Re: A nice puzzle: How fast does the E field move?

From: sal (pragmatist_at_nospam.org)
Date: 08/16/04 

Date: Mon, 16 Aug 2004 15:22:02 -0400

On Mon, 16 Aug 2004 18:20:38 +0000, Dirk Van de moortel wrote:

>
> "sal" <pragmatist@nospam.org> wrote in message
> news:pan.2004.08.16.16.42.44.517454@nospam.org...
>> On Mon, 16 Aug 2004 15:39:30 +0000, Dirk Van de moortel wrote:
>
> [snip]
>
>> > In the first case it takes time for some signal (carrying information
>> > about the location of a particle) to reach C. In the second case no
>> > signal is sent. C constantly 'knows' where to find the particle. No
>> > information must be sent.
>>
>> Quite true.
>>
>> But now, extend the particle's worldline backward. In the first case,
>> assume the particle started its journey 2 centuries ago, at a point 100
>> lightyears away, and it's been traveling at a uniform velocity ever
>> since. Does that affect the answer?
>>
>> With the line extended backward, just how, exactly, is the first case
>> different from the second case...?
>
> I see no difference with this first case (extended backward) and the
> previous first case (accelerated (almost) instantly). In both first cases
> a signal is sent at 2:00 restframe time, whatever happened before that. In
> the second case no signal is sent.
>
>
>>
>> > Note 1: At 2:00 in the restframe C must have stopped, or like you
>> > would express it "decelerated (almost) instantly to 0C", to see the
>> > E-field point to A. Otherwise there would be an aberration effect.
>>
>> Yup, and a magnetic field, too.
>>
>> But those angles are annoying, aren't they? Let's simplify things a
>> little.
>>
>> Start with the ships in an "L", keeping A-B and B-C fixed at 1 light
>> hour. This way, A and C start 1.4 light-hours apart:
>>
>> A------------------B
>> |
>> |
>> |
>> |
>> |
>> C
>>
>> Let C move parallel to the A-B line until it's directly across from A:
>>
>> A------------------B
>> |
>> |
>> |
>> |
>> |
>> C' <-------------- C
>>
>> Now, with C moving at 0.5c, an event on ship A and an event on ship C
>> which are simultaneous at 2:00 in the stationary frame will be
>> simultaneous at 1:30 in C's frame

Well, 1:44, not 1:30, actually.

> Huh?
> If C is still moving, these events will not (and never) be simultaneous in
> C's frame, if they are to be simultaneous in the stationary frame.

I beg your pardon! The two events are separated by a line which is
_perpendicular_ to the line of motion.

Put the origin at C's starting point in both frames. Distances in
light-hours. Motion is along the X axis. Separation of A and C at
the moment we're considering is along the Y axis. Coordinates of C and A
in A's frame at the moment when C "passes" the particle, given as (t,x,y),
are

  C = (2, -1, 0) and
  A = (2, -1, 1)

Coordinates in C's frame, moving to the left at v, will be

   t' = g*(t + v*x)
   x' = g*(x + v*t)
   y' = y

And so for C, with gamma=1.15, we'll have:

   t' = 1.15*(2 - 0.5) = 1.725
   x' = 1.15*(-1 + 1) = 0
   y' = 0

And for A:

   t' = 1.15*(2 - 0.5) = 1.725
   x' = 1.15*(-1 + 1) = 0
   y' = 1

and they are, indeed, simultaneous in both frames.

(But it's not 1:30 on C's clock -- it's 1:44; left out the square root in
gamma first time through. Oops.)

> If C has (almost) instantaneously stopped, his clock will be reading
> (almost) 2/gamma hours = 1.73 hours (not 1.50 aka 1:30)

Yeah, I was using gamma^2 by mistake

>, provided C's
> clock has been behaving (almost) nicely. In the latter case the events
> will be simultaneous at 1.73 in C's frame, which has of course become
> the stationary frame again.
>
>>, and a right-angle line with respect to
>> C's motion will be a right angle in both frames.
>
> Not while C is still moving, which obviously is what you have in mind...

See above -- check the computation. The events are, indeed,
_simultaneous_ in both frames, motion included.

You always get to have a plane of simultaneity containing any
particular event which is shared between two inertial frames; I oriented
it to include A and C at the moment I cared about.

Nothing deep, and it's not necessary to the problem, but it simplifies
things...

>> No more aberration to
>> worry about! At 1:30 on board C, and at 2:00 according to a stationary
>> clock, C will "see" the field pointing straight up, directly toward A.
>>
>> Right?
>
> Um no, not right I'm afraid.
> You forget aberration again ;-)

If it's not a 90 degree angle with the line of motion in one frame or the
other, care to compute what it actually is? :-)

>> (Imagine a test charge on C; it must accelerate directly toward A
>> at that point.)

Really, do imagine that. Tell me what direction C will see it accelerate
in, if it doesn't appear to accelerate directly up the Y axis.

>> But go back to the top, and start over with the negative charge moving
>> from A to B, using this new arrangement.
>>
>> What time will it be when C "sees" the E-field point to B, if the
>> particle leaves A at 12:00 and arrives at B at 2:00?
>
> 3 hours

AH! 3 hours! An hour of signal travel time, after it arrives at B.

Are you sure?

Quite, quite sure?

Check my calculations above, see if you can spot an error. If what I said
is correct, then when the ship moves and the particle remains stationary,
the field points straight up at the exact moment the ship is passing the
charged particle ... where "exact moment" is equally true in either frame,
since the plane of simultaneity includes the particle and the moving ship
at that moment. But ... that would correspond to a time of 2 hours in the
original problem, not 3 hours. (Time dilation is different in the two
cases, but that shouldn't really matter for this.)

Then tell me what the difference is between the ship moving, and the
particle moving, if the particle started a long, long time ago. Motion is
relative, after all -- only acceleration is absolute.

Transform the E field between frames, for good measure. I did that, and
as far as I can see, an E field perpendicular to the line of motion
transforms to ... an E field in perpendicular to the line of motion,
albeit with a B field thrown in (which we don't care about).

>> :-)
>>
>>
>> > Note 2: At 2:00 in the restframe C's clock will show less than 2:00.
>>
>> Yup, that's true; it'll read 1:30.
>
> 1.73 hours (decimal) actually.

Yeah, left out the sqrt in gamma.

>
> Dirk Vdm
> [a bit puzzled - everything okay over there?]

Give up? Makes no sense? Sal must be off his rocker?

;-)

MTW pp 110-111.

It's _not_ obvious. 

-- 
I can be contacted through http://www.physicsinsights.org

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