The correct thing to say is that photons are massless, even if they have momentum and energy, which is what explains that they interact with matter: that is what happens in the photoelectric effect, for example, where photons knock electrons out of a metal plate.
This poses, however, two difficulties for us students:
First, how to reconcile this with the definition of momentum as "mass times velocity"? Doesn't this mean that one of the two things is wrong: either light is massless, in which case it should not have momentum, or it does have momentum, in which case it should have "some sort of" mass...?
Second, the famous formula:
which people wear in T-shirts... doesn't it mean that light, which is energy, is mass?
Given this, some physicists answer: well, light does have mass, it has the so-called "relativistic mass", that is why it interacts with things and that is why it has momentum. To justify this, they follow this route:
- The fully-fledged formula is actually not the one that you see in T-shirts, but this longer one:
- In the theory of relativity, momentum is not simply m*v, but is preceded by the so-called Lorentz or gamma factor:
- If you replace p in the first formula with its relativistic value, after a few algebraic operations, you get:
- In this new formula, m with a subscript (mo) is "rest mass", that is to say, the mass that a particle at rest with you might have. A free photon cannot be at rest with you because you cannot travel at the speed of light, but it can have relativistic mass, which is rest mass times the gamma factor:
- But then... if v in this formula is the speed of the particle and c is the speed of light, when you speak of a photon, it turns out that v = c, so the gamma factor is 1/0... If you add to that that the rest mass of the photon is also 0, the expression looks as follows: (1/0)0. Anyhow, it looks like an undefined expression. How do you make sense of this?
Well, you can ask those physicists who defend this approach. There are several explanations around, but I simply don't grasp them. I will only mention that the "excuse" is often that v tends to c and you get the undefined only in the limit, when v = c....
In turn, the majority of physicists reject the above reasoning and the very concept of "relativistic mass".
For them the famous formula,
is only applicable for particles at rest with the relevant reference frame, not for free photons. In other words, for massive particles or for a photon that is trapped within a massive object. In effect, a photon can contribute to the mass of an object if, for example, it goes into a cavity with perfectly reflecting mirrors: in this case, if you could measure with an incredibly precise instrument, you would be able to detect a tiny increase in the inertia of the object. However, as long as the photon is free, it has no mass, full stop.
The photon does have momentum, although its value does not come from the usual formula (mass times velocity), but from quantum mechanics. The reasoning is in particular:
- In the above full formula, replace m with 0 and simplify as follows:
- As per quantum mechanics, the Energy of a photon is its frequency f times the Planck constant h:
- So the photon's momentum is:
Said this, what is my point? Well, I want to highlight that the mistake of those "some physicists" lies at the step of replacing p with the formula for momentum (relativistic momentum, in this case) and what this error means, because this is a very common issue in many fields of knowledge.
You cannot do this move (if we were playing, we would say that it is an "illegal move") when you are talking about a free photon, because, if you do, you are using the concepts at hand ("free photon", "mass" and "momentum") beyond their respective domains of applicability, which in turn have been defined based on the observation of how nature works.
It is not strange, therefore, that you arrive at an undefined, i.e. a dead-end street, which is proof that you have previously taken a wrong turn. An undefined is an expression that breaches the rules to make meaningful mathematical expressions, so whenever you end up with one is because you took the wrong path.
And it is not an excuse to say that, "in the context of calculus, you often encounter undefined expressions, but this difficulty is overcome through the use of limits". It is true that in calculus, limits enable you to get rid of the undefined, but that is because the undefined should never be there from the start and what you do by resorting to limits is solving that initial trouble!
I will explain myself briefly because the matter deserves a calmer discussion and now I just want to outline it. When you seek, for instance, an instantaneous speed (a derivative), you are frustrated because we define speed as the ratio between space traversed and time elapsed. This is a logical thing to do, because that is the way to physically measure speed. Hence if something is instantaneous, you should divide by 0, which is mathematically forbidden... But you don't despair, because you understand that one thing is the operative definition of concepts (how in practice you obtain their value) and another is the reality that the concepts aim at grasping. In this case, such reality is a "state of motion" that can be perfectly growing continuously and thus has a different value at each instant. This reality surpasses measurement and surpasses the algebraic expression of a ratio. So the trick cannot be easier: get rid of the ratio, i.e. make a number of algebraic operations, so that the time interval is not placed anymore at the denominator of a fraction. Then it is said that you make such an interval tend to 0 and eliminate it. I would simply say that you make it equal to 0, because the concept itself, in its purest form, does not need of any time interval at all.
In conclusion, in calculus we get an undefined because we ourselves created a language problem (we used a ratio to refer to what is not a ratio in nature). So it is legitimate that we sort out the problem by trying to extract the underlying reality that our language was hiding.
Instead here, if we get an undefined, it is because we are being inconsistent with nature and with the concepts that we have adopted to reflect nature.
The key assumption is that there are things that can travel at the speed of light (we call them light, among other things) and others that can't travel at the speed of light (we call them massive). [I would add that sometimes the former (light) becomes the latter (mass) when it gets trapped among the walls of something massive. And I would like to generalize the idea to other forms of energy, but this would be a speculation beyond the scope of this post.]
Given this, the momentum of massive things is defined as mass times velocity (preceded by gamma factor, if you want to account for relativistic effects). Hence this concept is designed from scratch as a one being applicable only to objects having mass, that is to say, being slower than light. Therefore, if you plug this definition (gamma *mass * v) into the energy formula, it must be because you are assuming that you are in face of something massive, that is to say, whose v is < than c. Otherwise, if you were thinking of a free photon, you would have simply made the wrong move. So forget about playing with limits for justifying your mistake.
Last but not least, under this light, if free photons' momentum is just hf/c, couldn't we dispense with this concept and talk only about light's energy, especially bearing in mind that, if we use natural units (whee c = 1), the numerical value of momentum of light equals that of its energy? I initially thought that, but then I realized that energy is a scalar (it has just magnitude), whereas momentum is a vector (it has direction)... Thus the concept of photon's momentum comes useful when analyzing its collision with another particle, like an electron ("Compton shift"): after the collision, each particle takes a direction (it is said that the photon is "scattered" and the electron "recoils"), but both directions are correlated, because momentum must be conserved. Thus by observing the direction of the photon's scattering, one could deduce the direction of the electron's recoil... were it not for the fact that... how can we guess the scattering angle of the photon? This remark paves the way for an interesting reflection about the uncertainty principle and randomness, but that will be another day.
Regarding the last point, I held an interesting conversation with ChatGPT, which seems to confirm my understanding: I will reproduce it:
ReplyDeleteMy question: Can we say that the utility of the concept of momentum of a photon, which justifies that we do not have enough with the concept of energy, is that momentum, being a vector, gives you information about the direction of the photon?
The answer:
You are correct in pointing out that energy is a scalar quantity and momentum is a vector quantity. The equation E^2 = (pc)^2 + (mc^2)^2 does not directly extract the vector nature of momentum or yield a vector quantity.
In the context of this equation, the momentum term (pc)^2 represents the magnitude squared of the momentum vector, which is a scalar quantity. The energy term (mc^2)^2 represents the square of the rest mass of the particle, also a scalar quantity. When added together, these scalar terms on the right-hand side of the equation yield the scalar quantity E^2, which is the square of the total energy.
While this equation does not explicitly provide the direction of momentum, it does incorporate the conservation of energy and momentum in relativistic systems. By considering the total energy and momentum, one can analyze various physical phenomena, such as particle interactions, where the total energy is conserved even if the individual energies and momenta change.
To fully capture the vector nature of momentum, one would need to consider the individual components of the momentum vector along with their respective directions. This is typically done by working with the vector components explicitly using appropriate mathematical formalisms, such as the three-momentum vector or four-momentum vector in the context of special relativity.