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How can photons have no mass and yet still have energy given that ?

In fact, is not the complete equation. The full version is:



Where is the energy of the particle, is its rest mass (in the case of the photon, ), is the speed of light in vacuum and is the particle's momentum.

So for a photon, or any other massless particle, the energy is .

For a massive particle, if its momentum is zero, we get the familiar . Even if its momentum is not zero, we can "absorb" the momentum into a coefficient, called the Lorentz factor, which is defined as:



And write the equation as . You can check and see (by squaring both sides and simplifying) that this is equivalent to the full version of the equation, given above.

That's correct. is the rest mass, the one you'd observe if you were in the same reference frame. A photon has no reference frame; it has zero rest mass.

The formula is actually an abbreviation for objects that aren't moving, i.e. objects with no momentum. This is sometimes written as , the rest energy, and for clarity emphasize that we mean the rest mass with :.

 While photons have no rest mass, they do nonetheless have momentum, given by . The momentum of a photon is kind of an unusual concept, since you're used to thinking of momentum as simply , which would be zero. Let's just say that photons have an inherent momentum that's different from the kind of momentum that massive objects have. (It's really exactly the same, but different from the kind you're familiar with in classical Newtonian physics). This is well-confirmed by experiment: it's what makes solar sails work.

For the full energy of a moving particle, with or without mass:



For non-moving objects, p=0, and you can see that it easily reduces to the form you're used to. For photons, with m=0, it reduces , and if you fiddle with that and the definition for above you'll see that you get the equation you listed in your question.

For objects with mass that are moving close enough to the speed of light for their momentum to make a difference, you do need the full form of the equation.

From relativity, we have the energy of an object , given by its mass and the speed of light, c.  According to this theory, mass depends on velocity v: , where is the "rest mass" - the mass at zero velocity.  So we can write the energy in terms of the rest mass and the velocity:



A photon has zero mass, but always travels at the speed of light, c.  As you can see, that makes both the numerator and denominator above equal to zero: 0/0 - so this equation cannot be used to determine the energy of a photon.  only applies to things with mass.

Edit 5/3/2013:
Though you can't get the energy of a photon in terms of frequency just from special relativity, you can get the energy in terms of momentum.  Relativistic momentum is given by , where again, m is relativistic mass. (yes, I insist on using relativistic mass where it suits me!)
Thus in terms of rest mass:
.
This has the same problem if we try to take the limit as m_0 goes to zero and v goes to c.  The solution to this problem is to combine the equations for E and p to eliminate either m_0 or v.  Then we only have to take the limit of one thing,  and there is no problem.

If you eliminate v and do some manipulation, you get
.

If you eliminate m_0, you get
.

In the former case, you can then set m_0=0, and in the latter case, you can set v=c, and in both cases you find E=pc.  This is consistent with the energy in terms of frequency discussed below, via the usual relations between momentum, wavelength, and frequency.

(End edit)

We actually have the energy of a photon from quantum mechanics: , where is the frequency of the photon.  (This was actually the first result of quantum mechanics, postulated by Planck at the beginning of the 20th century.) 

A tangent: Another interesting thing to note about is what happens when the velocity of an object is small compared to c.  In the expression for E in terms of the rest mass , the quantity is very small in this case, so we can get a good approximation using a Taylor series, ignoring any terms quartic or higher in v/c.  The first two terms of the Taylor series give .  This is just the rest energy plus the usual classical kinetic energy   That is, immediately gives you the classical kinetic energy, when v is much less than c.  In still other words, you can think of the classical kinetic energy of an object as simply the slight increase in mass that comes with relative velocity.

There are already many advanced answers here, so I would like to answer this from a more intuitive understanding.

What you are referring to must be this most famous equation:


When a particle has zero mass, its energy must be zero because E = 0 * c^2 = 0. So you are already concluding it right. A photon (light particle) that has its mass zero, could not have energy.

As many has answered, that E = mc^2 is not a complete formula. It's true, but we won't discuss it mathematically here. It's not really that interesting. But let me first to show its more complete formula, just for the sake of our discussion:


So, beside the usual mc^2 term, we have this new term: (pc)^2. Don't worry about the square to all those terms, it won't affect much to our discussion.

This "p" in the new term is called momentum. But wait, we do know momentum, it's mass times its velocity: mv. There's mass once again, and when it's zero, the momentum should be zero too! Nah, apparently we do NOT really know this momentum, because a photon could have momentum without necessarily having a mass! We back to the same question again: how could that be??

Momentum starts its career as a mathematical formula. We don't know what it is actually, but there's some quantity, in which we defined as: mass times its velocity, that when nature has done its tricks, this quantity stays the same. We may crash two cars, but after the crash, if we calculate the mass of those two cars times their velocities, we would get the same number we calculate before the crash. We may explode a bomb, but after that, if we collect all the debris, and calculate all their mass times their velocity, we would get the same number with the one we calculate before the bomb explode. Whatever we do, we would get this same number of momentum, mass times its velocity.

One of exciting moments in physics is when we get such a revelation moment. This is a such moment. It turns out that momentum is more REAL than mass! How could physicists say that? It's simple, we imagine something like this. When we are hit by a car, we would be bounced. Then we ask, is there something that could make us bouncing, without actually being hit by a thing such a car? The answer is: YES. And one of that something is photon.

We don't actually see a photon, it's just a hypothetical matter. What we know is electromagnetic radiation could excite electrons far away. When we could see stars, that's because the electromagnetic radiation / light from those stars travel a very far distance to reach our eyes, then excite the electrons in our eyes. If we could receive data through our wifi connection, that's because the electromagnetic radiation from the wifi router excites the electrons in our wifi receiver.  However when we observe and calculate the behavior of electrons that are being exposed to electromagnetic radiation, they behave in such a way of being hit by something!

We never found this "something", but if we pretend, imagine, guess, whatever, that this something is a momentum, all of our mathematical formulations involving momentum work perfectly well. Hah! Bingo! We have something that we never found what it is, but it definitely has a momentum, and for the sake of communication, we "pretend" to know it, and give it a name photon. That is why a photon could have energy or momentum.

But, is it for real? In physics, we don't really care about if something is real or not. As long as it agrees with experiment, it could be in the form of whatever it needs to be, we would say it's useful.

In this case, yes, we started with something easier to comprehend: mass. It's easier in a sense that we could see it directly with our own eyes, such a car, and say it has a mass. Then when we watch nature does its trick, through its mathematical formulation of course, we see quantity that we never actually see such these energy and momentum. We may wonder, is it real or just some result of a mathematical trick? But along with the progress of our understanding to the nature itself, we start to see some behaviors similar to these energy and momentum, without having something we could perceive directly. That makes us ask again, which are more real, energy and momentum, or mass? However as the experiment shows that energy and momentum always exist, where the mass is not necessarily so, we have to accept the fact, no matter how weird it is, that energy and momentum, somehow, are more real than mass. So the fact that photon could have energy and momentum, without having a mass, is something that we just have to accept.

They are not something that we could "touch" and "see", only their influences, but it's not us to judge which are more real. We could only say: it's amazing!

In its most general form, the in refers to the "relativistic mass" , where is the rest mass, and .  So then we have . 

It is specifically the rest mass of the photon that is zero.  To see why the rest mass can be zero without the energy being zero, you have to look at the value of .  Since a photon always travels at the speed of light (), that means that .  Plugging this and into the equation , we get .  This is neither zero nor infinity - it is undefined, so instead, we may wish to look at the limit of as the mass goes to zero and the speed goes to c. 

Imagine you have two knobs you can turn. One controls the mass of a particle, all the way down to zero.  The other controls the speed of the particle, all the way up to c.  Starting at some nonzero mass, as you increase the speed towards c, gets bigger and bigger, thereby increasing .  Now turn down the mass so that is reduced as low as you like.  Now increase the speed again to further increase , and bring back up.  You can continue this process as long as you like, getting closer and closer to and , while maintaining at any (positive) value whatsoever. The conclusion is that a photon with rest mass of zero and speed of c can have any value of energy.  This is what is observed (you can have a low energy radio-frequency photon or a high energy x-ray photon but they all have zero rest mass and travel at the same speed).

As a side note, Barak Shoshany's answer is entirely correct, but it doesn't answer the question directly in that it raises a similar question which is "Why is the momentum of a photon not zero if ?"  The answer to this question is the same as above - the product of and can be any value in the limit that v goes to c and goes to zero.  Indeed there is a relationship between that limiting value of E and the limiting value of p, but that is not relevant to the question here.

refers to the "rest energy" of a particle with mass - namely, the energy that the particle has just by existing.  It would have more energy if it started to move (kinetic energy).

A photon does indeed have no rest energy as it is massless.  However, that does not mean that it can't have energy by moving around (at the speed of light).   The energy of a photon is given by , where is the momentum of the photon.

The full equation is . The equation is only the case for nonrelativistic motion.   Thus for 0 mass a photon can still have energy. See further comments if you are interested in why the momentum is not 0 as well.

No. Photons have zero rest mass, but as they are never at rest and always moving with the speed of light, they do have energy and mass. m in the Einstein's formula is the relativistic mass, not rest mass.

Photons have no rest mass, and hence cannot be at rest. Mass is a "crystallised" form of mass/energy; photons are an "uncrystallised" form. If you transfer enough energy by creating, transmitting, and absorbing photons, the absorber will increase in mass. Because of the large value of c^2, this is hard to observe/

A portion of massive particle's energy is equal to rest mass. Now photon has no rest mass, it is always moving.

E=mc^2 is true for non-moving particles
E^2=m^2*c^4+p^2*c^2 is the full relationship. For a photon m=0 but p (the momentum) is not.

A photon doesn't have any intrinsic mass, hence its entire energy comes from its motion. By virtue of motion it carries a momentum .

The relation between energy and momentum is in special relativity. For a massless particle . So . Now, since , we see

In old literature, there used to be mention of rest mass or proper mass (as measured by an observer at rest with respect to the particle) and the relativistic mass ( as measured by an observer with respect to whom the particle has a velocity . They are related as follows



No modern book would call "relativistic mass". By "mass" it is always meant the proper invariant scalar mass (which was used be called rest mass).

You could look at it the other way round. Photons travel at the speed of light and yet clearly do not have infinite mass. The only reasonable explanation at the moment is that a photon at rest, if you could ever find such a thing, would have no mass. The only mass, according to the full equation which relates energy, momentum and mass comes from its energy, which is related to its frequency.
In short, if a photon existed at rest, it cannot have mass because as the energy was pumped in to accelerate it the mass would increase.

One simple way of looking at it is that waves transmit energy, and the photons are effectively oscillating electric and magnetic fields, in line with what Maxwell believed, except the wave packages have quantised action. Now the photon does not have mass, in the sense you mean, although in one sense it has inertia in that its path of least action is bent by gravitational fields.

Energy can be converted to mass, but is not the same as mass. Photons do carry momentum though; they can physically push things that they impact.

There are many other good responses to this, but since I was asked, I will try to give a concise answer.

Using the standard definition for mass , this equation gives us the energy of a particle at rest.  A photon at rest has no energy, therefore it has no mass.

I hope that is clear enough.

the total energy of a zero mass system is proportional to its' momentum, I've only covered this last semester so don't wish to come over as though I know everything, but I found this site helpful: http://galileo.phys.virginia.edu...

First of all let me clear you that E = mc^2 is the full equation. E^2 = (mc^2)^2 + (pc)^2 is as expanded form. Since m in expanded form is rest mass and in first equation m is relativistic mass. I could derive it if you want. The rest mass of a photon is zero therefore the equation reduces to E = pc where pc is the momentum of the photon. E = pc is derived from E = mc^2. Since this equation get expanded to relate it with momentum. m is what makes the difference. Therefore E = mc^2 = pc for a photon.

The rest mass of photon is Zero.

I believe that a photon's at-rest mass is zero only, and this is because photons are always moving.

When we say that photons are massless, we are referring to the invariant mass or the rest mass of the photon. The invariant mass is a Lorentz scalar, which means that it is the same as measured by observers in all reference frames. However, this does not mean that the Energy/momentum of the photon is zero. In fact, there exists a concept of relativistic mass (currently out of use), which is defined as the energy of the relativistic particle(in natural units). This is non-zero for the photon.

In fact, from a different perspective, we can consider processes where a photon is released or absorbed. This results in a reduction or increase in the mass of the system by E (in natural units). For instance, such effects are important in Quantum electrodynamics where the effective mass of particles is modified by radiative corrections.

P.S: A word of advice: Refrain from drawing physical inferences from layperson  statements such as "Energy is equal to mass". :)

Photons get something called relativistic mass and depends on their momentum. In fact they have a certain momentum depending on the velocity.

In a way it does. The energy of a photon is calculated as E=hf. but Einsteins theory of relativity (which uses little more than Pythagoras' theorem) can be used to show that photons can't have mass since they travel at the speed of light.

But photons are not matter. Bill Bryson (not a physicist) described light, or energy in general, as as the free form of matter, where as matter is a 'condensed' (not his words) form of energy.
Basically let it loose you have energy and compress it to get matter.

But E=mc² (for ² press ctrl+alt+2)  it shows how energy (the photon) can be "translated" into matter with a certain mass (m).

1. mass of photons is not 0, it's indeterminate. rest- mass of photons is 0( that is  is mass of an object at rest with respect to an observer. according to einstein's special theory of relativity mass increases with velocity, that is mass is not invariant, it is a relativistic quantity.)
2.mass of photons is indeterminate as  relativistic mass= rest mass/sq. root(1- v^2/c^2). for photons v=c. thus relativistic mass becomes indeterminate.
2.the total relativistic energy equation is given by - E^2 = (cp)^2 + (mc^2)^2. where p is the relativistic momentum, c is the speed of light, and m is the rest-mass. according to de-broglie's equation cp=hw/2pi and by planck's formula E=hw/2pi. thus, mc^2=0. since c is the speed of light which is not 0, hence m=0. the rest-mass of the photon is therefore 0.

Mass is a mathematical relation between external 'force' on a body and its acceleration. In physics, energy is an undefined term, which generally means ability to do work. As both these are functional entities, with sufficient deliberations any relation between them can be established.
An alternative view; Although photons are assumed as mass-less, it does not mean they have no 3D matter-content. Photons are the most basic 3D matter-particles. Certain work is done by universal medium to create and sustain a photon. Work is distortions in universal medium. Distortions have associated stress. This stress is the energy associated with the photon. For details, see: 'MATTER (Re-examined)'.