# Photon energy limits?

Is there a theoretical or practical limit to how little or how much energy a photon can have? Something below ELF radio, or above Gamma rays?

So far as I know, the energy of any photon is just going to be Planck’s constant times its frequency. For any single photon you can find a reference frame that makes the energy as small as you want (i.e., a frame where it’s frequency is redshifted towards zero), so it wouldn’t really make sense for there to be a lower limit. Likewise, you could increase the frequency (blue shifting), so the same argument suggests there isn’t a maximum (although I feel a little shakier talking about arbitrarily large energies – seems like some new physics could come into play at high energies).

There’s a practical limit, of a sort, where a photon has a high enough energy that it and a microwave background photon have enough energy in their center-of-momentum frame to create electron-positron pairs.

A quick Google for a cite failed me.

Good point, ZenBeam.

As ZenBeam says, it takes a couple photons to do this. (This is due to conservation of mass – a single photon doesn’t have mass, as evidenced by the fact that you can redshift it away to nothing, but a system of multiple photons can have mass). But as ZenBeam also notes, a second photon isn’t hard to come by.

It takes about 1.022 MeV to make an electron-positron pair. (Electron/positron mass = 511 keV/c[sup]2[/sup])

Mega high powered photons bring up an interesting, though probalistically equal to zero scenario.

Imagine the mother of all high energy photons created somewhere else in the universe. Its heading towards earth. When it hits us its like getting hit by the Death Star. And we would never even see it coming. We could even call it the OMG particle/photon.

Err…there already has been an OMG particle recorded. It was a proton with exceptionally high energy. That said it is still a proton and so, even with all that energy, it would amount to a brick falling on your toe. Huge for such a small particle but not Death Star like.

As for the blue shifting above I would assume there is a limit. You cannot move faster than the speed of light so, at some point, there is no more blue shifting to be done.

Errr yourself. I know about that particle. I SAID if there was no limit (or an absurdly high one) you actually could have a honest to goodness OMG! disturbance in the force level particle, not some namby pamby falling brick level omg particle.

There is no theoretical limit to the blue shift.

Check out the section “Motion along the line of sight” on this page, and the equation shown in the sentence “The ratio f_s/f_o = √(1+ß)/(1-ß) is called the Doppler Factor …” This is the ratio of a photon’s emitted frequency (at its source) to its frequency as measured by an observer moving toward or away from the source.

This ratio gets arbitrarily small, and therefore the blue-shift arbitrarily large, as ß goes to negative 1 (meaning you’re moving toward the source at a speed close to C).

Likewise, there’s also no theoretical limit to the red-shift, if you speed in the opposite direction.

Quantum gravity might also impose an ultimate limit, presumably somewhere in the vicinity of the Planck energy, but that’s pure guesswork, since we don’t know how quantum gravity work.

Huh? I thought it was mass-energy that was conserved these days - e=mc^2 and all that. A photon has energy, surely. Why can’t that convert to mass? (I guess there might be reasons why it doesn’t, but is it really because of mass conservation?)

Also, if one photon has no mass, how can “a system of multiple photons” have mass?

Yeah, I thought it was the case that a photon had no rest mass, but had a mass equal to h-bar * c / lambda, although I may have c and lambda inverted. IOW, any photon with an energy above 1.022 MeV could become an electron-positron pair which would immediately annihilate back into a photon (or more than one I guess). Please explain.

Thanks,
Rob

Mass and energy are not the same thing. In fact they’re not things at all, they’re properties of a system.

Einstein’s equation that relates energy, mass and momentum is:

m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup] (c = 1)

So if you have a system of photons that have a zero momentum frame then the momentum of the system (p) equals zero and m = E.

In other words: a photon has both energy and momentum. If it decays into a particle/anti-particle pair, the pair must have the same net energy (rest-mass energy + kinetic energy) and same net momentum (a vector) as the original photon. It’s been a while, but if I remember correctly, these quantities cannot be made to add correctly for a one-photo-to-two-particle decay. There must be another photon or particle to supply or absorb some of the momentum and energy.

Think of the reverse, the +/- pair combining to one photon, and change your reference frame so that the net momentum of the pair in zero (always possible). A single photon cannot have zero momentum without also having zero energy. Ergo, one other photon must be produced or some energy and momentum transferred to a fourth particle.

About the monster photon headed towards Earth: I don’t know if there’s a theory for the interaction of matter with extremely high-frequency EM radiation, but it’s possible that the Earth might be transparent to it.

Mass and energy ARE the same thing, that’s what E=mc^2 is all about. Total energy and rest energy (aka total mass and rest mass) are not the same thing. So, no, conservation of total mass does not preclude the decay of a single photon, and there is no such thing as the conservation of rest mass. Edward Lost has the correct argument, i.e. you cannot conserve both energy and momentum with a single photon decaying into an electron and positron.

The easiest way to see this is to think about the reverse reaction in the center of mass frame of the electron and positron. When they collide, their energy can produce a photon, but a single photon will travel in some direction at the speed of light and carry a momentum of E/c, while the momentum of the original electron-positron pair was zero.

Mass and energy ARE not the same thing. Mass is the magnitude of the energy-momentum four-vector and energy is its time component.

E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]

If you don’t agree with this I suggest you take it up with Einstein, and also explain why a single photon has energy but no mass.

There is only one meaning of mass and it’s what I stated above.There’s no need to speak of rest mass because mass is simply mass.

ETA you can say that mass is equalto the energy of a system that cannot be transformed away.

This is the way I usually define it. You could define “mass” in some other way, but it wouldn’t be nearly so useful to do so. Under the current definitions, which have proven to be very useful, Ring is correct.

You are using mass to refer to rest mass, I am using mass to refer either to E/c^2, or if I use the adjective “rest” to the rest mass. This is the conventional approach in the teaching of relativity at the elementary level, without four vectors. I am not familiar with the convention of forbidding the use of the term mass in expressing the quantity E/c^2, but that seems like another way to do things, and perhaps has some advantages, so I will concede the point, though pedagogically, I don’t think it is the best approach for explaining it to people for whom the conclusion is not already familiar.

We can all agree that the reason for two photons is either: simultaneous conservation of energy and momentum, or in the four vector terminology, conservation of four momentum.

So the “for dummies” answer is that you can’t get an e p pair out of a single photon because that would involve creating momentum out of nothing? That makes some sense to me. I guess you can’t get a vector out of a scalar (mass, or energy come to that). Is that basically right? (I don’t doubt that it is oversimplified.)

njtt

That’s the idea! The way to say it is that a single photon has a momentum and an energy, but there is no way to assign momentum and energy to the electron and positron so the total energy and total momentum is equal to that of the original photon.

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