Re: Gravity waves and "speed of gravity" questions
- From: carlip-nospam@xxxxxxxxxxxxxxxxxxx
- Date: Fri, 12 May 2006 19:00:34 +0000 (UTC)
ajiko <ajiko2004@xxxxxxxxx> wrote:
Binary star systems should be generating variations in gravity
where the gravity (toward us) is slightly different if the
stars are aligned perpendicular to us or aligned inline. This
would appear as a very slight dipole effect. The resulting
"waves" are possibly going to be detected in the relatively
near future.
No. The variation you're describing is very different from
gravitational radiation. A changing dipole moment of the
sort you're talking about gives a force that falls off as 1/r^3,
which is much too small to detect. Gravitational radiation,
on the other hand, falls off as 1/r.
Think about the electromagnetic case. When a charge accelerates,
it's true that the Coulomb force changes because the charge is
changing position. But that change is *not* the same as the
electromagnetic radiation the charge emits.
However, there doesn't seem to be anything special about
simply detecting them. For example, Newtonian gravity also
would have the same effect.
Newtonian gravity includes the (unobservably small) change in
dipole field that you describe. It does *not* have gravitational
radiation, which requires a field propagating at a finite speed.
So, my question has to do with the "speed of gravity".
Should the phase of the gravity effect exactly match the phase
of the visible positions of the two stars? In other words,
is the gravity effect expected to travel at a speed exactly
matching the speed of the light from the same stars?
You might start by asking the same question about electromagnetic
radiation. You will find a careful discussion in Volume II,
chapter 21, of the Feynman Lectures. Section 21-4 works out
the dipole case in some detail. The answer is that even though
electromagnetism propagates at the speed of light, the phase of
the electromgnetic force very nearly matches the "instantaneous"
phase of the charges, not the retarded phase. ("Retarded" means
"corrected for light propagation time," and is roughly what you
mean by the "phase of the visible positions.")
This is possible because electromagnetic forces depend not
only on the position of a charge, but also on its velocity.
The velocity-dependent term very nearly cancels the effect of
propagation time. For a charge moving at a constant velocity,
the net electromagnetic force points toward its "instantaneous"
position; for a charge whose velocity is not constant, it points
toward the "linearly extrapolated" position.
In fact, it is this effect that is responsible for electromagnetic
radiation. Consider a charge moving at a constant velocity that
abruptly stops at a point p, say at time t=0. Suppose you are
measuring the force due to that charge at another point a distance
d from the stopping point p. According to Maxwell's equations,
you will not know at first that the charge as stopped, and the
force you observe will continue to change direction as if the
charge were still moving. Then, at a time t=d/c, the information
that the charge has stopped will reach you, and the force will
abruptly swing back to point to p.
If you think about this, you will realize that this abrupt change
propagates out from point p at the speed of light. You can do
more -- using just Coulomb's law (or Gauss's law) you can compute
the energy and the power carried by this moving front. You will
find that it's just the right amount to match the calculation
for the power carried by electromagnetic radiation emitted by
the decelerating charge. In fact, it *is* the radiation --
this effect is the reason that an accelerated charge radiates.
If the force always pointed toward the instantaneous position
-- that is, if electromagnetism propagated at infinite speed --
this effect wouldn't occur, and we would have no radiation.
(You can find details of the calculation in Appendix B of
Purcell's undergraduate textbook, _Electricity and Magnetism_.)
For gravity, the situation is (according to general relativity)
similar, though more complicated because the field equations are
nonlinear. You will find the analog of the Feynman Lectures
computation for gravity in http://arxiv.org/abs/gr-qc/9909087.
Once again, the gravitational "force" will point almost exactly
at the instantaneous position of the source, as "extrapolated"
from the past motion; and once again, an acceleating source
will produce gravitational radiation.
Steve Carlip
.
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- From: ajiko
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