Introduction to Lagrangian Mechanics

[@fredg8328]

In french "kinetic energy" is "énergie cinétique". There is no "T" in that. But it could come from the word "travail" which means "work" in french.

[@joaovitorjungblut5225]

of course there is a 't,' it is right there: cineTique

[@HakingMC]

I thought it’s because T is before U, so they used T.

[@CliffSedge-nu5fv]

Whenever I see T used for kinetic, I usually also see V for potential. In Newtonian physics, U is used for potential and K for kinetic. V is used for electric potential per unit charge (voltage), so I appreciate the U to avoid that confusion. I still prefer K for kinetic, as T is often used for time period or for tension, depending on context.

[@cmmaslanka]

Exactly. Travail mécanique...

[@Xaver_44]

I always assumed it came from the word for work in European languages, which mostly start with "T", because there is a theorem that states that the total work is equal to variation of kinetic energy

[@richardrigling4906]

Nice reintroduction to the Lagrangian - it's been 50 years since I played with this.

[@rohitjha8626]

Still beautiful! isn't it?

[@richardrigling4906]

@@rohitjha8626 Yep!

[@richardrigling4906]

As you were going through the derivation, i could hear my classical mechanics professor's voice. Ghost from 50 years ago. Fascinating - to quote a certain Vulcan

[@randomdude9135]

So u never used it in 50yrs??

[@randomdude9135]

What's the purpose of knowledge is u rnt gonna use it?

[@RodrigodaMotta]

In my first year in physics 1 course, my teacher tried to give an overview on Lagrangian Mechanics, but he used a lot math tools that i didn't know at the time, for exemple a Taylor Series with two variables. So, I don't need to say that i didn't get anything that he said. But your video can give a very good overview of Lagrangian for a first year!. Good Job! I want a Series!

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐

[@suddhasattasaha4793]

Hey, if you want a complete course on Lagrangian Mechanics ( with problem solving sessions) , then here I am sharing a playlist with you, I created this series of videos to introduce LAGRANGIAN mechanics in a simple and exciting way, yet retaining the rigour of the subject. Hope you enjoy it ! THIS IS THE LINK TO MY PLAYLIST : kzbin.info/aero/PLZumX0a4ZsO0MSJ5Qujht8TOxm-w5XbBf

[@TelepathShield]

@@suddhasattasaha4793thank you

[@lookthesky9632]

After watching you video, I felt Lagrangian mechanics is a very smart way to do it. This feeling is like there is a house, front door is closed. Backdoor is opened but people don't know there has a back door in this house. When try to open the door and into the house, Newtonian mechanics try very hard use force to breaching the door, and Lagrangian mechanics just into the house from back door.

[@user-lb1ib8rz4h]

maths and even problem solving in general is poking on all sides until you see some give. then poke there more, rinse, repeat. and if you see similarities to another house, then it makes things even easier ofc every problems has tools that make it easier or harder. but often, differentiation is much easier than integration, hence the Lagrangian comes in handy.

[@dybydx31]

Absolutely right

[@CliffSedge-nu5fv]

Or can use brute force computational physics and take all the nails and bolts out and lift off the roof.

[@ysailokesh]

Your teaching levels of physics is very good and it makes me to listen to your teaching

[@Blake_47]

11:57 The Potential energy is mglsin theta since the restoring force is mg sine theta

[@davicruzpestana7329]

Really cool! I'm planning on getting a Physics degree and this just keeps me motivated.

[@bockminster7474]

Don’t do it until covid is cured. Trust me I’m dying

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐

[@moldysnek]

@@bockminster7474 glad to know I'm not alone xd

[@nurfitriyanaazizah3408]

just don't

[@DrAtomics]

@@bockminster7474 Facts man, me too... What year are you? (Senior coming into Fall semester for me). Things have been super tough during COVID

[@droher1344]

This is amazing man. I've always had a lot of problems with parametrization, and although that was not the core of the video you made me see it in a different way. Keep it up

[@donnymcjonny6531]

This is such a great breakdown of what's going on. I'm in a 700 level mechanics course and this is pretty much on par with what we've been reading (Goldstein)

[@kingplunger1]

700 level ?

[@ramsey55]

@@kingplunger1grad school

[@GalileanInvariance]

Nice introduction to LM ... An important point which was overlooked is the way in which LM can incorporate generalized forces (which would appear as extra terms in the E-L equation). Such forces must be taken into account when some physical forces acting on the system are not conservative (and therefore not expressible via potential energy). Such forces also are especially convenient/useful for assessing relevant constraint forces.

[@jiaxinl4036]

Thank you so much for posting this! The explanation is super clear and very helpful for me to understand this topic which is entirely new to me!

[@tornjak096]

should the y_double_dot on 10:37 be minus g because it is acting downwards or I am missing something

[@Vaderphobia]

You are right 👍🏽

[@user_2793]

Yep

[@Jacked_R_Us]

ye its -g

[@remavas7076]

He lost a minus when solving the last equation.

[@piotr_biły_1959]

I noticed this flaw too. Fortunately, someone had described this problem earlier, so my worries quickly disappeared.

[@pensivist]

At 13:14, the derivative should be positive because you defined y as -l*cos(theta). Although it doesn't affect the result as you are squaring it later!

[@tbatlas7243]

This video is really helpful, but I think the diagram at 4:45 could use a redraw of sorts ;)

[@DotPhysics]

I finally get this comment. I feel stupid now.

[@physicsiseverything15]

T can be calculated- using resolution of vectors resolute T in vectors components as Tcosx (theta=x) and Tsinx. Tcosx and mg gets neutralized and we are left with Tsinx. Since, bob is moving in circular trajectory it implies we can equate Tsinx= mR(dx/dt)^2 where R is radius Therefore,. T= mR(dx/dt)^2/sinx

[@brandonflorida1092]

Very clear, and this subject is often presented poorly. Thank you!

[@bwm71]

In your final solution for the pendulum problem you have upper case L in the denominator. It should be lower case.

[@khushilalit6551]

I wonder sometimes, how can something be explained this good. Thanksssss

[@DotPhysics]

Oh. That’s nice. Thanks!!

[@alex_ramjiawan]

10:27 I'm super sure that y'' is -g.

[@msergejev]

For people that can't simply get over using the equation without knowing where did it come from (god knows I can't), video from Eugen Khutoryansky gives a rather satisfying explanation of Euler-Lagrange equation derivation.

[@SolidSiren]

I can't let it go ever in any instance. I can't shutup and calculate. I need to know the derivation, and why, and how, and inside and out. How else will I fundamentally understand the things I'm doing?

[@msergejev]

@@SolidSiren I feel you brother. That is why I warmly recommend this one. Also 'Physics with Eliot' channel has amazing resources on this.

[@abhishekchatterjee7184]

This was a really good video, Physics is very interesting subject, although a little complicated.

[@kamalenduhaldar5805]

Every indian jee adv aspirant is kinda junior physicist lol... (me😭)

[@knotoftime9680]

@@kamalenduhaldar5805 🗿

[@scar6073]

Please keep making videos on Lagrangian mechanics 🙏

[@DotPhysics]

you know it.

[@richardrigling4906]

At time stamp 10:30, shouldn't it be mg = -my dot?

[@yiwang2537]

I just realized how important is this when I saw the "LAGRANGIAN" is so big. !

[@arnabc]

You missed a minus at 10:23. As a result you got y''=g, which is obviously wrong, as the acceln due to gravity is DOWNWARDS!

[@TschumiQu]

aha at 10:38 it's y''=-g not y''=g, since we defined the ground to be 0 and the sky to be positive.

[@sickboi11111]

This video deserves to be prefaced with links to the sources for related and assumed knowledge, so people can get the pre-requisite knowledge required to understand and appreciate this video.

[@tutrinhngoc4389]

Thank you so much 😘 From vietnam with love

[@youtubeuser1052]

It's been a long time since I studied multi variable calculus. Can you remind me why you can assume ydot doesn't depend on y? When taking the partial of ydot with respect to y my first thought is that ydot could possibly be equal to a function of y so I wouldn't assume the ydot term would be zero.

[@bingusiswatching6335]

I'm new to this but I think its more to do with the lagrangian itself, to predict an objects motion it requires an initial velocity and position which are not dependent on each other in configuration space?

[@user-lb1ib8rz4h]

they're taking the partial of L wrt y, and L is a function y, ydot, and t. so we ignore the ydot and t if taking the partial wrt y

[@alleksha]

Many thanks! There is an obvious mistake at 10:25 . You've lost te sign. There y double dots should be equal to minus g

[@alivia181]

Thank you so much Sir for this wonderful video. I am starting this topic in my 1st year. It is very helpful .

[@yansayidiyegrace854]

Are You Physics Student at University? I recommend this: kzbin.info/www/bejne/noi3YWlvr85qo68

[@quantummechanic8897]

The best playlist kzbin.info/www/bejne/noi3YWlvr85qo68&ab_channel=BiPhysiMath-Special

[@kristofferh2312]

I can see this is a BIG DEAL!

[@lloydbotway5930]

Excellent straightforward explanation

[@quantummechanic8897]

The best playlist kzbin.info/www/bejne/noi3YWlvr85qo68&ab_channel=BiPhysiMath-Special

[@1973jdmc]

Exceptional introduction- THANK YOU

[@quantummechanic8897]

The best playlist kzbin.info/www/bejne/noi3YWlvr85qo68&ab_channel=BiPhysiMath-Special

[@dr.emmettbrown7183]

12:05 Potential energy is not - mgl cos@ but mgl (1 - cos@)

[@martinullrich655]

Thanks for the super tutorial! (Last formular of the pendulum: instead of "L" should be "l" [length of pendulum])

[@HaliPuppeh]

T was chosen by the person who introduced the concept of kinetic energy. It stands for "translational" cause it's the energy of movement

[@discreet_boson]

Came here from the subreddit Awesome video, liked and subscribed

[@igorbarros3308]

simple and easy to understand, thank you!!!!

[@andrejburcev6023]

At 10:28 acceleration should be -g

[@silverlining6824]

Mewton: force LaGrange: energy

[@Imran52Feb]

Will it not be Theta double dot =- g sin Theta/ Lower case L rather than capital L in the denominator? Also del L/ Del theta dot is zero in the first term because Theta dot is a function of time and not Theta. Am I right?

[@sivavenkateshr]

Extremely wonderful video 👏.

[@nicolassanchez7259]

This is great! Thank you! Does someone learn this as an engineer? Because I study Civil but have never seen this.

[@DotPhysics]

For civil engineering - you might not get to this stuff.

[@diatlemaboe9942]

Mechanical engineering - vibration, yeah

[@nicolassanchez7259]

@@diatlemaboe9942 yess I'm taking a class in vibrations and you see this stuff

[@dwrd7356]

This is a really amazing video!. Actually I was a little bit confused in 13:11, why y' is d/dt(lcos(theta))? it shouldn't by d/dt(-lcos(theta))?

[@NateRiver-ph9co]

I´d like to know as well

[@user13rs258]

It gets fixed while adding their square if you have noticed it, even if it is fixed by mistake. 😂😂 I mean because of squaring and getting the right answer 😁

[@janovehansen9172]

At 17:19. Shouldn't it be -(g/l)*sin(theta) ... lower l as capital L is used as the Lagrangian?

[@arnabc]

Not sure about the right hand plot at 5:39. S, as defined by the integral, is a just a constant for each path, since y is a function of t, and we are integrating wrt t. So any plot S should be against "path". How is that represented by the horizontal axis?

[@miloszforman6270]

Sorry to say that, but it's an awkward BS diagram as well as an awkward BS explanation. Thumbs down.

[@classictutor]

Could you explain why L is defined as T - U? Also when you were doing the partials around @10:00, even though y dot and y are different variables, isn't y dot still dependent on y and vice versa? So if you are doing the partial of one variable, wouldn't the other be affected as well, in other words when you are doing the partial of y the y dot is moving and vice versa so that the results are not clean? Help me if I am over thinking or there is something to this.

[@maalikserebryakov]

Thats like asking why is newtonian Force defined as the product of mass and acceleration. These are just definitions invented to make calculation possible. you can invent a new definition yourself but the difficult part is to come up with a definition that is useful in calculation

[@classictutor]

@@maalikserebryakov Useful for calculation is good enough for me. I thought there was something more deeper.

[@maalikserebryakov]

@@classictutor yes thats all there is to it nothing deeper at all 👍

[@MrSidney9]

@@maalikserebryakov Net force being equal mass times acceleration is not “a definition invented “ it’s a fact about the natural world transcribed in mathematical term. L=T-U is however purely definitional, and useless considered by itself; However, the Lagrangian EQUATION, is a statement about the physical world.

[@speedbiker6132]

that path from 4:47 is something familiar !?

[@mohammadmahirareeb7513]

didnt know cheatcode existed irl💀💀

[@thevegg3275]

At minute 9:51 shouldn’t the derivative of why dot the 2MY dot?

[@navneet1533]

Mr physics explained make more video on Lagrangian principle it's so helpful for me👍🏻

[@anbudamodaran4162]

Interesting problem at 4:45, looks very familiar...

[@melchiortod29]

T stands for travaille i think. Which means work

[@clarysshow]

Afte practising so many questions, I have seen making mistakes due to writing x prime or x dot to show the velocity in x direction and it's not so efficient to always write like those old books, I do not make mistakes in differentiation or calculus, I make mistakes in distinguishing between velocity and coordinate. So I start writing Vx or Vy or Vz or Vi to show velocity XD

[@raunacsen179]

Thank u sir for this nice explanation 👍🏻👏🙏

[@AnuragGuptainspired]

How will we obtain the Lagragian for an object sliding under the influence of force 'F' on a frictional surface?

[@pritamroy3766]

Hi, Dot physics I have two question and im seeking my answer for long time never satisfied with any.1) why lagrangian are defined like L=T-U??? Why not any otjer form like L= sin T- log U or amy othe rkind of weired combination??? Dpes lagrange derrive this perticuler form from Anywhere or he got it in his dream??? 2) if a system is in motion with non conservative force, only like friction can we still define Lagrangian??? How???

[@grahamashton7210]

The Lagrangian takes different forms depending upon the theory in question - and yes he dreamt it in just the way Newton dreamt F=ma. It's a guess that can be compared with reality.

[@NicolasSchmidMusic]

I don't think the T has a particular meaning in french either...

[@eldarr0uge482]

Kinetic energy translates to "énergie cinétique" in french, so the T probably doesn't come from there ^^

[@usman6323]

Amazing intro to lagrangian.

[@DotPhysics]

thanks!

[@harshpandey7970]

So simply explained thanks ❤️

[@DotPhysics]

Glad it was helpful!

[@Lucky10279]

This video was really helpful!

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐

[@lanefaulhaber1824]

How does the problem change when you have a pendulum in 3D space? when I plug my z value in, is it also lcos(theta)(theta-dot)? Or would the angle be different when referring to z? And then how does that affect the rest of the process??

[@mihadbinislamtanim6267]

at 11:58 shouldn't the potential energy be U = mg(l - lcos(theta))

[@st_s3lios860]

It's actually the same. You just put the origine of potentiel to the bottom when he put it to the top. But whatever, the constant will diseapper when you will derive (sorry for my bad english)

[@atheroot]

@@st_s3lios860 no, it's not the same. Looks like the author don't understand the topic.

[@jujhaarsingh7889]

Good video and great explanation.

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐

[@dybydx31]

Great content

[@currenredman1864]

Great video, helped a lot. Also, would you happen to know why we use T-U as the lagrangian? I can't seem to find a good answer anywhere

[@gibbogle]

Same here. I'm guessing that was a lot of development by different mathematicians, maybe spanning years, to arrive at L = T - U. The teachers don't go into the history.

[@brighttech3470]

i have some doubt why we take the derivitive of the lagrangian function for the solutions ? and how does it work ?

[@JoeHynes284]

i love this channel

[@TimurChepiga]

@ 10:36 its negative g. great video!

[@ronaldjorgensen6839]

i prefer bance newtons cradle at each instance practice problem sets never tried double pendulum other than upside down newtons cradle at inverse forces

[@shreeniwaz]

@10:36 -mg-my'' = 0 is turned to mg = my'' This obviously doesn't happen..

[@ihbarddx]

Great! Where does it come from?

[@lykamontana9108]

then what should be the equation on the motion of the ball?

[@soundarrajankannan7692]

excellent teaching sir. But just one doubt, in potential energy, for Y we have to substitute (l - l cos(theta)) right??

[@dorol6375]

Ah yes I should clearly be learning this going to 10th grade

[@fargle2008]

At 9:35 you say that the kinetic energy does not depend on y. Surely from a physics point of view the kinetic energy is a maximum at the bottom, falls to zero at the top of the curve, then increases again on the way down - so it is continuously changing with y. I am confused.

[@mihadbinislamtanim6267]

The kinetic energy depends on the 'derivative of y' not y. At the bottom the y is 0 and it's derivative is at maximum, so kinetic energy is at maximum. At the maximum height y= max but it's derivative is zero, so there's no kinetic energy there. So kinetic energy is changing with derivative of y and not y

[@learnerman5712]

@@mihadbinislamtanim6267 But could not derivative of y itself be expressed in terms of y so that one could say that kinetic energy is indirectly a function of y?

[@fargle2008]

@@mihadbinislamtanim6267 Thank-you for taking the time to answer my question. You say "At the bottom the y is 0 . . . kinetic energy is at maximum". Then you say " height y= max . . . so there's no kinetic energy there [ i.e. KE = 0 ]". So you are repeating my assertion that KE does vary with y. The total energy is the sum KE plus PE. As PE increases linearly with y. KE must decrease linearly with y to keep the total energy constant.

[@mihadbinislamtanim6267]

@@fargle2008 Thank you for taking your time to point out my mistakes. Your right that kinetic energy does vary with y. But what I wanted to imply was that the kinetic energy term was depended on the value of y• and not y. Meter's and meter's per second aren't same. If we take the Taylor expansion and work for y• we will find a function which is dependent on y. I just wanted to be generic with the dimensions of the function. Sorry if my first comment was misleading 😅.

[@mohammaditani4558]

It's a partial derivative with respect to y_dot that's why it doesn't depend on y. I don't think he meant the kinetic energy doesn't depend on y.

[@thomasalgado15]

Should that derivative with respect y be my' * mdy'/dy - mg?

[@mrtienphysics666]

The Langrangian is the excess of kinetic energy over the potential energy of a system. ref: Lanczos

[@definty]

I like the introduction series

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 👍💐

[@100AngelBoy]

This might be a dumb question because I don't get the part from 12:20. I know that kinetic energy is (m*x*^2)/2 from newtonian mechanics but did you add an *y* because there is also the y coordinate? Or, if there would be an Oxyz space would we have to add also an *z* over there?

[@DotPhysics]

Since we are using a coordinate of theta for the motion, we need to get the kinetic energy in terms of theta. In general, that can be tough (yes, you could see it in polar coordinates). So, if we get an expression in terms of x and y (and z if you need it), then I know the kinetic energy is (1/2)m(x-dot^2+y-dot^2)

[@leoliu7492]

this is the velocity in 2D.

[@100AngelBoy]

@@DotPhysics I see, thank you! Dude, you are actually good 😁

[@roccotinitali1544]

I was following along until i got to the ball and shaft problem and i had to pause and reevaulate if i should laugh or cry

[@amudonchingangbam9256]

That 'identical' caught me off guard.. 🤣🤣🤣

[@yansayidiyegrace854]

Are You Physics Student at University? I recommend this: kzbin.info/www/bejne/noi3YWlvr85qo68

[@quantummechanic8897]

The best playlist kzbin.info/www/bejne/noi3YWlvr85qo68&ab_channel=BiPhysiMath-Special

[@magicstix0r]

5:00 Who else couldn't stop giggling at the diagram of his ball path?

[@rominaabadi]

@10:28 shouldn't be y double dot = -g?

[@quantummechanic8897]

The best playlist kzbin.info/www/bejne/noi3YWlvr85qo68&ab_channel=BiPhysiMath-Special

[@bernicenwabueze924]

Sir more examples 😢

[@DotPhysics]

Here's an older playlist - hope that helps kzbin.info/aero/PLWFlMBumSLSbX-yG0eUl5hBa_I6Oz5k7B

[@SolidSiren]

Wym there is no tension equation? T = mv^2/r + mgcosx

[@user-pb4jg2dh4w]

Aahh yeah this man needs more than just one like

[@matguenette1596]

Why is the derivative with respect to time of cosine theta not zero? Cosine theta would be a constant with respect to time, no?

[@DotPhysics]

Theta is a variable - it can change with time.

[@TheAmigodaniel]

What courses in mathematics do I have to have in my body before going inside the famous Lagrangian? Somebody can explain?

[@DotPhysics]

Really, you just need derivatives and partial derivatives. So, you would probably be fine with something like Calc II

[@hayatabbas4732]

Great lecture!! Thanks

[@DotPhysics]

Glad it was helpful!

[@Khutso1279]

Wow, u made it easier

[@markostojanovic5930]

Great video and great explanation!!! I'm becoming a fan.

[@bernicenwabueze924]

Perfect

[@djttv]

Very clear explanation of how to do problems using lagrangian dynamics, but I still wonder where it comes from. Why did someone define L=T-U in the first place, etc. Great video!

[@DotPhysics]

I think, like many concepts in physics, we use the Lagrangian just because it works. Lagrange showed that if you defined L, then the action integral would be the same as the actual trajectory. It's crazy.

[@miloszforman6270]

_"Great video!"_ _"but I still wonder where it comes from"_ Isn't that some kind of contradiction? If you can't understand it from the video, probably it's a bad video. We would not have to bother, but this guy is stealing our time.

[@BederikStorm]

It must be -g, not +g for the second derivative of y.

[@michaelli8523]

Anyone can tell me the playlist? I want to watch the other examples. Thanks!

[@DotPhysics]

Here you go - kzbin.info/aero/PLWFlMBumSLSbX-yG0eUl5hBa_I6Oz5k7B

[@michaelli8523]

@@DotPhysics Awesome, thank you for the excellent series!

[@281992pdr]

There ARE two forces not there is two forces. Conceptual clarity depends on prcision of expression.

[@michaelempeigne3519]

what is the difference between Lagrangian mechanics and Hamiltonian ?

[@DotPhysics]

This might help - but it doesn't fully answer your question. Here is the problem of a block sliding down an inclined plane. I solve it three ways: Newtonian, Lagrangian and then Hamiltonian. kzbin.info/www/bejne/g3vVn4iErKqYask

[@srallulrich]

Can proof that Newton laws are equal to Euler Lagrange equation