Register to reply

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

by Science>All
Tags: e=mc^2, einstien, light, mass, photons
Share this thread:
Science>All
#1
Jun2-10, 04:58 PM
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=(4.1356673310^-15)((299,792,458)/λ)
E=(4.1356673310^-15)(299,792,458)(1/λ)
E=(4.1356673310^-15)(299,792,458)(λ^-1)=mc^2
m=((4.1356673310^-15)(299,792,458)(λ^-1))/(c^2)
m=(4.1356673310^-15)(299,792,458)(λ^-1)(1/c^2)
m=(4.1356673310^-15)(299,792,458)(λ^-1)(c^-2)
m=(4.1356673310^-15)(299,792,458)(λ^-1)(299,792,458^-2)
m=(.00000000000000413566733)(299,792,458)(.0000000000000000111265006)(λ ^-1)
m=(1.3795101410^-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 (1.3795101410^-23) which I dub, were it to have any scientific ground to it, Demitri constant.
Phys.Org News Partner Physics news on Phys.org
'Squid skin' metamaterials project yields vivid color display
Team finds elusive quantum transformations near absolute zero
Scientists control surface tension to manipulate liquid metals (w/ Video)
ZapperZ
#2
Jun2-10, 05:24 PM
Emeritus
Sci Advisor
PF Gold
ZapperZ's Avatar
P: 29,242
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.
cragar
#3
Jun2-10, 07:05 PM
P: 2,468
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 .

arunma
#4
Jun2-10, 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 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!


Register to reply

Related Discussions
Absorbed photons, emmited photons Introductory Physics Homework 2
Universe is made of baryonic matter Cosmology 7
What happens to photons in a magnetic room? General Physics 1
Make photons into some material? General Physics 7
Are Virtual Photons Responsible For Real Photons? Quantum Physics 11