How can photon have energy, but no mass if “m=E/c^2” (E= mc^2)
In: 42
Because E=mc^2 isn’t the full equation.
The fuil equation is E^(2)=(mc^(2))^2 + (pc)^2 where *m* is mass and *p* is momentum. When momentum is zero (or near zero) then the mass portion is dominant because it is squared again (note that it is **E^(2)=** and not **E=**).
Photons have momentum even if they have no mass.
The equation is E^2 =(mc^2 )^2 +(pc)^2 where m is the *proper mass*
This E=mc^2 is only correct if the thing is at rest in the frame of reference (so that p=0), or if m is interpreted as relativistic mass.
Both p and E can depend on the frame of reference. The thing that is constant is m, also known as proper mass. The proper mass of photon is 0.
In that past, physicists put emphasis on the concept called “relativistic mass”, which is a different m that is *defined* by E=mc^2 , in other word, m is defined to be E/c^2 so that the equation is always true. Later, physicists realized this isn’t a good concept, but this concept is still remain in popular science. The relativistic mass of photon can be non-zero.
E=mc^2 only applies to stationary objects. Light is never stationary, thus is never described by this equation.
I dont understand any of this, I just wanna give a shoutout to all you mathematicians og physicist – You rock
Equations are not always true. Equations are true under certain conditions. 1+1 = 2 is not always true. 1 ft + 1 in. = 13/12 ft. 1 coin flip + 1 coin flip = 0 coin flips.
The equation you used works for massive particles at rest. The photon does not have mass.
In certain scenarios, a photon can behave as if it has (or had) a mass equivalent to E/c^2. For example if it hits a surface and is absorbed, like on a solar sail, the momentum of the object it hits will change as if the photon had mass.
One way to put it is like this. Things get mass by interacting with the Higgs field. But photons just don’t. It’s like it’s out of reach for them.
Because E=mc² is not the complete thing. We get these results form using special relativity (and the Lagrangian formalism) to calculate the motion of a charged particle.
I will show a few interesting results and add some context if you are interested. So the stuff in brackets are not that important just some extra.
The “full” result is: E=mc² / (1-v²/c²)^0.5
It comes from this equation:
d/dt mc²/ (1-v²/c²)^0.5 = qEv
E here is the electric field and q is charge. Now qEv is the power of the electric field. And power is the time derivitive of energy. (Change in energy over time is power by definition.) So that thing after d/dt is energy.
This 1/sqrt(1-v²/c²) shows up all the time it’s the gamma factor. I’m going to use y to label it.
So if we plug in 0 for v we get E=mc². As y = 1 here.
We can define the momentum vector not as p=mv but, p=ymv. (This again would require more context so: When we derive the stuff we get something that looks almost like F=ma for the electromagnetic case where F=q(E+v×B) The Coulomb and Lorenz forces. And the other side is the time derivitive of this new modified momentum. So F is still dp/dt it’s just that p also gets gamma factored.)
In special relativity we like the four momentum. (vectors with space and time like components, 1 time and 3 space of course.) p is the spacial momentum and we also need a time like thing into it. (ymc, p) = (E/c, p). As E=ymc².
#Energy is c× the first component of four momentum.
Lets calculate its lenght squared (We call the scalar product of a vector with itself its lenght squared. Here we have two kinds of vectors, and upper and lower index variant and the rule is that you multiply the upper and the lower together. The difference boils down to the lower index variant having -p not +p. So thats where the minus sign comes from.) I’m gonna use p_4 for the four momentum but this is the only time I’m using it:
p_4 p_4 = (E/c)² – p² = (m²c²-m²v²)/(1-v²/c²) = m²c²
#Mass is the lenght of four momentum. (per c)
Another rather useful way of writing things is:
E²=(mc²)²+p²c² (The square is important cause momentum alone is a vector and we can only talk about its length in terms of energy.)
If m=0, E=|p|c. (This is quite useful when you calculate stuff like Compton scattering, where you can treat light as little balls and calculate their momentum using this equation. And for light E=hf is also true and momentum is what we really care about. Or you can also use it to calculate radiation pressure. )
And as you can see E/c = |p|. So (E/c)²=p². So for light if we plug this into the lenght calculation we did, which defines mass, (thats the big deal) p_4 p_4 = (E/c)²-p² = p² – p² = 0. Mass squared is that per c² so m²=0 for light.
This might be a bit complicated but at least now we understand why saying stuff like mass depends on velocity is conceptually wrong and this is how formally you can introduce momentum for light and now we defined what mass is.
Let’s leave the formula aside. Actually everything have mass.
The gravity on the Earth is 9.81. That means every object with weight will be down to surface. But depends on the weight, the different objects have different weight and therefore they down to surface with different speed. If we go to the Moon, it has different gravity and the objects will down to surface with different speed versus Earth.
So, if the gravity be bigger the lightest objects will be down to surface.
Let’s leave the formula aside. Actually everything have mass.
The gravity on the Earth is 9.81. That means every object with weight will be down to surface. But depends on the weight, the different objects have different weight and therefore they down to surface with different speed. If we go to the Moon, it has different gravity and the objects will down to surface with different speed versus Earth.
So, if the gravity be bigger the lightest objects will be down to surface.
Latest Answers