# One has to wonder about photon interaction, and when to think of them

by Science>All
Tags: e=mc^2, einstien, light, mass, photons
 P: 1 One has to wonder about photon interaction, and when to think of them as a particle and when as simply a wave. A friend of mine told me to think of them as a wave, because they are without mass. But why should I view it that way, when even theoretical mass is mass, such as weak force. So how might one think of them? And what about this: E=mc^2 Energy of one photon: E=hv E=h(c/λ) E=(4.13566733×10^-15)((299,792,458)/λ) E=(4.13566733×10^-15)(299,792,458)(1/λ) E=(4.13566733×10^-15)(299,792,458)(λ^-1)=mc^2 m=((4.13566733×10^-15)(299,792,458)(λ^-1))/(c^2) m=(4.13566733×10^-15)(299,792,458)(λ^-1)(1/c^2) m=(4.13566733×10^-15)(299,792,458)(λ^-1)(c^-2) m=(4.13566733×10^-15)(299,792,458)(λ^-1)(299,792,458^-2) m=(.00000000000000413566733)(299,792,458)(.0000000000000000111265006)(λ ^-1) m=(1.37951014×10^-23)(λ^-1) m=(.0000000000000000000000137951014)(λ^-1) By this logic, and by no means do I claim it to be infallible, photons do have a theoretical mass inversely proportional to its wavelength and multiplied by the constant (1.37951014×10^-23) which I dub, were it to have any scientific ground to it, Demitri constant.
 Mentor P: 27,572 You make a mistake that is so common, we have a FAQ entry on it. Please review the FAQ thread first in the General Physics forum. Zz.
 P: 2,422 the full equation is E^2=(mc^2)^2 + (pc)^2 the mass of a photon is zero so it then becomes E^2=(pc)^2 then E=pc hc/(lambda)=pc then (lambda)=h/p which is the debroglie hypothesis .
P: 907

## One has to wonder about photon interaction, and when to think of them

Photons have a mass identically equal to zero. The equation E mc^2 cannot be applied to massless particles such as photons. The short explanation is that photons have momentum, but no mass.

The longer explanation has to come with the derivation of E = mc^2. As Cragar said, the actual equation for a massive particle is,

$$E =\sqrt{p^2c^2 + m^2c^4}$$

This can be rewritten as,

$$E = mc^2\sqrt{1+\dfrac{p^2}{m^2c^2}}$$

We can Taylor expand this in terms of p^2/m^2c^2,

$$E = mc^2(1+ \dfrac{1}{2}\dfrac{p^2}{m^2c^2}-\dfrac{1}{8}(\dfrac{p^2}{m^2c^2})^2+...)$$

When we assume a small momentum (in nonrelativistic cases the momentum is always much smaller than the mass times c), we can just take the first two terms,

$$E = mc^2+ \dfrac{p^2}{2m}$$

One can recognize the second term as the formula for kinetic energy. However, in the nonrelativistic limit we recover the peculiar mc^2 term, which seems to be momentum-independent. This is why we say tha E = mc^2. Note also that in the derivation, we must assume that p << m, and this certainly isn't true for a massless particle. E = mc^2 only works when you understand that a particle can have momentum but no mass.

Hope this helps!

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