Constrained Lagrangian mechanics: understanding Lagrange multipliers
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- Опубликовано: 12 дек 2022
- How to use Lagrangian mechanics to find the equations of motion of a system whose motion is constrained. Starting from Hamilton's principle of stationary action, we derive the Euler-Lagrange equations, then consider how to modify them to account for constraints using the method of Lagrange multipliers. We also consider the physical interpretation of the extra term that appears.
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #dynamics #calculus #calculusofvariations #differentialequations #lagrangian #mechanics #lagrangianmechanics #eulerlagrange #hamiltonsprinciple #action #stationaryaction #leastaction #lagrangemultipliers #maths #math #science #education
The introduction of Lagrange's undetermined multipliers was illuminating !!
Hello‼︎ I’m a freshman in Nagaoka university of technology. Now I study analytical mechanics in a phyics class.
I could understand how to derive "Euler-Lagrange epuation" and using "method of Lagrange multiplier for not only statics but also dynamics by watching this video.
Thank you for making very good video on youtube all the time‼︎
And I want to study physics very hard‼︎
Thanks for watching and I'm glad it was helpful!
The quality of your teaching is of the highest level. Thank you very much for these videos. I have watched 3 of them so far and I found all of them to be excellent. Please continue to spare some time and make more of these videos. I, for one, will definitely be watching all of your videos over time.
Thanks for your kind words, I appreciate your support! I have many more videos planned, but do let me know if there are any specific topics you'd like to see covered in the future.
@@DrBenYelverton Dr I am going to take you up on that offer. I am just getting back to studying physics all over again -trying to go beyond where I left it, and teachers like you are god's gift. Thank you for sharing your knowledge and insights. All the best sir! I will send you requests on interesting problems as and when I come across them. Thank you and all the best! You have a permanent sub over here!
@@shlokdave6360 Thank you and good luck with your Physics studies! That's great, it's always useful to hear what my viewers are interested in!
thanks for making the video. very nice, concise explanation of the concepts. I was reading the constrained equations in QFT and needed a reminder of this topic in a short manner. the video was perfect for me. so thanks again.
Thanks for saying so, I'm glad it helped!
Amazing video - minor comment, you should probably clarify a bit when you say “the lagrangian is by definition T - V”, which made that sound very general… it’s true for simple mechanics problems, but it isn’t general. Lagrangians in other regimes have lots of different forms. For anyone interested in a general look at this stuff I really recommend the Leonard Susskind lectures on classical mechanics, they’re available on RUclips
This was a super illuminating video! But are there any examples of actually doing this and solving the equations? When I tried it with a simple pendulum example (where L=½mx’²+½my’²-mgy and
the constraint is f(x,y)=x²+y²-l²=0), I got a super nasty expression for lambda that seems way worse than just restricting the coordinate system in the first place. Is this normal or am I going about it the wrong way?
Thanks, I'm glad it was helpful! Choosing coordinates that automatically satisfy the constraints is often less complicated and therefore a better choice, but the disadvantage is that you can’t get any information about the constraint forces this way. If you only care about how the system evolves over time then this doesn’t really matter, but sometimes you also need to know about the constraint forces - for example, you might want to work out when an object will leave a particular surface, which corresponds to setting the constraint force (and therefore λ) to zero. I'll be covering an example like this in my next video. Outside of physics problems, there are also practical cases where the constraint forces are important, e.g. in rigid body simulations (and presumably in engineering applications).
I haven’t worked through your particular example but I’m not surprised that the λ expression looks complicated using Cartesian coordinates. I’d suggest working through it using polar coordinates with the origin at the top of the pendulum, with the constraint r - l = 0, which should make things work out more nicely!
hiya Dr.Ben. I came across this topic when trying to learn Multi Body Dynamics. is that the main region of usage for this topic ? Also, is this a bachelors or a masters level topic ?
Thanks for watching. I think I first studied this in the final year of my bachelor's degree, but it probably varies between universities. It's useful whenever you need to calculate the forces required to maintain a constraint so I'd imagine that multi-body dynamics is an important application!
arigato ku saimas
Could I aplied this prescription to solve the shape of a cabe hanging from 2 poles (i.e. the catenary problem)?
If given constrained has generalized velocity terms then how to deal with that
Sir I want to ask how do we derive the equation for small variation in Lagrange at 3:18
We're basically just doing a first-order Taylor expansion of L about its stationary value. Intuitively, ∂L/∂qᵢ gives the rate of change of L with respect to qᵢ, so (∂L/∂qᵢ)δqᵢ is the change in L when qᵢ changes by δqᵢ. So we're just adding up all the small changes in L when each of the qᵢ (and its derivative) changes individually.
@@DrBenYelvertonThanks for the explanation
@3:45 is there a derivation or explanation of why delta of L is (partial L/wrt q) * delta q?
When one of the coordinates qᵢ changes by δqᵢ while the other coordinates remain fixed, L will change by some amount δL, and by definition the partial derivative with respect to qᵢ is the limit of δL/δqᵢ as δqᵢ -> 0. From this it follows that if δqᵢ is small (which it is in the video, as we're considering small variations to the paths through configuration space), then δL/δqᵢ ≈ ∂L/∂qᵢ and therefore δL ≈ (∂L/∂qᵢ)δqᵢ. Then since we're varying all of the coordinates we add all the changes due to each qᵢ together and turn the right hand side into a dot product. That's the intuitive way to understand this, or alternatively you can view it as a first-order Taylor expansion.
Hold on there is something that's really bugging me at around 17 minutes: at first before implementing the degee of freedom lambda we had n+1 equations et n-1 unknowns. So if we add lambda, we just have n unknows for n+1 equations so the system is still over constrained isn't it? Why don't we add another lambda to fill the gap?
Excluding λ, there are N unknowns since N is by definition the number of generalised coordinates, all of which are unknowns.
@@DrBenYelverton but haven't you said that having a constraint reduces the number of unknowns by one? Which means that before constraint: n unknowns and after constraint: n-1 unknowns
Edit: Or maybe I made a confusion between degree of freedom and unknown? It is not because a constraint decreases the number of degrees of freedom that it decreases the number of unknowns. Which means that even an imposed displacement is considered an unknown
@@qqn4531 Applying the constraint f = 0 reduces the number of degrees of freedom by one, but that's different from the number of unknowns. There are still N coordinates to solve for, it's just that they can no longer vary completely independently.
@@DrBenYelverton Ok thank you very much for the explanation. It really means that my misunderstanding stemed from the confusion between the amout of degrees of freedom and the amount of unknowns.
Thanks for the video, really appreciated as I have an exam soon
Thanks for watching and good luck!
Where is the lagrangian partial derivative by the time multiplied by the derivative of time? Am I missing something?
Hamilton's principle says that the system chooses the path that makes the action stationary, so we need to consider how L varies over different paths (i.e. consider variations in q and its time derivative) - but we can't vary time itself, which is why there's no (∂L/∂t)δt term.
why @9:50 is that the delta q's being independent means the coefficients must all be 0. Could the vectors not just be orthogonal and that is how the dot product equals 0?
It's because the δqᵢ are arbitrary - in order for S to be stationary we need δS = 0 for any combination of δqᵢ, not just for certain δqᵢ. The only way to guarantee that that's the case is to set all the coefficients to zero.
I agree that would guarantee the dot product to equal 0 but aren't we missing some solutions? When solving a problem, and you are given a set of delta qi would it not be possible to find a LHS of the dot product that does make the dot product equal 0? Hence the Euler Lagrange equation should be the entirety of the inside of the integral including the dot product as this would better encapsulate the certain delta qi that would make the integral equal 0. @@DrBenYelverton
for example if our δq vector is (δx, δy) (two independent variables) then we could set the LHS of the dot product to equal the vector (δy,-δx) and the dot product would be zero
@@alexmaciver-redwood3081 If you set the LHS of the dot product to (δy, -δx), then δS is indeed zero for that particular choice of δq, i.e. (δx, δy). However, the principle of stationary action requires δS = 0 for every possible δq. If you consider another δq, e.g. (δx', δy'), then the dot product will no longer be zero in general because the LHS of the dot product is a fixed quantity - we can't set it equal to something different for every possible δq. So, the action would not actually be stationary.
@@DrBenYelverton I wasn't making the case that δy, -δx would work in general. I was just demonstrating how in that particular case, setting the LHS of the dot product to 0 would miss some solutions. The general rule to guarantee that the dot product is equal to zero would be to make the vectors orthogonal as this includes the zero vector as well as all non-zero vectors that when dotted with δq equal 0. Surely this will make the inside of the integral equal 0 and hence the make the variation in the action also equal 0 in complete generality. Was a brilliant video. Much better than my university lecturer. Just want to tie up some loose ends.