Re: Electrolysing Steam
From: Bill Ward (bwardREMOVE_at_ix.netcom.com)
Date: 10/26/04
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Date: Tue, 26 Oct 2004 05:16:54 GMT
On Mon, 25 Oct 2004 20:52:41 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
>
>"Bill Ward" <bwardREMOVE@ix.netcom.com> wrote in message
>news:417cbc2a.5231492@localhost...
>> On 23 Oct 2004 17:58:13 -0700, hhc314@yahoo.com (Harry
>> Conover) wrote:
>>
>> >"G. R. L. Cowan" <gcowan@eagle.ca> wrote in message
>news:<417A8E83.135C3C0E@eagle.ca>...
>> >> Harry Conover included:
>> >> >
>> >> > hence I = E/R was likely the same type of
>> >> > oversimplification that led to F = MA.
>> >> >
>> >> > Just my 2-cents worth.
>> >>
>> >> You're asking too much ...
>> >>
>> >> I wonder what cases of 'F' being unequal to 'ma'
>> >> you find most interesting.
>> >
>> >There are two very obvious cases.
>> >
>> >The first is found in analysis of the force required for a rockets
>> >acceleration since the mass of the rocket varies with consumption
>of
>> >fuel mass.
>>
>> Harry, I'm out of my league here, but like R = E/I, isn't
>> A = F/M still true for instantaneous values?
>
>No, no and no again.
>The correct expression is
>F = dp/dt = m dv/dt + v dm/dt
>The first term is ma and the second term is an absolutely essential
>modification if the mass of the system varies.
Thanks for your response, Franz. I'm afraid I don't quite
understand why you're solving for the force, when I was
looking for the instantaneous acceleration at constant
force. Are you perhaps implying that there is no such
thing as acceleration?
>
>> If the thrust
>> stays constant, the acceleration will still increase in
>> inverse proportion to the instantaneous rocket mass.
>
>That is incorrect unless the mass is constant.
It's still not clear to me why, for instantaneous values. .
>
>> >The second is the acceleration of a charged particle in an
>accelerator
>> >where the force required to accelerate it uniformly varies with its
>> >relativistic gain in mass as it is accelerated.
>>
>> Doesn't it depend on the observer's frame of reference?
>
>Harry was assuming that the observer is in the frame in which he and
>the accelerator are sitting still on the lab floor. They are usualyy
>called the "lab coordinates"
>
>> Suppose I was riding on a photon propelled spaceship powered
>> by an Earth-based laser reflecting off a mirror on the back
>> of the ship. Assume all of the transmitted laser energy is
>> perfectly reflected, transferring its momentum to the ship,
>> thus exerting an initially constant accelerating force..
>
>
>> I start off at say 1G acceleration, which I can observe by
>> measuring the magnetic force required to balance a test
>> mass. All goes well, until I observe my acceleration to be
>> decreasing. I look back and see the laser is now red
>> shifted because the velocity relative to home has become
>> relativistic, and the photons now have lower energy.
>>
>> From my frame of reference, the force has decreased, because
>> the thrust from the red shifted photons is weaker. To the
>> homefolks, I must be getting more massive, because the same
>> photon beam is not able to accelerate me at the initial
>> rate.
>>
>> But the instantaneous A = M/F still holds, doesn't it?
>
>No, it does not. If I were you, I would not try to pursue this
>problem in the medium of a newsgroup on hydrogen.
I appreciate your comments, but if at any time you feel out
of patience, simply don't respond and I'll understand. The
ng does have a number of posters who are willing to teach
each other on various topics This is a subject on which my
interest exceeds my understanding, so please bear with me.
>The kinematic
>calculations actually get quite complicated, because you have an ever
>increasing Doppler shift and an ever increasing relativistic mass as
>the ship goes faster.
I'm specifying instantaneous values, The point of my
thought experiment was to show that the issue of lower force
due to Doppler shift or gain of mass due to velocity depends
on the frame of reference. Could you clarify why
instantaneous A = F/M doesn't apply in both frames of
reference?
>
>Strange as it may sound, the problem is probably slightly easier to
>solve by summing (integrating over) the succession of elastic
>collisions between each photon and the ship. Even so, it is
>impossible to obtain a closed solution for it.
Wouldn't conservation of energy/mass still apply, assuming
perfect reflection and no losses?
>
>> Either the mass is increasing, or the force is decreasing,
>> depending on the frame of reference. I think that may also
>> hold for particles in an accelerator.
>>
>> Way off topic, but WTH, it beats politics.
>>
>> I appreciate all comments, even if they're over my head.
>
>Franz
Regards,
Bill Ward
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