Is a photon truly massless? [duplicate]

Question

First of all, I am not a physicist or mathematician, not even a hobbyist but the following statements have always puzzled me:

1. $E=mc^2$
2. A photon is a desecrated particle of energy.
3. A photon is massless.

To me (as a complete novice), all three statements cannot be true at the same time because, if $E=mc^2$ then $m = E/c^2$. If $E$ is the energy (of a photon in this case), it is a 'positive number' no matter how small. Then $E >0$, $c^2 >0$ and therefore $E/c^2 >0$. Therefore, $m$ is a positive number, meaning a photon does have mass.

I saw another post here which asks the same question (again I am not a mathematician). If I read between the lines I believe it says $E=mc^2$ is not 100% true in that it misses some aspects (if I read between the lines correctly).

Can someone please give me an answer (in very very layman's terms) to my question above, i.e. am correct to say that give the three statements above and my basic logic of working this through there is a contradiction?

Answer 1

Although everyone has heard of the famous equation:

$$ E = mc^2 \tag{1} $$

few realise that this is a special case and applies only in limited circumstances. Specifically it applies only to a massive particle, and even then only in that particle's rest frame. The more general equation is:

$$ E^2 = p^2c^2 + m^2c^4 \tag{2} $$

where $p$ is the momentum of the particle. For a massive particle in its rest fram $p = 0$, and if we put this into equation (2) it simplifies to equation (1). That's why equation (1) applies only to massive particles in their rest frame.

You should now be able to see how the mass of the photon can be zero. If we substitute $m = 0$ into equation (2) it simplifies to:

$$ E = pc \tag{3} $$

For a photon the energy is related to its frequency by $E = hf$, and hence to its wavelength by $E = hc/\lambda$, and if we substitute this into equation (3) and rearrange to find the momentum we get:

$$ p = \frac{h}{\lambda} $$

which is indeed the equation for the momentum of a photon.

So your confusion arises simply because your point 1:

1. $E = mc^2$

does not apply to photons.

Answer 2

The correct version of your syllogism is:

• $E=mc^2$ for a particle at rest.
• For a photon, $E>0$.
• For a photon, $m=0$.

The correct conclusion is that a photon can never be at rest.

Answer 3

Light has inertia: it takes a force to change the direction in which they travel, and if you have a box with light bouncing around inside of it, it takes more force to change the speed of the box than if you had the same box with no light bouncing around inside of it. A given amount of energy E has an inertia $E/c^2$, as $E=mc^2$ suggests.

In the early 20th century, when $E=mc^2$ was popularized, mass and inertia were still used interchangeably. A new symbol, $m_0$ for "rest mass" was used to represent the amount of inertia that something has if the observer is comoving with it. This is also the least amount of inertia that something can have. Light does not have any of this, because the speed of light is the same in all reference frames.

Later in the 20th century, $m_0$ and the phrase "rest mass" was dropped from common use, and $m$ and "mass" were defined to mean the same thing that $m_0$ and "rest mass" had meant previously.

$E=mc^2$ is still true using the old meaning of $m$ (inertia) but it is not true using the new definition of $m$ (mass or "rest mass") except in the special case when $v=0$, which is never possible for light. For the meaning of $m$ in current use, Mike Stone's comment above is correct.

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notes:

Since you can be comoving with a box with light bouncing around in it, the energy associated with the light bouncing around in the box does contribute to the "rest mass" of the box a mass equal to $E/c^2$ where E is the energy of the light in the box. However, this is strictly the mass of the box because of the light, not the mass of the light in the box. If we put the observer in the box and asked him to get comoving with some of the light, it would be impossible.

I should also note that getting mirrors reflective enough to really have such a box for more than a tiny fraction of a second is not possible.

All of the $m$'s in Newtonian mechanics are inertia, not mass - although all of the masses you're likely to calculate with using Newtonian mechanics have masses which are so extremely close to their inertia that it doesn't matter.

Answer 4

I like pictures, and the relevant picture is:

You can ignore the formulae, they are just high school trig, tho, so nothing prohibitive. (The do look bad, though, I think it's because they are crammed in).

The only formula you need to grasp is the good ol Pythagorean Theorem:

$$ c^2 = a^2 + b^2 $$

which is invoked here as:

$$ (\rm Total\ Energy)^2 = (\rm Rest\ Energy)^2 + (\rm Momentum\ Energy)^2 $$

which is remarkably simple. Rest mass is the energy required to sit still and move forward in time. While energy from momentum is what it costs to move through space (always: in some frame).

The minimum energy is when you are motionless and just moving through time with the famous:

$$ E_0 = mc^2 $$

(I put the subscript on to avoid confusion, eventhough it is unsightly).

Meanwhile, the energy from motion is just:

$$ E_p = pc $$

and of course, it's all a photon ever has.

One thing that trips up our intuition (which is Newtonian), is that the concepts of momentum and kinetic energy were developed at low speed $v\ll c$, so we ignored the rest mass, and had this thing called kinetic energy (the purple part) which is just to the total energy minus the rest energy:

$$ T = E-mc^2 \rightarrow \frac{p^2}{2m} = \frac 1 2 mv^2 $$

where the RHS is in the low speed limit, and are the Newtonian formulae from Physics 1. Of course, when Newton did his thing, we were not splitting the atom and extracting kinetic energy from "mass" (not to be confused with "matter"), so the $mc^2$ was ignored, as it was effectively constant.

All we had to work with was kinetic energy, we did not know it was the difference between two relativistic energy concepts.

Kind of amazing the relativity has Newtonian physics as its low speed limit.

Also in the low speed limit we have:

$$ p = mv $$

which we intuitively associate with "inertia"--why being hit by a bus is worse than a smart car. Or a 400 pound lineman in a trot can level you as effectively as a 200 pound free safety at full speed (though the latter has more energy, and hurts more).

The problem is, we then associate momentum with mass in motion, but it's really energy in motion, and a photon is a pure chunk of energy, in maximal motion.

The irony is that if you look at momentum per total energy:

$$ \frac p E = \frac{\gamma mv}{\gamma mc^2} = v/c $$

you find that mass makes it smaller, and moving at the speed-of-light with no mass gives you the maximal momentum per unit energy.

Answer 5

Mass is not an inherent property of energy. It is a property energy can have, but does not always have. Specially, confined energy has associated mass, while free energy does not. A freely traveling photon has zero mass. A photon trapped in a mirrored box does increase the mass is the box by $E/c^2$.

It is roughly like saying the flame temperature of coal is $2200 \rm ° C$. A lump of coal sitting on a table certainly does not have that temperature, but it would have that temperature if it were ignited under the right conditions.