Do photons have mass? Why not?

[kuruman]

TL;DR

Why is the energy of a photon divided by the speed of light acceptable as the photon's momentum but the same energy divided by the speed of light squared is not acceptable as the photon's mass?

This thread was triggered by an earlier thread in "Classical Physics" asking what if the speed of light is not constant. Here, I bring up a "why not" question as expressed in the title of this current thread. The statement that got me thinking in the earlier thread is this,

The physics of a massive photon are understood, and we actually use this theory to measure the photon mass. So far it's indistinguishable from zero.

Right off the bat I admit that I am not one of the people who understand the physics of massive photons and how this theory, instead of an experiment, can be used to measure the photon mass and find it to be zero. My hope is that the discussion in this thread will be informative to me and others. Until this happens, I present below a simple heuristic argument that points to a definition for the photon mass as the ratio of its energy to $c^2$. It is an argument by analogy that relies on two separate experiments: (a) Measuring the speed of a mass after it has fallen from rest by height$H$ (frequently performed in undergraduate labs); (b) Measuring the frequency change of a photon emitted by a source at rest after dropping by height $H$ (Pound-Rebka experiment.)

Here we go.

First, consider releasing a rock from rest at the top of a tower of height $H$. Air resistance is ignored. In the proper frame of the tower, the total energy at the top is $$E_{\text{top}}=E_0 = m_0c^2 $$ where $m_0$ is the rest mass of the rock. The dropped rock will accelerate and reach the ground with total energy $$E_{\text{gnd}}=\gamma m_0c^2.$$ To a good approximation, $\gamma \approx 1 + \dfrac{1}{2}\left(\dfrac{v}{c}\right)^2~$ so that $$E_{\text{gnd}} \approx m_0c^2\left(1+\dfrac{v^2}{2c^2}\right)= E_{\text{top}}+\frac{1}{2}m_0v^2.$$When we perform a careful measurement of the rock's speed at ground level, we discover that $v=\sqrt{2gH}~$ which allows us to write $$E_{\text{gnd}}=E_{\text{top}}+m_0gH.\tag{1}$$
Now, consider a photon source at rest at the top of the same tower. The source is aimed towards the ground and air resistance is ignored. The photon's total energy at the top of the tower is $$E_{\text{top}}=E_0 = hf_{\text{top}} $$ where $f_{\text{top}}$ is the "proper frequency" of the photon, i.e. the frequency when the source is at rest with respect to the tower. The photon will propagate at constant speed $c$ and reach the ground with blue-shifted energy $$E_{\text{gnd}}=hf_{\text{gnd}}.$$ When we perform a careful measurement of the photon's frequency at ground level (Pound-Rebka experiment), we discover that $$f_{\text{gnd}}=f_{\text{top}}\left(1+\frac{gH}{c^2}\right).$$And when we multiply both sides of the last equation by Planck's constant $h$ we get $$hf_{\text{gnd}}=hf_{\text{top}}\left(1+\frac{gH}{c^2}\right)\implies E_{\text{gnd}}= E_{\text{top}}+\left(\frac{h f_{\text{top}}}{c^2}\right)gH.\tag{2}$$When we compare equations (1) and (2), it appears that the dropped photon acquires energy as if it were a dropped rock of inertial rest mass $m_0=\dfrac{hf_{\text{top}}}{c^2}.$ If it quacks like a duck . . .

So my question is, why do we routinely write the momentum of a photon as $p=E/c,$ but never write a "mass" for the photon as $m=E/c^2$ and affirm instead that the mass of the photon is zero? If this is a naive question or the heuristic argument is specious, I would like to discuss why.

 

[pines-demon]

This is not surprising. You must know at least that in general relativity, gravitational potentials change the passage of time, so a light ray changing frequency and thus energy is kind of normal. Also gravitational potentials can make light curve. None of this suggests anything related to light having mass or inertia, but energy is the key factor here.

 

[JimWhoKnew]

If it quacks like a duck . . .

The rock accelerates. The time it will take the rock to fall from top to midpoint is longer than the time from mid-point to ground. Light on the other hand...
So the quacking is not entirely the same.

 

[Sagittarius A-Star]

So my question is, why do we routinely write the momentum of a photon as $p=E/c,$ but never write a "mass" for the photon as $m=E/c^2$ and affirm instead that the mass of the photon is zero?

The mass of a moving rock is
${E \over c^2}\sqrt{1-v^2/c^2}=m$.

The photon is massless because it is moving with $c$:
${E \over c^2}\sqrt{1-v^2/c^2}= {E \over c^2}\sqrt{1-c^2/c^2}=0$.

 

[Ibix]

So my question is, why do we routinely write the momentum of a photon as $p=E/c,$ but never write a "mass" for the photon as $m=E/c^2$ and affirm instead that the mass of the photon is zero? If this is a naive question or the heuristic argument is specious, I would like to discuss why.

That's a matter of definition. In modern physics (last few decades, mostly), when we say "mass" we mean the invariant mass (a.k.a rest mass, but that's a terrible misnomer for light) defined via $m^2c^4=(E^2-p^2c^2)$, which is identically zero for light. What you are measuring here is the relativistic mass, $m_r=E/c^2$. In terms of four momentum, the former is the modulus of the particle's four momentum, while the latter is the inner product of its four momentum and the observer's four velocity.

It used to be perfectly normal to call $m_r$ a mass, but modern convention (thanks in large part to the late Lev Okun, known here as @levokun) is to reserve "mass" for the invariant mass and not use relativistic mass at all. Relativistic mass gets nasty if you start considering motion in more than 1d. It's also a minefield because particles no longer have a defined mass if you prefer their relativistic mass over their rest mass, just a minimum value that's unique to their species.

It has to be said that the relativistic mass does matter sometimes. A reflective box containing a photon gas has total mass equal to the invariant mass of the box plus the relativistic mass of the photons. This is because the four momenta add, and the modulus of a sum of vectors is not the sum of their moduli.

Right off the bat I admit that I am not one of the people who understand the physics of massive photons

My understanding is pretty limited, but the high level story is that if you quantise Maxwell's equations you get a massless spin-1 field. If you (or Alexandru Proca) write down the maths of a massive spin-1 field and deduce the classical limit you get a theory which deviates further and further from Maxwell the more massive (in the invariant mass sense) is the photon. In particular, it doesn't obey Gauss' Law and how badly it deviates depends on the photon mass. So you can measure how well reality obeys Gauss' Law, and depending on your experimental error and whether the photon really has a non-zero invariant mass you either get a measure of, or an upper bound on, the photon mass.

 

[bob012345]

So what effective mass is used when starlight is bent by the sun? Isn’t it just $m=\frac{E}{c^2}$?

 

[Ibix]

So what effective mass is used when starlight is bent by the sun? Isn’t it just $m=\frac{E}{c^2}$?

No mass is needed in GR - it's a theory of curved spacetime and particles in free fall follow geodesics (the generalisation of straight lines to non-Euclidean spaces) that look curved when you project them from 4d spacetime onto 3d space. Gravity isn't a force between masses in GR.

 

[bob012345]

No mass is needed in GR - it's a theory of curved spacetime and particles in free fall follow geodesics (the generalisation of straight lines to non-Euclidean spaces) that look curved when you project them from 4d spacetime onto 3d space. Gravity isn't a force between masses in GR.

Well something needs to have mass in order to curve spacetime.

 

[Matterwave]

Well something needs to have mass in order to curve spacetime.

The spacetime is curved by the source mass (e.g. the "M" in the Schwarzschild metric). Light itself is massless and will travel along null geodesics of the metric.

 

[Ibix]

Well something needs to have mass in order to curve spacetime.

Correct - as @Matterwave says, that would be the planet or the star or galaxy or whatever is doing the lensing. But the mass of a test object (something light, like an astronaut or a photon) nearby doesn't enter into the maths.

If the other object has significant mass (e.g. a binary star or an asteroid or whatever) then you need to take into account the curvature it generates too, sure. And you probably need a big computer.

 

[bob012345]

Correct - as @Matterwave says, that would be the planet or the star or galaxy or whatever is doing the lensing. But the mass of a test object (something light, like an astronaut or a photon) nearby doesn't enter into the maths.

If the other object has significant mass (e.g. a binary star or an asteroid or whatever) then you need to take into account the curvature it generates too, sure. And you probably need a big computer.

I wonder if that is just an approximation? I would think that in principle, the energy difference between a gamma Ray and a radio wave would change the spacetime curvature even if in the most minuscule amount to change to geodesic.

 

[Ibix]

I wonder if that is just an approximation? I would think that in principle, the energy difference between a gamma Ray and a radio wave would change the spacetime curvature even if in the most minuscule amount to change to geodesic.

A gamma ray is just a radio wave seen in a different frame of reference. The curvature must be the same for both!

The source of gravity in general relativity isn't relativistic mass. It is the stress-energy tensor, which is a symmetric 4×4 tensor, so has ten independent components. Some of them are non-zero for light and we do expect it to generate a gravitational field if we could get enough of it in one place, yes. But how the gravitational source term and field vary with reference frame (and hence energy of the light) is not simple.

Note that we don't have a quantum theory of gravity, and GR can't handle sources where quantum effects are important. We can talk about the gravitational field of a light pulse, but the gravitational field of a photon (or indeed any single particle) is not known.

 

[renormalize]

Note that we don't have a quantum theory of gravity, and GR can't handle sources where quantum effects are important.

How can this statement be true? A hydrogen atom in its ground state is stable against radiative decay precisely because of "quantum effects". Do you deny that the gravitational field outside such an atom in isolation is exactly that arising from the Schwarzschild metric of general relativity?

 

[pinball1970]

A gamma ray is just a radio wave seen in a different frame of reference. The curvature must be the same

Yikes. If I'm the guy it comes into contact with I'm hoping for radio!

 

[kuruman]

The rock accelerates. The time it will take the rock to fall from top to midpoint is longer than the time from mid-point to ground. Light on the other hand...
So the quacking is not entirely the same.

I agree, but it's still quacking and not barking.

The photon is massless because it is moving with $c$:
${E \over c^2}\sqrt{1-v^2/c^2}= {E \over c^2}\sqrt{1-c^2/c^2}=0$.

That is not what I asked. I asked why we must reject an effective, frequency-dependent, mass equal to $E/c^2$ when treating situations where a photon exchanges energy and momentum with the Earth (as presented in post #1) or an electron as in the Compton effect.

The closest answer so far is that

... modern convention (thanks in large part to the late Lev Okun, known here as @levokun) is to reserve "mass" for the invariant mass.

Are we saying, then, that the photon, which we all agree does not have a conventional invariant inertial mass,
cannot be considered to have non-zero mass $E/c^2$ even though there is experimental evidence that it gains energy $(E/c^2)gH$ when dropped from height $H$ near the surface of the Earth? Why?

Is there a simple answer that doesn't get into spacetime curvatures, metrics and 4x4 tensors?

 

[Ibix]

How can this statement be true? A hydrogen atom in its ground state is stable against radiative decay precisely because of "quantum effects". Do you deny that the gravitational field outside such an atom in isolation is exactly that arising from the Schwarzschild metric of general relativity?

Can you tell me the atom's momentum and location exactly so I can measure $r$ in its rest frame in order to check?

The obvious problem with quantum sources of gravity is superposition. Even if you can write down the gravitational field from an atom in some state, and its field in a different state, how do you superpose spacetimes when it's in a superposition of the two states?

 

[Ibix]

Yikes. If I'm the guy it comes into contact with I'm hoping for radio!

If I shoot a bullet at you, but do it while travelling away from you at several hundred meters per second, you can just catch the bullet as it falls straight down out of the barrel. Mythbusters actually did this with a cannon with a muzzle velocity of about 30mph mounted on the back of a truck doing about 30mph - it's on YouTube if you want to look it up.

Energy is frame dependent, both for EM and physical objects.

 

[Dale]

TL;DR: Why is the energy of a photon divided by the speed of light acceptable as the photon's momentum but the same energy divided by the speed of light squared is not acceptable as the photon's mass?

I am not one of the people who understand the physics of massive photons and how this theory, instead of an experiment, can be used to measure the photon mass and find it to be zero.

This is the idea of a “test theory”. If you are going to test a hypothesis (photon is massless) then you cannot use a theory that assumes your hypothesis (Maxwell’s equations). Instead, you use a “test theory” (Proca action) that has some parameter ($m$) that includes your hypothesis as a specific value ($m=0$). Then you design an experiment to measure that parameter.

 

[Ibix]

Are we saying, then, that the photon, which we all agree does not have a conventional invariant inertial mass,
cannot be considered to have non-zero mass $E/c^2$ even though there is experimental evidence that it gains energy $(E/c^2)gH$ when dropped from height $H$ near the surface of the Earth?

As long as you call it relativistic mass, not just "mass", you can use it as you are proposing. Most people will refer to it as the particle's energy (divided by $c^2$ if you're in units where that matters).

Okun (and others) persuaded the community that calling energy-divided-by-$c^2$ relativistic mass was confusing and pedagogically bad, so the term fell out of use. However, it isn't exactly wrong to use it - just not recommended.

 

[pines-demon]

Well something needs to have mass in order to curve spacetime.

What about energy

 

[renormalize]

The obvious problem with quantum sources of gravity is superposition. Even if you can write down the gravitational field from an atom in some state, and its field in a different state, how do you superpose spacetimes when it's in a superposition of the two states?

The answer is that you couple gravity to expectation values of the particles and/or fields. See, e.g.,
https://www.google.com/books/editio...in_Curved_Spacetime/Iud7eyDxT1AC?hl=en&gbpv=0:

 

[Ibix]

The answer is that you couple gravity to expectation values of the particles and/or fields. See, e.g.,
https://www.google.com/books/editio...in_Curved_Spacetime/Iud7eyDxT1AC?hl=en&gbpv=0:
View attachment 374157

As the screenshot says, that's a semi-classical model.

 

[Dale]

Well something needs to have mass in order to curve spacetime.

This is a bit of a misconception. First, there are vacuum spacetimes with curvature. So curvature does not require a gravitational source. Second, the source of gravity in GR is the stress energy tensor. Light has energy and momentum so it can act as a gravitational source even without mass.

 

[Dale]

Is there a simple answer that doesn't get into spacetime curvatures, metrics and 4x4 tensors?

Yes. @Ibix already gave it. We don’t call the quantity $E/c^2$ “mass”. We use the word “mass” to refer to $m$ in $m^2 c^2=E^2/c^2 -p^2$.

It is just a matter of words.

 

[pinball1970]

Energy is frame dependent, both for EM and physical objects

If I'm traveling away from the source approaching C will the gamma radiation still not travel towards me at C?

 

[kuruman]

As long as you call it relativistic mass, not just "mass", you can use it as you are proposing. Most people will refer to it as the particle's energy (divided by $c^2$ if you're in units where that matters).

Okun (and others) persuaded the community that calling energy-divided-by-$c^2$ relativistic mass was confusing and pedagogically bad, so the term fell out of use. However, it isn't exactly wrong to use it - just not recommended.

That's the answer I was looking for. Thank you.

 

[renormalize]

As the screenshot says, that's a semi-classical model.

Yes, but the left-side of eq.(4.6.2) is strictly classical general relativity. My point is simply that classical GR can be sensibly coupled to matter systems that do exhibit "quantum behaviors" in order to analyze certain situations of interest, e.g., the expected gravitational field external to a stable hydrogen atom, particle production in evolving spacetimes, the trace anomaly of the energy-momentum tensor and Hawking radiation outside a black hole. This "semi-classical model" has been an area of active research for almost 60 years.

 

[Matterwave]

First, there are vacuum spacetimes with curvature. So curvature does not require a gravitational source.

Could you clarify? There's a few different possible things you could mean with this, and I'm not sure which one (or all of them) you're talking about.

 

[bob012345]

A gamma ray is just a radio wave seen in a different frame of reference. The curvature must be the same for both!

Perhaps, but we are dealing with a gamma ray and a radio wave in the same reference frame. Consider what happens if both interact with a microscopic primordial black hole of comparable energy to the gamma ray for example?

 

[Matterwave]

There are research papers investigating the gravitational effects of light (as a source), see, e.g.:

https://arxiv.org/abs/1511.01023

The usually cited one is TEP: https://journals.aps.org/pr/abstract/10.1103/PhysRev.37.602

But it looks like it's behind a paywall even though it's from the 30's. Not sure why lol.