Re: There's work, and then there's work

[David Rutherford <drutherford@SOFTCOM.NET>]

Bob LaMontagne wrote:
> David Rutherford wrote:
>
> > The water flowing out through the side hole is analogous to the
> > particles in the environment outside the wire that have picked up energy
> > from the particles inside the wire. The particles inside the wire have
> > transferred the part of their energy that's due to the transverse
> > component of their motion to the particles outside the wire. Therefore,
> > the calorimeter experiment measures _only_ the part of the energy of the
> > particles inside the wire that is due to the _transverse_ component of
> > their velocity. Similarly, the current experiment measures _only_ the
> > part of the energy of the particles inside the wire that is due to the
> > _longitudinal_ component of their velocity. The reason that the two
> > experiments come out with the same result is that, on average, the
> > transverse and longitudinal components of the velocity are equal.
>
> The two experiments you refer to are actually one and the same. The purpose of
> measuring the current through the resistor is so one can CALCULATE i^2 * R,

The current, _by definition_, considers only the component of the
velocity in the longitudinal direction (through the wire or the
resistor). Therefore, the energy calculated using i only gives the
energy resulting from the longitudinal flow.

> integrate it over time, and PREDICT the result of the calorimetry experiment.

No more than the purpose of measuring the change in heat in the
calorimetric experiment, and calculating the energy from that, is so one
can predict the result of the current experiment.

> OTOH, you can actually PERFORM the calorimetry experiment. One of these is theoretical,
> the other is a measured result - which is why they give the same value.

Both are a combination of theory and measurement. In the current
experiment, you _measure_ the current, then you calculate the energy
using the _theoretical_ relationship between current and energy. In the
calorimeter experiment, you _measure_ the change in heat, then you
calculate the energy using the _theoretical_ relationship between the
change in heat and energy.

> You are attempting to count the same result twice.

No I'm not. I'm attempting to count the energies resulting from the
transverse component of the motion (calorimeter experiment) and
longitudinal component of the motion (current experiment), separately,
then adding them together. Each experiment, individually, gives only
half of the total energy. In the calorimetric experiment, the
contribution of the longitudinal motion to the energy is discounted, and
in the current experiment, the contribution of the transverse motion to
the energy is discounted.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf