# Photons have energy but zero mass?

1. Mar 26, 2006

### fatoomch

Photons have energy but zero mass??

hi,

Apologies if this is in the wrong message board.

Photons have zero mass.
If E=mc^2 then m = 0. So how can photons have energy?
Does this have anything to do with E=mc^2 only applying to matter at rest?

thanks

2. Mar 26, 2006

### Hurkyl

Staff Emeritus
Yes -- that formula tells you the rest energy of the object. It says nothing about other forms of energy, such as kinetic energy.

The formula for "total" energy, that is the rest energy + kinetic energy, is:

E² = (mc²)² + (pc)²

(p, here, is the magnitude of the 3-momentum)

3. Mar 26, 2006

### Astronuc

Staff Emeritus
Einstein's equation is an equivalence of matter (mass) with energy.

Energy does not have to coincide (simultaneously that is) with mass.

Photons 'always' move at the speed of light.

4. Mar 26, 2006

### rbj

but here, there is two different and incompatible (if the velocity is not zero) expressions for the same symbol, "E". if "m" is only rest mass, then fatoomch's "E" is not the same as Hurkyl's "E".

Hurkyl is the mentor here and i am of much humbler stature, but the unqualified language used in his answer is disputable and, essentially, a currently fashionable practice, but was not always so. somtime in the last 30 years, it became common practice among physicists to reserve the term "mass" for only rest mass. i do not know if the term "relativistic mass" is used much at all recently, it seems out of vogue, but it hadn't always been that way.

just as how time dilates between two inertial observers moving relative to each other, so does mass and the equations relating the two are identical:

$$t = \frac{t_0}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

$$m = \frac{m_0}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

so, just as a clock wizzing past an observer appears to be ticking more slowly than if it were resting in the observer's reference frame, the mass of that clock (and the parts inside) appears to having a larger value (by the same factor) than it would if it were at rest in the observer's frame of reference. so now, what i mean by $m$ is relativistic mass not rest mass (or "invariant mass") as is the usage of the term by Hurkyl (and, i guess, the textbooks currently out). for rest mass (or invariant mass), we used to used the symbol $m_0$ in relativistic contexts and that i am doing.

currently, it appears that rather than talk of the mass of an object changing as it's speed changes (relative to some observer), the leave that only to the expression for momentum which is commonly defined as

$$\mathbf{p} = m \mathbf{v}$$

but here $m$ is what i mean for mass, the total relativistic mass so the momentum an object, in terms of the rest mass, is

$$\mathbf{p} = \frac{m_0 \mathbf{v}}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

but here, Hurkyl and most other current physicists would say "$m$" instead of "$m_0$" meaning invariant mass and indeed "rest mass" or "$m_0$" simply is not very much in their usage and when they say "mass" they always mean what i would call "rest mass". but, using the corresponding symbols, we agree on momentum. as the velocity of an object increases, its magnitude of momentum increases faster or more than proportionally to the speed of the object.

when one derives an expression for relativistic kinetic energy, they get, using the "old" terminology:

kinetic energy $$T = m_0 c^2 \left( \frac{1}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}} - 1 \right)$$

which is $$T = m c^2 - m_0 c^2$$

which is interpreted as $$T = E - E_0$$

where $E = m c^2$ is the total energy of the particle (what Hurkyl calls "E" above) and $E_0 = m_0 c^2$ is the "rest energy" (the energy the particle would have intrinsically if it were just sitting in front of you, same as what Hurkyl mentioned but with a different symbol). note that when the particle is not moving, the total energy and rest energy are the same, so the kinetic energy $T$ is zero. in fact, you can rearrange the equation about to say the same thing as "total energy, that is the rest energy + kinetic energy":

$$E = E_0 + T$$ where

and that expression for total energy is totally compatible with the expression

$$E^2 = \left( m_0 c^2 \right)^2 + \left( |\mathbf{p}| c \right)^2$$

if you keep your symbols straight. note i am using $m_0$ instead of $m$ and i have to because, unless $\mathbf{v}=0$, there is no way for

$$E = m c^2$$

and

$$E^2 = \left( m c^2 \right)^2 + \left( |\mathbf{p}| c \right)^2$$

to both be true. if $m$ is the rest mass, then the top equation is only rest energy and not equal to the total energy in the bottom equation.

also it is not true (but currently in vogue) that the famous expression

$$E = m c^2$$

means only rest energy. if we understand $m$ to be the relativistic mass, then $E = m c^2$ means total energy of a particle and, if it is a photon, that energy is the same as

$$E = h \nu = \hbar \omega$$.

so, using the semantics i have above, photons do have mass because they have energy and the scaling factor to convert the energy to mass is $c^{-2}$. that is the effective mass of a photon is

$$m = \frac{E}{c^2} = \frac{h \nu}{c^2}$$

and this expression can be found in some textbooks, but they might be a few decades old. since the speed of a photon is definitively $c$, the momentum of the photon is

$$|\mathbf{p}| = m |\mathbf{v}| = \frac{h \nu}{c}$$

and that equation remains, even in todays texts.

now even if a photon has inertial mass, the rest mass is

$$m_0 = m \sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}$$

and since $|\mathbf{v}| = c$, then the rest mass $m_0$ must be zero.

Photons have mass, but they don't have rest mass. Their "invariant mass" is zero, and since it is the invariant mass that is a property of just the particle's nature and not a function of what speed it might have relative to any observer, then it is commonly said, particularly today, that photons are "massless particles".

that's the spin on this from a middle-aged electical engineer.

5. Mar 27, 2006

### ZapperZ

Staff Emeritus
I know that this has been talked about to death. However, you also need to keep in mind that whenever a question like this is asked, it is usually by someone who hasn't studied SR, and most importantly, someone who only has a concept of "mass" that is in terms of the familiar property, such as the mass of that object sitting on a coffee table. You seldom have to tell such a person "That object only has a rest mass, and it has no relativistic mass".

So one can safely assume that we are talking about the rest mass here, because that is the only concept that most people are aware of. If not, we run the risk of creating this confuson ("Oh look ma, even the scientist can't agree about mass!") to a question that should have been handled in a straightforward manner. Hurkyl's answer would be the same answer that I would give simply because it introduced to most people the more GENERAL energy equation that most don't know about that allows something with zero mass to have an energy. We may have to explain how something with no mass can have a momentum, but that's a different matter (no pun intended).

Zz.

6. Mar 27, 2006

### da_willem

Because there are no such objects...;). But I agree on your point!

7. Mar 27, 2006

### skywolf

yeah photons have no energy, when they're not moving! (eg. black hole horizon)

8. Mar 27, 2006

### rbj

i dunno, i think that

$$E = m c^2 = \frac{m_0 c^2}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

is pretty general.

that, to me, seems to be the more difficult (because p = mv is commonly known) than to explain the difference between relativistic mass and rest mass to someone who is already aware of E = m c2.

9. Mar 27, 2006

### Hurkyl

Staff Emeritus
That reasoning is circular, though -- you show that the photon has relativistic mass by first showing it has momentum! I don't see, off hand, how else to do it.

Incidentally, I really prefer using 4-vectors. In that case, you have p=mv where m is the rest mass. And the magnitude of the 4-momentum of a lightlike particle is, in fact, zero.

10. Mar 27, 2006

### krab

How can you say that? it's not general because it doesn't apply when m_0=0 and v=c.

11. Mar 27, 2006

### rbj

the other way to do it is show that the photon has relativistic mass by showing it has energy. you get energy from Planck's equation.

i dunno what a 4-vector is. sounds like something GRish.

you're right, krab, about the second equality. it isn't general to that important limiting case. that's why i kept the first equality. in the case that $m_0=0$ and $|\mathbf{v}|=c$, then you need to use $E = m c^2$.

Edit: actually, krab, you can use

$$|\mathbf{p}| = \frac{m_0 |\mathbf{v}|}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

to solve for $m_0$ in terms of $\mathbf{p}$ and stuff that into the $m_0$ of

$$E = \frac{m_0 c^2}{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

and get

$$m_0 = \frac{\sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}{|\mathbf{v}|}$$

$$E = \frac{|\mathbf{p}| \sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}} c^2}{|\mathbf{v}| \sqrt{1 - \frac{|\mathbf{v}|^2}{c^2}}}$$

$$E = \frac{|\mathbf{p}| c^2}{|\mathbf{v}| }$$

i guess that's good for nothing. it just repeats $E = m c^2$. it doesn't mean that the equation doesn't apply, but you just have to resolve the 0/0 limit correctly.

Last edited: Mar 27, 2006
12. Mar 29, 2006

### pmb_phy

It depends on what one means by the term "mass." The relativitistic mass (aka inertial mass) of a particle is the m in p = mv.

When the particle is a tardyon (a particle which always travels at speeds < c) the mass is a function of speed, i.e. m = m(v). The rest mass is given by m0 = m(0).

For a luxon (particles for which v = c) the mass is given by m = p/c = E/c2 (where E = pc for luxons). The rest mass is found from

E2 + (pc)2 = (m0c2 )2

Replace E with pc then you get m0 = 0.

Pete

http://www.geocities.com/physics_world/mass_paper.pdf

Last edited: Mar 29, 2006
13. Mar 29, 2006

### rbj

i think there's a typo. should be:

E2 = (pc)2 + (m0c2 )2

otherwise it's about what i was trying to say.