I believe that is because dS = 0 only when the Taylor expansion is truncated at first-order terms. For example, consider dS = S[r(t) + ŋ(t)] - S[r(t)], where ŋ(t) represents an infinitesimal deviation from the stationary action path. Therefore, S[r(t) + ŋ(t)] can be expanded as S[r(t)] + ∇S[r(t)]·ŋ(t) + high-order terms. Here, ∇S[r(t)] = 0 because it lies on the path of least action. Thus, when ignoring high-order terms, dS = 0. If we add (∂L/∂t)*dt to dL, we will have a term with the integral [ ∫(∂L/∂t)dt ]*dt, resulting in a high-order term. I'm not sure if this is correct, but it occurred to me. I would appreciate it if someone could confirm if this makes sense.
A couple of cirrections: 1. on Min. 37:13 - the derivative od the potential should be w.r.t. the angle \theta. 2. on min. 41.40 - z should probably be y, for consistency.
Nice clear derivation. But at 24:20 did you justify that the sum of the integrals equal to 0 implies that each individual integral = 0 and that each integrated term = 0? This is the hard part for me.
Good question: it's a subtle point. The idea is that if we start out with the path with least action then the action will not change (to first order) if we change the path a bit. But there are many ways we can change the path: we can vary each of the several generalized coordinates q_i at any time between the start and end points. The change in the action, dS, is zero for *any* change we make to the path, changing any of the coordinates at any time, dq_i(t). But how can dS always be equal to zero for any dq_i(t)? It can only happen if the function itself being integrated is zero. Then multiplying by the arbitrary function dq_i(t) for each i won't affect the integral -- it's always just zero!
Good lecture, but the application of EL EoMs to the pendulum around 43:30 is somewhat faulty: the constraint r(t)=l should be taken into account from the very start, but here it is used only after writing out the EoMs. This does not matter for the EoM in theta, but if the EoM in r is written out as on the preceding slide, then it will falsely say that mr-dbldot=centrifugal force + radial component of gravity, instead of r-dbldot=0 in reality (the radial gravity term comes from differentiating -V = mgrcos(theta) w.r.t. r, with V=0 at the pivot). This says that the bob flies off the rod. The mistake comes form leaving out the reaction force of the constraining rod, which in reality balances both the radial component of gravity and the centrifugal force (or, from the inertial frame PoV, provides the centripetal force). In other words, the EL EoMs are applied to the problem with two coordinates (r, theta) as though there were no constraints except at the endpoints. The simplest remedy is, of course, to eliminate r from the start and consider the problem with just a single coordinate, theta. (Or, if one insists on keeping both coordinates (r, theta) then one has to apply the version of EL EoMs with additional holonomic constraints (which here would be r(t)=l for each t).)
I believe that is because dS = 0 only when the Taylor expansion is truncated at first-order terms. For example, consider dS = S[r(t) + ŋ(t)] - S[r(t)], where ŋ(t) represents an infinitesimal deviation from the stationary action path. Therefore, S[r(t) + ŋ(t)] can be expanded as S[r(t)] + ∇S[r(t)]·ŋ(t) + high-order terms. Here, ∇S[r(t)] = 0 because it lies on the path of least action. Thus, when ignoring high-order terms, dS = 0. If we add (∂L/∂t)*dt to dL, we will have a term with the integral [ ∫(∂L/∂t)dt ]*dt, resulting in a high-order term. I'm not sure if this is correct, but it occurred to me. I would appreciate it if someone could confirm if this makes sense.
This lecture would improve a lot IMHO if the speaker focused on the concept of "action" before stating the Lagrangian. Explain what is going on with all the trajectories, what is the problem mathematically. As it is action remains something arbitrary and mystical, while the Lagrangian pops up "deos ex machina" of sorts.
I’ve been loving your channel since I came across it recently. Keep up the good work. One note: The armchair physicists in the comments of your videos are almost as entertaining as the videos. They are theoretical physicists, because their physics degrees are theoretical 😂
since you're setting delta_A to be 0, is that enough to say you are deriving the principle of least action (as in there's a global minimum at the Euler-Lagrange equation). Isn't it more like the principle of stationary action? I've heard that phrase somewhere but I'm not sure.
i guess if you're finding the only situation in which the delta(S) is zero that must be either a global maximum or a global minimum. is it possible to take the second derivative of the functional w.r.t. the path to prove that it is in fact a minimum?
Why aren't we considering ∂L/∂t*dt in the differential of L at 17:26 ?
I believe that is because dS = 0 only when the Taylor expansion is truncated at first-order terms. For example, consider dS = S[r(t) + ŋ(t)] - S[r(t)], where ŋ(t) represents an infinitesimal deviation from the stationary action path. Therefore, S[r(t) + ŋ(t)] can be expanded as S[r(t)] + ∇S[r(t)]·ŋ(t) + high-order terms. Here, ∇S[r(t)] = 0 because it lies on the path of least action. Thus, when ignoring high-order terms, dS = 0.
If we add (∂L/∂t)*dt to dL, we will have a term with the integral [ ∫(∂L/∂t)dt ]*dt, resulting in a high-order term. I'm not sure if this is correct, but it occurred to me. I would appreciate it if someone could confirm if this makes sense.
Great lectures! Sad to see that there are few views as soon as the physics get really interesting.
A couple of cirrections: 1. on Min. 37:13 - the derivative od the potential should be w.r.t. the angle \theta. 2. on min. 41.40 - z should probably be y, for consistency.
37:14 EOM in theta: Left side should be - partial derivative of V w.r.t theta instead of w.r.t. r.
Great job, thank you.
Nice clear derivation. But at 24:20 did you justify that the sum of the integrals equal to 0 implies that each individual integral = 0 and that each integrated term = 0? This is the hard part for me.
Good question: it's a subtle point. The idea is that if we start out with the path with least action then the action will not change (to first order) if we change the path a bit. But there are many ways we can change the path: we can vary each of the several generalized coordinates q_i at any time between the start and end points. The change in the action, dS, is zero for *any* change we make to the path, changing any of the coordinates at any time, dq_i(t). But how can dS always be equal to zero for any dq_i(t)? It can only happen if the function itself being integrated is zero. Then multiplying by the arbitrary function dq_i(t) for each i won't affect the integral -- it's always just zero!
Good lecture, but the application of EL EoMs to the pendulum around 43:30 is somewhat faulty: the constraint r(t)=l should be taken into account from the very start, but here it is used only after writing out the EoMs. This does not matter for the EoM in theta, but if the EoM in r is written out as on the preceding slide, then it will falsely say that mr-dbldot=centrifugal force + radial component of gravity, instead of r-dbldot=0 in reality (the radial gravity term comes from differentiating -V = mgrcos(theta) w.r.t. r, with V=0 at the pivot). This says that the bob flies off the rod. The mistake comes form leaving out the reaction force of the constraining rod, which in reality balances both the radial component of gravity and the centrifugal force (or, from the inertial frame PoV, provides the centripetal force). In other words, the EL EoMs are applied to the problem with two coordinates (r, theta) as though there were no constraints except at the endpoints. The simplest remedy is, of course, to eliminate r from the start and consider the problem with just a single coordinate, theta. (Or, if one insists on keeping both coordinates (r, theta) then one has to apply the version of EL EoMs with additional holonomic constraints (which here would be r(t)=l for each t).)
WOW.
This lecture was such a pleasure to follow. Thanks a ton, Dr Andrew Mitchell, ur the best.
sir can you pls tell me in 19:55 you expanded q and q' from L=L(q,q',t) but why t was not written kindly can you answer this doubt??
I believe that is because dS = 0 only when the Taylor expansion is truncated at first-order terms. For example, consider dS = S[r(t) + ŋ(t)] - S[r(t)], where ŋ(t) represents an infinitesimal deviation from the stationary action path. Therefore, S[r(t) + ŋ(t)] can be expanded as S[r(t)] + ∇S[r(t)]·ŋ(t) + high-order terms. Here, ∇S[r(t)] = 0 because it lies on the path of least action. Thus, when ignoring high-order terms, dS = 0.
If we add (∂L/∂t)*dt to dL, we will have a term with the integral [ ∫(∂L/∂t)dt ]*dt, resulting in a high-order term. I'm not sure if this is correct, but it occurred to me. I would appreciate it if someone could confirm if this makes sense.
Simply a great teacher 👏🏿👏🏿👏🏿 I take my heart off t you Sir
This lecture would improve a lot IMHO if the speaker focused on the concept of "action" before stating the Lagrangian. Explain what is going on with all the trajectories, what is the problem mathematically. As it is action remains something arbitrary and mystical, while the Lagrangian pops up "deos ex machina" of sorts.
I’ve been loving your channel since I came across it recently. Keep up the good work. One note: The armchair physicists in the comments of your videos are almost as entertaining as the videos. They are theoretical physicists, because their physics degrees are theoretical 😂
Dr. Mitchell's presentations are clear and fresh, giving a view that really enforces ones understanding if you have a earlier introduction.
Many thanks! I am very glad the lectures are helping people with their studies!
20:22 time derivative of (q) --> time derivative of (dq)
Thank you for the correction, well spotted!
@@drmitchellsphysicschannel2955 Thank you for uploading your lectures. They are fantastic!
14:10 Principle of Least Action
is there still a cash prize for simplifying LaGrange you simplified it for me poetic fluid 1/137
41:40 z is l*costheta, it's not dz. But anyway, with derivation it does not matter in the end.
Such good presentation
Brilliant.
Brilliant. Thank you.
Good video, congratulations.
Could you do a video explaining Euler Lagrange using Cosines? Also within an algebraic setting.
since you're setting delta_A to be 0, is that enough to say you are deriving the principle of least action (as in there's a global minimum at the Euler-Lagrange equation). Isn't it more like the principle of stationary action? I've heard that phrase somewhere but I'm not sure.
i guess if you're finding the only situation in which the delta(S) is zero that must be either a global maximum or a global minimum. is it possible to take the second derivative of the functional w.r.t. the path to prove that it is in fact a minimum?