Comment on The speed of light

• ttayh@lemmy.zip ⁨2⁩ ⁨days⁩ ago

  They have momentum, but not mass (ignore the other explanation, like yes, E=sqrt(m0^2c^4 + (pc)^2), but so what? m0=0 for photons)

  As you can see, momentum, p, is p=E/c, and we know that the energy for light is proportional to its frequency, f, E=hf (h is Plank’s constant). So, p=hf/c. When light is absorbed by a material momentum (and energy) conservation apply and it imparts p onto the object. If light is reflected it imparts 2*p, showing this is left as an exercize to the reader

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• BlackLaZoR@lemmy.world ⁨2⁩ ⁨days⁩ ago

  They have mass. Everything that has energy has mass. They don’t have inertial mass but it’s just part of the equation

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  • Zink@programming.dev ⁨2⁩ ⁨days⁩ ago

    I think you are stating that backwards.

    You can definitely say that everything with mass has energy. And yeah the two are kind of interchangeable.

    But that does not mean that a photon HAS mass. It just means that we can calculate how much mass its energy is equivalent to.

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  • porous_grey_matter@lemmy.ml ⁨2⁩ ⁨days⁩ ago

    They have momentum, not mass, in relativistic physics you need something more complex than p=mv to describe momentum.

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    • BlackLaZoR@lemmy.world ⁨2⁩ ⁨days⁩ ago

      Energy=mass always. Theoretically you can make a black hole out of light, or you can turn that photon energy into inertial mass by running light in a closed loop.

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      • porous_grey_matter@lemmy.ml ⁨2⁩ ⁨days⁩ ago

        No lol, there’s a momentum component to that equation which is just conveniently 0 for massive objects at rest, photons don’t have a rest reference frame and are governed by E=pc.

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      • dangrousperson@feddit.org ⁨1⁩ ⁨day⁩ ago

        nope, the full relativistic energy relation is:

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

        p is the momentum and for a (massive) particle at rest the momentum p=0, taking the square root if both sides you get the more familiar:

        E = m c^2^

        Now a photon (and any other massless particle) can’t be at rest, it is forced to always travel at the speed of light and since it is massless m=0 and the energy becomes (again taking the square root):

        E = p c

        When a particle has both mass and isn’t at rest (but not traveling anywhere close to the speed of light) the E = m c^2^ is much, much larger than the E = p c part (ignoring that a square root isn’t linear {Hello Dirac}). Because the speed of light c is such a a huge number that squaring it makes it even bigger. It is usually fair to say that mass is ‘equivalent’ to energy, but it isn’t strictly true and actually false for massless particles (or particles traveling close to the speed of light [velocity v~c] -> p=m v~m c -> E = pc ~ m c^2^ , which has close to c^2^ in it).

        So photons have energy, not because they have mass (where massive particles have most of their energy), but because they have momentum (p).

        You bring up the theoretical black hole from photons, which are called ‘Kugelblitz’ black holes, iirc. They (theoretically could) exist, not because photons have some sort if mass, but because Spacetime curves because of Energy, not mass. Again, for most regular objects, the vast majority of its energy comes from its mass and the momentum doesn’t play a huge role. But for photons all.the energy comes from its momentum, since it doesn’t have any mass.

        Source: my bachelors degree in physics, I suppose.

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