3 ms·

The gravitational field points at where the object would be, given its current velocity. This is not to say that the object will actually end up there. If the

• jimmahoney 13y ago
  Yes. This is a case when a picture (well, a GIF animation) is worth lots of words. Look at http://jimmahoney.net/pulse_acceleration_field_lines.gif http://jimmahoney.net/pulse_acceleration_field_lines.gif . (I made this from the "Moving Charge" java physics applet at http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/Mo... . It's a demo of the physics of electric charges, not general relativity, but the essence of the idea is the same even if the field equations are different. This follows directly from the central idea of relativity: the physics of uniform constant motion is the same as no motion, if you're moving along at the same speed.) The point charge (mass for gravity) in the center is first moving slowly to the right, then suddenly changes direction to move slowly to the left. (The small red vector indicates velocity.) The white lines represent the direction of the field, that is, the direction an object would be pulled. Near the point source, the field lines point to where it is. Far away, the news hasn't arrived yet that it's motion has changed, and so they point to where it would have been. During the brief acceleration when the source changed direction, the field lines connecting these two regions are "kinked" strongly - that's the radiation, which propagates outward at the speed of light. A second object would feel a sideways pulse as the wave (kink) passed by.