I never understood these Feynman diagrams but after watching your videos everything is crystal clear now. Thank you very much for making these videos :)
2:32 Aniha-whOOps ani-oop. He's a VSCO guy!!! sorry I had to, btw great video, really helps for revision. And your methods for remembering are not lame, they're just hilarious, which helps u remember it better!
Thank you so much. Exams are in a week and this helped me understand it so well. Even when I wanted to quit your jokes at the begging helped me stay motivated and complete all the videos.
I'm so glad to help! Actually, it looks like photons are bosons! physicsworld.com/a/ultra-precise-test-confirms-photons-are-bosons/ When solving exam questions, whenever there's an interaction, you'll need one of these particles (normally you should expect to use either a photon, Z, or W+ or W-).
Ahh, but no, it can't be a Z, since at the point in time (on the x-axis how I drew it), there is ONLY the boson in existence. And since as you wrote correctly the left side has +1e charge, and the right side has +1e, then the boson in the middle must carry that exact +1e charge. You see, charge needs to be conserved at every moment in time, not just on the left or the right (there they're the same). It also needs to be +1e in the middle, when there's ONLY the boson existing. That's why we know we need a boson with a +1e charge (and therefore why we know it has to be a W+). Hope that helps! Well done for double checking. Not many students are so brave to ask a question :)
Hi, I understand the conservation rule in this case but with the beta minus isn't the left side just -1/3e. Why does the boson then become W - if the left side is only -1/3e. Am I missing something?
Hi, but that's beyond the syllabus of what we need for the IB diploma. Although calculating probabilities using Feynman diagrams is actually super useful. It allowed Feynman to be much quicker than calculators at the time!
Thanks for the video. I have a question: in the exam question, could the answer be a z0 boson, a tau neutrino and a positron, all of them going out of a single junction? The conservation laws are conserved if I'm not wrong.
So long as the conservation laws work out, anything is possible I suppose for an exam. However, there are actually many more factors to consider in real-life interactions. For the purposes of IB exams at least, it will be kept nice and simple. Just look at conservation of charge, baryon number, strangeness, lepton number. That's all you need for the exams.
In the previous video, with the beta minus decay, you said that tha charge of the left side (which is quark d) is -1/3e and for the right side (quark u) its 2/3e. So you put a W- boson so 2/3 - 1 = -1/3. I understood this, you searched the charge of the left side ans did the sum of the charge of the boson with the right side and you had the conservation of both of the charges. But in this video, in the last feynman diagram you said that tha boson is W+ because in the left side we have a charge of 1e and sale for the right side. So you didn't do the sum of boson with right side here. I DIDNT UNDERSTAND WHAT I HAVE TO DO SO IF SOMEONE COULD HELP ME WITH THIS INFORMATION IT WILL SAVE MY YEAR. IB finals are in 3 days so please I need help. My english is maybe bad cause I'm doing IB in french. Thank you in advance I hope someone will help me like this amazing teacher did
You need to make sure the charge is conserved at ALL points in the diagram with respect to time. So this means at any given snapshot of time, the charge should be the same. That's why I needed to add the W+ boson. Cheers, Mitch
If it's something to do with just electrons or positrons, then it's a photon as your 'wingman'. If it's with quarks, then it's either the Z^0, W^+ or the W^- particle. Depends on the charge
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Dude, I wonder if you sometimes think about how many students you saved with your videos.
Awww, thanks. I am just happy to help, seriously. Your comment made my day :) Cheers, Mitch
sir, you are literally a legend
Thank you so much for the kind words. You made my day :). Cheers, Mitch
I never understood these Feynman diagrams but after watching your videos everything is crystal clear now. Thank you very much for making these videos :)
That makes my day - thanks for the lovely comment :)
been binging your videos and solving questions for the upcoming exams. Thank you for all the content that you've uploaded
I'm so happy to help! Good luck with your upcoming exams - you can do this!
@@OSC1990 thank you!
2:32
Aniha-whOOps
ani-oop.
He's a VSCO guy!!!
sorry I had to, btw great video, really helps for revision. And your methods for remembering are not lame, they're just hilarious, which helps u remember it better!
You're "the best teacher I never had" :--) Thank you for all the videos, it's life saving
Hah, thanks! I'm so glad you like the videos. Keep working hard on the subject - you can do it!
Thank you so much. Exams are in a week and this helped me understand it so well. Even when I wanted to quit your jokes at the begging helped me stay motivated and complete all the videos.
I'm so glad to be able to help you! Good luck with the upcoming exams - you've got this!
sir...i cant explain how thankful i am
You're too kind. I'm happy you can benefit from the videos! Cheers, Mithc
Thank you so much Mr.Mitch, this was very helpful! Just want to ask when do you us the photon as a bozon? Thanks!
I'm so glad to help! Actually, it looks like photons are bosons! physicsworld.com/a/ultra-precise-test-confirms-photons-are-bosons/
When solving exam questions, whenever there's an interaction, you'll need one of these particles (normally you should expect to use either a photon, Z, or W+ or W-).
Shouldn't the boson for the last question be Z^0? Since both sides have +1 charge, there should be no charge changes.
Ahh, but no, it can't be a Z, since at the point in time (on the x-axis how I drew it), there is ONLY the boson in existence. And since as you wrote correctly the left side has +1e charge, and the right side has +1e, then the boson in the middle must carry that exact +1e charge. You see, charge needs to be conserved at every moment in time, not just on the left or the right (there they're the same). It also needs to be +1e in the middle, when there's ONLY the boson existing. That's why we know we need a boson with a +1e charge (and therefore why we know it has to be a W+). Hope that helps! Well done for double checking. Not many students are so brave to ask a question :)
Hi, I understand the conservation rule in this case but with the beta minus isn't the left side just -1/3e. Why does the boson then become W - if the left side is only -1/3e. Am I missing something?
Nevermind got it, thanks!
@@Anukritlol Hi! I am a little bit confused about it. Could you kindly explain it? I appreciate that!
ohhhh I got it! thx
You're the best
Thank you so much for the kind words. Cheers, Mitch
Do we need to memorize the charge and information about quarks and leptons? Or will we be given information in our data booklet?
No, they give you these on the Data Booklet :)
Hi can you make a video about Fayenman reules to calculate probability of this diagrams
Hi, but that's beyond the syllabus of what we need for the IB diploma. Although calculating probabilities using Feynman diagrams is actually super useful. It allowed Feynman to be much quicker than calculators at the time!
Thanks for the video. I have a question: in the exam question, could the answer be a z0 boson, a tau neutrino and a positron, all of them going out of a single junction? The conservation laws are conserved if I'm not wrong.
So long as the conservation laws work out, anything is possible I suppose for an exam. However, there are actually many more factors to consider in real-life interactions. For the purposes of IB exams at least, it will be kept nice and simple. Just look at conservation of charge, baryon number, strangeness, lepton number. That's all you need for the exams.
okay, thank you very much!@@OSC1990
In the previous video, with the beta minus decay, you said that tha charge of the left side (which is quark d) is -1/3e and for the right side (quark u) its 2/3e. So you put a W- boson so 2/3 - 1 = -1/3. I understood this, you searched the charge of the left side ans did the sum of the charge of the boson with the right side and you had the conservation of both of the charges. But in this video, in the last feynman diagram you said that tha boson is W+ because in the left side we have a charge of 1e and sale for the right side. So you didn't do the sum of boson with right side here. I DIDNT UNDERSTAND WHAT I HAVE TO DO SO IF SOMEONE COULD HELP ME WITH THIS INFORMATION IT WILL SAVE MY YEAR. IB finals are in 3 days so please I need help. My english is maybe bad cause I'm doing IB in french. Thank you in advance I hope someone will help me like this amazing teacher did
How do I know when to use a Z boson or a photon boson if no charge is required from the "wingman"?
I think it is because charge needs to be conserved, therefore it must remain the same at every point of the diagram
yes, I m also confused with when to use the Z boson and a photon boson??
@@arwa519 i dont know either, but i think the question will say "photon is produced" so if it doesnt ig just assume it is a Z boson
@@CR-ri9gq oh makes sense, thank youu, btw are you M22?
@@arwa519 yes i am! And I'm entirely unprepared for physics.
Why does he use a W+ Boson if the charges on each side of the equation are same?
You need to make sure the charge is conserved at ALL points in the diagram with respect to time. So this means at any given snapshot of time, the charge should be the same. That's why I needed to add the W+ boson.
Cheers, Mitch
@@OSC1990 Much appreciated!
7:15 how do we know that the wingman particle is boson instead of something else?
If it's something to do with just electrons or positrons, then it's a photon as your 'wingman'. If it's with quarks, then it's either the Z^0, W^+ or the W^- particle. Depends on the charge
@@OSC1990 What do you mean if it has to 'do' with electrons? Do you mean if something transforms into an electron?
@@kellyjin2469 2/3 e + 1/3 e = e. W+ has a charge of e. Hence e=e, so we use W+