[time-nuts] Thunderbolt stability and ambient temperature
magnus at rubidium.dyndns.org
Fri Jun 12 08:32:38 UTC 2009
> J. Forster wrote:
>> It has nothing to do with this.
>> A long (length >> width) bar can simply be modeled as a long ladder of
>> series resistors's and capacitors to ground:
>> ---zzz---zzz---zzz---- ... ---zzz---
>> _|_ _|_ _|_ _|_
>> ___ ___ ___ ___
>> ----| ----|-----|----- ... ----|-----
>> If you put a rectangular pulse in the left end, it will emerge later and
>> very much rounded at the right end.
>> Either do the math or simulate it in Spice or with a handful of R's and
>> C's and a pulse generator and scope.
>> No inductors needed. PERIOD. That model fully accounts for your
>> observations with the bar heated at one end.
> Ok, but isn't that propagation rate constant? Obviously, the heat from
> the hot end will eventually propagate with some attenuation to the cold
> end. My observation was that shoving cold into the hot end seems to
> accelerate the propagation of heat toward the cold end. This model won't
> show that effect, will it?
> This would be a double step something like this:
> | \
> ---| \ ----> + time
> | /
> That 2nd opposite step won't make the first pulse propagate faster or
> with more apparent intensity, will it?
No, you just perceive it to be related. If you had waited it would have
happend anyway without the cooling period. Your cooling has nothing to
do with the already heated wave. While the speed does depend on
temperature, it is not that distinct effect. The cooling will also take
that time to get through.
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