# If light has no mass, how can it have energy according to e=mc2?

1. Jan 16, 2008

### Meatbot

If m=0, then e=0(c^2), so e=0

2. Jan 16, 2008

### f95toli

E=m2^2 is not the complete equation.
There is also a momentum term and since light carries momentum there is no contradiction.
The complete equation is
E=sqrt(mc^2+(pc)^2)

sqrt=square root

In most cases the momentum term can be neglected, but obviously not for light.

3. Jan 16, 2008

### Staff: Mentor

The e in e=mc^2 is the "rest energy". The total energy of a particle is the rest energy plus the kinetic energy. So your observation about the equation implies that a massless particle does not have any energy at rest (or in fact at any speed < c) or equivalently that all of a photon's energy is kinetic energy.

4. Jan 16, 2008

### dst

Because mass is defined after not before.

e = hf for a photon. And you can get a virtual mass for it through e = mc^2, useful or not. It comes from the relativistic equation of mass/kinetic energy.

e^2 - p^2 = m^2c^4

But since RHS is 0, then e^2 = p^2.

Edit: I seem to post very slow? 2 replies...

5. Jan 16, 2008

### Meatbot

What is the point of saying a photon has zero mass if that's only true when it's not moving (which can't happen (can it? (I love nested parentheses.)))? Why do we not say that the photon has the "virtual" mass in actuality?

6. Jan 16, 2008

### dst

Because of pair production, it can spontaneously become real mass, i.e. electron/positron pair. Only when the photon is colliding with something to transfer momentum, however. So a photon with a certain energy can create 'real mass', in a way.

7. Jan 16, 2008

### Meatbot

What's the difference between virtual mass and real mass?

8. Jan 16, 2008

### rbj

there is also different qualifications of the word "mass". photons, if they really do travel at the speed of c (relative to any observer), cannot have rest mass a.k.a. invariant mass (what i'll denote as m0). but if you think of momentum as simply

$$p = m v$$

then the mass of a body that is moving past an observer (at a relative speed of v) appears to that observer to be heavier than the rest mass:

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

but that is for a regular old object or particle with rest mass m0. note that the relativistic mass approaches infinity as v approaches c. if you use this (relativistic) m in $E = m c^2$, then the E is the total energy of the particle (in the reference frame of the observer that the particle is whizzing past) and, for photons, is the same energy as

$$E = h \nu$$

so, if we're talking relativistic mass, photons have a mass of

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

but if you turn the previous equation around a little:

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}} = \frac{h \nu}{c^2} \sqrt{1 - \frac{v^2}{c^2}}$$

you will see that for any finite m, the rest mass, m0, must go to zero if the speed of the photon, v, is c. photons are called "massless", because their rest mass is zero. but their relativistic mass is not zero and is, indeed, proportional to their energy which is proportional to the frequency of radiation.

9. Jan 16, 2008

### ZapperZ

Staff Emeritus

Zz.

10. Jan 16, 2008

### Meatbot

Ok, I understand that now. But why do we even bother to use the concept of rest mass when everything is in motion relative to something else? Nothing is ever at rest. Or is it used in calculating the apparent mass to different observers?

That also brings to mind the perspective of the photon. From its perspective, can't it "see" itself as being at rest and everything else is moving at c? Then, if it's massless at rest AND not moving, then it doesn't have mass OR momentum and thus has no energy at all. If you have no mass and no momentum and no energy how can you even be said to exist at all?

11. Jan 16, 2008

### rbj

an inertial object is at rest in its own frame of reference.

yes. but i have to be careful here.

it is true that the terminology differentiating rest mass (the apparent mass in an objects own frame of reference) from the apparent inertial mass (a.k.a. "relativistic mass") of an object whizzing past an observer at 0.99c, has been "falling out of favor".

contemplating the "perspective" or frame of reference of a photon gets one in trouble. we really can't deal with a reference frame (ostensibly with an observer) that whizzes past another at a speed of c. the physics doesn't work out.

we get into arguments here about this at PF. i'm a relatively old electrical engineer who had a college level modern physics course in the 70s that had no problem with terminology of relativistic mass and of the masses of photons. back then we said:

total energy of a body:

$$E = m c^2$$

where

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

where m0 is the "rest mass" or what they now call "invariant mass" of the body. nowadays they just like to use the symbol "m" for that, but, to keep our terminology straight for this discussion, let's not do that. now the total energy of a body was the energy that it had at rest (used to be called the "rest energy") plus the additional energy it gets from being in motion (a.k.a. "Kinetic Energy"):

$$E = m c^2 = E_0 + KE$$

and since the kinetic energy of a body (from the perspective of some observer) is zero if it's not moving (relative to that observer) and the apparent inertial mass of the body not moving is the same of the rest mass (m = m0), then the rest energy of the body is

$$E_0 = m_0 c^2$$

and then, in general, the kinetic energy of the body (the total energy less the rest energy) is

$$KE = E - E_0 = m c^2 - m_0 c^2 = m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right)$$

now we all agree with the last two equations since they are expressed with no reference to this "relativistic mass" that has fallen out of favor, and if you allow the velocity v to be much much smaller than the speed of light c, then you will get the more familiar classical expression for kinetic energy:

$$KE = m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \approx \frac{1}{2} m_0 v^2$$

another difference in nomenclature or expression i have with the current trend is that while we all agree that this expression relating (rest) mass, momentum, and total energy is:

$$E^2 = E_0^2 + (p c)^2 = m_0^2 c^4 + p^2 c^2$$

i would disagree that this comes as a first principle and that the expression for relativistic momentum, p, is derived from it. i would say that the apparent momentum of a body is derived first from the kinematics and the knowledge of what relativistic time-dilation does, we get

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

and if we continue to interpret momentum as the product of the inertial mass, m, and velocity, v,

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

then we get the above expression for the inertial mass or "relativistic mass":

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

but this mass is not only a property of the body or particle itself, and i think that is why its use is commonly discouraged. but it is, in my opinion, a useful quantity and concept to keep. this relativistic mass, m makes the famous Einstein equation work, not only for rest energy, but for the total energy of a body or particle:

$$E = m c^2$$

that can validly represent the total energy of a body or particle if the relativistic mass is used for m. but if m means only the "rest mass" (what i've been calling "m0") then the above equation means only the "rest energy" (what i've been calling "E0"), devoid of kinetic energy, not the total energy of the body.

now, about photons, i have always qualified the term "massless" in regard to photons. photons have mass, but they do not have rest mass. the photon total energy is:

$$E = h \nu$$

and if that is equated to the relativistic total energy

$$E = h \nu = m c^2$$

then it is clear that the photon has a relativistic mass of

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

which is neither zero nor infinite. people here don't like me saying this, but what we do agree on is that, assuming the speed of the photon is c (for any observer), the momentum of the photon is

$$p = \frac{h \nu}{c}$$.

but i would say it is because

$$p = m v = m c = \frac{h \nu}{c^2} c = \frac{h \nu}{c}$$.

the other guys (that discourage talk of "relativistic mass") would say it's because

$$E^2 = m_0^2 c^4 + p^2 c^2$$

and because photons are "massless" (as a first principle), then m0 = 0 and

$$E^2 = (h \nu)^2 = (0)^2 c^4 + p^2 c^2$$

and solving that you get

$$p = \frac{h \nu}{c}$$.

I would not say that photons are massless as a first principle. i would turn the relativistic mass equation around and express the rest mass in terms of the relativistic mass (and v and c):

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

but if the apparent inertial mass or relativistic mass m is finite (as it is for a photon), then the rest mass must go to zero if v = c. so photons, although the have inertial mass (which i think is a necessary component to have a non-zero momentum), do not have any rest mass. nor do they have rest energy, all of their energy is "kinetic" (that might not be the best word for it, so i am bracing for pervect or someone to slap me down for saying it).

this expression of the concepts makes sense to me, and i believe is pedagogically more accessable than how most of the real physicists here like to express it (without the use of relativistic mass, so then given the only mass they recognize, rest mass or "invariant mass", then photons are massless). but, again, i am not a physicist. i am an electrical engineer who also does math for a living. i get into other pedagogical disputes here regarding the dirac delta "function", but that's a different topic.

12. Jan 16, 2008

### Staff: Mentor

Have you learned about the spacetime interval? In special relativity certain concepts that we once thought were independent and absolute, like time, were found to be frame variant. And yet, we were able to find a spacetime "distance" that is not frame variant and that all observers can agree on. That is the length of the spacetime interval.

Similarly for energy and momentum. They are combined in the same way that time and space are combined, and the length of the energy-momentum 4-vector is the invariant mass. It is something that all observers can calculate and agree on and it is the same as the rest mass. For light the length of that energy-momentum 4-vector is 0 so it has 0 invariant or rest mass. It is the only mass that all observers will agree on.

13. Jan 17, 2008

### Meatbot

Why has it been falling out of favor? I was under the impression that this mass gain was real and not virtual. If its real then you have to deal with it as if it were. Is it just because it's harder to do the math if you have to keep in mind that the mass is different for different observers?

Apparently, as I was shown yesterday. You get division by zero in the Lorentz equation. But my question is why can't we deal with it sonehow? Is our equation wrong? Is a photon doing something qualitatively different than other moving objects so it requires a different equation altogether? Perhaps its irrelevant and a photon does not really move at c or perhaps there are no photons. What new physics is required to resolve this?

14. Jan 17, 2008

### lightarrow

Also, but even because you would have a different mass in the motion direction and in the perpendicular direction; also for other reasons. Rest mass is the only meaningful mass in SR.

If you think that rest mass is meaningless for photons, think about the fact that "two" photons not travelling in the same direction, have a total rest mass =/= 0:

E^2 = (cp)^2 + (mc^2)^2

If the two photons don't travel in the same direction, you can find a ref frame in which they travel in opposite directions, so that total momentum p = 0, so E^2 = (mc^2)^2 --> m = E/c^2 =/= 0.

Edit: you can compute mass in a different ref frame because mass is invariant = doesn't change with ref frame.

Last edited: Jan 17, 2008
15. Jan 17, 2008

### lightarrow

The problem is that "photon's perspective" doesn't exist because to assign a physical meaning to that concept means to find a meaningful ref frame co-moving with a photon and that frame cannot be find in physics. So you can't do that reasoning.

16. Jan 17, 2008

### twinsen

What exactly does this mean? that we cant travel as fast as light so there is no way we can create a refference frame?

Im also interested in how the energy of a photon is derived as E=h\nu
Has this come from experimentation?

Alex

17. Jan 17, 2008

### CJames

rbj, I'm a fan of using relativistic mass as well, even though it can be confusing for some, because it's so much easier to work with mathematically (for me anyway) than relativistic kinetic energy

18. Jan 18, 2008

### chooserslain

No, we can travel with the speed of light. But even when we travel with the speed of light in the same direction with the photon, the photon, with respect to us, still travel with the speed=c.

There is no reference frame that light, in that frame, travels with speed different from c.
That because the space-time is distorted. As we try to travel faster, in our reference frame, the more space is stretched and time is condensed. So, the light, due to the fact that distance is stretched and time is condensed, has it velocity= distance/time still remaining c.

You can use the Lorentz fomula of velocity addition to verify:

A body travels with speed vx in a rest reference frame RF1. The speed of it when we observe it while we are resting in a reference RF2 which is travelling with speed v0 with respect to the RF1 is vx':

vx'=(vx-v0)/(1-vx.v0/c^2)

Let's consider the body as a photon, so vx=c. Let's try to travel with the speed of light, so our speed v0->c, then the speed of the photon in our reference frame, RF2, is
vx'=lim (as vx->c) (vx-c)/(1-vx.c/c^2)=c

In general, no matter what we are travelling, following the photon, with the speed of c/100, c/10, c/2 or 0,9999c, or even c, the space will stretched, time will condensed, so as that the photon still travel with its speed=c with respect to us. That the weird characteristic of photon.

Last edited: Jan 18, 2008
19. Jan 18, 2008

### Staff: Mentor

Eh? How do we accomplish that?

20. Jan 18, 2008

### jostpuur

This is incorrect use of the velocity addition formula.