From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Mon Jun 09 2003 - 14:13:40 MDT
Randall 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.
>
> No doubt there's some bit I haven't caught. :)
IANAP. With that in mind, my understanding goes something like this: The
instrument is attracted to where the star's projected position is,
according to the star's velocity in the past. This is because of the way
the star's velocity bends the potential well generated by the star. If
you're one light-minute away, you are not attracted to where the star is
now. You're attracted to where, one minute ago, the star would have been
in one minute, had it continued on at the same velocity it had one minute
ago, from the position it had one minute ago. (I'm not quite sure if
acceleration is supposed to be taken into account as well, or maybe it was
some other property of the star's behavior aside from or in addition to
the velocity, but I think this is roughly how it goes...) This is quite
distinguishable, behaviorally and experimentally, from an infinite
propagation speed. It also produces the same behavior if we shift
reference frames to regard the instrument as moving and the star as
motionless.
-- Eliezer S. Yudkowsky http://singinst.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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