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

[Science>All]

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.

 

[ZapperZ]

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]

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]

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!

 

[sardar]

I would say that the concept of photons can be understood in both ways - as particles and as waves. The wave-particle duality of photons is a fundamental principle in quantum mechanics and it is important to understand that they can exhibit properties of both particles and waves depending on the experimental conditions.

When thinking about photons as particles, it is important to note that they do not have a rest mass like other particles such as electrons or protons. However, they do have a theoretical mass, as shown in the equations provided. This theoretical mass is related to the energy of the photon and is inversely proportional to its wavelength. This is why photons with shorter wavelengths (such as gamma rays) have more energy and therefore, a higher theoretical mass.

On the other hand, thinking of photons as waves helps to explain their behavior in phenomena such as interference and diffraction. These are characteristics of waves and can be observed in the behavior of photons as well. This is why your friend suggested thinking of photons as waves.

Ultimately, both perspectives are necessary to fully understand the nature of photons. They are both valid and can be used to explain different aspects of their behavior. As scientists, we must continue to study and learn about photons in order to gain a better understanding of their true nature.