Two body problem reduced to One body problem | Classical Mechanics Lectures 21 | Central Force

[@ProfSivakumarRajagopalan]

MSc level Classical Electrodynamics Playlist th-cam.com/play/PLtUBquogGkRv5CME__m2oudBqW3_ffv33.html
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[@V15H4L_4RJUN]

Best explanation for solving centre force problems , thank you sir

[@theerdhathulasi6875]

Well explained sir! Thank you

[@asifalis9741]

Clearly understood sir

[@adwaitm6871]

Thank you, sir!

[@abhisheks1723]

Thanks a lot sir!

[@sushreemonalishasahu9320]

sir, while calculating for the kinetic energy part you have cancelled the terms, i think it can only be cancelled when m1= m2 but in this case m1 is not equal to m2. so we can't cancel it. kindly check that part and please correct me if i am wrong.

[@ProfSivakumarRajagopalan]

At 19:50, we are canceling the 2ab terms of (a+b)^2 and (a-b)^2. Both terms have 2*m1*m2 as the common factor. I forgot to write the masses in these terms. Thank you for identifying.

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Thank you sir. ✨

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Very good and we understood well.

[@ProfSivakumarRajagopalan]

Glad to hear that.

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Thank You sir.

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Thank you Sir

[@vishwa9713]

Very well explained sir. Thank You👍🏻

[@ProfSivakumarRajagopalan]

Best wishes.

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Thank u sir❤️❤️

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thank you sir

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Thanks a lot sir❤️

[@ProfSivakumarRajagopalan]

Most welcome.

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Nice lecture sir

[@ProfSivakumarRajagopalan]

Thanks for liking.

[@sampadsaha3277]

Thank you, Sir.

[@saadhassan8813]

Thank you sir. I have a question, when we translated our co-ordinate, we converted our problem from 6 DOF to 3 DOF. But we developed another formulation, parallel to r_1 and r_2. If we did translation in that formulation, we would again get 3 DOF. What was the need for formulation with respect to centre of mass vector and that small curly r vector.
Thanks once again for great lecture.

[@ProfSivakumarRajagopalan]

The need for re-formulation is to reduce the number of degrees of freedom. In case if the center (CoM) of mass formulation is not capable of reduction of CoM, then there is no need of new formulation.

[@saadhassan8813]

@@ProfSivakumarRajagopalan Thanks Sir

[@anjalim5135]

thanku sir

[@anaghavinayan7143]

Tnq sir

[@criticalthinking575]

Sir why can't we. Just use polar coordinates in which there just be r1 and r2 .. So 2 degrees of freedom

[@blackhole1222]

Why not from the beginning we define the coordinate system such that it's centered at m1 without the need to introduce the center of mass and all the other stuff? ?

[@ProfSivakumarRajagopalan]

At the beginning, we do not know that choosing a COM (center of mass) frame is more easier than otherwise. First we perform our calculations with origin at some arbitrary place. Then repeat the calculations with origin at COM. After doing all the calculations, we discover that COM leads to simplified calculation steps.

[@tayyabahaider3872]

Sir..why you did not translated it directly to the point of centre of mass?

[@ProfSivakumarRajagopalan]

The equation (6) is the Lagrangian in the center of mass coordinates. Further continuation is in the next lecture, which is Lecture-22. See at 36 min 42sec of the lecture-22 where we get the equivalent one-body problem. Ultimately we are obtaining the Lagrangian in the center of mass frame of reference. So, what is your question/doubt? I could not understand your question.
Lecture #22 th-cam.com/video/A8fExeJY40Q/w-d-xo.html
Full Classical Mechanics Playlist th-cam.com/play/PLtUBquogGkRukyAAWOmq2LmQx8Yf9H_3E.html

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@@ProfSivakumarRajagopalan Yes sir I have taken the next lecture and got it thank you for response

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