Physics students learn this better than math students!! 

Well noticed, I wanted to mention it too. Now, alpha doesn’t have to be “small”; it can be simply any function on the function space, because the perturbation is parametrized by the number epsilon, of which the limit to zero will be taken. So any non-asymptotic function alpha will suffice as any value in alpha will when multiplied with epsilon, in the limit, go to zero. But an asymptote in alpha between a and b would exclude this pseudo alpha because it’s domain is not compatible with the definition of the function space. Thus, ANY alpha in the function space can be considered.

Ah! thank you so much. That clears up my question with later at 22:31 he said that if y minimizes the condition then the euler lagrange equation must be satisfied. That didn't make sense to me at first with what was said just before that because I was thinking that euler lagrange equation being true implies the integral is zero not necessarily the other way around. but with what you just said, it's clear now that the only way that the integral is zero is if the euler lagrange equation is true. Thank you for taking the time to comment and point that out.

yeah this video has some logical problem

it's a running gag, that every michael penn video has to have crucial mistakes but end with a good result 😂

Correct. This is a crucial point in the whole derivation of the EL equation.

That is actually a paradox, because which of these comes closer to the concept of clicking: Physicist: "It does not matter what your intuition tells you, the only thing that matters is the outcome of reproducible experiments." Mathematician: "Give me a set of consistent axioms, and using my own logic I will build a set of theorems from it."

It's not just because of our familiarity with the laws if physics but also because the laws of physics are the fundamental laws of nature.

@@koenth2359 I can say that mathematicians sometimes get lost in their proofs. For example, if we have some equations (integrals, derivatives etc.) mathematician would sometimes unneccessarily take extra steps, for example they would prove that the solution exists, then they would prove that the solution is unique etc. On the other hand a physisit would see that some equations represent the output of a physical system in nature, so he knows that the output must exist (because the world exists) and that there is only one output.

@@srki22 Except that those aren't unnecessary steps, you're just working in a different system.

I kinda view it a bit like this: The physicist usually works with observations that are (usually) continuous over time and if not the discontinuities can be treated separately as indeed they should as events govern the observed information The mathematician usually works in an abstract space and is equally content with what happens in regions of continuity and additionally of discontinuities also exploring theoretical dangers or events that are inherent within such a constructed structure. Identify anomalies, groups of anomalies, classes of solutions and establishing a framework in which physicists can calculate error bounds confidently based on explorations of theoretical nature of math being used. In short?: physics happens in real world of natural observables whereas math happens in abstract world of thought and conceptual variables? Examples: Euclidean geometries, projective geometries, injective geometries, classification of these and patterns and/or relationships that arise from such attempted classifications.

Same!!!

Yes! It is really a good subject!

Thank you, when he got to this part, i got really confused as to why he said the inside of the integral was 0 and not a constant!

Ah! thank you so much. That clears up my question with later at 22:31 he said that if y minimizes the condition then the euler lagrange equation must be satisfied. That didn't make sense to me at first with what was said just before that because I was thinking that euler lagrange equation being true implies the integral is zero not necessarily the other way around. but with what you just said, it's clear now that the only way that the integral is zero is if the euler lagrange equation is true. Thank you for taking the time to comment and point that out.

Completely agree. If the Euler-Lagrange equation isn't satisfied then I think it is sufficient to choose α(t) equal to the other function factor inside the integral so that we have the integral of a function squared and not identically 0. This integral must be strictly positive, thus giving a contradiction.

@@MarcoMate87 The only subtlety is that α(t) must satisfy α(a) = α(b) = 0 But the main idea is like you constructed it. Another approach is to take a non negative bump function as our α(t), with a compact support in which the rest of the integrand is strictly possitive / strictly negative. This will indeed ensure us a non zero integral

And on a side note, the reason why Physicists learn it (quite early on) is because apparemtly any problem in physics can be expressed in terms of such an extremum function problen. Instead of working with clumsy vector analysis, this approach is much more elegant and allows to solve much more difficult problems. It also has a great theoretical value

Perhaps in another video. I don't think Bernoulli posed it just to Newton?

The expression involving alpha(t) was obtained after using the chain rule. He took the conclusion for the d/dt terms and went back to d/dx terms, but it wasnt explicitly stated

Yes, I that alpha part is critical in helping students understand that this approach works for all possible alphas. In other words, each alpha is a differently shaped deviation from the critical curve and by setting the other part of the integrand to zero you show that you've found the critical curve for all deviations. I get frustrated with a few Classical Mechanics texts that only show it for very simple deviations rather than indicating that the part he left off at 20:30 is the key to having it work for all deviations.

Yepp, he forgot it and skipped an important step. Rest is good though

Yes it requires the fundamental lemma of calculus of variations if he had correctly included alpha, which is a pretty big step to skip tbh

Okay, i got the alpha part now. For anyone having trouble understanding that too: If we factor out alpha in the integral and keep it when we set the condition, that the integrand should equal 0, we get a formula of the form 0 = (df/dx + d/t * df/dx')*alpha. But now we want that to be 0, regardless of which pertubation alpha we choose. So for any alpha this should be 0, meaning, the left-hand-side should be 0. From this we get the Euler-Lagrange-Equation. I would still be happy for an explanation of the first question though.

I think it was because if the integral of the (product of alpha and stuff) is always zero, since alpha is arbitrary, it means that the stuff is zero.

He missed a crucial step, the one described by @sqohapoe (if the integral is zero for every alpha(t) it's because alpha is multiplied by zero). As he said it, the statement is false.

It's not a straight line because it's the shortest distance between two points, it's a straight line because that's what the curve looks like between two close points in the limit as a consequence of the concept of differentiation. The curve length formula doesn't rely on the assumption that the shortest distance between two points is a line. It only assumes differentiability

Not true. ds^2=dx^2+dy^2 results from Pythagoras' theorem applied to a plane. ALL smooth curves approach a straight line for sufficiently small dx. This is the basis for Taylor's theorem.

@@byronwatkins2565 and Pythagoras can be applied because the curve must be differentiable. Differentiability necessitates a curve being a straight line in the limit. No assumptions about minimizing distance here

Exactly. And the easiest way to make such an alpha is alpha(t) = f(t, x,(t) x'(t)) itself, then the integrand becomes f^2 which is always positive and never integrates to zero (except when it is always zero, the boring case).

From what I understand, we are not trying to find all functions X that minimise distance (if more than one of them even exist). You just need to prove that at least one exists. And the integrand being 0 is a solution you can have which proves that. And by definition, if more functions exist, they will all evaluate to that same minimum distance. It can't be greater since they are all assumed to be *the* minimum, and they can't be smaller because nothing is below the minimum of tat first "trivial" function. The scope of the video is not interested in finding all functions themselves but rather *a* function that finds the minimum distance, so I guees that's why we didn't care to find them all.

For the chain rule, we need to list all functions involved and only the variables these functions depend on. While x'(t) is obviously related to x(t), we effectively only need the information that it depends on t in order to properly apply the chain rule. Put otherwise: it only takes t as its input, and not x or another variable. So that is why we (are allowed to and therefore) list it separately. Hope this helps!

@@boristerbeek319 can we prove something like dx'/dx (I am not sure how to define this but I saw that we can differentiate a vector with another vector) is equal to 0 to prove this assumption?

f depends on three inputs, and the partial derivatives just indicate with respect to which input we differentiate

My understanding is that the function inside the integral is really just a function of three variables: f(t, u, v). That's what makes them independent as far as the function is concerned. We can plug in whatever we want for those parameters, but to get the right answer we choose x, y(x) and y'(x). What makes it confusing is when the partials are written like df/dy. This is an abuse of notation and it should really be something like df/du (rate of change with respect to the second parameter, nothing to do with y(x)). Then it becomes much more obvious that something like d(uv)/du = v and d(uv)/dv = u, whereas d(yy')/dy' is not so clear.

Yes. In the case of a spherical surface you can check that the minimal length lines are arcs of circumference.a

You don't really need to pick out the pieces. What you need to have started with is a functional which really includes y and y', and thus is defined on any differentiable function y which is defined in the suitable domain. If you get a closed form of an integral which looks correct, with y and y', and the form defines it for all suitable y-s, this is going to work. If you're starting from something that is not an integral, or something more complicated, then you may have multiple forms like that, but that's OK. As long as it is possible to write down the partial of f wrt y and partial of f wrt y', this will work. If you have a product yy', treat this as completely independent variables and just write down the partials like the other variable is a constant amd this works.

i feel a similar unease i deal with it by thinking of the kernel as taking three independent arguments h(x, f(x), g(x)), doing all the fancy calculus first, and _then_ saying to myself "only _now_ shall i impose the constraint g(x) = f'(x) and see what are the consequences of that" because all of the calculus up to the Euler-Lagrange equations works whether g(x) = f'(x) or not

One way how to look into it is to think of the integral in the Riemann sense. In every partial Riemann sum, integrand is always evaluated in a discrete set of points (in this case for discrete set of values of t). When the value of t is already given, function value y(t) and its derivative y'(t) can be chosen independently, as they influence only an infinitesimal surrounding of a given t and we can always make sure that size of that surrounding is always smaller than a given partition. In other words, for a given set of unique discrete points one can always find a smooth function which have both values and their derivatives prescribed over the given set.

the derivation started with the assumption that f is a FUNCTION, not a functional... nothing to wory about ;)

@@Cazolim The function inside the integral is really just a function of three variables: f(t, u, v). That's what makes them independent as far as the function is concerned. We can plug in whatever we want for those parameters, but to get the right answer we choose x, y(x) and y'(x). What makes it confusing is when the partials are written like df/dy. This is an abuse of notation and it should really be something like df/du (rate of change with respect to the second parameter, nothing to do with y(x)).

That's exactly how the Physics Nobel laureate Richard Feynman saw his mathematical physics. And if you've read his biography you'll know he also knew how to have fun in other ways (art, lock picking, bongo drums among other things) so he was far from being boring outside the lab.

@@trueriver1950 oh I didn’t know his. But my professors and some PhD students play piano, drums, guitar and so on pretty well, and they played like a real band sometimes when we had a party together. I’m a junior right now. I’m learning piano, and oil painting is my major hobby. The universe is so fantastic. So many things to learn, no time to be bored.

It does - but generally, you do not have \epsilon as a parameter as the curves x need not map to a linear space. Either you endow the target space (preferrably a smooth manifold) with a metric (finsler), or you consider sections of vector bundles, equipped with some bundle metric.

Physicist: Did you know that multiplication is commutative? That means that you can write the items in the integral in any order you like.

Does it have to do with how he later evaluated it at epsilon =0 maybe? at time 17:18. or.. shoot, I think I need to go back and review my multivariable calc stuff on the multivariable chain rule. or perhaps things will become more clear if we go back and instead of writing it x_epsilon = x(t) + epsilon*alpha(t) if we call the function x_epsilon something else like u which is itself a function dependent upon x, epsilon, and alpha (with x and alpha each depending on t).

@@nucreation4484 evaluating at 0 later might clarify it but at the time didn’t look right to me, I thought the bottom term and the top ones of the next “fraction” were supposed to be the same, so both x_epsilon but not sure. Been a while since I took multivariable calc. Doing more abstract stuff in grad school now and this is more applied lol

I think it’s because f is a function that depends on t, x, and x’ as it originally appeared. It’s just that in that part we have substituted x_ε for x, like we’re evaluating at x_ε.

@@divisix024 ahh. Thank you

People identifying with the tag 'engineer', 'physicist' or 'mathematician' are very childish.

​@@michaelpieters1844not so, in my opinion. It's an expression of where your motivation comes from in doing the maths. A mathematician from this perspective does it simply because it's fun in and of itself. A physicist does it because they really want to understand how the world works. The engineer does it in order to create new stuff in that world. The interplay among those three motivations is important, and all of us have a bit of all three of them; but most of us veer more to one of them than the other two. For me, as a physicist, it's understanding that Einstein's General Relativity says something about the real universe was my main motivation for learning it, and for later teaching it in my career. What IS childish is to think that any of us is better than those motivated by the other three corners of this triangle of motivations.

We need not put any restrictions on \alpha‘ (except for smoothness, bit that follows from smoothness of alpha). Note that, e.g., for \alpha(t)=(a-t)(b-t), we have \alpha(a)=\alpha(b)=0, whereas both \alpha‘(a) and \alpha‘(b) are non-zero.

@johannesmoerland5438 That was helpful, thank you.

Thanks for that comment: it shows insight in a surprising way. In physics there are cases where the extreme value you want is a maximum, so it doesn't always matter which it is As for the local/global thing, actually often it doesn't matter if it's only a local extreme value. So we don't want to close the door on your future learning by limiting ourselves to just global mimina. Example 1: Look into the corner of a rectangular fish tank and you'll sometimes see the same fish twice, once through each glass panel. The rays of light from the fish to your eye travel by locally minimised time delays. You get two images because the universe does not censor the longer path through the other pane of glass, it only censors the partners that are locally suboptimal. Example 2: you are with a friend next to a mirror. You can see your friend's face in the mirror and in real life. Again, the universe allows the rays of light to get to you via the mirror, which must take longer than the direct path, even though the optimisation is one of minimum time delay Example 3: in short wave radio, you can sometimes hear a transmitter with an echo: there are two paths round the planet for the radio signal: the most direct possible path and the so called "long path" all the way around the planet Example 4: when we see a far off galaxy twice due to gravitational lensing by a black hole that's almost on the direct line of sight, then just like the fish in the tank, it's because there are two different paths of local minimum times between the source and ourselves. Examples 1 and 2 show the exceptions to the straight line rule when reflections or refractions are involved. Both effects can be shown to be consistent with the calculus of variations. Unlike the way Maths works, physicists discovered the rules for reflections and refractions long before we knew them to be results from the deeper truth found by calculus of variations. Example 3 is interesting because the fact that there are only two paths show that great circles take the place of straight lines on the surface of a sphere (in this case a spherical shell between planet and ionosphere) and again can be shown by calculus of variations Example 4 is special here: we could figure out what's going on in the first three because cases simply be assuming that might travels in straight lines (plus Snell's law for the aquarium or the great circles rule for radio). These empirical rules turn out to be handy short cuts to doing the full 😢analysis each time. And being easier to write down they got discovered first, before physicists understood why straight lines were important to light, and why light apparently breaks that rule when crossing from one medium to another. But in the general relativity (GR) case we have to use similar maths to this, but with the GR tensors, in order to figure out the mass of that black hole. The kind of maths shown in this video is not used by cosmologists to derive straight line rules, but to actually find out something about how that black hole is distorting the spacetime continuum, and from that they can tell us how many times more massive than the sun that black hole is. In those multiple images we can often see that the images are out of sync ( like if they contain the flashing lights that cosmologists call pulsars). So clearly only one of those paths is globally optimized. So physicists are great fans of locally extreme values. 🎉🎉🎉🎈

@@trueriver1950 Wow, that's a long answer. Thanks "In physics there are cases where the extreme value you want is a maximum, so it doesn't always matter which it is" I don't get it: if you want a maximum and the method gives you a minimum, you will be disappointed. So it does matter. As it stands, from the arguments of the video, you cannot if it is an (incomplete) answer to the question "Find the shortest path" or to the question "Find the longest path". The second one is stupid, since there is no longest path, but we know it only because of additional mathematical or physical arguments. "As for the local/global thing, actually often it doesn't matter if it's only a local extreme value." Sure, in real-life problems, it is not easy to find the absolute extreme values (if any) and we are often satisfied with local extreme values obtained by numerical methods. But the goal of the video is to find a global minimum, so their argument should provide that. I don't understand what your examples are bringing to the table. The purpose of the video is to find the shortest path between two points (see the title and the introduction): it should bring all elements necessary and sufficient for that goal. Or at least mention when some part is omitted. Finding situations where the global maximum and the global minimum are of interest (and it does not remove the burden of proving that they are actually global maximum and minimum) does not change this.

@@TaladrisKpop and thank you too for your long response. Yes, I take your point that for the specific purpose of the task to find the shortest distance between two points, a proper solution would need you to prove that the result is both a minimum and a global minimum. I was answering from the standpoint of being a retired Physics lecturer. I would be likely to present the students with this example as an introduction to calculus of variations (though not as gracefully as Michael does). Thinking ahead to what my students would be doing with the subject in the next few years (those who stick with physics) I would not want to mislead them into thinking that all minima are global, nor that we always know whether we want a max or a min. Back when Snell's law was first explained we didn't know if light travelled faster or slower in water. If light were a wave, we could drive Snell's law from the assumption that the refractive index was the ratio of the speed in the medium to the sled in free space. Ditto is light was a particle. Interestingly, the ratios were the opposite ways up depending whether light turned out to be a wave or a particle. (Waves turn towards the slower medium because of the effect on the wave front, but particles turn towards the faster one of you assume the particle accelerates only in the direction normal to the boundary) So the same analysis could be used to prove Snell's law whether light was a classical wave or a classical particle, leaving it for future determination whether the turning point we could was a max or a min. Neat eh? So then, we now know that light is sometimes one and sometimes another, as is anything else. Why did the wave-theory win out in this case? Well light is actually neither a classical wave now a classical particle: it's a quantum wave like anything else. Those quantum waves coincide with the waves predicted by non-quantum electrodynamics, but we didn't know that till within my lifetime... Feynman (you'll maybe spot that I'm a big Richard Feynman fan) showed how this "coincidence" between the quantum waves and the EM waves is not actually a coincidence after all. That's called QED, quantum electro dynamics. So to sum up: whether you do the job as would be required in an assignment on this question depends whether the teacher is aiming to just teach the example as literally presented, or is instead creating a stepping stone to future topics that may be in the same course; or may be for future study that some of the group will never take. From a teaching perspective it's always difficult to know how much to plan ahead in physics: it annoys me that high schools teach as facts things from Newtonian mechanics that are contrary to either relativity or to quantum, or both. But equally, to give your weaker students a fair chance to pass high school physics exams you need to stick within the Newtonian model or you leave some students floundering.

Functional, linear form, distribution, etc

@@hybmnzz2658 yeah I'm stupid. I've looked up on the internet and it has many names x)