# Prove that light has no mass

1. Mar 3, 2004

### taylordnz

could you out there prove to me that light has no mass because i read a lot and it always says that light has no mass and no one has told me why it has no mass plz help

2. Mar 4, 2004

### Haelfix

I can give you a proof in quantum field theory, with one assumption (that matches experiment to ~11 significant digits).

Namely, that the global U(1) guage symmetry of electromagnetism, is also local. That will directly imply a massless photon.

The proof is a little involved, but I can handwave it out.

Or do you want something simpler?

3. Mar 4, 2004

### suyver

The fact that the electromagnetic interaction is a long-range interaction with a 1/(distance^2) dependence is considered a proof of the masslessness of the photon.

4. Mar 4, 2004

### LURCH

Also the fact that photons travel at light speed without their mass becoming infinite.

However, the absence of mass is something that cannot be directly measured, since "0" is an amount we cannot observe. So as better equipment and methods develope, our ability to detect smaller and smaller amounts of mass enables us to set a "lower limit" that is progressively less.

5. Mar 5, 2004

### wisp

Some proof that light has no mass.
If it had mass then photons travelling to us from far reaches in the universe would clump together because of gravitational attraction.
There is no evidence that this happens and so we can assume light has no mass.
But it does have momentum and so there must be some sort of "mass" associated with its speed. But this does not have a gravitational effect like normal mass.

My own view is that normal matter causes spherical distortion to the ether that causes the mass from the ether to be attributed to matter. Light distorts the ether in a different way to normal matter and as such doesn't cause the gravitational effect. It does have mass, but not mass that has a gravitational effect.

6. Mar 5, 2004

### ZapperZ

Staff Emeritus
You need to be a bit careful here in drawing parallels between "momentum" and "mass". This is especially true if something can have a larger and more general definition, which is the case for the quantity we call "momentum".

While the definition of momentum being "p = mv" is popular (and correct in many instances), it isn't the ONLY expression in common use in physics. I could also point out that p=hk is very common expression also, especially in solid state/condensed matter physics, where h is Planck constant, and k is the "wave number". In CM physics, we sometime call this the crystal momentum. Whatever it is, notice that it is missing one thing - mass! It neither has gravitational mass, nor inertial mass, which you alluded to.

But here's another fascinating thing. The idea that light has a momentum did not originate when photons came into existence as a viable model. Even when light was thought to be E&M wave based on Maxwell equations (i.e. no one thought light could be "particles"), you can still see the concept of "radiation pressure", etc within this classical theory of light. So even then, there is such a thing as momentum for something which was thought to be purely a classical wave!

But here's the kicker. You can detect light's momentum most efficiently when it impinges upon a metallic surface. On non-metallic surfaces, it is more difficult to detect this momentum. It is why sci-fi books always imagine using solar sails made of mylar, or the likes, as a means of propulsion, etc.

I know I haven't exactly given you an answer on the origin of this momentum without the need of having a mass. But I've given you plenty of clues! :)

Zz.

7. Mar 6, 2004

### Hypercase

I hav been confused now for the past couple of weeks regarding light and its momentum.
Here below I hav complied my thoughts and I would like to request u guys to read it and correct me.

First I think that the word mass can be asscociated with diffrent definitions.
first :
1) First there is the newtonian concept, the rest mass a.k.a the proper mass.
2) the mass a body is said to have which is equivalent to its energy content( calculated using E=Mc^2)
I think that just as one should'nt confuse ma as a force present in a free body diagram in Newtonian physics, in a similar manner 'this 'mass should'nt be confused with the one stated in 1).
3) then there is the mass a body gains as it gains speed.[from m=m/sqrt(1-(v^2/c^2)]

In the case of the photon the rest mass is zero, or else it would require infinite energy to propel it to light speed.
However we notice that light has momentum.(photoelectric effect, use of light sails.)(zapper the are not scifi anymore, plz refer to scientific american 2003 nov issue.)

the newtonian formula for momentum is p=mv
using this light has no momentum.
However if i calculate mass accordin to the second definiton above,
I get m=E/c^2
now energy of a photon is
E=hf
substituting this above I get m=hf/c^2

now again using the newtonian notion
I get p=(hf/c^2)*c
and I end up with p=hf/c
which is the formula given by de broglie fo the momentum of a photon.
therefore i hav ended up with the conclusion that the in the formulap=mv
the m is not the 'rest mass', but rather themass equivalen to its energy content as was defined in 2).

If I am wrong in my thoughts above( as is in most cases.)
plz clear out my confusion by answering the following:-
1) What is momentum?
2) why or how can a massless particle hav momentum.
3) If it does hav momentum, why is momentum, defined as m*v

8. Mar 7, 2004

### ZapperZ

Staff Emeritus
(i) I don't actually consider something published in Scientific American as acceptable. :) It doesn't mean it isn't true, but new work in physics doesn't appear in SciAm first. They appear and get reviewed in a peer-reviews journals.

(ii) I only brought examples of solar sails in sci-fi books since that is where most people would encouter it. I didn't say photon momentum is a "sci-fi". And I know all about photoelectric effect since I've done extensive experimental work in photoemission spectroscopy. The photon momentum here is negligible to explain the photoelectric effect.

Unfortunately, this is a very common mistake. The full relativistic energy equation is

E^2 = (pc)^2 + (mc^2)^2.

By definition, m=0 for light, and so E=pc is the only expression left behind. Most people are only aware of the more "famous" part of the equation and erroneously equate the energy that light has with it having a mass.

Zz.

9. Mar 22, 2004

### Grim

so Zapperz you seem to know a bit about this,

if light has no mass would that mean that it is pure energy.. also how does this fit in with light being called a dualility?

10. Mar 22, 2004

### ZapperZ

Staff Emeritus
Notice the REASON why we think light has a "dual" behavior. It is due to our insistance that if object A behaves in so-and-so way, it must be a particle, while if object B behaves in so-and-so way, it must be a wave. If you look closely, these definitions are built on top of our classical ideas on the properties of matter and waves. There are no a priori reasons that the universe should follow such rules.

Secondly, there is NO duality in the description of light in the QM formulation. This may surprise a lot of people, since the whole origin of "duality" came along side the history and advances of QM. But if you really, really look at the QM description of light/EM interaction, there is only ONE, and ONLY ONE, consistent set of equations/descriptions. Not two, not three... but ONE! Now, if we mean by "duality" as in something that has two completely independent and incompatible sets of description, then I would say that there are no duality of light (or matter) within the QM description. I can as easily describe the interference pattern of light as easily as the photoelectric effect and compton scattering using the same, identical formulation.

A lot of the issues and so-called problems in QM are not really with QM itself, but with our insistance that objects and descriptions at the QM level MUST satisfy and follow all our classical notions. When this is the case, OF COURSE we are going to get weird answers. It's like trying to force a square object into a round hole! Once we start reexamining what it is that we are insisting and relax those insistance, then things are not as troubling as one would expect.

Zz.

11. Mar 24, 2004

### pmb_phy

Regarding the definition of mass - There are many old threads in this forum which cover this topic so you might consider doing a search and browsing through them.

There are three aspects to mass - inertial mass (the "m" in p = mv), passive gravitational mass (that which gravity acts on) and active gravitational mass (that which is the source of gravity.

When people say that light has no mass what they mean is that light has zero proper mass. Light does have inertial mass since it has momentum. Proper mass is an inherent property and for a tardyon (particles which travel at speeds less than light) the inertial mass is a function of speed, i.e. m = m(v). The proper mass, m0, of the tardyon is then given m(0) = m0. For a tardyon the proper mass is related to the inertial mass as

$$m = \frac {m_{0}}{\sqrt{1 - v^2/c^2} }$$

For proof please see
http://www.geocities.com/physics_world/sr/inertial_mass.htm

The kinetic energy, K, of the tardyon is given by

$$K = (\gamma - 1)m_0c^2 = \gamma m_0c^2 - m_0c^2$$

or rearranging terms

$$\gamma m_0c^2 = K + m_0c^2$$

This sum is given the name "total energy" but I prefer to call it "inertial energy". Label it E

$$E = \gamma m_0c^2 = K + m_0c^2$$

the momentum of the tardyon p = mv and so when this is substituted into the expression above for E and rearranged once more we obtain

$$E^2 - (pc)^2 = (m_0c^2)^2$$

Now that we know the proper mass in terms of the inertial energy and the momentum we can find the proper mass once we know E and p. Now consider what happens when we measure E and p for light and substitute in the above expression: It can be shown that the momentum of an electromagnetic wave is related to the energy if that wave by E = pc. Substituting this above gives m0 = 0

However please note that the above relation was derived with the assumption that v < c and therefore m0 > 0. Its for that reason why I think its a bad idea to think of mass as being defined by $$E^2 - (pc)^2 = (m_0c^2)^2$$. It's impossible to measure mass in this way for a tardyon. To know E we must know both K and m0 and that's impossible since it's m0 that we assume we don't have.

There are better ways to measure the proper mass if light and that is through other means such as EM methods (i.e. precise measurements of the Coulomb force being a 1/r^2 force etc)

re -
1) What is momentum?

It's defined as p = mv. m is defined such that the quantity mv is conserved in all particle collisions. See above link for details.

2) why or how can a massless particle hav momentum.

Anything which can collide with another particle and change its motion must have a non-zero value of "m" as defined in #1 above. Therefore since a photon can change another particles velocity (scattering off it) it has momentum and it therefore has mass. Momentum defines inertial mass. You're thinking of proper mass.

3) If it does hav momentum, why is momentum, defined as m*v

Because we observe in nature that the quantity mv is a conserved quantity. Hence the reason for giving it a name.

RE - "Unfortunately, this is a very common mistake. The full relativistic energy equation is.."

Actually its not a mistake at all. Its very well known that Light does have a non-zero value of inertial mass. You're thinking of proper mass. In cosmology one often assigns mass to light. E.g consider Alan Guth's lecuture notes from his course "The Early Universe"

http://www.geocities.com/physics_world/guth.gif

See Eq. (7.3) in that page

It's also well known that light has both gravitational and inertial mass since (1) it is acted on by gravity and (2) it generates a gravitational field. See example at

http://www.geocities.com/physics_world/grav_light.htm

Anything which has energy has inertial mass, active gravitational mass and passive gravitational mass. However it does not mean that it has proper mass. However a gas of disordered radiatons has proper mass in that it has enery and a zero-momentum frame call that "invariant mass"

http://www.geocities.com/physics_world/sr/invariant_mass.htm

Last edited: Mar 24, 2004
12. Mar 24, 2004

### Grim

ohh I see so the term dualility is used because we created rules, the universe doesn't follow our rules, and it doesn't really apply to the QM way of working things....

pmb_phy, sorry man I only understood half of what you said and had no idea about those equations i'll try re-reading it a bit later when I can focus more.

thanks guys, I dont know who started this thread but it's useful i'll post more questions later if they pop up

13. Mar 24, 2004

### Loren Booda

I seem to recall that a photon's zero mass, constant speed of c, and infinite lifetime are directly interrelated.

14. Mar 24, 2004

### Janitor

If you buy the idea that every particle in the universe was created and will at some point be destroyed, then every particle can be called "virtual." And of course virtual particles don't have to be "on mass shell." That means a virtual photon doesn't have to be massless, doesn't it? But for a photon that manages to cross 5 billion light years of space before it is absorbed by a material object, I suppose it's really splitting hairs to say it was virtual and not real! The longer-lived a virtual particle, the closer it has to be to its "ideal" mass, if I can use that term.

15. Mar 25, 2004

### pmb_phy

Regarding electrical means - Proca was the first to take into consideration a non-vanishing photon (proper) mass in the Lagrangian density of the electrmagnetic field. This Lagnangian density reads

$$\mathcal{L}_{Proca} = -\frac{1}{16\pi}F_{\alpha\beta}F^{\alpha\beta} + \frac{\mu^2}{8\pi}A_{\alpha}A^{\alpha} - \frac{1}{c}J_{\alpha}J^{\alpha}$$

where

$$\mu = m_{\gamma}c/\hbar}$$

where $$m_{\gamma}$$ is the photon's proper mass. The Coulomb potential would then be

$$\Phi(x) = q \frac { e^{-\mu r} }{r}$$

Then one simply measures the elecric force on a charged particle, i.e. $$\mathbf{F} = -q\nabla \Phi(x)$$ and one can, in principle, determine the photon's proper mass.

Last edited: Mar 25, 2004
16. Mar 25, 2004

### kheorman

Okay. Let's try this on for size.

Light has no mass? Must we have numerous definitions of mass to satisfactorily explain physical phenomenon, or have we just not discovered the true nature of matter and energy? Remember that every theory is just a means for us to explain observed phenomenon and nothing more. The more a theory can explain, the better the theory is received. Theories are not absolutes, and to believe such is to violate the very essence of physics research.

So light has no mass? But light is effected by strong gravitational fields. It bends around the sun and provided evidence that Einstein was correct. It can be trapped by the event horizon of a black hole according to astrophysics. Is it just light (visible spectrum) or are all forms of electro-magnetic radiation effected by strong gravitational fields? If so, are higher energy photons more 'massive' than lower energy ones, and therefore would 'bend' further when passing through a strong gravitational field? Has any astronomical research been conducted to determine these answers?

Maybe part of our problem with the concept of photon mass is that we are limited by current theories in such that they may only predict macroscopic gravitational effects, and therefore only allow confirmation of gravitational interactions when at least one object is super massive. It is possible that our current gravitational attraction equations do not hold true between two objects at what could be photon mass levels.

So if light has mass, why isn't it's mass infinite at 'c'? Again, I believe we are limited by the current equations. Recall that relativity is based on inertial frames of reference and not an absolute frame of reference. Does this mean that an absolute frame of reference does not exist? No. It only means that we are unable to knowingly make observations and measurements in the absolute frame of reference. In fact, the only plausible absolute frame of reference would be the center of the Big Bang for the universe. Maybe Newton was correct and we just didn't know where to look? Now there's an interesting math problem.

So if the speed of light in a vacuum is a constant regardless of the inertial frame of reference, then why would is be so far fetched to believe that the mass of light is constant also?

In order for us to truly progress beyond our current limitations, we have to stop thinking that the earth is the center of universe and start transforming everything to an ultimate reference frame. If reality started with the Big Bang, then reality should be explained within the Big Bang reference frame.

One could only wonder, and as Einstein noted, "Imagination is more important than knowledge."

17. Mar 25, 2004

### pmb_phy

Please note: If I use the term "mass" unqualified below or anywhere else in this forum then I mean inertial mass aka "relativistic mass".

Light has no proper mass (aka "rest mass"). That doesn't mean it doesn't have "mass" i.e. inertial, passive gravitational and active gravitational mass.

There are inertial aspect of matter - hence the term inertial mass. There is the passive gravitational aspect of matter, i.e. that which gravity acts upon - hence the term passive gravitational mass. There is the active gravitational aspect of matter - hence the term active gravitational mass.

Like momentum, inertial mass is a function of velocity. The particle's proper mass is simply the inertial mass at low speeds. For light it's zero.

Thus light as a non-zero passive gravitational mass.
Light is just electromagnetic radiation of very high frequency.

Last edited: Mar 25, 2004
18. Mar 26, 2004

### Antonio Lao

Previous posts already shown the following equation for the relativistic mass m(v) of a photon.

$$m(v) = \frac {m_0}{\sqrt{1- \frac{v^2}{c^2}}}$$

transposing

$$m_0 = m(v) \sqrt{1- \frac{v^2}{c^2}}}$$

if v=c then $$m_0 = 0$$

The rest-mass of the photon is zero but the relativistic mass of the photon is embedded inside its momentum p which is

$$p = \frac {E}{c}$$

so that its relativistic mass is

$$m(c) = \frac{E}{c^2}$$

19. Mar 26, 2004

### pmb_phy

Actually you forgot to mention that m(v) goes to infinity as v -> c so you have an indefinite form of infinity * zero. It's better to start with the definition or relativistic mass. Relativistic mass is defined as the ratio of the magnitude of the particle's momentum to speed of the particle. For a photon each is a well defined quantity.

20. Mar 27, 2004

### Antonio Lao

pmb_phy,

Actually you forgot to mention that m(v) goes to infinity as v -> c so you have an indefinite form of infinity * zero.

Your point is well taken.

I always think that when I calculate a quantity, I always have to make the assumption that there are unknowns. If all quantities are known, then there is no point in trying to calculate a known value.

I made the assumption that I dont know the m(c) when I try to find m0.

When I found m0, then I can use it to find m(c) which in this case I have to use the relativistic energy formulation.

$$E^2 =c^2 p^2 + m^2_0 c^4$$