# Gravitational potential of an approaching mass

Let $$\Phi(0, t)$$ be the gravitational potential at the origin and $$t$$ the clock ticking at infinity, i.e. not influenced by any gravitation.

Let a mass $$m$$ approach the origin with constant velocity such that its distance to the origin is $$r(t)$$. What is $$\Phi(0,t)$$?

The naive answer would be $$\Phi(0,t) = -Gm/r(t)$$. But this would mean instantaneous influence. My next best guess is $$\Phi(0, t+\frac{r(t)}{c}) = -Gm/r(t)$$
which assumes that the influence of the mass on the origin can only travel at the speed of light.

But for the observer at infinity (that's were the $$t$$ clock ticks), the gravitational influence travels not with $$c$$ in empty space, but with a $$c_{\Phi}$$ that depends on the gravitational potential or (even worse) the metric along the path which, for the observer at infinity, is less than the constant $$c$$.

While bringing in the word 'metric' I immediately concede that I have no idea in which metric to measure $$r$$ anyway.

# Short Version

What is the time dependent gravitational potential at the origin when a mass approaches it with constant velocity in terms of General Relativity. (It may turn out that gravitational potential is not the right thing to talk about, but rather metric or stress-energy tensor. I am happy to get answers as such.)

• What if the mass is stationary while the origin is approaching with the same speed? – safesphere Nov 25 '18 at 12:06
• It's not obvious to me that it makes sense for an observer at one point in space to ask about the gravitational field at another point in space without specifying how the field there is to be measured. For example, is there a clock at the origin sending out radio "ticka" at intervals that are, from its perspective, at regular intervals? – S. McGrew Nov 25 '18 at 12:22
• @safesphere this should hopefully be symmetric. Fell free to elaborate. I am happy to accept a solution that shows that there is no $\Delta t$ involved. – Harald Nov 25 '18 at 14:08
• @S.McGrew feel free to define the way to measure it. Sending radio is clearly an option. But this is not about an experiment, but rather what the theory predicts. Then creating an experiment without tripping over our own feed by messing up coordinate transformations is secondary. – Harald Nov 25 '18 at 14:12
• A problem here is the clock at infinity. This question is essentially about the relativity of sumultaunety. You need to define specific reference frames for your measurements. – safesphere Nov 25 '18 at 14:51

If the velocity is relativistic, and we do not restrict ourselves to a linear approximation, then the gravitational field of an isolated mass is best described by a metric known as a boosted Schwarzschild solution. It is simply the Schwarzschild metric written in a reference frame where this mass is moving with a constant velocity in a flat background.

Note, that since there is no point masses in general relativity, this metric either describes the spacetime outside of a moving extended body or a moving black hole. This last interpretation makes this metric quite useful in the field of numerical relativity where such solutions serve as initial data for numerically simulating black hole collisions. For example, the paper:

• Matzner, R. A., Huq, M. F., & Shoemaker, D. (1998). Initial data and coordinates for multiple black hole systems. Physical Review D, 59(2), 024015, arXiv:gr-qc/9805023.

contains the explicitly written forms for such metrics, and so the following equations are mostly lifted from there.

We start by noting that Schwarzchild metric could be written in Kerr–Schild form: $$$$ds^{2} = \eta_{\mu \nu} dx^{\mu} dx^{\nu} + 2H(x^{\alpha}) l_{\mu} l_{\nu} dx^{\mu} dx^{\nu} \label{1}$$$$ where $$\eta_{\mu \nu}$$ is the usual flat space Minkowski metric, $$H$$ is a scalar function of position and time and $$l_{\mu}$$ is a (position dependent) null vector (null both in the background, and in the full metric), $$\eta^{\mu \nu} l_{\mu} l_{\nu} = g^{\mu \nu} l_{\mu} l_{\nu} = 0,$$ For Schwarzschild metric specifically, the function $$H = M/r$$, where $$M$$ is the mass of the particle and $$r=(x^2+y^2+z^2)^{1/2}$$ is a normal expression for radial coordinate. The null vector is $$$$l_{\mu} = \left(1, \frac{x }{r}, \frac{y}{r}, \frac{z}{r}\right).$$$$

Transformation into a “boosted” reference frame consist in applying the usual (Minkowski space) Lorentz transformation to the cartesian coordinates $$x^{\mu}=\{t, x, y, z\}$$ which also straightforwardly transforms $$H$$ and $$l_{\mu}$$: $$$$\begin{array}{rcl} x'^{\beta} & = & \Lambda_{\alpha}^{\beta} x^{\alpha} \\ H(x^{\alpha}) & \rightarrow & H(\Lambda^{-1 \alpha}\,_{\beta} x'^{\beta}) \\ l'_{\delta} & = & \Lambda^{\gamma}\,_{\delta} l_{\gamma}(\Lambda^{-1 \, \alpha}\,_{\beta} x'^{\beta}) \\ g'_{\mu \nu} & = & \eta_{\mu \nu} + 2H l'_{\mu} l'_{\nu} \end{array}$$$$ Note, that since the new vector $$l'_\mu$$ is also null, the metric $$g'_{\mu \nu}$$ also has Kerr–Schild form.

For example, for the motion along the $$z$$-direction ($$v \equiv v^z$$), with new coordinates $$(x, y, z, t)$$ we have $$$$\begin{array}{lcl} r^{2} & = & x^{2} + y^{2} + \gamma^{2}(z - v t)^{2} \\ l_{t} & = & \gamma(1 - v \gamma (z - v t)/r) \\ l_{x} & = & x/r \\ l_{y} & = & y/r \\ l_{z} & = & \gamma(\gamma (z - v t)/r - v) \\ H & = & M/r \\ \end{array}$$$$ where $$\gamma = 1/\sqrt{(1-v^2)}$$.

As an illustration here is a slice through this spacetime (also from the abovementioned paper). The lapse scalar and shift vector plotted could be viewed as analogues of scalar and vector potential.

Figure caption:

Contours of lapse and shift vector $$\beta_{i}$$ for a boosted Schwarzschild (Eddington-Finkelstein) black hole. This is a cut through the equator; the hole is boosted at 0.5 upward in the figure. The shift is smaller ahead of the hole than behind, which allows the black hole to move through the computational grid. The ovoid figure is the horizon (distorted in these coordinates). The cardioid curves are contours of constant lapse $$\alpha$$, with values from the topmost contour down of 0.91, 0.86, 0.78, 0.71, 0.64, 0.58, 0.52. Along the axis of motion the lapse for this $$v=0.5$$ case is $$\sqrt{3}/2$$ at the leading point and $$1/2$$ at the trailing point.

• I always get confused about which coordinate system is which, so apologies for asking something silly: the unprimed coordinate system is used when "sitting on the mass, moving with it" and the primed one is the one of the outside observer seeing the mass approaching the orgin? Or just the other way round (grmmmmmlllll:-/ ? – Harald Dec 2 '18 at 14:21
• Somewhat inconsistently, the paper used $g'_{\mu\nu}$ for “boosted” metric (in frame where the mass is moving) yet renamed that same coordinates back to unprimed for the last set of equations, starting with $r^2={}$. – A.V.S. Dec 2 '18 at 15:41