Elementary Particle Questions...fundamentals of Higgs Boson Theory

[fivemack]

Originally Posted by Dubslow

That's ~90 billion Joules, or ~25GWhrs. That sounds like a lot, and all at once it is, but (I can't find figures to back this up) it wouldn't power Chicago for a month. (And it took 40 years to make it all.)

You've lost several orders of magnitude there (and are using 'billion'=10^12 which is by now archaic; just use scientific notation). Americans use about 1.5 kilowatts of electricity (counting industrial use: that's electricity-consumption-of-US divided by population-of-US); there are three million of them in Chicago, so they use 4.5 gigawatt-hours hourly, or 4.5*10^9 joules per second.

One gram of antimatter annihilates with one gram of matter to give 0.002c^2 joules, 1.8*10^14J, which is the electricity used by Chicago in half a day.

[fivemack]

Originally Posted by Christenson

What happens if those electrons have many kEV of energy when annihilated?

I believe that the annihilation of a fast electron with a stationary positron just gives two more-energetic gamma-ray photons - momentum is conserved, all that's left is photons, so the momentum must end up in the photons.

So if you look at the spectrum of the sum of the two gamma-ray energies it starts at 2x511keV and goes up.

I'm not quite sure how you measure gamma-ray energies in this range; I think higher-energy gamma rays make scintillators sparkle more brightly.

[retina]

Originally Posted by Dubslow

Also keep in mind that over the 40 year history of Fermilab, they produced about enough antimatter to make a dollar bill

A negative dollar. Or -$1.

So if I buy something from them that costs $10 then I pay them $9 and they give me change of -1$.

[Christenson]

Originally Posted by retina

A negative dollar. Or -$1.

So if I buy something from them that costs $10 then I pay them $9 and they give me change of -1$.

Exploding change, that is!!!!

[lavalamp]

Originally Posted by fivemack

I believe that the annihilation of a fast electron with a stationary positron just gives two more-energetic gamma-ray photons - momentum is conserved, all that's left is photons, so the momentum must end up in the photons.

So if you look at the spectrum of the sum of the two gamma-ray energies it starts at 2x511keV and goes up.

Supposing you had a positron at rest relative to your frame of reference, and an electron zipped in at 0.936c (chosen for aesthetic reasons), collided and annihilated with the positron.

The energy to be conserved is therefore 511 + 2.84*511 ~= 1963 keV, and the momentum to be conserved is 2.84*511*0.936 ~= 1359 keV/c. The energies of the photons can be calculated by solving a couple of simultaneous equations.

Therefore the two gamma rays emitted in this case, with respect to your frame of reference, would have energies 302 keV and 1661 keV.

Notice that one of the photons has an energy LESS than the rest mass of a positron/electron.

Although I didn't include the maths, the solution is easy to verify:

1661 keV + 302 keV = 1963 keV
1661 keV/c - 302 keV/c = 1359 keV/c

[davieddy]

A tadge one-dimensional perhaps?

[Dubslow]

This is physics. Remember the spherical cow?

Seriously speaking, you can consider all these numbers to be the magnitudes are arbitrary-dimensional vectors. That is, his statements are true, even if the direction is unknown.

[davieddy]

Originally Posted by Dubslow

This is physics. Remember the spherical cow?

Seriously speaking, you can consider all these numbers to be the magnitudes are arbitrary-dimensional vectors. That is, his statements are true, even if the direction is unknown.

But there are udder things to consider.

David (Hilbert)
x

[Random Poster]

Originally Posted by lavalamp

Supposing you had a positron at rest relative to your frame of reference, and an electron zipped in at 0.936c (chosen for aesthetic reasons), collided and annihilated with the positron.

The energy to be conserved is therefore 511 + 2.84*511 ~= 1963 keV, and the momentum to be conserved is 2.84*511*0.936 ~= 1359 keV/c. The energies of the photons can be calculated by solving a couple of simultaneous equations.

Therefore the two gamma rays emitted in this case, with respect to your frame of reference, would have energies 302 keV and 1661 keV.

Notice that one of the photons has an energy LESS than the rest mass of a positron/electron.

Although I didn't include the maths, the solution is easy to verify:

1661 keV + 302 keV = 1963 keV
1661 keV/c - 302 keV/c = 1359 keV/c

That is the only possible solution in one-dimensional space. In spaces with more dimensions (like the one we happen to live in) there is an infinity of other solutions, because the photons aren't restricted to move parallel to the electron's trajectory.

[Dubslow]

Originally Posted by Dubslow

This is physics. Remember the spherical cow?

Seriously speaking, you can consider all these numbers to be the magnitudes are arbitrary-dimensional vectors. That is, his statements are true, even if the direction is unknown.

I partially take this back. The magnitude of the energies of the photons would be the same -- not necessarily so for the momentum. (Thanks to RP for getting me thinking properly.)

[davieddy]

Originally Posted by Dubslow

I partially take this back. The magnitude of the energies of the photons would be the same -- not necessarily so for the momentum. (Thanks to RP for getting me thinking properly.)

RP?