Re: Is LIGO just observing viscosity of the vacuum?

Jonathan Thornburg [remove -animal to reply] <jthorn@xxxxxxxxxxxxxxxxxxxxxxx> wrote:

[...]
More precisely, John Baez's (excellent!) web page which you cited
is referring specifically to the case where we have two masses in
orbit about each other, and where that system is "nearly isolated"
(in the sense that we can neglect outside influences on it), and
where the two bodies are orbiting relatively slowly compared to
the speed of light.
[...]

If you consider Newtonian mechanics, then the orbit is stable
(it doesn't decay), essentially because the Newtonian gravitational
force is always central, i.e. parallel to the (instantaneous) vector
between the two bodies, and thus perpendicular to the instantaneous
velocity vector. Thus the dot product of the force and the velocity
vector (which gives the work done by the force) is zero.

Suppose you naively try to insert a finite-speed-of-light correction
into Newtonian gravity, by saying that the gravitational force points
in the "retarded" direction, i.e., in the direction where the other
body was a time r/c ago (where r is the distance between the two bodies,
i.e., the orbital diameter). Then it's easy to see that you've
effectively introduced a drag proportional to v/c into the Newtonian
orbital dynamics. This would make orbits decay very rapidly.
(This is not observed: the Earth has been in roughly its present
orbit around the Sun for 4+ billion years. So we conclude that
the retarded-Newtonian-gravity theory doesn't accurately describe
orbital motion.)

In general relativity, it turns out that this v/c drag cancels out.
I think (but am not quite sure) that the reasons for this are somewhat
similar to those for a similar effect in (flat-spacetime) Maxwell
electromagnetism, where the electric field from a charge which is
moving at a uniform velocity, and has *always* been moving at that
same (vector) velocity, points towards/away from the *current* position
of the charge, not the retarded position of the charge. [The "has
always been moving at that same (vector) velocity" part is essential
here!]

In general relativity, it turns out that any drag proportional to
(v/c)^2 also cancels out. [Alas, I don't have a nice intuitive
explanation of why this is so -- it comes out of a long & tricky
calculation.]

Even powers of v/c don't give drag -- they're time-reversal invariant.
For this particular question, you only have to worry about odd powers.

Remarkably, the (v/c)^3 and (v/c)^4 terms also cancel out.
[Once again, I don't have a nice intuitive explanation for why.]

Roughly speaking, it's because electromagnetism allows dipole radiation,
while gravity allows only quadrupole and higher radiation. By energy
conservation, "drag" forces have to be balanced by radiated energy,
and are restricted by how fast energy can be radiated away.

Here's an argument that's basically dimensional analysis. For E&M,
radiated power goes as the square of the second derivative of the dipole
moment, i.e, as e^2a^2, where a is the acceleration. For two charged
bodies in a circular orbit, this gives a power
P ~ (e^2/r^2)(v^4/c^3) = F_0 v (v^3/c^3)
where F_0 is the Coulomb force. For this power to balance orbital
energy changes due to "drag," we need corrections in F_0 to first
appear at order v^3/c^3.

For gravity, though, the mass dipole moment is constant -- it's first
derivative is just total momentum, which can be set to zero by going
to a center of mass frame. The first contribution to radiation comes
from a changing quadrupole moment, and by dimensional analysis,
one needs an extra time derivative. This gives a radiated power
P ~ F_0 v (v^5/c^5)
where F_0 is now the Newtonian attraction.

These arguments don't explain exactly *how* the v/c terms for E&M
and the (v/c)^3 terms in gravity are cancelled, but they show that
such a cancellation must occur. I've worked out the details for gravity,
if anyone wants to look at the math, in gr-qc/9909087.

Steve Carlip
.

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