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What is the Matter with Matter?

by FredJR
Tags: anti matter, hep, particle physics
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FredJR
#1
Jan23-13, 02:18 AM
P: 4
Physicists often remark about the absence of antimatter in the universe. My question here sounds simple, but the answer has proved elusive to me. Perhaps someone here can shed some light.

The question is: What is the physical property that "matter" has that distinguishes it from "anti-matter"?

The universe around us is made up of what we call "matter" which is essentially a couple of quark types (u and d) and a lepton type (e). So my question can be rephrased as:
"What physical property does the lepton 'e' share with the quarks 'u' and 'd' that gets them all in the same group (matter) and their anti particles classed as anti-matter?"

Particle Physics says we have 24 fermions (spin half particles). Each fermion has an anti-fermion. Each fermion also has its weak isopsin partner (the particle you get if you add or remove a W+/- as appropriate. Eg u->d when you emit a W+ or e->neutrino when you emit a W-. etc.). But that doesn't say which of the fermions is matter and which is anti-matter.

The fermions are split into those that experience the strong force (quarks and anti quarks) and those that don't (leptons and anti-leptons). The grouping of the electron with the up and down quarks into "matter" spans this. But what is the property?

The Standard Model of Electro -Weak interactions says that only left handed fermions and right handed anti-fermions experience the weak force. I've heard this used as an arguement to group left handed electrons and left handed up and down quarks etc. However theory applies to zero mass particles. I can accept that this can be used in the lepton regime where the neutrinos are effectively massless so even though the charged leptons are massive there is one component -the neutrino- which is massless. However in the quark regime all the fermions have mass.

So can anyone explain to me what is the property that the electron and its associated leptons have that gets them grouped with the 'up' quark rather than the anti-up quark? Other than its statistically large representation within the observed universe, which is not a property of the particle itself.
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Bill_K
#2
Jan23-13, 06:18 AM
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So can anyone explain to me what is the property that the electron and its associated leptons have that gets them grouped with the 'up' quark rather than the anti-up quark?
It's because, contrary to appearances, QCD and electroweak theory are not completely independent of one another. This requires the particles to be related. A world with leptons without baryons could not exist. The standard model is renormalizable only thanks to anomaly cancellation. Fermions come in three generations, and for each generation certain conditions must be satisfied. One is that the sum of the charges in each generation must be zero. (A quark counts as three particles, due to the three colors.)

0 = 3(2/3) + 3(-1/3) + (0) + (-1)

The second condition is quadratic:

0 = 3(2/3)2 - 3(-1/3)2 + (0)2 - (-1)2

Thus the charged lepton that goes with the up and down quarks must be negatively charged.
Simon Bridge
#3
Jan23-13, 06:27 AM
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A particle is an antiparticle if it annihilates with another particle.

The distinction of calling one kind matter and the other antimatter is a historical accident.
A distinction has been found in terms of CP violations... it's just not normally how we tell them apart.

Strictly speaking it is accurate to say that an electron is the antiparticle of a positron.

I don't think many people outside science fiction refer to a positron as "antimatter" though we may say that it is an anti-electron... I cant think of a reason you cannot call an electron an anti-positron.

http://en.wikipedia.org/wiki/Antiparticle

FredJR
#4
Jan24-13, 01:27 AM
P: 4
What is the Matter with Matter?

Thanks Simon Bridge for your answer. I don't think you quite got the bit I was stuchk at, maybe my question wasn't well worded. However Bill_K managed to work out what I was looking for and provide and answer. I'll read up on my QCD.

Thanks to both of you for your answers.

Regards
Simon Bridge
#5
Jan24-13, 03:33 AM
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Yeah - I was fishing ;)
Glad Bill_K could help.


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