@Fausto Saporito asks a great question (below), which raises an important point: the final reaction is called a "flavor-changing neutral current"--something that has never been observed but may be theoretically possible. This Wikipedia page is a good starting place if you want to read more: en.wikipedia.org/wiki/Flavor-changing_neutral_current
8:25, why are we referring to the down quark as the "primary" quark we base our charge on, the arrows go in the other direction so I thought we would see what we need to do to 2/3 charge to get a -1/3 charge, which would be subtract one
Good questions!! You're thinking about it in terms of an equation (which is correct), but you're just using the wrong equation. What you wrote is "what do I need to do to 2/3 to get -1/3" which is equivalent to the equation: 2/3 + x = -1/3. This equation is wrong because it implies that x is an initial charge (co-occurring with 2/3)--that's incorrect; x is a final charge that co-occurs with -1/3. So the equation should instead be: 2/3 = -1/3 + x. There's an easy way to see what the equation should be, because it's always the same equation: (total initial charge entering the vertex) = (total final charge exiting the vertex) The way we figure out if a particle is "initial" or "final" is by (a) looking at the time axis over on the side, (b) ignoring the arrows on the particles themselves, and (c) looking at whether each particle touches the "initial" side of the vertex or the "final" side of the vertex. In this example, the time axis is pointing upward. Thus, anything touching the bottom of the vertex is an "initial" particle, and anything touching the top of the vertex is a "final" particle. The u quark touches the bottom of the vertex and thus it belongs on the "initial" side of this vertex's equation. The d quark and the exchange particle both touch the top of the vertex and thus both belong on the "final" side of this vertex's equation: (total initial charge entering the vertex) = (total final charge exiting the vertex) (charge of u quark) = (charge of d quark) + (charge of exchange particle) (+2/3)e = (-1/3)e + ? +1e = ? So when I'm solving this, I don't think of any of these particles as primary, and I think of the calculation as being based on both the u charge and on the d charge. There are two tricky things to keep in mind: (a) there are 2 different points/vertexes in our picture. The first vertex is what we've been discussing and shows up earlier in time. The second vertex shows up a little higher / later in time. When we consider a different vertex, we have to re-do the analysis, and particles that are "final" for the first vertex could end up being "initial" for the later vertex. Specifically, the W+ boson (1) is a "final" particle for the first vertex because it's touching the top of the first vertex, and (2) it's also an "initial" particle for the second (higher) vertex because it's touching the bottom of the second vertex. In general, for these Feynman diagrams, there will always be 3 particles at every vertex: an initial particle, a final particle, and an exchange particle. Sometimes the exchange particle will be on the "initial" side, and other times it will be on the "final" side. So the conservation of charge equation will always be one of these two: (charge of initial particle) + (charge of exchange particle) = (charge of final particle) (charge of initial particle) = (charge of final particle) + (charge of exchange particle) and whether we need the top equation or the bottom equation just depends on the picture and which side of the vertex the particles are touching, relative to the time axis. The arrows we draw on a particle itself doesn't tell us whether it's an initial particle or a final particle. That just tells us whether the particle is matter (the arrow points with the time axis) or antimatter (the arrow points against the time axis). We don't need to show this using arrows. If you had colored pencils, you could remove the arrows altogether and just draw matter with blue and antimatter with red. There is an interesting historical reason for using arrows (physics.stackexchange.com/questions/391/is-anti-matter-matter-going-backwards-in-time), but it's not really anything students need to know.
Very interesting video. I have two questions, in the last diagram about strange quark, the Z boson could be also a gluon, and in this case is there also a color change ? Could you explain this ? thanks.
This final reaction (mediated by a Z boson) would a flavor-changing neutral current, which hasn't actually ever been observed, to my knowledge. According to the Standard Model, these flavor-changing neutral currents are suppressed and can only occur via certain QM processes. So, as I understand, the diagram really should have a loop to reflect this suppression. I would like to re-do this video, remove this last example, and replace it with something that actually occurs. I don't believe it's possible to replace the Z boson with a gluon, because strange quarks cannot decay via the strong interaction--they can only decay via the weak interaction. In general, strong interactions conserve flavor.
Yes, that would be a nice thing to add--I created this video for an IB Physics high school course, and color isn't covered at all in that course, so unfortunately it probably won't be featured in the next iteration of this video. I'm also not an expert at all in this topic, and I would be a little wary to go very in-depth on it. I bet you could get great info from Fermilab's channel: ruclips.net/user/fermilabvideos
I believe Feynman diagrams can be used for any reaction. I think this must be true, since Feynman diagrams actually represent *mathematical equations* that are used in Feynman's path integral formulation of quantum mechanics. Here's a great video about that topic from the true experts: ruclips.net/video/hk1cOffTgdk/видео.html
![FNAF Into the Pit has A LOT of Unused Graphics | LOST BITS [TetraBitGaming]](http://i.ytimg.com/vi/6YhZCRdipl4/mqdefault.jpg)
@Fausto Saporito asks a great question (below), which raises an important point: the final reaction is called a "flavor-changing neutral current"--something that has never been observed but may be theoretically possible. This Wikipedia page is a good starting place if you want to read more: en.wikipedia.org/wiki/Flavor-changing_neutral_current
Thanks!!! I
Best explanations I've ever seen for beginners...THANK YOU!!
8:25, why are we referring to the down quark as the "primary" quark we base our charge on, the arrows go in the other direction so I thought we would see what we need to do to 2/3 charge to get a -1/3 charge, which would be subtract one
Good questions!! You're thinking about it in terms of an equation (which is correct), but you're just using the wrong equation. What you wrote is "what do I need to do to 2/3 to get -1/3" which is equivalent to the equation: 2/3 + x = -1/3. This equation is wrong because it implies that x is an initial charge (co-occurring with 2/3)--that's incorrect; x is a final charge that co-occurs with -1/3. So the equation should instead be: 2/3 = -1/3 + x.
There's an easy way to see what the equation should be, because it's always the same equation: (total initial charge entering the vertex) = (total final charge exiting the vertex)
The way we figure out if a particle is "initial" or "final" is by (a) looking at the time axis over on the side, (b) ignoring the arrows on the particles themselves, and (c) looking at whether each particle touches the "initial" side of the vertex or the "final" side of the vertex.
In this example, the time axis is pointing upward. Thus, anything touching the bottom of the vertex is an "initial" particle, and anything touching the top of the vertex is a "final" particle. The u quark touches the bottom of the vertex and thus it belongs on the "initial" side of this vertex's equation. The d quark and the exchange particle both touch the top of the vertex and thus both belong on the "final" side of this vertex's equation:
(total initial charge entering the vertex) = (total final charge exiting the vertex)
(charge of u quark) = (charge of d quark) + (charge of exchange particle)
(+2/3)e = (-1/3)e + ?
+1e = ?
So when I'm solving this, I don't think of any of these particles as primary, and I think of the calculation as being based on both the u charge and on the d charge.
There are two tricky things to keep in mind: (a) there are 2 different points/vertexes in our picture. The first vertex is what we've been discussing and shows up earlier in time. The second vertex shows up a little higher / later in time. When we consider a different vertex, we have to re-do the analysis, and particles that are "final" for the first vertex could end up being "initial" for the later vertex. Specifically, the W+ boson (1) is a "final" particle for the first vertex because it's touching the top of the first vertex, and (2) it's also an "initial" particle for the second (higher) vertex because it's touching the bottom of the second vertex.
In general, for these Feynman diagrams, there will always be 3 particles at every vertex: an initial particle, a final particle, and an exchange particle. Sometimes the exchange particle will be on the "initial" side, and other times it will be on the "final" side. So the conservation of charge equation will always be one of these two:
(charge of initial particle) + (charge of exchange particle) = (charge of final particle)
(charge of initial particle) = (charge of final particle) + (charge of exchange particle)
and whether we need the top equation or the bottom equation just depends on the picture and which side of the vertex the particles are touching, relative to the time axis.
The arrows we draw on a particle itself doesn't tell us whether it's an initial particle or a final particle. That just tells us whether the particle is matter (the arrow points with the time axis) or antimatter (the arrow points against the time axis). We don't need to show this using arrows. If you had colored pencils, you could remove the arrows altogether and just draw matter with blue and antimatter with red. There is an interesting historical reason for using arrows (physics.stackexchange.com/questions/391/is-anti-matter-matter-going-backwards-in-time), but it's not really anything students need to know.
@@danielm9463 oh that makes a lot more sense, thank you for your time and the quick response!
@@Ptoki1 Good luck with your studies!! Feel free to ask questions any time.
@@danielm9463 thanks :)
Very interesting video. I have two questions, in the last diagram about strange quark, the Z boson could be also a gluon, and in this case is there also a color change ? Could you explain this ? thanks.
This final reaction (mediated by a Z boson) would a flavor-changing neutral current, which hasn't actually ever been observed, to my knowledge. According to the Standard Model, these flavor-changing neutral currents are suppressed and can only occur via certain QM processes. So, as I understand, the diagram really should have a loop to reflect this suppression. I would like to re-do this video, remove this last example, and replace it with something that actually occurs. I don't believe it's possible to replace the Z boson with a gluon, because strange quarks cannot decay via the strong interaction--they can only decay via the weak interaction. In general, strong interactions conserve flavor.
This should be very nice. Maybe with something describing the colour change.
Yes, that would be a nice thing to add--I created this video for an IB Physics high school course, and color isn't covered at all in that course, so unfortunately it probably won't be featured in the next iteration of this video. I'm also not an expert at all in this topic, and I would be a little wary to go very in-depth on it. I bet you could get great info from Fermilab's channel: ruclips.net/user/fermilabvideos
Daniel M I understand, no problem :-) I just didn't understand when the color change happens. Anyway, I'll check the Fermilab channel, thanks!
Can we make feynman diagram for every reaction??
If charge, lepton, baryon, strangeness, isospin conserve.???
I believe Feynman diagrams can be used for any reaction. I think this must be true, since Feynman diagrams actually represent *mathematical equations* that are used in Feynman's path integral formulation of quantum mechanics. Here's a great video about that topic from the true experts: ruclips.net/video/hk1cOffTgdk/видео.html
Can we make feynman diagram???
If strangeness or baryon or laptop not conserved but charge conserved??
If I'm understanding the question correctly, then yes, we can draw Feynman diagrams for reactions where certain quantities are not conserved.