Why is the photon's mass zero in classical equations for light?

[JulianM]

Mentor's note: this thread has been forked from https://www.physicsforums.com/threads/why-the-photon-has-no-mass.899792/#post-5912751

The correct equation, that is, the one which is always correct in special relativity, is:
E^2 = (mc^2)^2 + (cp)^2 (1)
p = momentum.
We know from classical electrodynamics that light's energy E is equal to cp. Then, from (1) you see that m must be zero.

lightarrow

But momentum = mass x velocity so that part of the equation is E2 = c.(mv)

If the mass is zero then that formulation also yields zero energy (which we know is not true)

 

[DrStupid]

But momentum = mass x velocity

This definition is based on another concept of mass. Today "mass" means invariant mass and that results in another equation for momentum.

 

[PeroK]

But momentum = mass x velocity so that part of the equation is E2 = c.(mv)

If the mass is zero then that formulation also yields zero energy (which we know is not true)

The momentum of a photon is given by $p = hf/c$. Not by $p = mv$.

In fact, $p = mv$ is a non-relativistic approximation of the momentum of a massive particle. The correct equation is:

$p = \gamma mv$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

 

[JulianM]

The momentum of a photon is given by $p = hf/c$. Not by $p = mv$.

In fact, $p = mv$ is a non-relativistic approximation of the momentum of a massive particle. The correct equation is:

$p = \gamma mv$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

So lightarrow was wrong?

 

[DrStupid]

So lightarrow was wrong?

No, he was and is still right.

 

[JulianM]

So lightarrow was wrong?

Regardless of whether you include gamma this equation still contains mass.
Anything multiplied by zero is still zero.

 

[PeroK]

For a photon $E = cp$. So, he is right and you are not.

$p = mv$, as I have already explained, is a classical equation for the momentum of a particle that is not valid in relativity - for any particle, massless or otherwise.

 

[DrStupid]

He said E = cp
I said p = mv
therefore E = c (mv)

Again, this "m" is an outdated concept of mass. It dates back to classical mechanics and is not used anymore because it is frame dependent in relativity. With the modern concept of mass (invariant mass) the equations for momentum and energy are

p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

Of course they cannot be used for photons (because v=c), but they result in equations like

p = \frac{{E \cdot v}}{{c^2 }}

and

E^2 = m^2 \cdot c^4 + p^2 c^2

which are valid for v=c and m=0.

 

[PeroK]

Regardless of whether you include gamma this equation still contains mass.
Anything multiplied by zero is still zero.

That equation is for a massive particle. It doesn't apply for a photon. In fact, it is not $0$ for a photon, it is undefined. Since the photon has zero mass and speed $c$ you would get:

$p = \frac{mc}{0} = \frac00$

Which is undefined. So, you have to look elsewhere for the equation that gives the momentum of a photon.

 

[JulianM]

This is the problem with asking something like "was lightarrow wrong"? How was I supposed to know what you thought was wrong? So, I read his post and found an error. Which I quoted above. He said:
Which is clearly a typographic error. That's where $p = cp$ came from.

So, back to your real question.

For a photon $E = cp$. So, he is right and you are not.

$p = mv$, as I have already explained, is a classical equation for the momentum of a particle that is not valid in relativity.

So what is the equation in relativity? It can't be just multiplying momentum by gamma as PeroK did (above). That still gives zero energy.

 

[PeroK]

So what is the equation in relativity? It can't be just multiplying momentum by gamma as PeroK did (above). That still gives zero energy.

The general equation for momentum is:

$p^2c^2 = E^2 - m^2c^4$

That works for both massive and massless particles.

For a massive particle the energy is $E = \gamma mc^2$

And, for a photon the energy is $E = hf$.

 

[DrStupid]

Since the photon has zero mass and speed $c$ you would get:

$p = \frac{mc}{0} = \frac00$

Which is undefined.

But the limit for $v \to c$ (and therefore $m \to 0$) is defined.

 

[JulianM]

The general equation for momentum is:That works for both massive and massless particles.

For a massive particle the energy is $E = \gamma mc^2$

And, for a photon the energy is $E = hf$.

How does

$p^2c^2 = E^2 - m^2c^4$

work for a massless particle if p= m.v

That would give (for v = c)
( mc^2).c^2 = E^2 - m^2.c^4
is equivalent to
E^2 = m^2.c^4 - m^2.c^4

There's something wrong, surely it cannot apply to a massless particle which posses energy

 

[PeroK]

How does

$p^2c^2 = E^2 - m^2c^4$

work for a massless particle if p= m.v

That would give (for v = c)
( mc^2).c^2 = E^2 - m^2.c^4
is equivalent to
E^2 = m^2.c^4 - m^2.c^4

There's something wrong, surely it cannot apply to a massless particle which posses energy

What will it take for you to accept that $p = mv$ is not a valid equation?

 

[Orodruin]

work for a massless particle if p= m.v

Again, as you have been told repeatedly, p = mv is not a valid equation relativistically. Repeating the same mistake does not make it true.

 

[JulianM]

Again, as you have been told repeatedly, p = mv is not a valid equation relativistically. Repeating the same mistake does not make it true.

Well now you have me very confused. How do we know when to apply relativistic equations or not?

 

[phinds]

Well now you have me very confused. How do we know when to apply relativistic equations or not?

For a massless particle, you have no choice. For massive particles, the relativistic equation is always true to but the slower the particle is, the less difference there is between the results of the classical equation and the relativistic equation. At human-scale speeds, the classical equation always gives results that are useful.

 

[PeroK]

Well now you have me very confused. How do we know when to apply relativistic equations or not?

In principle, the relativistic equations are always valid and the classical equations are an approximation. If that approximation is good enough, then you can apply the classical equations.

If the velocities involved are a significant proportion of the speed of light, then you definitely need the relativistic equations.

In practice, it depends how accurate you need your answer. Time calculations for GPS satellites must be so accurate that relativistic equations are needed despite the modest speeds. But, normal engineering calculations are usually more than accurate enough with classical equations.

 

[ZapperZ]

But momentum = mass x velocity so that part of the equation is E2 = c.(mv)

If the mass is zero then that formulation also yields zero energy (which we know is not true)

No, in THIS case, momentum is defined via the wavenumber, i.e.

p=ħk

So there is no requirement for a "mass" here.

Zz.

 

[Nugatory]

Well now you have me very confused. How do we know when to apply relativistic equations or not?

You can always apply the relativistic equations, and you will always get the right answer. They're more exact than the non-relativistic ones.

However, the relativistic equations are also more complicated and using them is more work, so we don't use them when the relativistic effects are small enough to ignore. For example: what is the kinetic energy and the momentum of a one-kilogram mass moving at ten meters per second? Try calculating that using the classical formulas and using the relativistic ones. Which was more work? And what difference did it make?

The classical formulas can be used any time the speeds involved are small compared with the speed of light (equivalently, $\gamma$ is so close to 1 that you're OK with the approximation $\gamma=1$). Clearly this is not the case for particles traveling at the speed of light, so you know that the classical formulas can't be used here.

 

[JulianM]

In principle, the relativistic equations are always valid and the classical equations are an approximation. If that approximation is good enough, then you can apply the classical equations.

If the velocities involved are a significant proportion of the speed of light, then you definitely need the relativistic equations.

In practice, it depends how accurate you need your answer. Time calculations for GPS satellites must be so accurate that relativistic equations are needed despite the modest speeds. But, normal engineering calculations are usually more than accurate enough with classical equations.

What should I use then for light which has been slowed to "non-relativistic" velocities? I presume that only relativistic equations would still have to apply?

 

[PeroK]

What should I use then for light which has been slowed to "non-relativistic" velocities? I presume that only relativistic equations would still have to apply?

The speed of light is always $c$. You can't slow it down.

 

[JulianM]

The speed of light is always $c$. You can't slow it down.

Sure you can. C is only for a vacuum.

 

[Orodruin]

Sure you can. C is only for a vacuum.

You are trying to open an entirely different can of worms here. The dispersion relation for light in a medium is not the same as that in vacuum. You really cannot say that light in a medium is massless.

 

[PeroK]

Sure you can. C is only for a vacuum.

That's a very different scenario. In a medium photons interact with the medium, are absorbed and may be re-emitted. That's what causes the apparent change in speed. You can't directly use equations for undisturbed motion in this case.

 

[Orodruin]

In a medium photons interact with the medium, are absorbed and may be re-emitted.

I must say that I really dislike this quite popular heuristic. I do not think it adequately describes what is going on and leads to misconceptions and misunderstandings.

Either way, I generally recommend staying away from ”photons” when discussing classical relativity and instead look at the dispersion relation of plane EM waves in the medium.

 

[Herman Trivilino]

So what is the equation in relativity?

$\vec{p}=\left(\frac{E}{c^2}\right) \vec{v}$

Well now you have me very confused. How do we know when to apply relativistic equations or not?

Your confusion is over when we can use equations that apply to massive particles versus equations that apply to massless particles. The equation I wrote above applies to both, as does the one previously mentioned in this thread: $E^2=(pc)^2+(mc^2)^2$.

The equation $\vec{p}=\gamma m\vec{v}$ applies only to massive particles. For low speeds $\gamma \approx 1$ and we can write $\vec{p}=m \vec{v}$.

Sure you can. C is only for a vacuum.

Up until now we have been discussing a single massless photon moving at speed $c$ through a vacuum. Once you start looking at collections of photons you see that the collection can and usually does have mass. Having a beam of light scatter as it passes through a medium involves the photons interacting with the atoms that make up the medium. You can try to understand that, but modelling the light as a collection of massive particles moving at speeds less than $c$ will not work in the sense that the model will not be consistent with the way we see light actually behaving.
.

 

[ZapperZ]

Furthermore, the CHANGE in the apparent speed of light in a medium is because we tend to measure the group velocity of light! This automatically means that we measure not the speed of a photon, but rather the speed of a collection of LARGE number of photons, which we collectively call "light" or EM wave. This is what we typically measure as the speed of light.

Zz.