Re: How can this work in relativity?

In sci.physics.relativity, dseppala@xxxxxxxxxxxxx
<dseppala@xxxxxxxxxxxxx>
wrote
on Wed, 03 Aug 2005 14:29:48 GMT
<42f0d3d3.4206208@xxxxxxxxxxxxxxxxxxxxxxxxx>:
> On Mon, 01 Aug 2005 15:00:08 GMT, The Ghost In The Machine
> <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
>
>>In sci.physics.relativity, dseppala@xxxxxxxxxxxxx
>><dseppala@xxxxxxxxxxxxx>
>> wrote
>>on Mon, 01 Aug 2005 11:51:40 GMT
>><42ee0c2a.59179836@xxxxxxxxxxxxxxxxxxxxxxxxx>:

[snippage]

>>> I have an extremely long line of people holding a rod that is
>>> light-years in length. At time t0, they all step onto a conveyer belt
>>> that is moving at 3 meters / second. The conveyer belt is also many,
>>> many light-years in length. Tell me what happens.
>>> David
>>
>>Erm... they all suffocate and die because nobody provided for air? :-P
>>
>>Of course, there's the little problem of in which reference frame t0 is
>>specified.
> The reference frame is the initial rest frame of the long line of
> people prior to them stepping on the conveyer belt.
>>
>>If it's at one end, the person 1 light-year from that end won't get
>>the signal until 1 year has elapsed (as measured from the endpoint of
>>the conveyor-belt). This can be worked around by sending the signal
>>from the far end, and engineering a countdown so that the near end
>>steps on first. An observer at the rod's midpoint might even
>>see the endpoints step on simultaneously under such a scheme.
> The way this works is that everyone is at rest with respect to each
> other and everyone has a synchronized clock. True, it takes years to
> synchronize the clocks, but once they are synchronized, no signal has
> to be transmitted along the 10**8 light-seconds of length. Instead,
> the long line of people simply all step on to the conveyer belt at the
> same designated time, using the time displayed on their synchronized
> clocks.
> David


OK. But are those clocks synchronized *in the reference frame of
the moving belt*?

Let's assume for giggles that two lightning strikes occur at both
ends of the belt. In O's space (he's standing at one end);
he sees the one flash immediately, and the other flash 3 years
later (we assume 3 light-years for specificity).
However, were O midway between he'd see them simultaneously -- 1.5
years later.

But they're still "simultaneous", if one uses the definition you're
suggesting, in the still-frame.

Now let's have A sit on the conveyor belt moving 3 m/s (10^-8 c).
We assume A sees the one flash immediately as he's standing next to O
at the time.

An event E can be represented as a coordinate-point (x,t).
Hence:

E_1_O = (0,0)
E_1_A = (0,0)

The other event can be represented

E_2_O = (3 l-y, 3 y)

The Lorentz of course applies, leaving us with:

E_2_A = ( (3 - (10^-8 * 3)) * 1/sqrt(1-10^-16),
(3 - (10^-8 * 3/1^2)) * 1/sqrt(1-10^-16))

Since x = t in O's observations, the two flashes are simultaneous.
(In other words, all of the time difference is accunted for because
of lightspeed traveling at 1 light-year per year.) Can A say the
same thing?

It turns out he can. So maybe we can perform your experiment after all.

In any event, what did you expect? The people step onto the belt,
1 meter apart each. Their distance will change to sqrt(1-10^-16) m
as observed from the belt, because of the Lorentz spacetimetwist.

(The reason it's sqrt(1-10^-16) and not 1/sqrt(1-10^-16) is because
of that twist. The issues are similar to measuring a moving rod
(moving at v) by waiting for the endpoints to pass under one's camera;
the calculations I leave for the reader -- or you can search for
an earlier post of mine which calculates this result.)

However, they're 1 meter apart *on the belt itself*; each one
can verify this using a ruler carried along for the occasion.

There's a few incongruities -- the persons at the very end of
the belt walk off almost immediately, for example.

[.sigsnip]

--
#191, ewill3@xxxxxxxxxxxxx
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