oneself in a stationary frame in a static curved spacetime, and to deduce what Newton's gravitation law would look like in that frame. Static spacetime is needed in order to reproduce what we experience on Earth's or Sun's neighborhood, where we observe bodies moving in a static gravitational field and explain their motions, after Newton, as due to a gravitational force proportional to their (passive gravitational) mass.

Of course in non-relativistic limit $\bF=m\ba$ is used. I would take instead a SR approach, using $\mathbf F=d\bp/dt$. You should not think I'm trying a SR theory of gravitation: mine is simply a phenomenological approach. I see objects are falling and try to find the force law.

To be definite, I used the only curved static spacetime I'm familiar with: Schwarzschild's. A stationary frame is one placed at fixed space coordinates, in particular at $r=\br$. Local means of infinitesimal extension. As a preliminary step, a precise meaning had to be given to time, speed, acceleration in that frame. This is done in an obvious way: time is what I read on a clock standing still in my frame. So it's not coordinate time; instead, calling local time $\bt$ we have $$d\bt = dt\,\sqrt{1 - {2GM \over c^2 r}}.$$ Distance means proper distance $d\ell$: $$d\ell = {dr \over \sqrt{1 - \dss{2GM \over c^2 r}}}.$$ Definitions of radial velocity and acceleration follow: $$v = {d\ell \over d\bt} \qquad a = {dv \over d\bt}.$$

The rest is standard. Law of radial geodesics is well known. After some manipulations we arrive at $$a = -{GM \over \g^2\br^2}\,\sqrt{c^2 \br \over c^2 \br - 2GM}.$$ From acceleration we obtain force by $F=m\g^3 a$: $$F = -{G M m \g \over \br^2}\,\sqrt{c^2 \br \over c^2\,\br - 2GM}.$$ In the weak-field limit $(\br\gg 2GM/c^2)$ Newton's law is recovered, but with a passive gravitational mass $m\g$.

This not the whole story however: we may not be content with radial fall. It could happen that for transversal motion a different law results. So I had to study transversal motion, and this is a more difficult thing. Just a few words to explain the hardest point. It is that angular motion comes into play, so that the problem becomes two-dimensional. Unfortunately in Schwarzschild geometry space sections are not euclidean, so that it is far from trivial to disentangle curvature of space geodesics from additional curvature due to gravity. (BTW, this is the same conundrum we face when we try to compute light deflection. It is well known that a "newtonian" calculation gives one half the value of GR calculation.) I'll limit to give the result, which is easy to say as it is the same as for radial motion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.