fatoomch said:
If E=mc^2 then m = 0. So how can photons have energy?
Does this have anything to do with E=mc^2 only applying to matter at rest?
Hurkyl said:
Yes -- that formula tells you the rest energy of the object. It says nothing about other forms of energy, such as kinetic energy.
The formula for "total" energy, that is the rest energy + kinetic energy, is:
E² = (mc²)² + (pc)²
(p, here, is the magnitude of the 3-momentum)
but here, there is two different and incompatible (if the velocity is not zero) expressions for the same symbol, "E". if "m" is only rest mass, then fatoomch's "E" is not the same as Hurkyl's "E".
Hurkyl is the mentor here and i am of much humbler stature, but the unqualified language used in his answer is disputable and, essentially, a
currently fashionable practice, but was not always so. somtime in the last 30 years, it became common practice among physicists to reserve the term "mass" for only
rest mass. i do not know if the term "relativistic mass" is used much at all recently, it seems out of vogue, but it hadn't always been that way.
just as how time dilates between two inertial observers moving relative to each other, so does mass and the equations relating the two are identical:
t = \frac{t_0}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}
m = \frac{m_0}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}
so, just as a clock wizzing past an observer appears to be ticking more slowly than if it were resting in the observer's reference frame, the mass of that clock (and the parts inside) appears to having a larger value (by the same factor) than it would if it were at rest in the observer's frame of reference. so now, what i mean by m is
relativistic mass
not rest mass (or "invariant mass") as is the usage of the term by Hurkyl (and, i guess, the textbooks currently out). for rest mass (or invariant mass), we used to used the symbol m_0 in relativistic contexts and that i am doing.
currently, it appears that rather than talk of the mass of an object changing as it's speed changes (relative to some observer), the leave that only to the expression for momentum which is commonly defined as
\mathbf{p} = m \mathbf{v}
but here m is what i mean for mass, the total relativistic mass so the momentum an object, in terms of the rest mass, is
\mathbf{p} = \frac{m_0 \mathbf{v}}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}
but here, Hurkyl and most other current physicists would say "m" instead of "m_0" meaning invariant mass and indeed "rest mass" or "m_0" simply is not very much in their usage and when they say "mass" they
always mean what i would call "rest mass". but, using the corresponding symbols, we agree on momentum. as the velocity of an object increases, its magnitude of momentum increases
faster or more than proportionally to the speed of the object.
when one derives an expression for relativistic kinetic energy, they get, using the "old" terminology:
kinetic energy T = m_0 c^2 \left( \frac{1}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}} - 1 \right)
which is T = m c^2 - m_0 c^2
which is interpreted as T = E - E_0
where E = m c^2 is the total energy of the particle (what Hurkyl calls "
E" above) and E_0 = m_0 c^2 is the "rest energy" (the energy the particle would have intrinsically if it were just sitting in front of you, same as what Hurkyl mentioned but with a different symbol). note that when the particle is not moving, the total energy and rest energy are the same, so the kinetic energy T is zero. in fact, you can rearrange the equation about to say the same thing as "total energy, that is the rest energy + kinetic energy":
E = E_0 + T where
and that expression for total energy is totally compatible with the expression
E^2 = \left( m_0 c^2 \right)^2 + \left( |\mathbf{p}| c \right)^2
if you keep your symbols straight. note i am using m_0 instead of m and i have to because, unless \mathbf{v}=0, there is no way for
E = m c^2
and
E^2 = \left( m c^2 \right)^2 + \left( |\mathbf{p}| c \right)^2
to both be true. if m is the rest mass, then the top equation is
only rest energy and not equal to the total energy in the bottom equation.
also it is not true (but currently in vogue) that the famous expression
E = m c^2
means only rest energy. if we understand m to be the relativistic mass, then E = m c^2 means total energy of a particle and, if it is a photon, that energy is the same as
E = h \nu = \hbar \omega.
so, using the semantics i have above, photons
do have mass because they have energy and the scaling factor to convert the energy to mass is c^{-2}. that is the effective mass of a photon is
m = \frac{E}{c^2} = \frac{h \nu}{c^2}
and this expression can be found in some textbooks, but they might be a few decades old. since the speed of a photon is definitively c, the momentum of the photon is
|\mathbf{p}| = m |\mathbf{v}| = \frac{h \nu}{c}
and that equation remains, even in todays texts.
now even if a photon has inertial mass, the rest mass is
m_0 = m \sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}
and since |\mathbf{v}| = c, then the rest mass m_0 must be zero.
Photons have mass, but they don't have rest mass. Their "invariant mass" is zero, and since it is the invariant mass that is a property of just the particle's nature and not a function of what speed it might have relative to any observer, then it is commonly said, particularly today, that photons are "massless particles".
that's the spin on this from a middle-aged electical engineer.
