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One has to wonder about photon interaction, and when to think of them 
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#1
Jun210, 04:58 PM

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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. 


#2
Jun210, 05:24 PM

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P: 29,238

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. 


#3
Jun210, 07:05 PM

P: 2,466

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 . 


#4
Jun210, 07:44 PM

P: 906

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, [tex]E =\sqrt{p^2c^2 + m^2c^4}[/tex] This can be rewritten as, [tex]E = mc^2\sqrt{1+\dfrac{p^2}{m^2c^2}}[/tex] We can Taylor expand this in terms of p^2/m^2c^2, [tex]E = mc^2(1+ \dfrac{1}{2}\dfrac{p^2}{m^2c^2}\dfrac{1}{8}(\dfrac{p^2}{m^2c^2})^2+...)[/tex] 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, [tex]E = mc^2+ \dfrac{p^2}{2m}[/tex] 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 momentumindependent. 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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