Can relativistic momentum and its conservation be derived fromconservation of Newtonian momentum and the Lorentz transformations?

On Apr 26, 8:09 am, Daryl McCullough wrote:
wrote:

The conservation of momentum (in either Newtonian or Minkowski
space-time) is an easy consequence of the classical Noether
theorem and the homogeneity of space(time).

That's an incomplete answer. If you know the Lagrangian,
then you can apply Noether's theorem to prove that the
momentum (defined as the variation of the Lagrangian with
respect to velocity) is conserved. But first you have to
find the Lagrangian. How do you derive a Lagrangian?

Since the first equation to start the derivation of the Lagrangian
method is:

** Action = integral(t1, t0)[L dt]

So, all you have to do is to look for something that will fit in.
shrug

Luckily, the Lorentz transform can be written into just one single
equation.

** c^2 dt”^2 – ds”^2 = c^2 dt^2 – ds^2

Where

** ds”^2 = dx”^2 + dy”^2 + dz”^2
** ds^2 = dx^2 + dy^2 + dz^2

If “ frame is observing itself, the above equation can be simplified:

** c^2 dt”^2 = c^2 dt^2 – ds^2

Or

** dt” = sqrt(1 – (ds/dt)^2 / c^2) dt

So, the Lagrangian is very obviously to be:

** L = sqrt(1 – (ds/dt)^2 / c^2)

Where

** Action = Time elapsed at “ frame

In the case of GR, this method of finding the Lagrangian also works.
Of course, this is the Lagrangian of geodesics. It should not be
confused with the Lagrangian that Hilbert pulled out of his ass to
derive the field equations. To this date, there is no understanding
of how Hilbert’s Lagrangian can be derived and how the Eisntein-
Hilbert action really means and why it has to be extremized. It is
all bull****. shrug

Daryl McCullough wrote:
Koobee Wublee wrote:

Since the first equation to start the derivation of the
Lagrangian method is:

** Action = integral(t1, t0)[L dt]

So, all you have to do is to look for something that will
fit in. shrug

Luckily, the Lorentz transform can be written into just
one single equation.

** c^2 dt”^2 – ds”^2 = c^2 dt^2 – ds^2

Where

** ds”^2 = dx”^2 + dy”^2 + dz”^2
** ds^2 = dx^2 + dy^2 + dz^2

If “ frame is observing itself, the above equation can be
simplified:

** c^2 dt”^2 = c^2 dt^2 – ds^2

Or

** dt” = sqrt(1 – (ds/dt)^2 / c^2) dt

So, the Lagrangian is very obviously to be:

** L = sqrt(1 – (ds/dt)^2 / c^2)

Where

** Action = Time elapsed at “ frame

In the case of GR, this method of finding the Lagrangian
also works. Of course, this is the Lagrangian of geodesics.
It should not be confused with the Lagrangian that Hilbert
pulled out of his ass to derive the field equations. To this
date, there is no understanding of how Hilbert’s Lagrangian
can be derived and how the Eisntein-Hilbert action really
means and why it has to be extremized. It is all bull****.
shrug

I consider what you've written as, at best, a heuristic. It's a way
of *guessing* a lagrangian. I don't really consider that a derivation
of the lagrangian.

Daryl doesn’t know what it is talking about as usual. shrug

Oh, come on, Koobee. Do you really not understand what it means
to give a derivation? Suppose the tables were turned, and *I*
had been the one who wrote your supposed derivation of the
relativistic langrangian. Would you have said: "Good job!"

Of course, Koobee Wublee would say “good job” if Daryl has shown
ingenious derivations and discloses amazing insights and revelations,
and this is one of the differences between Daryl and Koobee Wublee.
shrug

After Koobee Wublee has correctly identified Daryl’s idol Einstein the
nitwit, the plagiarist, and the lair as merely nobody but a nitwit, a
plagiarist, and a liar, the Einstein Dingleberries started to call
Koobee Wublee anti-Semitic. The forensic evidences are all in the
mathematics involved. Daryl has rejected any competence in Koobee
Wublee and tried to discredit Koobee Wublee every chance Daryl has,
but with scientific method and analysis, Daryl is striking out
everywhere. It is a pity that Daryl would even call himself a
computer scientist. What a fvckign joke, eh? shrug

In fact, Daryl would behave and has behaved childishly bitching about
the following.

“[Daryl has] called [Koobee Wublee’s derivation] a piece of garbage.
But for some reason, [Daryl has] much lower standards for something
[Koobee Wublee has] written than for something that someone else has
written.”

But even if you somehow knew that the lagrangian for a free
particle is

L = -mc^2 square-root(1-(v/c)^2)

No, the Lagrangian is unitless. Each Lagrangian relates the local
flow rate of time of the observed to each observer. Thus, the action
involved is none other than the elapsed time of the event. His
derivation of the Lagrangian in the content of SR is impeccable.
shrug

For the ones who are interested, the famous equation

** E = m c^2 / sqrt(1 – v^2 / c^2)

Can trivially be derived after arriving at the conserved quantity

** (v / sqrt(1 – v^2 / c^2))

Which is interpreted as the conservation of momentum. Well, that is
SR. Under GR, it is slightly more complicated. It is a little bit
tricky to get to the following. shrug

** E = m c^2 [g]_00 / sqrt([g]_00 – v^2 / c^2)

Where

** v^2 = (ds/dt)^2
** [g]_00 = Function of s

Under low curvature of spacetime (low speeds and weak gravity), the
above equation simplifies into orbital energy of a Newtonian system.

** Total energy = kinetic energy – sqrt((potential energy)^2

For reference, the Lagrangian under GR is

** L = sqrt([g]_00 – (ds/dt)^2 / c^2)

This exercise demonstrates how the conservation of four-momentum can
be derived if one is willing to get himself demystified. If not, the
haunted soul will be trapped in the limbo world of knowing the
Lagrangian method is the key but failed at delivery as Tom put it the
best in the second post of this thread.

“I doubt very much this (derivation of the conservation of four-
momentum) is possible (under the content of SR or GR), but it's
irrelevant (Tom is totally mystified) -- see below (the Lagrangian
method).”

shrug

What you wrote is not a derivation. You know that. You're just
being a jerk. An arrogant one at that.

Koobee Wublee has derived the relativistic momentum which Tom is
doubtful even that can be done. If you have found something wrong,
just point it out instead of bitching about something you do not
understand. shrug

The units have no bearing on the equations of motion.

Yes, if you understand the subject at all, the Lagrangian has no
restriction on what units it takes. shrug

However, the definition of canonical momentum is

p = @L/@v

It is not a definition but without Koobee Wublee’s derivation, it is
only a fvcking assumption. shrug

So if p is to have the normal units (mass times velocity)
then the Lagrangian must have dimension mass * velocity^2.

Wrong, again. shrug

The classical free particle lagrangian is L = 1/2 mv^2. Taking the
derivative with respect to v gives: p = mv.

Yes, but it is more involved than that. shaking head

The relativistic lagrangian is L = - mc^2 square-root(1-(v/c)^2.

No, it is not. The Lagrangian with no potential energy is simply.

** L = sqrt(1 – v^2 / c^2)

That satisfies as the density to the following action

** T’ = integral(L dt)

Where

** T’ = Total elapsed time of a remote event
** dt = Observer’s flow rate of time

[rest of ignorant rant snipped]

Get a life, Daryl. Relativity appears to be something you can never
understand. Quit fudging this and that. Take up on something less
mentally challenging. shrug

After all, Koobee Wublee has correctly identified all Einstein
Dingleberries as zombies under Orwellian ideology whe

** FAITH IS LOGIC
** LYING IS TEACHING
** NITWIT IS GENIUS
** OCCULT IS SCIENCE
** FICTION IS THEORY
** PARADOX IS KOSHER
** FUDGING IS DERIVATION
** BULL**** IS TRUTH
** MYSTICISM IS WISDOM
** BELIEVING IS LEARNING
** IGNORANCE IS KNOWLEDGE
** PRIESTHOOD IS TENURE
** CONJECTURE IS REALITY
** HANDWAVING IS REASONING
** PLAGIARISM IS CREATIVITY
** FRAUDULENCE IS FACT
** MATHEMAGICS IS MATHEMATICS
** INCONSISTENCY IS CONSISTENCY
** INTERPRETATION IS VERIFICATION

shrug

why do you put such a simple "norm" on PE?

unforunately, evn though "four-momentum" can
be mathematically conserved, the whole
idea of spacetime is an obfuscation, nothing
but mere phase-space. momentum cannot usefully be thought of,
without the concept of time, and this is clear
from the use of quaternions in SR,
where the "real, scalar" part is time,
clearly completely different from the "pure,
imaginary vector" part of space.

** *Total energy = kinetic energy – sqrt((potential energy)^2

For reference, the Lagrangian under GR is

** *L = sqrt([g]_00 – (ds/dt)^2 / c^2)

This exercise demonstrates how the conservation of four-momentum can
be derived if one is willing to get himself demystified. *If not, the
haunted soul will be trapped in the limbo world of knowing the

in other words, the modification of quaternions,
due to Gibbs, that engineers use
as "vector mechnics," wasn't really needed, although
it seems to have a pegagocical role!

where the "real, scalar" part is time,
clearly completely different from the "pure,
imaginary vector" part of space.

thank you.