Why photon is mass-less?
- Thread starter Saint
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A photon has zero rest mass, but of course it never is at rest. It's energy is a result of its momentum. SR tells us that anything moving at "c " does not have non zero rest mass.
Janus58
Valued Senior Member
That formula assumes that Newtonian physics strictly holds. However, we don't live in a strictly Newtonian universe, but a Relativistic one .
Even for something with a non-zero rest mass, it is not correct (though in most everyday cases it gives a good enough answer to be useful in practice.)
For photons, the momentum is found by h/wavelength, with h being Planck's constant.
One way to think about it is that the momentum of something is mv + a "bit more" and that "bit more" is due to the energy of the object.
While a photon has a m of zero, and thus no mv component to its momentum, it still has energy, and thus that "bit more" component of momentum.
exchemist
Valued Senior Member
That equation is ONLY for objects at rest relative to the observer. A photon isn't. The full formula is: E² = (mc²)² + p²c², in which p is the momentum of the object, as seen by the observer and m is the rest mass.
If p=0, as it does for an object at rest, this reduces to E=mc².
If however the object is moving and has zero rest mass, which is the case for a photon, it reduces to E=pc*. So the "strange", that is to say non-Newtonian, thing about a photon is that it has momentum even though it has zero rest mass. But this is what we observe, so it seems to be right.
*Applying de Broglie's QM relation for momentum and wavelength (p=h/λ) and the relation between frequency, wavelength and velocity for any wave (c=νλ), you get that p=h/(c/ν) =h.ν/c.
So using that to replace p, E=pc becomes E = h.(ν/c).c = hν. E=hν is Planck's relation, one of the fundamental relations of QM, which describes the connection between the energy in radiation and its frequency.
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Photons aren't energy. Neither is mass. But let's not get into that here.
You already know that there are types of energy that aren't associated directly with mass, right? So, it turns out that photons have energy that isn't dependent on them having mass. Enough said.
Only approximately, and only for things that have mass. It doesn't apply to photons.
It's just the way that nature works. The aim of science is to describe nature. Science isn't prescriptive, it's descriptive. If you ask questions like "why are there photons?" or "Why can massless things have momentum?", there are no "final" answers. It's just the way the universe is.
Typically, photons disappear when they are absorbed by matter. Their energy is transferred into some other form in the process.
Yes.
Halc
Registered Senior Member
It seems the better way to put it is that momentum is really mv, and photons have mass, and are thus attracted by gravity.
What they lack is proper mass. Energy and mass are different forms of the same thing and there cannot be one without the other.
The mass of any object (rock photon) is frame dependent, and thus its momentum is also frame dependent.
Photons/light simply follow geodesics in curved/warped spacetime. I'm fairly sure you have been told this before?
I don't think there's much to be gained from introducing the concept of "relativistic mass" or any velocity-dependent mass. It leaves less room for unnecessary confusion if we just talk about rest mass, because rest mass is a frame-independent quantity.
Momentum is a frame-dependent quantity in relativity, but it is already frame-dependent even in Newtonian/Galilean physics. If the "relativistic momentum" is $\gamma m\vec{v}$, nothing much is gained by defining $m_R=\gamma m$ and calling it "relativistic mass". It just tends to obscure the real reason that $\gamma$ is there.
Halc
Registered Senior Member
But a photon doesn't have proper mass (same as rest mass). It isn't that it has zero proper mass, but rather that there is no valid frame in which a photon is stationary, hence the proper mass not even being defined. If momentum as mv is actually (relativistic, or inertial) mass and not proper mass, it works fine, and no need for complications like $\gamma m\vec{v}$. I admit this goes against the comment by Janus who says that p=mv works only under Newtonian physics, but only if proper mass is used for m. The alternate p=h/wavelength (per Janus again) is the more proper way to measure momentum since m for a photon cannot be directly measured. It does have inertial mass since light can be used to exert force on an object. A photon can be accelerated (change in wavelength) by exerting force (probably gravity) on it.
Janus58
Valued Senior Member
Again, the idea the gravitation interaction is just due to mass(rest mass), is a Newtonian physics idea. Under Relativity, gravity is due to the Stress-energy tensor, and this includes energy as a participant in the gravitational attraction. Photon's have energy, and thus are effected by gravity.
Yes.
In Newtonian gravity, the relevant equation is the force law:
$F=\frac{GMm}{r^2}$
In general relativity, the relevant equation(s) are the Einstein field equations:
$G_{\mu \nu}+\Lambda g_{\mu \nu} = \frac{8\pi G}{c^4}T_{\mu \nu}$
The $G$ on the right-hand side of the Einstein equations is the same constant that we find in Newton's equations - the universal gravitational constant. The $G$ you see on the left-hand side of the Einstein equations is the Einstein tensor, which (roughly speaking) describes the geometry of spacetime, while the $T$ on the right-hand side is the stress-energy tensor that describes all the things that cause spacetime to curve. Oh, and the second term on the left-hand side has something to do with dark energy, or else it is a "cosmological constant", depending on which side of the equals sign you prefer to put it.
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Not directly. Energy and momentum are both related to Newtonian ideas of force, though. In many ways, modern physics doesn't really use the notion of force directly.