Hi,
I was wondering whether you could suggest a site that derives E =
mc2 ... or whether you yourselves could derive it...I
was wondering why this particular proposition holds true...
Thanks
It can be derived from the law that the
mass of a particle at velocity v with rest mass m0
is:
m = m0 /sqrt(1 - v2 /c2 )
provided you allow yourself to make a few other assumptions (e.g.
conservation of energy). See the discusion Monkey on
a Rope , about half way down for two derivations (dated April
13 2000).
[Or read on... - The Editor]
I'm not sure I follow the step where integration occurs from u
= 0 to u = V/sqrt(1-V2 /c2 )
Thanks
I'll let Michael get back about the
integration in his proof, one thing I think should be mentioned
however, is that the famous formula is really only a special case
(at rest, so the momentum, p=0) of the result
E2 - c2 p2 = m2
c4
which can actually be proved fairly quickly but it uses
four-vectors, which takes a while to set up, so for the moment I
would concentrate on the kind of proof Michael gave in the monkey
discussion.
Sean
OK, might as well go from scratch. We
have a particle with rest mass m which is travelling in a
straight line, and has position vector x at a time t. It is acted
on by an external force F which is constant.
Its relativistic mass at time t is therefore m/sqrt(1 -
v2 /c2 ) where v = dx/dt. Applying
Newton:
F = d/dt(m/sqrt(1 - v2 /c2 ) x v)
Multiply both sides by dx/dt. Noting that F dx = dE where E =
total energy of rocket we get:
dE/dt = v d/dt(mv / sqrt(1 - v2 /c2
))
Now let u = v/sqrt(1 - v2 /c2 ):
dE/dt = mv du/dt
We also know that:
u2 = v2 / (1 - v2 /c2
)
So:
u2 - u2 v2 /c2 =
v2
v2 = u2 /(1 + u2 /c2
)
And as v> 0:
v = u/sqrt(1 + u2 /c2 )
So our equation is:
dE/dt = mv du/dt = mu/sqrt(1 + u2 /c2 )
du/dt
Then using the fact that (dE/dt)/(du/dt) = dE/du we get:
dE/du = mu/sqrt(1 + u2 /c2 )
Integrate this indefinitely:
E = mc2 sqrt(1 + u2 /c2 ) +
D
where D is a constant. So:
E = mc2 sqrt(1 + v2 /(c2
-v2 )) + D = c2 m/sqrt(1 - v2
/c2 ) + D
Now note that if we call M the total mass of the particle
then:
M = m/sqrt(1 - v2 /c2 )
so:
E = Mc2 + D
where D is constant. In other words the total energy of the
particle is in direct proportion to its mass, and the constant of
proportionality is c2 . This is, essentially, exactly
what E = mc2 says.
Then by using conservation of energy and conservation of mass you
can show E = mc2 holds on all occasions, not just when
you have a particle under a constant force. Actually, come to
think of it I haven't actually assumed F is constant above. Of
course it doesn't matter anyway.
Yours,
Michael
Also one thing worth pointing out about
Sean's equation:
E2 = m2 c4 + c2
p2
is that here m is rest mass. If you're dealing with particles
then it is always true to say:
E = mc2 where m is TOTAL (not rest) mass. So you don't
have to think of E = mc2 as a special case of
E2 = m2 c4 + p2
c2 because in fact the m's mean different things in
the two equations. I guess Sean's version is more useful if
you're dealing with light as well as just particles, but it is
nice to know that the total mass and the total energy are always
connected by E = mc2 even when p is not 0.
Although Michael is right, it turns out
in the long run that it is generally more useful to work with a
constant rest mass and a momentum 4-vector (E,p) than it is to
work with gamma's and total masses. For example, gamma doesn't
generalise well to GR but the momentum 4-vector does. And it is
conceptually more satisfying too, for instance the proper way to
write the equation is in fact
E2 - c2 p2 = m2
c4
where the left side is in fact the norm of the momentum 4-vector
(exactly as x2 - c2 t2 is the
norm of the position 4-vector (t,x) ). The right side is the norm
in the rest frame where it turns out that the 4-momentum is
(mc2 ,0). So all the equation is saying is that the
norms is conserved under Lorentz transformations which is in fact
the definition of a 4-vector, exactly as the norm of the position
4-vector is conserved under Lorentz transformations.
Sean