You are invited to Log in or Register a free Frihost Account!

# The photon and E=mc^2

ocalhoun
I'm having a little problem with understanding the 'massless' photon...
If we use the equation E=mc^2, on a photon, using zero for the mass, we get E=0*c^2, reducing to E=0.
So, if a photon has no mass, then it must also have no energy.
But, suppose we start with the other variable known (I'll assume the value of it is '2', since I don't know the real value, and it doesn't really effect my point.) Then we have 2=mc^2, after solving that, we'll end up with the result of the photon having a tiny, but definite mass. The problem with that, though, is that if it does have any mass at all, the mass of it should effectively be infinitely large (according to relativity), since it is traveling at the speed of light.

Anybody have any insights to help me out of this conundrum?
Indi
Yes, you're using the wrong equation. It's actually E² = m²c⁴ + p²c² where p is momentum.

The square of the total energy of a particle (E) is the sum the square of of its energy due to its mass (mc²) - it's rest energy - and the square of its kinetic energy (pc). Or, E² = (mc²)² + (pc)².

The rest energy of an object with mass is:
E² = (mc²)² + (0 * c)² (the momentum of the object at rest is 0)
E² = (mc²)²
E = mc²

The total energy is:
E = √((mc²)² + (pc)²)

For a massless object, m is 0, so it has no rest mass, and its total energy is:
E = pc

Now, the momentum of a photon is related to its frequency (or wavelength).
p = ℎf/c (ℎ is Planck's constant)

So the energy of a photon is:
E = ℎf
Chris65536
If the equation is E² = m²c⁴ + p²c², then why do we all remember E=mc²? I can see that it's the simple case for objects at rest, but when relativity is introduced in physics class, we still use E=mc² for a given mass or energy to work out the other variable for high-speed particles.
Indi
That depends on who "we all" is. ^_^;

If you mean the general public: because E = mc² is much more thrilling. mc² is invariant in all reference frames, so it's kind of a universal constant for any object. An electron's total energy can change depending on its velocity - you know the equation from high school: KE = ½mv² (which is technically wrong, by the way, but good enough at low velocities). The total energy also changes: E = mc² + ½mv². But that rest energy never changes. If you want to get poetic, then for a massive particle, ½mv² is the energy it has due to its motion, and mc² is the energy it has due to its existence. mc² is also the glue between matter and energy - if you convert a massive particle to energy, you get mc² worth of energy, or to create a particle of mass m it will require mc² worth of energy. That's what happens in nuclear reactions (more or less). Really, it was the big paradigm shift of Einstein - everyone knew a particle had energy when it was moving long before he came along, but this idea that a particle is energy, mc² worth of energy, was new.

If you mean your physics class: you'd have to ask the teacher. Either you didn't understand the lesson, or he was teaching it wrong.
E ≠ mc² for high-speed particles.
E = mc² for all particles (note the subscript)
E = mc² for particles at rest
E = E₀ + KE = mc² + ½mv² for low-speed particles
E = √((mc²)² + (pc)²) for high-speed particles
It's probable he wasn't dealing with high-speed particles at all - he was only talking about particles at rest. If you didn't deal with Lorentz, he wasn't dealing with high-speed stuff.
quasar
The popular formula E = mc^2 cause confusion. The correct expression is
E = mc^2 / sqrt[1 - (v^2 / c^2)] ---> sqrt: square root
where m is independent of velocity.
Chris65536
I did A-level physics. E=mc² was basically the last topic we covered, and only occasionally showed up in questions (and like I say, we were usually told that it has "x joules of energy; what is the mass?")

Now I understand though. E² = m²c⁴ + p²c² says that moving particles have energy, E=mc² says that particles are energy. Plus it's shorter!
Indi
 Chris65536 wrote: I did A-level physics. E=mc² was basically the last topic we covered, and only occasionally showed up in questions (and like I say, we were usually told that it has "x joules of energy; what is the mass?") Now I understand though. E² = m²c⁴ + p²c² says that moving particles have energy, E=mc² says that particles are energy. Plus it's shorter!

Ah, ya, i don't remember doing anything noteworthy in terms of relativity in A-level physics either - the high school kids here do more than we did.

Yes, shorter. ^_^; And more culturally gripping - "Matter and energy related by the speed of light? Shocking! But what does it mean?" But keep the longer forms in mind for the more complex stuff. They all reduce down to E = mc² very easily. A particle at rest has no momentum so p = 0, so E² = m²c⁴ + p²c² reduces to E = mc². Or in the form quasar gave: E = mc² / √(1 - (v² / c²)). At rest v is 0, so E = mc² again.

Incidentally, the form quasar gave is identical to the form i gave - you just need to do some manipulation using p = mv / √(1 - (v² / c²)), the momentum for a massive particle (note how it reduces to the standard formula for momentum when v is much smaller than c), and you can switch between the two.

Of course, the form quasar gave doesn't work for massless particles, but it does fail in informative ways:
If m = 0 and v < c, E = 0... so a massless particle travelling at less than c can have no energy... which means it can't be travelling, which means it can't exist.
If m = 0 and v = c, then E = 0/0, which is indeterminate, which means it can be anything at all... but at least we know it is possible... so a massless particle can theoretically exist if and only if it is travelling at c.
And if v = c, then E = mc / 0, which goes to infinity, which is nonsense unless m is also 0... so the only things that can travel at c must be massless.

And there you have it:
A massless particle is possible if and only if it is travelling at c.
The only thing that can travel at c is a massless particle (a massive particle travelling at c would have infinite energy).

Of course, this doesn't prove that massless particles exist, just that they can exist (and if they do exist, they must be travelling at c).

A whirlwind tour of special relativity. ^_^;
knight_frost
 Quote: In physics, the photon is the elementary particle responsible for electromagnetic phenomena. It is the carrier of electromagnetic radiation of all wavelengths, including in decreasing order of energy, gamma rays, X-rays, ultraviolet light, visible light, infrared light, microwaves, and radio waves. The photon differs from many other elementary particles, such as the electron and the quark, in that it has zero rest mass;[1] therefore, it travels (in a vacuum) at the speed of light, c. Like all quanta, the photon has both wave and particle properties (“wave–particle duality”). Photons show wave-like phenomena, such as refraction by a lens and destructive interference when reflected waves cancel each other out; however, as a particle, it can only interact with matter by transferring the amount of energy E = {hc\over\lambda}, where h is Planck's constant, c is the speed of light, and λ is its wavelength. This is different from a classical wave, which may gain or lose arbitrary amounts of energy. For visible light the energy carried by a single photon is around 4×10–19 joules; this energy is just sufficient to excite a single molecule in a photoreceptor cell of an eye, thus contributing to vision.[4]

Source - http://encarta.msn.com/encyclopedia_761574512/photon.html

[MOD - source and quote tags added - Bikerman]
infinisa
 quasar wrote: The popular formula E = mc^2 cause confusion. The correct expression is E = mc^2 / sqrt[1 - (v^2 / c^2)] ---> sqrt: square root where m is independent of velocity.

Surely you mean "where m is the mass of the particle at rest", let's say m₀.

So we have:
E = m₀c^2 / sqrt[1 - (v^2 / c^2)

But a particle travelling at velocity v has a relativistic mass of m₀/ sqrt[1 - (v^2 / c^2)]
So we get:
E = mc^2
where m is the relativistic mass

And so the simple version of the equation is right after all!

I think the version involving momentum is needed only in the case of a massless particle - which is of course the reason for this thread in the first place!
victornumber
great post guys. I thought this would be a simple conversation on the forum when I came to help to solve it but it turned really really complex. I didn't know there were so many smart people on this forum. Kudos to all of you guys ...............X100
Rohedi
I thing infinisa's explanation is correct, we must distinguish the total mass m with the rest mass m0. I love E=mc^2 forever, hehehehe...
Dennise
You can compute a photon's energy from it's 'wave' frequency.

i.e. E=hxf

where h = Plank's constant.