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?

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    $\begingroup$ 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! $\endgroup$
    – Danu
    Nov 28 '13 at 18:24

The answer to your question:

Would it be correct to assume that the particle has a stronger gravitational field [...]?

is no, it would not be correct. Here is why.

Comparing gravitational field in special relativity to its Newtonian limit means trying to take an ill-defined limit. If one wants to include relativistic corrections such as relativistic expression for the energy, gravitational interaction has to be described by the general relativity. There is a weak field limit of GR, linearized gravity, but even in this limit the gravitational field no longer can be described by a single Newtonian potential, so we need a new framework to define the 'strength' of gravitational field.

If we were talking about moving charge, it would have been wrong to say "the electric field of a moving relativistic charge is weaker in comparison with the stationary one", the right statement would have been "moving charge has qualitatively different field: the electromagnetic field, which has magnetic component, different dependence on the angles...".

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.

  • $\begingroup$ Your answer is reasonable,however it is misleading to compare a charge to mass...a gravitational field behaves less similar to an electric field,in string theory gravity appears to be a loose string,unlike an electric string which appears to be a band...if the two were similar it would be wrong to think of gravity as a an extension of some electromagnetic field(and this could only be true under certain circumstances,and not always) $\endgroup$
    – user34793
    Nov 29 '13 at 8:55
  • $\begingroup$ @user34793: The analogy is only illustration for the fact that relativistic corrections are not simply 'stronger' or 'weaker' fields. $\endgroup$
    – user23660
    Nov 29 '13 at 9:05

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