From: Hal Finney (hal@finney.org)
Date: Mon Jun 09 2003 - 13:02:22 MDT
Randall wrote:
> On Sunday, June 8, 2003, at 11:25 PM, Hal Finney wrote:
> > It turns out that gravity is not unique in this respect. Electric fields
> > work the same way. If the star carried an electric charge, and you had
> > an instrument to detect it, the instrument would be attracted towards
> > the current position of the star, not to where the star was in the past.
> > It's for the same reason, that the electric field moves along with
> > the star.
>
> If this were true, wouldn't it follow that one could build an FTL
> communications device consisting of a sufficiently precise detector
> and a movable highly charged object? For this reason alone, it seems
> implausible that electric fields behave as perfectly rigid objects, as
> you describe.
Sorry, I wasn't completely clear. The electric and gravitational fields
move along with the star only while it is moving uniformly. If the star
gets bumped, the *change* in the field propagates outward at c. At time
t after the bump there is an expanding sphere of size c*t, where inside
the sphere the field has changed due to the bump, but outside the sphere,
nothing has changed yet. Outside of the sphere, the field points to
where the star would have been at that moment if it hadn't been bumped.
This part of the field doesn't know yet about the bump.
This can perhaps be most easily understood if you think of the simplest
case, where the speed of the star is zero. Then it is not surprising
that the field is spherically symmetric around the center, right?
Now bump the star, and the field will change; but the change will
propagate outward at speed c. Outside of the sphere of size c*t, the
field is undisturbed and still points placidly at where the star was
before the bump.
Think about how this would look from the perspective of a moving observer
(which is the same as looking at a moving body+field from a stationary
observer) and it will hopefully be easier to see how it works.
Hal
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