Re: Mass

[Hugh Logan <hloganPHY@CFL.RR.COM>]

Edmiston, Mike wrote:

> And what exactly does E=mc^2 mean?
It is difficult to say until E and m are defined. There is a formulation
of special relativity, often used in popular treatments or in chapters
of modern physics texts largely intended for non-physics majors, in
which E is total energy and m is "relativistic mass." In this
formulation, the rest mass is m_0 and the rest energy is E_0. Then
E_0=m_0*c^2. With gamma=1/sqrt(1-v^2/c^2), m=gamma*m_0 and
E=gamma*E_0=gamma*m_0*c^2. Any energy in excess of the rest energy is
kinetic energy K, so that
E=E_0+K. If relativistic momentum is defined as a vector with magnitude
p=gamma*m_0*v, a little algebra allows one to arrive at the well-known
equation

E^2=p^2*c^2+E_0^2=p^2*c^2+m_0^2*c^4  .

This formulation has fallen out of favor in deference to the 4-vector
formulation in which the mass is invariant. (I grew up thinking that
mass increased with speed. This was something that almost everybody
knew.) This mass is the same as the rest mass in the older formulation,
but the subscript "0" is usually dropped, since this is the only mass in
town. Thus E_0=m*c^2. Some newer modern physics texts that don't use the
4-vector treatment refrain from using relativistic mass, writing the
total energy as
E=gamma*m*c^2, m being mass (which is rest mass). Neither can I find
"relativistic mass" in Einstein's
popular book, _Relativity, The Special and General Theory_, Crown
Publishers. His "m" is rest mass. He discusses how the rest mass of a
body is changed by an amount E_0/c^2 when the body gains or loses energy
E_0 (not the rest energy in his notation.) Einstein gives the total
energy mc^2/sqrt(1-v^2/c^2), expanding it  to compare it with the
classical kinetic energy, only later interpreting the rest energy, mc^2.
He often writes only the right side of the equation, so there is no
notation for rest mass or total mass in this book

In an essay, "E=MC^2," in _The Theory of Relativity and Other Essays_,
Crown Publishers, 1950,
p.14, Einstein explicitly states, "E is the energy that is contained in
a stationary body; m is its mass."

In the 4-vector formulation, a timelike displacement is given by the
4-vector, (c*dt, dx, dy, dz), the square of  whose norm is
c^2*dt^2-dx^2-dy^2-dz^2>0. If the object is at rest in this frame,
dx=dy=dz=0, so that the square of the magnitude is clearly  c^2*(d
tau)^2 where tau is proper time. Since the magnitude of (c*dt, dx, dy,
dz) is invariant under the Lorentz transformation, it is always true that

c^2*(d tau)^2=c^2*dt^2-dx^2-dy^2-dz^2. We will need to get dt/(d tau)
from this equation. Dividing by dt^2 and letting v_x=dx/dt,v_y=dy/dt,
and v_z=dz/dt ,

c^2*[(d tau)^2/dt^2]=c^2-v_x^2-v_y^2-v_z^2=c^2-v^2  or

[(d tau)/dt]^2=1-v^2/c^2. Inverting and taking the square root, dt/(d
tau)=1/sqrt(1-v^2/c^2)=gamma

Thus c* __d tau__=(c*dt, dx, dy, dz) where __d tau__ is a 4-vector with
magnitude (d tau).

The 4-velocity __u__ is defined as

__u__=__d tau__/(d tau)=[c*dt/(d tau), dx/(d tau), dy/(d tau), dz/(d
tau)]  so that

__u__=(gamma*c, gamma*v_x, gamma*v_y, gamma*v_z)=gamma*(c, v_x, v_y, v_z) .

Note that  norm ( __u__)=gamma*sqrt[c^2-v^2]=gamma*c*[1-v^2/c^2]=c

The momentum-energy 4-vector is defined as

__P__=m*__u__=[gamma*m*c, gamma*m*v_x, gamma*m*v_y, gamma*m*v_z], so that

norm(__P__)=m*norm(__u__)=mc . This could also have been seen by
considering a frame in which the object was at rest, so that gamma=1,
v_x=v_y=v_z=0, giving (mc, 0, 0, 0). The norm is obviously
mc, and it is invariant under a Lorentz transformation. We have

__P__=(gamma*m*c, p_x, p_y, p_z) where p_x=gamma*m*v_x, p_y=gamma*m*v_y,
and  p_z=gamma*m*v_z are the components of relativistic 3-momentum.

gamma*m*c=E/c as noted above. There are several ways to derive this from
scratch, and that
E=gamma*m*c^2 is true can be made plausible by expanding it with the
binomial expansion.

E=(1-v^2/c^2)^(-1/2)*m*c^2=(approx)=[1+(1/2)*v^2./c^2+(3/8)*v^4/c^4+...]*m*c^2
E=approx=m*c^2+(1/2)*m*v^2+ (3/8)*m*v^4/c^2+... . For small velocities,
this reduces to

E=m*c^2+K where K=(1/2)*m*v^2 in agreement with classical mechanics.

The 4-momentum can now be written as

__P__=(E/c, p_x, p_y, p_z). Knowing that norm(__P__)=mc, we have

m^2*c^2=E^2/c^2-p^2 or

E^2=m^2*c^4+p^2*c^2

as in the older formulation of special relativity.

>  Doesn't it mean mass and energy are
> equivalent?
Yes, if one regards c^2 as a conversion factor. As a matter of fact,
many authors use a system of units in which c=1 (dimensionless) and in
which length and time have the same units. Similarly, energy, momentum,
and mass have the same units. With these units, the above equations
could be simplified by setting c=1. This is done in _Spacetime Physics_
by Taylor and Wheeler and also in _The Meaning of Relativity_ by
Einstein.  In the latter Einstein gives the 4-vector treatment in which
the (total) energy appears as a component of the momentum-energy vector
:  E=m/sqrt(1-q^2) where q is the speed.
He considers the case in which q=0, leading to E_0=m, which he converts
to E_0=mc^2. But nowhere have I so far found Einstein to use
"relativistic mass" which increases with speed. Perhaps because it would
mess up the 4-vector formulation? From E_0=mc^2 he concludes (p. 47),
"Mass and energy are therefore essentially alike; The mass of a body is
not a constant; it varies with changes in its energy. ..."

Rick pointed out that Phil Morrison prefers to write E=mc^2  as E=m (as
does Einstein). I presume this is the equation for a body at rest. I
have seen so many elementary modern physics texts with m denoting
"relativistic mass" that I am never sure.

>
Jim Green wrote:

> Jack, inertial mass is just fine for basic intro problems, but it doesn't
> give me much insight into E=mc^2 and complimentary ideas.  Neither
> does it
> help me much with GR.
Einstein, in his book, _Relativity, The Special and the General Theory,
15th ed._ is explicit about this.
On p.47, he writes"If a body takes up an amount of energy E_0, then its
inertial mass increases by an amount E_0/c^2; the inertial mass of a
body is not a constant, but varies according to the change in the energy
of the body. The inertial mass of a system of  bodies can even be
regarded as a measure of its energy. The law of the conservationof the
mass of a system becomes identical with the law of the conservation of
energy, and is only valid provided that the system neither takes up or
sends out energy."
It seems reasonable that the relevant mass is inertial mass, since the
kinetic energy of special relativity reduces in the classical limit to
(1/2)*m*v^2 in which m is clearly inertial mass. I believe John Denker
mentioned that the norm of the momentum-energy 4-vector is inertial
mass. (My value above, mc, would agree in a system of units in which c=1.)

Hugh Logan