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[Phys-L] OLD --Re: conservation of momentum



Pair production:

On Aug 31, 2016, at 9:00 AM, John Denker <jsd@av8n.com> wrote:

In the early going, the electric field of the two leptons, as
observed from far away, is zero. That's because they have opposite
charges, at initially the same position. The reaction on the proton
is initially zero, as it should be.


The leptons overlap 100% ?

When they separate?


If you do it right, momentum is conserved. Always. Everywhere.
Strictly. Locally.

The leptons are not point objects? therefore, locally means at their edges?



Does the “effect” travel at C or sl. slower thru the leptons?

bc uneducated


I think I understand the description of the effect on the proton.

=============================

Executive summary:
-- Momentum is conserved.
-- Momentum is always conserved.
-- Momentum is strictly conserved.
-- Momentum is locally conserved. That means no matter where you
look, momentum is conserved right here, right now.

++ You have to account for all forms of momentum.

On 08/30/2016 07:48 AM, Moses Fayngold wrote:

"action-reaction" concept is most simple and fruitful only in
classical and static situation, in which case the notion of their
starting and ending together is meaningless.

The action/reaction concept is synonymous with conservation of
momentum. If it doesn't appear synonymous, you're doing it wrong.

In more realistic situations involving dynamics this notion does not
work or is, at best, not straightforward.

It does work. It always works.

Consider, for instance, an electron-positron pair production from
collision of 2 neutral particles in the field of a distant proton.

That's a complicated setup. It's neither simple nor straightforward,
for reasons having nothing to do with conservation.

Each member of the newly-born pair immediately feels the field of
the proton and experiences the corresponding action,

OK....

but the proton will start feeling their field (and respective
reaction) much later.

In the early going, the electric field of the two leptons, as
observed from far away, is zero. That's because they have opposite
charges, at initially the same position. The reaction on the proton
is initially zero, as it should be.

Later on, as they move apart, there is a changing electromagnetic
field. This must be analyzed using tools appropriate to the task.
In particular, there is RADIATION, and there is MOMENTUM IN THE
FIELDS. This is small, as it should be. It is just sufficient
to account for the small reaction on the proton.

It is a fundamental fallacy to focus attention to one small effect
while neglecting other equally-small effects.

If you do it right, momentum is conserved. Always. Everywhere.
Strictly. Locally.