# Gravitational field of a particle in SR

According to special relativity, what is the gravitational field due to a particle moving with a constant velocity v? Would it be correct to assume that the particle has a stronger gravitational field because of the relation $M=\gamma m$, where m is the rest mass and M is the relativistic mass? How is the spacetime affected?

• Nothing happens to the spacetime: in SR, one assumes that the Minkowski metric is fixed, so nothing affects it. It is only in general relativity that the metric can change, and that spacetime is no longer necessarily flat. The field due to the moving particle would be stronger, yes, although the concept of relativistic mass is no longer used by most physicists. It can simply be seen from the fact that the energy increases: apply $E=mc^2$ and voila!
– Danu
Nov 28 '13 at 18:24
• Nov 28 '13 at 18:43

Such is also situation for the gravitational field of a relativistic mass: It is different from Newtonian limit, not just stronger or weaker. To illustrate, let us take the first post-Newtonian approximation for the gravitaional dynamics: Einstein–Infeld–Hoffmann equations which contains $1/c^2$ correction to the point particles accelerations. Taking the equation from Wikipedia page: \begin{align} \mathbf{a}_A & = \sum_{B \not = A} \frac{G m_B \mathbf{n}_{BA}}{r_{AB}^2} \\ & {} \quad{} + \frac{1}{c^2} \sum_{B \not = A} \frac{G m_B \mathbf{n}_{BA}}{r_{AB}^2} \left[ v_A^2+2v_B^2 - 4( \mathbf{v}_A \cdot \mathbf{v}_B) - \frac{3}{2} ( \mathbf{n}_{AB} \cdot \mathbf{v}_B)^2 \right. \\ & {} \qquad {} \left. {} - 4 \sum_{C \not = A} \frac{G m_C}{r_{AC}} - \sum_{C \not = B} \frac{G m_C}{r_{BC}} + \frac{1}{2}( (\mathbf{x}_B-\mathbf{x}_A) \cdot \mathbf{a}_B ) \right] \\ & {}\quad{} + \frac{1}{c^2} \sum_{B \not = A} \frac{G m_B}{r_{AB}^2}\left[\mathbf{n}_{AB}\cdot(4\mathbf{v}_A-3\mathbf{v}_B)\right](\mathbf{v}_A-\mathbf{v}_B) \\ & {} \quad {} + \frac{7}{2c^2} \sum_{B \not = A}{ \frac{G m_B \mathbf{a}_B }{r_{AB}}} + O (c^{-4}) \end{align} we see a lot of new terms. If we try to test the field of relativistic object by measuring acceleration of a small test particle at rest in our reference frame ($v_A=0$) we will notice that indeed, there is a term proportional to $v_B^2$, which could be interpreted as additional force from added relativistic mass, but we also see its dependence on the angle between $\mathbf{n}$ and $\mathbf{v}_B$, terms proportional to acceleration $\mathbf{a}_B$, $1/r^3$ terms. And this is just the first relativistic correction.