# Is the relativistic increase in (relativistic) mass same for a freely falling body as compared to body which is accelerated to the same velocity?

For a similar mass either falling toward a black hole or accelerated by a force is there a comparable increase in mass for similar velocity? Suppose a mass of one kilogram is accelerated to 99% of the speed of light by application of a force, or falls toward a black hole, reaches 99% of the speed of light. Is there an increase in mass similar for both or there is no increase in mass for a falling body. I believe falling body does not increase in mass as no force is being applied. What do you think?

Suppose a mass of one kilogram is accelerated to 99% of the speed of light by application of a force, or falls toward a black hole, reaches 99% of the speed of light. Is there an increase in mass similar for both or there is no increase in mass for a falling body.

In both cases a force is applied. In the case of a falling body the force is gravity. There is after all only one way of changing an object's velocity: by applying a force (see Newton's Laws)! In both cases its final momentum $$p$$ will depend on its rest mass and its velocity $$v$$ (see below).

Even if the body falls in Earth's gravitational field, say over a small height $$\Delta h$$, its mass will slightly increase because its (kinetic) energy increases by $$\Delta E$$ (if we observe the falling body alone):

$$\Delta E=mg\Delta h$$

And:

$$\Delta E=\Delta mc^2$$

So that (approx. for small $$\Delta E$$):

$$\Delta m=\frac{mg\Delta h}{c^2}$$

Because $$c^2$$ is huge, $$\Delta m$$ is of course negligibly small.

However when $$v \geq 0.99 c$$, it's no longer advisable to look at relativistic mass:

The word mass is given two meanings in special relativity: one (rest or invariant mass, and its equivalent rest energy) is an invariant quantity which is the same for all observers in all reference frames; the other (relativistic mass or the equivalent total energy of the body) is dependent on the velocity of the observer. The term relativistic mass tends not to be used in particle and nuclear physics and is often avoided by writers on special relativity.1 They do, however, talk about the (total) energy of a body, which is the equivalent to its relativistic mass, rather than the rest energy equivalent to its rest mass.

Rather the momentum $$p$$ is considered at high $$v$$:

$$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$

where $$m_0$$ is its (invariant) rest mass.

$$\let\g=\gamma \def\half{{\textstyle {1 \over 2}}} \def\bg{{\mathbf g}} \def\bp{{\mathbf p}} \def\bu{{\mathbf u}} \def\const{\mathrm{const.}}$$ As far as I can understand, the question has to be answered in a GR framework. No other relativistic theory of gravitation is available and generally accepted. Therefore answers which assume a different viewpoint shouldn't be considered.

It is well known that professional physicists (not only high-energy physicists) do not use the so-called relativistic mass, for several reasons. Nor I'll do it myself. Both in SR and in GR invariant mass only is a useful (and fundamental) concept. In both cases the answer is no. Mass doesn't change.

The answer will be different for energy, which in SR is a better substitute for relativistic mass (a factor $$c^2$$ apart). In SR the kinetic energy theorem of Newtonian mechanics still holds true:

work done by a force equals increment of kinetic energy.

Only expression of KE as a function of velocity changes: $$K = m\,c^2 (\g - 1)$$ instead of $$K = \half\,m\,v^2$$ (I'm assuming expression of $$\g(v)$$ is well known). Total energy is simply $$E = m\,c^2 \g = m\,c^2 + K.$$

And what about GR? A general answer can only be given for a static spacetime. Shortly, it's a spacetime where coordinates $$(t,u,v,w)$$ exist with the following properties:

• $$g_{tu} = g_{tv} = g_{tw} = 0$$
• no component of $$\bg$$ depends on $$t$$.

Free fall of a test body (i.e. a motion under no forces - recall that in GR gravity isn't a force) has a worldline which is a timelike geodesic. If it's parametrized with proper time $$\tau$$, then its tangent vector $$\bu$$ (the 4-velocity) and the 4-momentum $$\bp = m\,\bu$$ obey $$u_t = \const \qquad p_t = \const$$ We could define $$c\,p_t$$ as the body's energy, and it's constant along the body's worldline. It can be shown that in weak field and slow motion approximation $$c\,p_t$$ can be written as the sum of three terms:

• rest energy $$m\,c^2$$
• gravitational potential energy
• kinetic energy

so that energy conservation of Newtonian mechanics is recovered.

In both cases the mass of the body increases. Moreover the increase in mass from it's rest mass is dependent on the speed rather than what caused the speed.

Also, as mentioned by @Gert, gravitational force is also a kind of force. So you can't say there is no force acting on a freely falling body.

• An accelerating body feels a force while a falling body does not feel a force. That is the difference. – Humayoun Khan Dec 2 '18 at 19:27
• Please refer to Newton's first and second law of motion. There can not be acceleration without force. – mayank1513 Dec 3 '18 at 1:12