Gamma waves and visible light?

[BoJangles2]

Ok it’s been a while since I’ve asked a question and I'm probably waaay overdue.

We know light has mass (light is energy, m=e/c^2) do different waves have different energy?

I.e. do gamma waves weigh more than visible light? It seems it must?

Or am I just getting confused here?

 

[MeteorWayne]

Not an easy question to answer in the proper terms. This sounds like one for the good Dr.

Hes higher energy (UV, Xray, Gamma etc) do contain more energy than lower energy EM radiantion (visible, infrared, radio)

And yes, since mass=e/c^2 the equivilent mass is less.

That's about as far as I can go without getting in trouble, though

 

[drwayne]

Photons of different wavelength have different energy (and momentum), since they do not have a rest mass*
per se, they do not have different "rest masses".

The energy relationship looks like this

E = h * nu = h * c / lambda

where

h is Planck's constant
nu is the frequency
lambda is the wavelength.

*or very small rest mass. Interesting experiments have been done over the years to establish "upper limits" on zero here

 

[DrRocket]

Photons of different wavelength have different energy (and momentum), since they do not have a rest mass*
per se, they do not have different "masses".

The energy relationship looks like this

E = h * nu = h * c / lambda

where

h is Planck's constant
nu is the frequency
lambda is the wavelength.

*or very small rest mass. Interesting experiments have been done over the years to establish "upper limits" on zero here

From that formula you can also apply E-mc^2 to calculate a "mass". When astrophysicists catalog the available mass in the universe, for purposes of attempting to find the curvature for instance, the mass tied up in photons is counted. There is quite a bit in that form.

 

[drwayne]

I am aware of what some refer to as "energy equivalent mass" - I tried to put "rest" in front of my masses
(I missed one) to distinguish what I was trying to say. I'll put the other one to rest here in a bit.

Wayne