# Photon mass - clarification request

1. Jul 10, 2013

### brenan

Am I correct in thinking it is currently believed that

a single photon at rest has no mass
a single photon in motion does have mass ?

2. Jul 10, 2013

### Staff: Mentor

There are no photons at rest.

A single photon does not have mass (=invariant mass, rest mass). It has energy (sometimes described as "relativistic mass", but that concept is not used any more in physics).

3. Jul 10, 2013

### DrGreg

Last edited by a moderator: May 6, 2017
4. Jul 11, 2013

### harrylin

5. Jul 11, 2013

### Staff: Mentor

I fixed it, thanks.

6. Jul 12, 2013

### brenan

Thanks guys -
Looking at the info it seems it's not photons I have a problem with it's the definition of "mass"

I'm not sure if its me missunderstanding or poorly phrased explanations:

On the ucr.edu page above:
"Photons are traditionally said to be massless. This is a figure of speech that physicists use to describe something about how a photon's particle-like properties are described by the language of special relativity."

Mass is a figure of speech? To describe behaviour by analagy is one thing - but this doesnt seem right.

Then you have mass = energy. equivalence exists.
OK I understand that - a given mass can be turned into a certain energy.
No problem - but mass is not energy. Are they not 2 different things?

Reading your own faq underlines there is a difference between a mass at rest and one in motion. Static and invariant mass - ok the concepts make sense execept the mass in motion appears to always be treated as energy. Its as if we are saying that we treat it as one thing but then in other situations the math doesnt work so we have to treat it as something else. Its like the real world is being fudged to make the math work.

At this point perhaps I should stop and say that my confusion then arises from
E=mc^2

i.e. energy = mass x speed of light.
or using my own missunderstanding
energy = energy x c
That doesnt make sense.

A canon ball at rest is said to have mass. I dont understand the definition of "mass" there.
In motion the cannon ball has kinetic energy and mass. not energy and energy.

Obviously I've missed something fundamental - can anyone straighten my head there ?

What exactly is the definition of mass. I think a clear explanation of that may sort me out?

7. Jul 12, 2013

### Staff: Mentor

That is an equivalence between [mass] and [energy at rest].
The energy contained in that mass can get released. It is just a conversion between different types of energy.
They are different things.
No it does not.
That is the formula for a particle at rest. It is better to write this formulas as $E=\gamma m c^2$ with the relativistic gamma-factor - this is valid for moving particles, too.

8. Jul 12, 2013

### Staff: Mentor

We aren't saying that mass is a figure of speech - you're right, that would be absurd.
We are saying that "photons are massless" is a figure of speech; people say something is "massless" when they really mean it has zero rest mass.

Stop thinking in terms of $E=mc^2$ for a moment, instead use the less famous but more useful $E^2=(m_{0}c^2)+(pc)^2$ where p is the momentum and $m_0$ is the rest mass. (There's an interesting history around why the first equation gets all the good press when it's just a special case of the second).

[edit: or as mfb says above, use $E=\gamma{m_{0}c^2}$, another equivalent but less publicized form that can be used for things with non-zero rest mass]

Last edited: Jul 12, 2013
9. Jul 12, 2013

### Staff: Mentor

There is no single definition of "mass", in relativity. Instead, a variety of different kinds of "mass" have been defined, over the years. In connection with special relativity, there are (at least) the following kinds:

invariant mass (also known as "proper mass" and "rest mass")
relativistic mass
transverse mass
longitudinal mass

In general relativity there are more definitions:

Komar mass
(and probably some others)

Unfortunately, for historical/traditional/conventional reasons, there is no universal agreement about which one people mean when they say simply "mass."

For most physicists who work with relativistic objects or particles nowadays, "mass" = "invariant mass". They define it via $mc^2 = \sqrt{E^2 - (pc)^2}$, which gives the same result in any inertial reference frame (hence the name "invariant mass"). It reduces to $mc^2 = E$ in the object's rest frame, if it has one (hence the name "rest mass"). It also agrees with the mass that you measure classically when you put an object on a scale or balance beam, at rest.

The "relativistic mass" that most popular treatments of relativity, many introductory textbooks, and some physicists talk about, is usually defined as the "m" that makes the classical definition of momentum, p = mv, continue to "work" in relativity.

"Longitudinal mass" and "transverse mass" are defined as the "m's" that make F = ma "work" when the force is parallel to or perpendicular to the object's motion. "Transverse mass" turns out to be the same as the "relativistic mass" defined above.

In the early days of relativity, physicists used "relativistic mass" (and "transverse mass" and "longitudinal mass") more than they do now. But this gradually changed, whereas popular and introductory treatments did not. There's even a quote from Einstein in the 1940s which disparages the use of "relativistic mass." Someone will probably provide the exact reference.

Last edited: Jul 12, 2013
10. Jul 12, 2013

### harrylin

You are perfectly right, that is sloppy language, suggestive of a lack of physical understanding. Mass cannot be turned into energy; energy is conserved! Funny enough, one of the first papers on energy-mass equivalence expresses it accurately:

"If a body gives off the energy E in the form of radiation, its mass diminishes by E/c². [..] the energy withdrawn from the body becomes energy of radiation [..] The mass of a body is a measure of its energy-content."
(I took the liberty to replace L by E for your convenience).
- http://www.fourmilab.ch/etexts/einstein/E_mc2/www/
There is no fudging, only different new definitions of "mass". The classical definitions lead to inconsistencies.
On that issue you missed nothing fundamental - quite the contrary.

11. Jul 12, 2013

### Staff: Mentor

It's the other way around - we tweak the math until it describes the real world (although this can be obscured by the way physics is often taught).

Look at that equation for total energy again: $E^2=({m_0}c^2)^2 + (pc)^2 (1)$

It basically says that in the most general case two things contribute to the energy: the $(pc)^2$ bit, which depends on the momentum and the ${m_0}c^2$ bit which does not. Now let's consider a few different problems and how the math applies to then:

1) Billiard balls rolling around on a table. We know that in these classical problems, the sort that you come across in introductory mechanics, the rest mass doesn't change so ${m_0}c^2)$ will always be the same throughout. So we don't need to worry about calculating it at all, we can just work with the kinetic energy given by ${E_k}^2 = E^2 - {m_0}c^2$; substitute that into equation 1 and do some algebra and you'll see that the ${m_0}c^2$ term disappears.

Furthermore, if v is small compared with c, it's easy to show that ${E_k}^2 = \frac{m_{0}v^2}{2}$ - this is just the standard Newtonian kinetic energy and it applies just fine for objects of non-zero rest mass moving at non-relativistic speeds.

2) Two masses of uranium moving towards one another at a few hundred meters per seconds. When they meet some of the uranium atoms will fission and $m_0$ before will not be the same as $m_0$ afterwards, so I can't apply the math the way that did in the previous case. Instead, I ignore the $(pc)^2$ term because the kinetic energy is well and thoroughly negligible and I'm looking at $E=mc^2$ and calculate the energy released in the collision from the change in mass. In practical terms, when you're calculating the damage done by nuclear bomb, you don't bother to add in the kinetic energy it picks up as it falls towards its target..

3) A proton and and an antiproton collide head-on at relativistic velocities. Now I have to consider both parts of equation 1 above. If they are moving really fast, p may be so large that I don't need the rest-mass term, in which case we have....

4) If the second term is really large compared with the first, then I can leave out the rest-mass term. This is always the case for a photon because its rest mass is zero. hence, $E^2 = (pc)^2$.

12. Jul 12, 2013

### Staff: Mentor

No, $E_k = E - m_0 c^2$ or $E = m_0 c^2 + E_k$.

This is true for $E_k$, not for $E_k^2$.

I suspect you accidentally got in the habit of "squaring" everything. :-)

13. Jul 12, 2013

### Staff: Mentor

14. Jul 12, 2013

### Ookke

Is a photon able to carry anything else than energy and momentum, like charge or elementary particles (quarks etc)? I wonder how conservation laws would hold, if we drain all energy from a photon and the photon "particle" just vanishes...

15. Jul 12, 2013

### PAllen

Every photon has arbitrarily low energy in some frame, and arbitrary high energy in some other frame. Thus there are not different energy states of a photon intrinsically.

I do not believe there is any fundamental principle that prevents massless particles from carrying electric charge. It does not occur in any existing theories resembling reality. However, gluons are massless and carry color charges.

16. Jul 12, 2013

### Bill_K

A photon is not just an energy packet, it's a full fledged elementary particle, on an equal footing with all the others. In the electroweak theory it's a first cousin of the W and Z bosons.

It carries angular momentum, as well as space parity -1, charge parity -1. Particles are often distinguished by their interactions, and the photon is uniquely the particle that couples to the electric current Jμ.