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Re: A question on inelastic relativistic collisions



Hi Ed,
I have not looked into the relativistic case, but in my Mech
class I always show that in the Newtonian model the amount of kinetic energy
lost or gained (the Q) in an inelastic collision is an invariant with
respect to a Galilean transformation to another inertial frame. (That's
why, for example, in all inertial frames the totally inelastic collision is
that in which the two final velocities are equal - and in the CM frame all
the kinetic energy is lost and the final velocities are zero.)
You have provoked me to look into the relativistic case, when time allows :)

Bob Sciamanda
Physics, Edinboro Univ of PA (ret)
trebor@velocity.net
http://www.velocity.net/~trebor

-----Original Message-----
From: Ed Schweber <edschweb@IX.NETCOM.COM>
To: PHYS-L@LISTS.NAU.EDU <PHYS-L@LISTS.NAU.EDU>
Date: Wednesday, February 17, 1999 11:06 AM
Subject: A question on inelastic relativistic collisions


Ed Schweber (edschweb@ix.netcom.com)
Physics Teacher at The Solomon Schechter Day School, West Orange, NJ
To obtain free resources for creative physics teachers visit:
http://www.physicsweb.com

Hi gang:

The problem with the week long February break many schools have, besides
that it wastes class time, is that it gives us teachers too much time to
think.

Anyway, I was wondering: Since in special relativity, momentum and
energy
are merged into a single four vector whose components will be different for
different moving observers, does that mean that the amount of heat lost in
an inelastic collision is relative to the observer?

If so, what does that do to specific heats and latent heats. Do these
also become relativistic quantities?

What about Kelvin temperature. Is that absolute or relativistic? Kelvin
temperature is proportional to the average KE per molecule and would not
that average KE be relative to the motion of the observer?

Thanks in advance for any input and I promise in the future that I will
stop trying to think.

Ed Schweber