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Suppose we have a mechanical balance, with two identical particles kept in the two sides. Now the balance does not show any deflection. Now, one of the particles is given some constant horizontal velocity. Will the balance show the moving particle to be heavier (that side will move downward )or not?

(There is no friction between the balance and the moving particle)

Archisman Panigrahi
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  • This question is not clear. Do you want to know whether the so called "relativistic" mass is actual and mass increase with increasing speed, or the gravity effect on moving bodies? Using scales to measure mass in the relativity framework is not so straight forward. – J. Manuel May 09 '18 at 06:38
  • I wanted to know whether the relativistic mass increase is actual (also if gravitational force on it is the relativistic mass times g) – Archisman Panigrahi May 09 '18 at 09:41
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    The objection to relativistic mass has never been that it is wrong, but that it is not covarient (and so has to be dispensed with in covarient formulations), that it doesn't mean what students think it means, and that (as a consequence of the second point) it encourages and enables sloppy thinking. See https://physics.stackexchange.com/q/3436/520, https://physics.stackexchange.com/q/180698/520, and https://physics.stackexchange.com/q/362545/520. – dmckee --- ex-moderator kitten May 09 '18 at 16:38
  • What does "actual" mean in this context, by the way? – probably_someone May 09 '18 at 20:15
  • By actual, I meant whether the mechanical balance will deflect. If it is just a formal convention, then the balance would not show anything. My language was not precise, though. – Archisman Panigrahi May 10 '18 at 03:23

2 Answers2

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Yes, it would -- the elementary pre-general relativity answer is "because gravity (which is what is measured by your balance) depends on the energy (mass plus kinetic energy) of an object, not the rest mass". So although the mass remains the same, a special relativistic correction to Newtonian gravity would be to consider the total energy instead of invariant mass.

This answer is unsatisfactory, however, because it doesn't seem to make sense to consider energy individually when calculating gravity, when it's just the time-like component of the four-momentum $(E, p_x,p_y,p_z)$ -- it would actually seem more natural to use mass (the norm of this vector) in your calculation than to use energy.

This is actually the theoretical motivation for general relativity, which explains that this force which depends on energy, $\Gamma_{00}^i$, is just one of the components (the "time-time component") of the sixteen components of the gravitational field tensor, albeit the most significant component we see at weak gravitational fields and low speeds. This is seen, e.g. in the Einstein field equation, where each of the sixteen components of the Einstein tensor depends on a corresponding component of the energy-momentum tensor, such as energy, momentum, pressure and shear stress.

Abhimanyu Pallavi Sudhir
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  • When you say norm of this vector, what exactly do you mean? $p^2-\frac{E^2}{c^2}$? That is the negative of the inertial mass. – Archisman Panigrahi May 09 '18 at 05:34
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    @ArchismanPanigrahi Just to be sure we're on the same page, the norm of that four-vector is the rest mass. The meaning of inertial mass can be ambiguous depending on whether you're using the invariant-mass framework or the relativistic-mass framework. – probably_someone May 09 '18 at 06:42
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Your experiment doesn't actually distinguish between the two definitions, because the two definitions give equivalent dynamics.

In the "rest-mass" framework, the gravitational force between two objects, where one is stationary and the observer's frame and one is moving, is

$$F=\gamma \frac{GMm_0}{r^2}$$

where $m_0$ is the rest mass, because forces perpendicular to the velocity (as is the case here) transform as $F\to \gamma F$ under Lorentz boosts in this framework.

In the "relativistic-mass" framework, we get to keep $F\to F$ under Lorentz transforms, because $m=\gamma m_0$ is no longer Lorentz-invariant. So the force is still

$$F=\frac{GMm}{r^2}=\gamma \frac{GMm_0}{r^2}$$

The difference is merely convention. In one case, you associate the required $\gamma$ with the way that force (or more fundamentally, momentum) transforms under Lorentz boosts. In the other case, you associate $\gamma$ with the way that mass transforms under Lorentz boosts. The two formulations give consistent results.

probably_someone
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