Derive E = MC^2
Jonathan
For anyone that hasn't gone through this exercise, this
is a great example of combining two concepts, Newton's
Second Law, and the Lorentz transformation into a single
concept that ...changed the world.
Force equals mass times acceleration
F = MA
The work done on a body equals the force integrated
over the distance it moves from xo to x1.
W = | Fdx (from xo to x1)
That work turns into the body's kinentic energy
K = | Fdx
But force changes momentum
F = dp/dt
Substituting
K = | (dp/dt) dx
Then
K = | (dx/dt) dp
But
dx/dt = v
So
K = | vdp
And momentum P depends on mass
P = Mv
Which also changes with speed
If the Lorentz factor is 1/(1 - v2/c2) ^1/2
Then
P = Mo v/(1 - v2/c2) ^ 1/2
Gives
dp/dv = Mo/(1 - v2/c2) ^ 3/2
Then
dp = Mo/(1 - v2/c2) ^ 3/2 dv
So Kinetic Energy may be expressed as an integral
over changing speed, from rest until the force stops
pushing.
K = | Mo v /(1 - v2/c2) ^3/2 dv
Anti-differentiating gives
K = Mo c^2/ (1 - v2/c2) ^1/2
(from zero to one)
K = Mo c^2/(1 - v2/c2) ^ 1/2 - Mo c^2/(1 - 0^2/c2)^1//2
K = Mo c^2/(1 - v2/c2)^1/2 - Mo c^2
At low speeds
K = 1/2 Mv^2
But at high speeds, the curve for increasing energy begins
to look just like the curve for increasing mass.
K = Mo (1/(1 - v2/c2)^1/2 c2 - Mo c^2
Since the Lorentz factor is
1/(1 - v2/c2)^1/2
Then
K = Mo 1/(1 - v2/c2)^1/2 c^2 - Mo c^2
At rest
Mo = M
So
K = Mc^2 - Mo^2
At any speed the kinetic energy is equal to the
change in mass times c^2.
K = ( M - Mo) c^2 is the kiinetic energy at any speed
Since Mo is the mass of the quantity as rest, then
Eo = Mo c^2 Is called the rest-mass energy
Adding the kinetic energy to the rest-mass energy
gives the ...total energy of the body.
K = (M - Mo) c^2
+ Eo = Mo c^2
--------------------
= E = M c^2
s
The Mechanical Universe
http://www.malcolmgin.com/blog/2008/10/07/caltech-the-mechanical-universe-series-on-google-video/
Jonathan
s
Arindam Banerjee
Herein the bullshit.
Cheers,
Arindam Banerjee
> The Mechanical Universehttp://www.malcolmgin.com/blog/2008/10/07/caltech-the-mechanical-univ...
>
> Jonathan
>
> s
Zinnic
wrote:
kinetic energy (1/2MV*2) in order to 'invent' a perpetual motion
machine.
>
Ray Vickson
> For anyone that hasn't gone through this exercise, this
> is a great example of combining two concepts, Newton's
> Second Law, and the Lorentz transformation into a single
> concept that ...changed the world.
>
> Force equals mass times acceleration
>
> F = MA
No, not in relativistic mechanics. Force = rate of change of momentum;
that is, F = (d/dt)[ m*u/sqrt(1-\u|^2/c^2)], where u = velocity. If
you set M = m/sqrt(1-\u\^2/c^2), this is F = M*du/dt + u*dM/dt. The
first term is Ma, but the second one is not.
R.G. Vickson
> The Mechanical Universehttp://www.malcolmgin.com/blog/2008/10/07/caltech-the-mechanical-univ...
>
> Jonathan
>
> s
Jonathan
"Ray Vickson" <RGVi...@shaw.ca> wrote in message
news:b0123e3b-d9ed-4b6c...@o21g2000prn.googlegroups.com...
On Mar 2, 7:40 pm, "Jonathan" <Em...@Yahou.net> wrote:
> >
> > Force equals mass times acceleration
> >
> > F = MA
> No, not in relativistic mechanics. Force = rate of change of momentum;
Right! A couple of steps down you'll see how he
got from F = MA to F = dp/dt.
Jonathan
"Arindam Banerjee" <banerjee...@gmail.com> wrote in message
news:abeead9d-b379-4821...@d26g2000prn.googlegroups.com...
> > So
> >
> > K = | vdp
> >
> > And momentum P depends on mass
> >
> > P = Mv
> >
> > Which also changes with speed
> > If the Lorentz factor is 1/(1 - v2/c2) ^1/2
> Herein the bullshit.
> Cheers,
> Arindam Banerjee
Can you be more specific? The only pause the Lorentz
transformation gives me is that I believe as one appoaches
the speed of light, the body must be in the process
of changing states from matter to energy.
At the speed of light the body must become as much
matter as it is energy, acting as both a particle and
wave at the same. And wouldn't that transition essentially
be a paired property? When the more one knows about
one property, the less can be known about the other
as in the Uncertainty Principle?
So how can it be possible for a single deterministic equation
to fully describe the transition state between two entirely
different states, matter and energy, at the same time?
I just think it's incomplete though, not wrong.
Arindam Banerjee
> "Arindam Banerjee" <banerjeeadda1...@gmail.com> wrote in message
>
> news:abeead9d-b379-4821...@d26g2000prn.googlegroups.com...
>
> > > So
>
> > > K = | vdp
>
> > > And momentum P depends on mass
>
> > > P = Mv
>
> > > Which also changes with speed
> > > If the Lorentz factor is 1/(1 - v2/c2) ^1/2
> > Herein the bullshit.
> > Cheers,
> > Arindam Banerjee
>
> Can you be more specific?
Please check out:
http://www.adda-enterprises.com/MMInt/MMInt.htm
and if you search in usenet using keywords as Arindam, mass, energy,
Newton, IFE, etc. you will find lots and lots of articles I have
written over the past 11 years.
Cheers,
Arindam Banerjee
> > s- Hide quoted text -
>
> - Show quoted text -
Wayne Throop
: Can you be more specific? The only pause the Lorentz transformation
: gives me is that I believe as one appoaches the speed of light, the
: body must be in the process of changing states from matter to energy.
Except of course that the whole point of relativity is that the
body doesn't change squat. At any time it's the selfsame object,
in the selfsame physical state, but just described from two different
perspectives, in one of which it is motionless, and the other one of
which it is moving near lightspeed. Since it's the same object, and
the only thing that changed is who's looking at it, there's no
"converting" going on (in terms of the actual object's state,
as opposed to the relationship of the object to the coordinates
it's described in).
: At the speed of light the body must become as much matter as it is energy,
Nah, that's pretty much nonsense. Just near-random word-salad.
Tossed with a light vinaigrette dressing.
In special relativity, no matter how much an object accelerates,
it's as far away from the speed of light as it was when it started.
Light is always zipping past it at exactly c, no matter how hard
it runs the red queen's race.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
rasterspace
is it true that Ariandam is alike to Neinsitein,
in proferring momentum as force?
:ScrollinG:
positive cloud-feedback results in the transport of water,
ultimately to the poles. the question is,
how does the heat actually dissipate from them,
in line with the actuality that they are not warming
per the spherical geometry-ignorant GCMers?
isn't most soot from crop-residue & forest fires?...
the supposed damaging effects of "gronudlevel ozone" may
be based upon LD-50 tests on helpless lab critters;
I love the smell of ozone, whenever!
in distinction to my Congressman's old '91 cap&trade nostrum,
"let the arbitrageurs jack the price of energy,
as much as they can; free beer, free market, free trade, freedom!"
Rock Brentwood
> > Force equals mass times acceleration
>
> > F = MA
>
> No, not in relativistic mechanics. Force = rate of change of momentum;
> that is, F = (d/dt)[ m*u/sqrt(1-\u|^2/c^2)], where u = velocity.
This is also wrong. The correct answer is: "No, not in relativistic
mechanics NOR in non-relativistic mechanics." The correct statement of
the law is independent of paradigm and applies across the board to
both. The form which Newton originally stated the law, in fact, was
what we now write as F = dp/dt, calling p the quantity of motion and
stating the 2nd law as equating the force to the rate at which the
quantity of motion changes with respect to time.
More generally, for any system in any setting (relativistic, non-
relativistic; classical or quantum; elementary or composite system)
the correct statement is given in terms of its formulation as a
Lagrangian or Hamiltonian system, namely that for each degree of
freedom (indiexed by a = 1, 2, ..., N) the corresponding momentum p_a
and force F_a are related by F_a = dp_a/dt. For a Lagrangian system
the total derivative of the Lagrangian is given by dL = sum(p_a dv^a +
F_a dq^a) in terms of the configuration coordinates q^a and their time
rates of change v^a; and for a Hamiltonian system it's given by dH =
sum(v^a dp_a - F_a dq^a), with the first order laws v^a = dq^a/dt, F_a
= dp_a/dt applicable in both types of system.
maxwell
Newton's genius was to realize that he could create new concepts, like
momentum P, by conflating two simpler ideas: in this case, velocity V
& Galileo's inertial mass M and map this combination using DesCartes'
new algebraic notation; eg P = M * V. His physics (Law II) then
defined impulse DI as the REASON (external cause) for the change in
momentum; DI -> DP or using algebra: DI(t) = DP(t) = P(t + Dt) -
P(t). He then proposed that a third concept (force F) could now be
defined: DI = F * Dt. For calculational purposes he ASSUMED that the
series of impulses occurred so closely together that a continuous
limit (Dt -> 0) could be defined. There's an awful lot of math hiding
behind this approach, hence his title: The MATHEMATICAL Principles of
Natural Philosophy. Unfortunately, these ideas were mainly picked up
& developed by mathematicians & not natural philosophers, who were
much more comfortable with words, hence the explosion of new math,
like calculus etc.
willia...@mokenergy.com
hanson
<willia...@mokenergy.com> wrote:
> <http://www.osti.gov/accomplishments/nuggets/einstein/speedoflight.html>
>
Here are some more nuggets:
<http://tinyurl.com/E-mc2-existed-before-Einstein>
<http://tinyurl.com/How-Einstein-stole-E-mc-2>
<http://tinyurl.com/Kwublee-views-Einsteins-Theft>
< http://www.italian-american.com/depretreview.htm>
and --- Einstein arrested, twice ---
<http://tinyurl.com/Einstein-wife-beater-arrested>
<< http://tinyurl.com/Blouse-Plumber-Einstein >>
