Question

how can photons have no mass but momentum bcuz p=mv?

Answer 1

hiii mate

hope this helps you



OK, that question is disingenuous, you were probably told that by your physics teacher. At least you were introduced to momentum with an equation such as p⃗ =mv⃗ p→=mv→ and perhaps told that this is what momentum is. But that is only how to calculatemomentum. Sometimes ‘how to calculate X’ is the best way to define X… but this is not one of those times. Let’s go back a bit to the original motivation for the concept of momentum and a basic understanding.

Momentum is, roughly and conceptually speaking, the amount of motion something has. I sometimes introduce it as how much ‘oomph’ there is. For regular objects it is pretty easy to see where that equation comes from then…

amount of motion = amount of stuff ×× how that stuff moves
p⃗ =mv⃗ p→=mv→

But mass is ‘the amount of stuff’ only if the stuff we have in mind is matter, and matter is not the only kind of stuff. Light (and other ‘massless’ things) are ‘stuff’ too, but we must estimate the amount of ‘stuff’ present another way. The answers that talk about ‘relativistic mass’ etc. are trying to do this, but most physicists would acknowledge that this is a rather rough way to do this. It gets the job done, but it is a little iffy.

First, let’s understand that it is perfectly reasonable to expect light to have momentum if we can understand that the old p=mv is only a special case, that just summarizes the idea that momentum is the amount of motion. And I think we can all agree that light has motion!

So let’s see if we can look at this another way, still through the lens of relativity. In relativity motion, and hence momentum, in the usual sense of spacial displacement / time is not a complete concept… just as space is not a complete concept and needs to be linked with time into spacetime. We can have displacements in the x-direction, the y-direction, the z-direction and the time-direction. This defines something called the ‘4-velocity’:
ux=dxdτ,uy=dydτ,uz=dzdτ,ut=−dtdτux=dxdτ,uy=dydτ,uz=dzdτ,ut=−dtdτ

Since we can have motion in these different directions, so too we can have an ‘amount of motion’ in the x-direction, the y-direction, the z-direction, and the time-direction. The ‘amount of motion in the time direction’ is the energy of an object. (For this reason we sometimes call the 4-vector that describes this the ‘momenergy’!)

But if we look at light we hit a roadblock again… because light follows a ‘null’ geodesic, it is different from everything else because it does not have a ‘proper-time’ with which to determine motion... because it moves equally in space and time the two cancel out! So the ττ used above becomes uniformly zero, and dividing by zero is just yucky (to use a technical term.)

However we can get around this because that very fact means that the energy and momentum (adjusted by constants) are the same, and we know the energy of light. So we just need to convert the energy appropriate units and that will be the momentum!

To do this conversion we just need to use the number of metres there are in a second, in other words to use a conversion factor! And this conversion factor turns out to be one we know well… there are 2.99752458×1082.99752458×108 metres in one second… this is what ‘c’ really is (not a velocity but a conversion factor!) So the units of momentum are just the units of energy / c

Since light has energy and momentum equal we can find the momentum by just converting the energy to space units: 
p = E/c

TL/DR: It turns out that mass is not the only way that things can have momentum… because momentum is the ‘amount of motion’, and therefore light has momentum because it has motion. We can find the momentum of light using its energy and the fact that light follows a path equal in space and time (a ‘null geodesic’)