The Principle of Least Action: Derivation of Newton's Second Law

This video provides an introduction to the principle of least action and shows how Newton's Second Law emerges as a constraint for particle moving in one dimension. This video is based on Richard Feynman's lecture on the Principle of Least Action, which I highly recommend.
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[@michaeldamolsen]

@1:00 I love how all the example paths involve traveling backwards in time :)

[@Lectrikfro]

Time travel, not even once

[@deeceeshadelaydelux1112]

Exactly my thought ^^

[@gamerkaue88]

I was going to point that out. I guess he thought it was an x-y graph instead of an x-t graph.

[@Luke-me9qe]

My point exactly

[@DJ-mj5pq]

mentioned as the basis for particle physics at the end

[@leoads]

Wow!!! I’ve been studying this for a long time, and I could understand more in this 25 minutes than the whole course with a bad professor. Congratulations you have a good teaching skills. I subscribed.

[@PhysicsExplainedVideos]

Glad it was helpful! Thanks for the feedback

[@pritamroy3766]

​@@PhysicsExplainedVideos hi ​ Physics Explained , I have two questions, (1)why do we always define Lagrangian as L =T-U ? Could we have L = f(T) - f (U)+constant ( 2) as newtons second law naturally imply that we are dealing with inertial frame, and you showed that from the least action principle we can arrive at newtons second law easily , so can we say that least action principle defines inertial frame ??? if the frame is not a inertial, we can use the least action principle ??? Or can we say excluding the higher order term we are making a general frame forcefully inertial, that's why least action principle implies newtons second law ??? is least action principle frame dependent at all ??? And thank you for your videos, a wonderful series, I'm watching all of those. Keep educating us, bro. A wonderful effort

[@joelmayer1018]

@@pritamroy3766 I have an answer for your first question. In the video he showed that principal works for T-U. The Lagrangian is nothing else than our definition of that term L = T-U so it must be equivalent. And when you *add* an constant to your Lagrangian you change it to a new one, you only can *multiply* it with an constant [ L` = L * constant ]

[@pritamroy3766]

@@joelmayer1018 hi thanks for you ans. then if I define lagragian L = exp T- Sin u and do the math will we get same answer ? actually my question is why Lagrangian is defined as L== T -U ? Why not any arbitrary way ?

[@taiwuchiang8]

T is the kinetic energy, -P is the potential energy (is 0 at infinity). I guess if the function you want to apply to them is continuous and monotonically increasing, the minimizing L will probably produce the same result.

[@SuperNovaJinckUFO]

"Every possible path" *draws three impossible paths*

[@ps200306]

Stueckelberg and Feynman showed that antiparticles could be interpreted as particles moving backward in time. And it's interesting that least action derivation in the vid works regardless of the path.

[@SuperNovaJinckUFO]

@@ps200306 Interestingly enough, applying the principle of least action was actually critical in the discovery and development of quantum field theory

[@ps200306]

@@SuperNovaJinckUFO , indeed that's what brought me here -- reading analytical mechanics as a precursor to QFT. I don't think any of my text books explained as cogently as this video that least action is equivalent to Newton's 2nd law. Quite neat!

[@SuperNovaJinckUFO]

@@ps200306 Agreed :)

[@xhawkenx633]

@@ps200306 however that particle in the video would have to travel booth forward and backward in time. Which would mean that particle would switch to being anti and then switch back to being normal and that multiple times, which is as far as I know impossible

[@DoubleA44]

At 22:15, you can't say that what's inside the square brackets is equal to 0 unless you hypothesize that eta is either always positive or always negative (which does not hamper the reasoning because you can still achieve any path if you add only positive or negative etas to your original path). If you forget to make this hypothesis, then the integral can cancel itself out even if what's inside the square brackets is not 0.

[@maelitodeutschmann4891]

In the third term of the Taylor series, the second derivative, 1/2! is missing as a factor.

[@Dario01101]

Maelito Deutschmann I guess that factor gets absorbed in epsilon squared, since its an infinitesimal value

[@nyoron39]

@@Dario01101 nah, he missed it, since he did write it in the \eta expansion, but it doesn't matter too much since he only wanted to show that the expansion was zero at first order.

[@ernestschoenmakers8181]

@@nyoron39 He also forgot to write down dt behind the integral.

[@canberksonmez3146]

@1:00 It's amazing to see how some of those paths require existing at several different positions simultaneously!

[@tonydai782]

Yes, to an observer the paths going backward would appear as anti-particles and the paths moving forward as particles. To an observer this would mean that a particle anti-particle pair suddenly appears and the anti-particle and one of the particles annihilate.

[@theultimatereductionist7592]

THANK you for doing this derivation at this level of generality. Michel van Biezen does fantastic calculations, but he plugs numbers for all the variables early on, so we don't see the general result.

[@AA-gl1dr]

You just helped me make sense of maths in a way that’s been nearly impossible for my entire life. Nobody told me you don’t need numbers to do math.

[@michaelwang1730]

24:24 a "short introduction to a fascinating topic," and it was nearly 30 min of maths. But I have learned a lot. Thank you so much!

[@tryingsmall]

Very clearly explained...thank you

[@michaelzumpano7318]

I’ve watched two of your videos and they we’re both excellent! Don’t change a thing. I’m subscribing and I hope to see many more from you in the future.

[@braedenlarson9122]

Thank you! I’ve been reading through the Feynman lectures archive and stumbled upon this and he makes it really vague what the concept is. This makes a lot more sense :)

[@alquinn8576]

i follow the principle of least action in my daily life; here is a detailed write-up on how to do this:

[@Xaminn]

I was going to write this. RIP

[@lPlanetarizado]

now you have an excuse to being lazy (when I learned about this that was my first thought )

[@unorthodoxpickle7014]

I've been waiting 🙏

[@rostamnasibullin8633]

You have a real talent to explaining these things. Thank you.

[@kaidenschmidt157]

I’ve watched two videos on this channel and they’ve both been of such top tier quality. I sincerely hope you produce more

[@jinks908]

Unbelievable quality and clarity. You, sir, have a gift!

[@theultimatereductionist7592]

5:15 Math Nazi here: the Taylor series expansion requires division by factorials, which you omitted: i.e. (1/2!)*epsilon^2 * d^2 f/d epsilon^2

[@PhysicsExplainedVideos]

You are right, well spotted!

[@joeo3377]

Please direct me to your other comments where you also point out that, as the Taylor series expansion "requires" division by factorials, the expansion should have read f(x+ε)=(1/0!)*f(x) + (1/1!)*ε*df/dx + (1/2!)*ε^2*d^2f/dx^2 + ... Being such a "math nazi", you would have pointed out those glaring errors as well. You wouldn't be so stuck up your own butt that you would point out something which isn't an error while failing to apply the same logic consistently, yourself, would you?

[@bobstevens3203]

Joe O he just forgot dividing by 2! on the second order term, the factorials are generally left off of the first two terms since they’re just 1 anyway

[@joeo3377]

@@bobstevens3203 When I checked the time code, I swore that he had divided by 2 and that he was being busted that he didn't divide by 2!, which is of course the same as 2. The factorial is generally left off this term because 2! is just 2 anyway.

[@dojinho]

@@joeo3377 Most importantly, the Math Nazi wrote : (1/2!)ε² · d²f/dε² instead of (1/2!)ε² · d²f/dx²

[@xPrism0x]

Absolutely loved this video! Great explanation!

[@FlavioLanfranconi]

Wonderfully clear and concise! Brilliant reference. From here on out, I will refer to this video whenever I mention the principle of least action or the equivalence of Lagrangian and Newtonian physics in general. 😃👍🏼🙏🏼

[@Real46]

Very clear and good explanation!

[@rgudduu]

Loved the quality, the content, topic, depth, voice, accent. Subscribed looking for more

[@PhysicsExplainedVideos]

More to come!

[@learnmore7692]

Love your work man. Subbed!

[@sghrd]

Really well explained! I liked it very much

[@bird9]

A HUGE THANKS ! I was going to give up and go watch tv before I found your video !! So Thank you you cheered me up!!

[@qidirbinu5640]

very good explanation...

[@maurosobreira8695]

Excellent Explanation, Thanks!

[@PhysicsExplainedVideos]

Glad you enjoyed it!

[@algorithminc.8850]

Thanks … another great explanation ...

[@cv21a]

Really well done.

[@hp127]

I learned this from Erik Verlinde and I remember my fascination. My second physics explained and its wonderful. Compliments and thanks.

[@PhysicsExplainedVideos]

Glad you enjoyed it!

[@Jim-be8sj]

Really nice. Reminds me of the time I took a classical mechanics course.

[@ShivamPhysics1]

Wonderful! Delightful!

[@XgamersXdimensions]

Amazing video: extremely clear derivation and quite intuitive! Could you do a similar video on deriving the Euler-Lagrange equation (Lagrangian Mechanics)?

[@PhysicsExplainedVideos]

Great suggestion. I will see what I can do. I am currently working on a series of quantum mechanics lectures and once these are done I will get onto Euler-Lagrange as a route into Quantum Field Theory

[@mayaschoenfeld1383]

Great video! Very intuitively explained and clear. Thank you! Any idea where I can find a recording of Feynman's lecture on the Principle of Least Action on which you based this lecture and which you highly recommend? :) Thanks!

[@n0n3st0p]

that pen looks so smooth to work with... nice vid:)

[@CatzNGaz]

Wonderful explanation. May I suggest 'squiggly equals' when you approximate or drop higher order terms for us pedants?

[@dinomonaco2665]

Brilliant explanation! Please make more videos.

[@mplaw77]

Well done, have seen this for decades, long before technology more advanced than a grease pencil on glass.

[@trnfncb11]

Eccellent illustration. One slight oversight, in the digression on Taylor series expansion, the 2nd order term should be halved.

[@swamihuman9395]

Nicely done. Thx.

[@franks42]

Great video! At 18:04 you may have missed to add a "dt" on the left-side of the equal sign... (also when you rewrote that same formula at 21:07) (without the "dt" you cannot tell over what variable you're integrating)

[@sageofsixpack226]

22:00 I see a lot of comments about this. This is called The fundamental lemma of calculus of variations. It's not an obvious fact, indeed. You can Google the proof

[@OfficialYunas]

This is beauty.

[@rezasorna]

Brilliant! thank you!

[@mustafabayulgen4357]

That is a great explanation

[@macfrankist]

Loved it!

[@mohak8368]

Please do more of these video.

[@fminc]

Excellent.

[@MrJatinderseehra]

Great videos, nicely explained. Where do I find your video on Taylor series?

[@alyasker2194]

Mind= blown that was just amazing

[@driggou]

Playslist with more videos like that. Thanks.

[@csvaughen]

Such beautiful work! What is your camera and recording process?

[@dr.mohamedaitnouh4501]

Very nice presentation! are you using a document camera for this presentation? Thank you sir!

[@lm58142]

Great video, thanks for sharing. Just one small remark if I may: at 3:36, the coefficient of the quadratic term should be divided by 2.

[@thedevansaini7634]

Loving your Chanel

[@GastroenterologyPINNs]

Thank​ you!

[@sinekonata]

"The negative gradient of the potential energy function, which as you know is the Force" Me : ....

[@dalegriffiths3628]

For gravitational fields -dV/dx equals g (gravitational field strength) but that is just Force per unit mass!

[@attheratehandle]

Force is rate of change of kinetic energy w.r.t. displacement (alternative definition to rate of change of momentum), the kinetic energy is being supplied by some form of potential energy. As you add kinetic energy you lose potential energy, so your rate of change of potential energy is really negative and equal in magnitude to the force (for a gain in k.e.). Therefore -m*dV/dx=F. (P.E.=mV, V is potential function, i.e. GM/r) There are a lot of 'ifs' here, some things have to be taken as axiom or left undefined, for example you may choose to leave force undefined and directly write the rate change of potential energy (which we understand as force, potential energy is an axiomatic law for any field). Even in situations where fields may not be directly involved, say two objects pushing away from each other in free space, there is some potential energy that has been lost to create the kinetic energy, therefore the law shall still hold. An example of this is a vertically falling object, as it falls it is losing GPE (=-GMm/r). -dV/dx=-d/dr(-GMm/r)=-GMm/r^2=m (d2 r/d2 t)=-ma [:. r is decreasing, both sides are negative] => a=GMm/r All quantities are treated like a scalar, direction is determined by the applied sign.

[@chaijulut4143]

And that's at the very tip of the Physics iceberg, let me tell you.

[@attheratehandle]

@Kelvin Dale isn't incorrect as dV/dx=GM/x^2=F/m.

[@rogersommers872]

"Use the negative gradient of the potential energy function, Luke!"

[@admiral_xanadu]

At 1:48, the integral looks a little dodgy to me. Shouldn't there be a dx/dt inside, since you are integrating over the path of the particle in 1D?

[@andrewk.7065]

This video is magnificent

[@danilo.magister]

I love his handwriting and his british accent

[@abrown6539]

A quibble: At 22:28 you require that the thing in brackets must equal zero to make the expression equal to zero. BUT, don't forget that eta is a function of time and can itself take on zero values over the range being considered; that is, the "alternate" path can cross the original path (as it does in every example diagram you drew). At those points the value of eta goes to zero; this means that the thing in brackets need not go to zero at those points and the overall expression will still give the path of least action. So you can't logically declare that the thing in brackets is simply equal to zero. The exceptions would involve discontinuous behavior in which the bracketed object is equal to zero only everywhere that eta is not. Really you have to make an explicit assumption of continuity, I believe, for the reasoning to be valid. Note that I'm not saying it is the PATH that has to be continuous, it is the KE and PE functions of time that must be. Exceptions might be a little contrived, but it's not wholly impossible to imagine. Perhaps to nail it down, as I think about it, you should emphasize that the idea is that for ALL possible alternate paths the values of t where eta is zero will be found over the whole stretch of the original path, which forces the thing in brackets to be zero everywhere in that range if we are choosing the path of least action from among all possible paths. There. It's the same conclusion (not surprising, since you DID successfully get Newton's law out of it!), but that step, at least for me, needed a bit of elaboration to be sound. So how would all this be affected if, say, the potential energy function were discontinuous? Seems like it would be relevant in QM.

[@jekoddragon6227]

In the process of taking an integral over a continuous time period, it's not a problem if eta reaches zero at some point. In fact, it could be jumping discontinuously to zero at countably many points and it wouldn't change the result.

[@angelmendez-rivera351]

As Jeko said, there is no need for an explicit assumption of continuity, because bounded discontinuities do not affect Riemann integrals or Lebesgue integrals. Also, in QM, the principle of least action does not at all apply in this format, since QM uses operators instead of path variables.

[@ozzyfromspace]

Nice work! I think I finally understand why the Lagrangian density function is defined as it is. It’s defined in such a way as to spit out Newton’s laws on demand. But then the principle goes deeper by affirming that all systems have a Lagrangian. Is this point of view correct?

[@surenmoodley7744]

Really cool stuff.

[@mitchellmiller9026]

At 22:20 after factoring out eta, why do the terms inside the brackets have to be equal to zero rather than eta itself? I thought the point of eta was to represent the deviation from the function, which would ultimately be (its limit to) zero? Not sure if my question is well-worded

[@ryanashi]

I can’t seem to find your video on Taylor Series that you refer to at 3:19

[@pritamroy3766]

hi ​ @Physics Explained @Physics Explained , I have two questions, (1)why do we always define Lagrangian as L =T-U ? Could we have L = f(T) -f (U)+constant ( 2) as newtons second law naturally imply that we are dealing with inertial frame, and you showed that from the least action principle we can arrive at newtons second law easily , so can we say that least action principle defines inertial frame ??? if the frame is not a inertial we can use least action principle ??? or can we say excluding the higher order term we are making a general frame forcefully inertial, that's why least action principle implies newtons second law ??? and thank you for your videos, a wonderful series, im seeing all of those. Keep educating us bro. an wonderful effort

[@ki-ka]

Thank you for the derivation. But why is the integral (KE - PE, dt, [t2, t1]) stationary for the path taken by a physical particle?

[@Robinson8491]

Great. Wonder how this relates to Keplers square law: is this also about action and action being conserved? And is it necessarily least action?

[@DoNotBeASIMP]

Amazing video. I have a question though. Since the concepts of energy depends on Newton's 2nd Law (ie, you can prove that the kinetic energy plus the potential energy in an isolated system are conserved, and such proof requires assuming Newtown's second law is valid) wouldn't this somehow be considered a cyclic argument?

[@DavidVonR]

That's really cool.

[@mehrdadmohajer3847]

Thx. & well done🌷

[@MakingVidsBreh]

You have an excellent way of explaining concepts and building. I would be indebted to you if you could explain a physical concept that has always eluded me: what is a potential drop or potential difference across a resistor? I understand how it can be calculated by Ohm's law, but it does not agree with the intuition built by electrostatic potential difference. Are there more electrons on one side of the resistor vs the other? This is how a point charge going though a potential difference works in electrostatics.

[@phasma6669]

Hope this will help: KZbin search: Electricity 101 - Potential Difference ( Part 5 of 6 ). I really like the Analogy with the skier and GPE to a simple DC circuit and Potential Difference... used this video series in my classes.

[@tissuepaper9962]

Think of electrons in a resistor as the pucks in Plinko, where the pegs stand in for the atoms of resistive material, and gravity stands in for the electric field driving the electrons through the resistor. Every time a puck hits a peg, it loses some of its kinetic energy to the peg, vibrating it and making the namesake "plink" sound. Now, instead of pucks hitting pegs, think of electrons "hitting" atoms. Every time an electron "runs into" (that's in air quotes because I don't feel like fighting off the physicists who will undoubtedly critique the simplification) an atom, it transfers some of its kinetic energy to the atom, vibrating it. On the atomic scale, that vibration isn't experienced as a sound, but as an increase in temperature in the resistor. It's very important to remember that this isn't a static system. It is in "steady-state", but that is not the same as static, i.e. a lot of what you may have learned in electrostatics may not apply without modification. The electrons are moving around, they just happen to be moving around at a constant rate. Another thing you can think about is a drinking straw. If you blow through a straw gently, it feels like the straw isn't even there, there isn't much of a pressure difference between one end and the other. As you blow harder and harder, more and more air molecules are trying to fit through the same straw in the same amount of time, and a larger pressure difference builds between your mouth (the positive potential in this analogy) and the outside air (ground). We can say that the pressure difference across the straw is proportional to the rate of air flow, with a constant of proportionality (i.e. "resistance") which depends on the exact construction of the straw. That's basically just Ohm's law but for air instead of electrons.

[@magnushelliesen]

4:19 What happened to the 2 factorial in the denominator of the second order term...?

[@sisirpynda5875]

Yeah even I hav the same question

[@nyoron39]

He clearly missed it, but that doesn't affect what he wanted to show in that calculation.

[@poge6192]

He did the calculation to show the orders of epsilon, which are independent of any constants you multiply them by

[@magnushelliesen]

Andrew 2 yes I realize that the point was to show that only the second, third, ... order terms matter at the minimum. But it’s sometimes a bit confusing when stuff doesn’t match what you expect... even though it doesn’t matter...

[@jeffreykalb9752]

Your Taylor series (3:56) is missing the factorials...

[@titobaba5362]

I thought so too

[@bangscutter]

He realised his mistake and added factorials in another Taylor expansion. It's moot point anyway, since he ignores the higher order terms.

[@ishanhamid691]

@8:09 So where is your video on the Taylor Series? I could not find it in your channel! Did you ever make one? Or have you removed it? PLEASE respond!

[@aidanbrophy9947]

Brilliant.

[@deeptochatterjee532]

Unless you can derive the principle of least action, I wouldn't call it a "derivation" of Newton's Second Law. It's just showing that an alternate axiom is consistent with nature. It's like the well-ordering principle and the principle of induction. One can be used to derive the other, but you must assume one of them to be true.

[@goldengeek3320]

Shhh, let the physicist have his fun. Don't expect any sort of mathematical (logical really) grace in a physical context. It's nice when it happens but that won't always be the case.

[@deeptochatterjee532]

@@goldengeek3320 I mean physicists should (and do) acknowledge this too. The fact is that the principle of least action is just an insanely helpful formalism that is experimentally verifiable and emergent from quantum mechanics.

[@goldengeek3320]

@Sthaman Sinha Not all physicists, just ones who approach any subject in an unsatisfactory way (something that goes for mathematicians and just about any other academic you can think of as well). I feel like it shouldn't be said as it is kind of by definition. My original reply had more the intention of humor than actual substance in light of that. (yes deepto chatterjee I am aware that physicists should acknowledge it, it was a joke and yes I am also aware of what the principle is, that explanation was unnecessary and I feel, was unprovoked ;D) I think that what I said is true, however. I happen to notice in my experience with physicists (in general) that they are often sloppier or "practical" in their work. Maybe its because 95% of my study is mathematics and I guess I just appreciate that grace over raw results / explanations. It would have been nice if the guy in the video explained the experimental evidence for the principle of stationary action and acknowledged how circular reasoning can be easily applied here between the principle and newton's law. Have a good day :)

[@goldengeek3320]

@Sthaman Sinha I totally agree with you. I never said that physics needs to be rigorous or mathematically beautiful, I said that it is nice when it happens and that we can't always expect that. REREAD CAREFULLY WHAT I WROTE INSTEAD OF GETTING TOUCHY OVER YOUR DEFENSE OF PHYSICS. Physics is beautiful on its own (both nature and the scientific process for discovering and describing it) BUT WHEN it happens to have a mathematical grace to it, for me it holds an even stronger appeal. Having said that, I maintain that it is not too bothersome for the guy in the video to explain the experimental origin of the principle and discuss how the equivalence of the principle and newton's law can lead to circular reasoning. This happens to be a logical/mathematical concept inherent to the physical world and beyond. The original commentator gave an example with well-ordering and induction. Sorry but if a physicist can't do that then I believe they don't fully understand the physics themselves. Physicists should have some mathematical fluency regardless of if their work turns out "mathematically beautiful". It would have added to the educational value of this video period. Sorry it took so long to write, I wrote the whole thing then deleted it by accident and had to retype it. Too bad, so sad.

[@matheussales4861]

It looks more like to me that he used Newton's 2nd law to "derive" the least action principle since that for K=(1/2)*M*V^2 you assume that F=M*a.

[@bennicholl9855]

@12:10 how is (dn/dt)**2 considered second order? It seems as though its just a squared first order derivative.

[@ernestschoenmakers8181]

You wrote down Newton's 2nd law for conservative forces. The Lagrange's equations that follows from the principle of stationary action are based on this, in order to bring in non-conservative forces like friction you need to add the Rayleigh dissipation function.

[@workerbee2003]

how do the particles go back in time? How do the occupy two positions in time?

[@siamsama2983]

I'm from the UK and I'm due to start Undergrad Physics this fall and my maths ability is up to A-level Further Maths. I was able to follow everything in this video wow, I'm actually surprised coz I was expecting harder maths. That final result is surprising, for something at first glance that seems unrelated to Newton's 2nd law.

[@chrisallen9509]

In general classical mechanics math isn't that bad, but the problems can still be quite difficult

[@yasirmuslim1859]

Simply wow

[@ThatDrummerFrank]

Can you make a video about QFT that jumps off from here?

[@eskilolsen3783]

What happens if you also consider the higher terms? Or is QM a better model at that level?

[@basicsofmathematics3799]

. You explained it very well. Thank you so much. It helped. Can you also describe derivation on Lagrang's equation using principle of least action? .

[@randymartin5500]

Thank you! This has helped me understand the principle of S much more clearly with the integration by parts. Is this also referred to as the Lagrangian or Path Integral? I see these used in Prof. Lenny Susskind's lectures.

[@Robinson8491]

Why do you call him Lenny, is he your friend?

[@randymartin5500]

@@Robinson8491 apparently you don't know much about the physics community. All who know this, call Leonard "Lenny" as a token of deep respect and affection because he is such a down to earth easy going guy . Dont be a sanctimoneous asshole , and stick to learning something.

[@benliu7801]

I am lost at 22:30; why that term has to be 0; can the integration of eta over dt = 0? which means at any particular time, eta !=0, but overall all (integration) = 0?

[@sageofsixpack226]

The fundamental lemma of calculus of variations. Google the proof

[@willnotsmith5058]

I want to know how we figured out that action principle holds. How do we know that (T-V) dt should be minimum? Please explain this.

[@danielgordon9444]

How do I project this over a Mandlebrot set?

[@markbakke471]

22:43 am I missing something here? Constraining the integral to be zero does not necessarily mean that the integrand is zero at all points in the interval, because you could have positive and negative areas of the sum that cancel out. If the integrand was always positive or always negative, I could see this argument holding, but clearly the x-acceleration is not always one sign, and the change in potential energy is not always one sign either.

[@DrDeuteron]

it has to hold for all values of eta.Yes it could cancel for a cherry picked eta, but for all possible eta: F=ma.

[@sageofsixpack226]

The fundamental lemma of calculus of variations. Google the proof

[@des1n5ekt]

@@DrDeuteron Wouldn't that mean that there might be a cherry picked eta, which describes a valid path while breaking F=ma?

[@manskiptruck]

At 21:05 why is there not a 3rd dt term in the left hand side of the equation?

[@junkim5954]

When you write the Taylor series at 3:20, where do all the factorials go?

[@ronaldjorgensen6839]

thank you

[@RD2564]

You use the same pen as me Pilot Hi-tecpoint V7 Grip. I use blue and put the cap on the back because I find it balances nicely that way, but you use yours to do the math 10x better than me ... Nice vid, thanks.

[@pishengding3984]

At the very beginning of the video, three "paths" of motion in the t-x plane are drawn. A "path" in the t-x plane is merely the graph of a position function x(t). But, none of the three paths shown is the graph of a function.

[@milasudril]

How did you derive mv^2/2 ? I guess this has to be an axiom in this formulation of classical mechanics.

[@jekoddragon6227]

You can think of the kinetic energy as the integral of the object's momentum. The more "push" it has amassed, the more energy it has

[@hrperformance]

Does anyone know where/what his Taylor series video is? I can't seem to find it...or any other good Taylor series video? Great channel. Thanks for making this fantastic videos.

[@reintsh]

The kinetic energy T = (1/2).m.(dx/dt)^2 results from integrating W=Fdx where F=m.d2x/dt2, so Newton's 2nd law is put in on beforehand. Around 22:25 he says: for the integral to be zero WE REQUIRE the integrand to be zero. Hmmm.

[@tissuepaper9962]

For the integral to be zero over any arbitrary interval, I'm pretty sure the integrand has to be zero. There are plenty of functions that integrate to zero over one particular interval, but I only know of one whose integral is always 0 for any arbitrary interval, f(t) = 0.

[@sageofsixpack226]

1) for lagrangian formalism we simply define T to be that expression. The fact it represents energy can be derived then from our postulate. 2) The fundamental lemma of calculus of variations. Google the proof

[@dr_ned_flanders]

Love it

[@leonig100]

I am unhappy with the paths with loops at 1.10. This suggests that time is going backwards. Am I wrong?

[@ryanyoon6772]

hello, can you derive the principle using d’alembert’s principle?

[@pharaohgarmar5611]

I’m not a physicist, but I think that the final step where the conclusion is that the integral equals zero implies the integrand is zero needs further explanation. It is because it holds for arbitrary eta that implies this is true. This is glossed over somewhat.