- #1
vincentchan
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Is there a systematical way to count all the possible feynman diagram with vertices less than or equal to n...
Counting Feynman Diagrams with n Vertices
- Thread starter vincentchan
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AI Thread Summary
There is no systematic method to count all possible Feynman diagrams with vertices less than or equal to n in quantum field theory (QFT). To determine the number of distinct diagrams, one must compute all possible unconnected Green functions using the generating functional approximation. The discussion also clarifies that two specific Feynman diagrams related to Compton scattering are indeed distinct contributions. This highlights the complexity of diagram counting in QFT. Overall, the conversation emphasizes the challenges in quantifying Feynman diagrams systematically.
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dextercioby
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They're not.There are the 2 (distinct) contributions to the Compton scattering of electrons (if i saw well,or positrons in the other case).
To the first question:nope,in general u cannot tell how many distinct Feynman diagrams a QFT has in a certain order.U have to compute all possible unconnected Green functions,using the approximation for the generating functional.
Daniel.
To the first question:nope,in general u cannot tell how many distinct Feynman diagrams a QFT has in a certain order.U have to compute all possible unconnected Green functions,using the approximation for the generating functional.
Daniel.
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.
Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
F1=(5000+200)g=5200*10=52000N --> F2=(5000+100)g=5100*10=51000N ?
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