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## Does mass increase in SR mean higher gravity in GR?

It's not necessarily unrealistic to envision a spaceship with an external power source - rockets are so bad for reaching relativistic velocities (except perhaps for anti-matter ones) that a laser driven light-sail is quite competitive. It's extremely unrealistic to imagine accelerating a planet to light speed, but significantly less so to ask what the gravity of a planet would "look like" as seen from such a viewpoint.

I think I'm finally beginning to understand why most of the analyses I've seen covers only the low velocity case, though, and why I should probably do likewise.

The problem is fairly basic, but hard to see under all the math. Gravity acts on everything - so in order to determine the "acceleration of gravity", one needs an external reference system. With such an external coordinate system, one can impose the condition that an object moves "in a straight line with constant velocity", (this will be a different path than the natural geodesic that the object would follow on its own, of course), and then measure the acceleration required to force the object to follow this path. (One can measure this acceleration from any of several possible viewpoints, a convenient one would be an accelerometer mounted on the body itself).

Such a procedure requires that space be reasonablly close to being "flat", however. At first it is not obvious why high velocities should be an issue with respect to the flatness of space.

But if we look at the connection coefficients in more detail, some of the concerns become apparent.

Take the value of d^2 x / d tau^2 in a Newtonain gravity field (g_xx = g_yy = g_zz = 1+2U, g_tt = -1+2U, where U is the additive inverse of the Newtonian potential and assumed not to be a function of time).

Let the velocity be in the x direction. The connection coefficients that will influence this are

$$\Gamma_{xxx} ={\partial U}/{\partial x}}$$
$$\Gamma_{xxt} = \Gamma_{xtx} = 0$$
$$\Gamma_{xtt} = -{\partial U}/{\partial x}$$

Then
$$\frac{d^2x }{ d \tau^2} = g^{xx} ( \Gamma_{xxx}(\frac{dx}{d\tau})^2 + \Gamma_{xtt}(\frac{dt}{d\tau})^2)$$

thus for a velocity $$\mbox{\beta}$$ in the x direction

$$\frac{d^2x }{ d \tau^2} = g^{xx} ( \Gamma_{xxx} \frac{\beta^2}{1-\beta^2} + \Gamma_{xtt} \frac{1}{1-\beta^2})$$

When $$\mbox{\beta}$$ << 1, the time component of the curvature provides the largest contribution to the acceleration - but for large $$\mbox{\beta}$$, approaching 1, the space curvature term is just as important, and of opposite sign. Thus the notion that space is "flat" is suspect, because the contribution of the purely spatial curvature to the acceleration is comparable to the time curvature.
 Recognitions: Science Advisor Staff Emeritus I'm pretty sure everyone's probably very tired of this thread by now, but I can't resist One More comment. It seems to me that the best approach to considering what happens to gravity at high velocities is to go back to the idea of exploring "tidal gravity", the Riemann tensor, at high velocities, due to the aforementioned problems of dealing with the traditional notion of gravity as a "force". This approach has the definite advantage that it can be done at a point - one does not need any reference to the outside world or dependence on an external global coordinate system to know what tidal forces one is experiencing, one can measure the tidal forces directly. (Except for the problem of rotation, which I'll get into). The biggest stumbling block I have here is the issue of how to deal with rotation. Some relativly simple calculations can give the tidal forces on a body moving at relativistic velocities relative to a large mass. One needs to compute the Riemann tensor rather than the connection coefficients, then the tidal forces can be neatly summed up by the following matrix: $$E^a{}_b = R^a{}_{bcd} u^b u^d$$ where u^x is the four-velocity of the moving object. The main difficulty in making this approach rigorous is dealing with eliminating rotation from the coordinate systems used, so that "centrifugal forces" from a rotating coordinate system don't get confused with the actual components of the tidal force. Thus, we can directly measure the tidal forces we experience due to passing close to a body moving at relativisitic velocities directly if, and only if, we have zero rotational angular momentum - the later is not a very stringent condition, but it does mean we can't quite ignore the rest of the universe, we have to pay enough attention to it to be able to say that we aren't rotating. Another thing which one can calculate in principle is the total amount of momentum transfered to a body by an object "flying by". So if one was initially at rest, and an object came whizzing by at ultra-relativistic speeds, then left for infinity again, it's reasonable to ask what velocity one has after the body has left. Space and space-time should be perfectly flat when the massive body has "passed through", so there shouldn't be any ambiguity in this question.
 It was mentioned earlier in the thread that if you didn't account for the Energy required to reach that velocity you would violate thermodynamics. What if this relative mass was seen as adding mass from one side, and subtracting mass from the other? So essentialy a fast moving body would pull things stronger then a rest mass body from one side and weaker, (or possibly negative) from the other side?
 Also, from the first example, that would mean the ball on a stick would have no effect on the moon if it was being swung parallel to the moon. If the ball was being swung so that on one side of the orbit it was moving directly towards the moon and the other directly away, then the overall result would still be null because it would pull one direction and push the other.
 Maybe this question has already been answered but: Assuming two particles (X1 and X2) moving at relativistic velocities relative to each other. X1 ---> | d | <---X2 Is there some distance d and some less than c velocity that X1 and X2 orbit each other? Cheers, Bert
 Mentor Not due to SR effects.
 Bert: Maybe this question has already been answered but: Assuming two particles (X1 and X2) moving at relativistic velocities relative to each other. X1 ---> | d | <---X2 Is there some distance d and some less than c velocity that X1 and X2 orbit each other? Cheers, Bert DaleSpam: Not due to SR effects. Bert: What about GR effects increases mass increases the gravity?

Mentor
 Quote by hcm1955 What about GR effects increases mass increases the gravity?
Mass isn't the source of gravity in GR. The source of gravity is the stress-energy tensor:
http://en.wikipedia.org/wiki/Stress-energy_tensor

As you can see from the link, energy (proportional to mass) is only one component of 10 independent components. As you add energy to a particle by accelerating it not only do you increase the energy component, but you also increase the momentum component. The net effect is that you don't get increased gravity.

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