Imagine there are two observers $A$ and $B$ and a particle $P$.

$A$ and $B$ are at about the same point, $P$ is some distance away.

From $A$'s point of view, $P$ has velocity $V$ and $B$ has velocity $-V$.



Suppose $A$ and $B$ have the same rest mass $m$ and $P$ has rest mass $m_P$.

It seems to me that $A$ will observe a gravitational force between $P$ and $A$ of $F_a$ and between $P$ and $B$ of $F_b$ with $F_a\neq F_b$ because the relativistic masses of $A$ and $B$ are different.

However, B will observe forces $F_a'$ and $F_b'$ and it seems to me that $F_a' \neq F_a$ and $F_b' \neq F_b$ because $A$ and $B$ observe different relativistic masses for everything involved.

I've tried demonstrating $F_a=F_a'$ on paper and gotten into a tangle. Intuitively, I think this equality is false and expect the difference to be resolved elsewhere.

I have a qualitative understanding of general relativity, but can't handle tensors etc. Can this be explained with high-school maths?


2 Answers 2


The quick way to answer this is to point out that the Riemann curvature tensor is co-ordinate invariant. This means the increase in the relativistic mass does not change the curvature i.e. if $\gamma$ is 2 that does not mean the gravitational force is twice as strong. The same argument prevents bodies collapsing into black holes at speeds near the speed of light.

The confusion arises because people tend to consider gravity as being a function of mass, but life is more complicated than that. The Einstein equation states:

$$G_{\alpha\beta} = 8\pi T_{\alpha\beta}$$

where $G_{\alpha\beta}$ is the Einstein tensor that describes the curvature and $T_{\alpha\beta}$ is the stress-energy tensor. So the curvature is not determined by mass, it's determined by the stress-energy tensor.

As normally written, the stress-energy tensor is a $4 \times 4$ matrix with 10 independent entries because it's symmetric and the mass/energy density only contributes to the $T_{00}$ component. The $T_{0\beta}$ components (and by symmetry the $T_{\alpha 0}$ components) are momentum densities, so for the moving bodies in your example these are not zero and will affect the curvature.

I must admit I have not attempted to calculate what the stress energy tensor looks like for systems like the one you describe, and how the momentum components balance out the mass/energy component in your system. I think this would be a hard calculation.

  • $\begingroup$ Well, the trace $T_\alpha^\alpha$ needs to stay constant under boosts, and equals the particle mass (times a delta function!) in the rest frame. In a boosted frame there will be diagonal flows of momentum (i.e. flow of $p_x$ along $x$), which means that $T_{00}$ will change. $\endgroup$ Jun 6, 2012 at 17:24

If mass is defined as a ratio of force and acceleration, it depends on the angle between force and velocity, and there would be a parallel mass, a transverse mass, or even a 30-degrees mass. So the standard (and I think the only) definition of mass is the rest mass measured at zero velocity.

Well, I'm confused about "relativistic" mass. Is it a mass calculated from energy-momentum or strength of a gravitational pull?

  • $\begingroup$ Now I found it's a 2 yrs old question :-) $\endgroup$
    – user40602
    Dec 11, 2014 at 9:02

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