Actually a good question arising from that false proposition....While all mass is energy, not all energy is mass. eg: kinetic/potential energy.
Rest mass of a photon
- Thread starter Magical Realist
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Magical Realist
Valued Senior Member
Isn't the energy of a photon kinetic since it is a moving object?
And why it has no rest mass.
Magical Realist
Valued Senior Member
K.E. = 1/2 m v2
Again, while all mass is energy, not all energy is mass.
Well, yes and no. Look....
The state of a quantum system is given by a state vector, $$\psi$$ (say). Any potentially measurable property of this system is described as an operator $$O$$ acting on the state vector. Momentum is one such, as you say - in itself it requires no operand to be potentially measurable.
The actual measurements - actual momenta - are given as the eigenvalues of the operator acting on the state vector. As in $$O\psi= \epsilon \psi$$ where $$\epsilon$$ is a spectrum of possible of possible momenta of the system.
exchemist
Valued Senior Member
Yes, the momentum is the eigenvalue, not the operator.
Janus58
Valued Senior Member
That's the Newtonian expression for KE. We don't live in an Newtonian universe, but a relativistic one.
A Relativistic expression for KE is
KE = mc^2 (1/sqrt(1-v^2/c^2) -1)
But again with the stipulation that m is the rest or invariant mass, which a photon doesn't have.
The point is that science has moved on since Newton developed his equation. Our notions of mass, energy, time, and space have all been modified.
Of course we still teach Newton, as in most cases it gives good enough answers to be practical. If I want to calculate the KE of a baseball, Newton suffices because the uncertainty in the mass of a baseball would make more of a difference than that between using the Newtonian or relativistic equations.
No, that is not at all what I said. Momentum is a property of the system and is given by the momentum operator (in QM). The eigenvalues are the possible numerical values that this property is allowed to take.
DaveC426913
Valued Senior Member
Yeah. Newton hasn't been replaced, he's just had some new terms added. For most real-world applications (such as sub-relativistic speeds and sub-black hole gravity) the new variables effectively reduce to zero, so the simpler Newtonian equations work just fine if you set them to zero and drop them out.
Quarkhead:
I mean, for starters, if you wanted to start with Newtonian kinetic energy and then naively replace the velocity with the speed of light, then you'd end up with $E=\frac{1}{2}mc^2$ rather than $E=mc^2$. And besides, there's the whole problem with $m=0$ for photons.
In the case of momentum, the substitution you mentioned above is the momentum operator. Without a state vector to operate on, that does nothing.
I don't see how it you get from $p=mv$ to $E=mc^2$, according to how you describe it. It looks like you're taking a Newtonian formula for momentum and sort of waving your hands vaguely until you produce the relativistic rest energy formula. I'd say the actual derivation is not so obvious.
I mean, for starters, if you wanted to start with Newtonian kinetic energy and then naively replace the velocity with the speed of light, then you'd end up with $E=\frac{1}{2}mc^2$ rather than $E=mc^2$. And besides, there's the whole problem with $m=0$ for photons.
You're agreeing with exchemist, there, apart from the bit about needing no operand. You said yourself that the operator $O$ operates on a state vector $\psi$. That state vector is the operand that is required. The result of $O$ operating on $\psi$ is that you get a number multiplied by $\psi$, provided that $\psi$ is an eigenvector of the operator, of course.
In the case of momentum, the substitution you mentioned above is the momentum operator. Without a state vector to operate on, that does nothing.
Correct.
The momentum of a system is the outcome of a measurement made on that system. Until you measure it, the system may not have a definite momentum.
Correct.