Ambiguity With Partial ∂ Notation, and How to Resolve It

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  • Опубликовано: 30 ноя 2024

Комментарии • 255

  • @EpsilonDeltaMain
    @EpsilonDeltaMain  7 месяцев назад +180

    Of course the issue stems from mathematicians and physicists each consolidating on different conventions!!
    Mathematicians
    d/dt Ordinary Derivative
    ∂/∂t (Total) Partial Derivative
    Physicists
    d/dx Total Derivative
    ∂/∂x (Explicit) Partial Derivative
    Mathematicians stuck around using d/dt for one variable case and ∂/∂t for just a multivariable version of it.
    So ∂/∂t means the "total" partial derivative with respect to an independent variable.
    Since 1800s, mathematics was all about about functions, not equations,
    so they got around the issue by defining explicit functions.
    Physicists on the other hand use d/dt as total derivative to mean every chain leading to that terminal independent variable,
    and use ∂/∂t to mean the "explicit" partial derivative leading directly to that variable.
    I left a link in the video description detailing the difference in convention.

    • @AndDiracisHisProphet
      @AndDiracisHisProphet 7 месяцев назад +2

      that explains a lot. thx

    • @markusklyver6277
      @markusklyver6277 7 месяцев назад +7

      Just write down the limit definition before using a notation - no confusion arises. For example, what you call "total derivative" is just
      lim_{h to 0} 1/h[u(x(t+h), y(t+h), t+h) - u(x(t), y(t), t)],
      and what you call the "direct partial" or "the explicit partial" is just
      lim_{h to 0} 1/h[u(x(t), y(t), t+h) - u(x(t), y(t), t)].

    • @Alan-zf2tt
      @Alan-zf2tt 7 месяцев назад

      ahhh ... (1) sound of hand slapping forehead, ... contextual definitions rather than absolute definitions -> repeat (1)
      Thank you for sharing

    • @TJ-hs1qm
      @TJ-hs1qm 7 месяцев назад

      I'll add a bit of Gerald Jay Sussman
      ruclips.net/video/arMH5GjBwUQ/видео.html
      ruclips.net/video/2MYzvQ1v8Ww/видео.html
      and
      Programming for the Expression of Ideas @ InfoQ

    • @88coolv
      @88coolv 7 месяцев назад +2

      There is no differencies here. Mathematicians use ∂/∂t as partial derivative only, there are no exceptions

  • @EneldoSancocho
    @EneldoSancocho 7 месяцев назад +192

    I knew something was wrong with the notation!! For years I have been working my way around this pesky partial symbols without thinking about the root problem.
    Excellent video!!

  • @Geovani1642
    @Geovani1642 7 месяцев назад +186

    I have a phd in math, and i still hadn't understood some partial derivative notations in thermodynamics... until i watched this video. Thank you!

    • @cpiantes
      @cpiantes 7 месяцев назад +18

      A good notation is immensely helpful. If anything, contrary to the other posts in the comments section, I think that a better notation for differential calculus would need just _one_ symbol: ∂, and a subscript for partials.

    • @cpiantes
      @cpiantes 7 месяцев назад +6

      For example, a function f(x,y,z) has the differential ∂f = ∂_x f ∂x + ∂_y f ∂y + ∂_z f ∂z; if x,y,z are themselves dependent on a variable t, then ∂f =(∂_x f ∂_t x + ∂_y f ∂_t y + ∂_z f ∂_t z) ∂t.

    • @jonathan3372
      @jonathan3372 7 месяцев назад +7

      @@cpiantes although very neat, I think with this method it would also be hard to avoid the ambiguity mentioned in the video. I.e. ∂_t u could mean the explicit partial derivative or the total partial derivative of u=f(x(t),y(t),z,t) w.r.t. variable t.

    • @Zxv975
      @Zxv975 6 месяцев назад +5

      I literally came here to comment the same thing (I had a PhD in theoretical physics though). I've never understood the motivation behind why thermodynamics explicitly stated which variables are being held constant until this video.

  • @paulcho7898
    @paulcho7898 7 месяцев назад +384

    "This is why we cant have nice things" 😂

  • @alexatg1820
    @alexatg1820 7 месяцев назад +202

    As a math students, I would say it is just a bad habit not to think about what the actual function is. It’s actually f:U⊂R^3->R, f(x1,x2,x3), g:R->R^3, g(t)=(x(t),y(t),t), and u(t)=f(g(t)), and the problem is solved automatically. du/dt is well defined, ∂g/∂xi is well defined, not ambiguity. Often people use the notion without care, this is just the consequence of such carelessness, not of the notation.

    • @alexatg1820
      @alexatg1820 7 месяцев назад +34

      As for the case if there is another variable z, f:R^4->R, f(x1,x2,x3,x4), g:R^2->R^4, g(t,x)=(x(t),y(t),t,z)
      u(t,z)=f(g(z,t))
      ∂u/∂t=Σ∂f/∂gi•∂gi/∂t, note that ∂g4/∂t=0.
      ∂f/∂g3 is well defined, no ambiguity involved.
      It is just that people use short hand notation inappropriately (abuse of notation), which twisted the meaning of the original notation, that caused the problem

    • @alexatg1820
      @alexatg1820 7 месяцев назад +37

      I know that using such rigorous notation, especially when applied in physics, would be rather tedious. I am just sharing my thoughts on such notation, and i think physics having its shorthand notation for clarification could be a rather good solution as well.

    • @derickd6150
      @derickd6150 7 месяцев назад +19

      I don't think the problem is that it's ambiguous, I agree that is the result, but rather it's that we don't have a notation now for the partial derivative only with respect to changing t, and not taking into account the intermediate derivatives of x and y with respect to t

    • @assassinosoldato92
      @assassinosoldato92 7 месяцев назад +3

      Vero good comment on a very good video

    • @98danielray
      @98danielray 7 месяцев назад +3

      ​@@derickd6150you can use f. that is the point of op. that said, it may be impractical to carry the composition symbol around

  • @nathanisbored
    @nathanisbored 7 месяцев назад +84

    i knew there was a reason i didnt like this notation but i could never quite put my finger on it

  • @geekjokes8458
    @geekjokes8458 7 месяцев назад +13

    wow, thank you for finally making me understand the parenthesis+subscript in thermodynamics classes... and why one fluid dynamics book was very serious (but very unclear) about calling it material derivative

  • @mr_hxid
    @mr_hxid 7 месяцев назад +49

    Physicist here, this is the way I learnt it.
    Say we have u(x(t), y(t), z, t) where u is a function of x, y, z and t and where x and y both depend on t but z doesn't.
    Then the total derivative of u with respect to t would be written as
    du/dt = ∂u/∂x * dx/dt + ∂u/∂y * dy/dt + ∂u/∂z * dz/dt + ∂u/∂t
    Here ∂u/∂[.] refers to the explicit partial derivative where all variables except [.] are interpreted as constants. In physics we say dz/dt = 0 which then gives the correct formula. This removes any ambiguity that mathematics has.

    • @kimchi_taco
      @kimchi_taco 7 месяцев назад

      If u(z,t), do you suggest to write partial derivative as du/dt? 😮

    • @mr_hxid
      @mr_hxid 7 месяцев назад +10

      @@kimchi_taco In that case du/dt and ∂u/∂t are the same. I would still write it in partial notation to be more explicit.

    • @welcomeblack
      @welcomeblack 7 месяцев назад +3

      +1 This is how I think the notation should work

    • @GeodesicBruh
      @GeodesicBruh 7 месяцев назад +13

      Physicist here, I agree and was shocked that the video didn't mention this as a solution.
      You can always pretend that Everything is a function of a function of a parameter, t for example, and take a total derivative with respect to that; if you then wanna hold some parameter, say x, as constant then you just put dx/dt=0 as you said.

    • @deltalima6703
      @deltalima6703 7 месяцев назад

      Omg. I have lost my mind. That actually made sense!

  • @ChaoticNeutral6
    @ChaoticNeutral6 7 месяцев назад +24

    This was so good it deserved money

  • @Minecraftster148790
    @Minecraftster148790 7 месяцев назад +14

    Something I don't understand about your f(x, y, z, t) example at 2:32 is that if we take the total derivative with respect to t then we add a df/dz dz/dt term (with partial ds). dz/dt is 0 so it evaluates to the same thing as before. This seems to be working as intended in my opinion.

  • @TheMaginor
    @TheMaginor 7 месяцев назад +26

    This used to confuse the hell out of me when I studied physics. From a pure mathematical perspective, we are really talking about two different (but related) functions. The first one is the function (x, y, t)->u(x, y, t), where x, y and t are free variables, while the second one is the composition t -> u(x(t), y(t), t), where x and y are now functions of their own. The latter function may be called g := u \circ (x, y, id) . The confusion comes from the fact that we are treating x and y both as functions and as free variables depending on the context, and that context is what you have to keep in mind when doing it. Now (\del u / \del t) is just d_3(u) (i.e. the differentiation of u along the 3rd variable), while what is called (du/dt) is actually d_1(g). Of course, it would be too tiresome to write this out each time, so it is fine to use non-rigorous shorthands as long as the context is implicitly understood, but I still wish I learned to think about it this way earlier. Edit: I see somebody already pointed this out in another comment.

    • @robvdm
      @robvdm 7 месяцев назад +3

      Same here. I seriously hate this notation. If you want an equally confounding notation you can look at Bayesian notation where p(x) and p(y) are not equal when x=y.

    • @88coolv
      @88coolv 7 месяцев назад

      > and as free variables depending on the context
      There is no such object, if we look from mathematics perspective. So, those x,y are always functions, in all contexts. Derivatives - that's what can be different

    • @quantumsoul3495
      @quantumsoul3495 7 месяцев назад +1

      That is what bothered me with physics, all variables are also functions

  • @thegozer100
    @thegozer100 7 месяцев назад +5

    This has cleared up so many things in just 10 minutes! Ive always felt the notation was slightly off but I did not know why

  • @theultimatereductionist7592
    @theultimatereductionist7592 7 месяцев назад +17

    THANK YOU FOR ADRESSING THESE FRUSTRATING AMBIGUITIES!

  • @tedsheridan8725
    @tedsheridan8725 7 месяцев назад +19

    Cool video. I remember how the total and partial derivative w.r.t. time used to confuse the hell out of me in fluid mechanics.

  • @paulcho7898
    @paulcho7898 7 месяцев назад +12

    A new upload from epsilon delta, was waiting for this! Yay! Love your work ❤

  • @eqwerewrqwerqre
    @eqwerewrqwerqre 7 месяцев назад +17

    Instantly subscribed. This is god tier, and not just the same regurgitated lesson I've seen a million times! I've had partials explained like 3 times now and it's never been so clear how to truly utilize this notation. I'd become aware of functionals when learning Lagrangian mechanics a couple months ago and this effortlessly cleared up a serious tension i hadn't even been able to understand yet.
    God tier video, i thank god and you that you made it

    • @geekjokes8458
      @geekjokes8458 7 месяцев назад

      yeah! it makes it a lot more clear on why we can take the partial with respect to x-dot and not care about the rest

  • @martingibbs8972
    @martingibbs8972 7 месяцев назад +1

    I’ve always had an issue with the apparent ambiguity of partial derivatives. You explained it beautifully.

  • @zacklee5787
    @zacklee5787 7 месяцев назад +20

    I don't see the problem, if z doesn't depend on t, then you can take the total derivative and dz/dt is just 0. Whether or not all paths lead to t is irrelevant.

  • @sdsa007
    @sdsa007 5 месяцев назад

    OMG! I struggled for sooooooo long trying to upgrade my high-school math beyond the 17th century, so I can learn scientific maths to understand climate change, only to learn now that it is sooooo incomplete and we don't have 'nice things'... because 17th people were being to carelessly vague about differential calculus terminology! THANK YOU FOR EXPLAINING, I love the arrow diagrams! I hope we get nice things soon, so we can understand how to model nature better and avoid bad things! I have looked at more recent work like the work from Élie Cartan... but what an odessy to understand!

  • @ratfuk9340
    @ratfuk9340 Месяц назад

    Omg, I was searching for information on this for so long. Thank you.

  • @Strategies2010
    @Strategies2010 6 месяцев назад

    Your visualizations honestly helped cement these kinds of concepts more than whole classes that I’ve taken… great job! I’ve always been left wanting to know more about the math intuition behind things like wavefunctions or heat transfer 😊

  • @CT-pi2gl
    @CT-pi2gl 6 месяцев назад +4

    For u(x(t), y(t), t, z) I would have said du/dt (total) still encompasses t, x(t), y(t), with z being irrelevant. And ∂u/∂t (partial) identifies the direct dependence on t, without x or y

  • @Nerdimo
    @Nerdimo 6 месяцев назад

    I’m not quite into material science, however, I’m a fan of machine learning in which partial derivatives come up often. I never considered the ambiguity of partial derivatives, but this video did a great job laying them out. I’ll try to follow the scheme you shared for whatever work I share, so thanks for awesome content.

  • @OrWeinstein
    @OrWeinstein 7 месяцев назад +1

    That's amazing. Always found it so confusing. As a mathematician studying physics, the most confusing step is Legendre transformation.

  • @yqisq6966
    @yqisq6966 7 месяцев назад +1

    I think the most general and reliable solution is to always consider the computational graph. Algebraic notations can only take you so far before it becomes overencumbered. It's precisely how modern backpropagation algorithm is implemented.

  • @armagetronfasttrack9808
    @armagetronfasttrack9808 7 месяцев назад +5

    Not necessarily a correction, but to add some context to the temperature function: the "original" temperature function u(x,y,t) is dependent on 3 independent variables and there is no meaning to the idea of (non-zero) dx/dt or dy/dt since x and y are independent variables. A different scenario you can consider is taking a _path_ (x(t),y(t)) through the 2d space and consider the temperature along that path through time. This temperature function through time would just be u(t) (or use a different name for the function like @alexatg1820 said, let's call it U(t)).
    We can then say U(t) = u(x(t),y(t),t). And we can also make sense of total vs partial derivatives as d/dt U = (δ/δt u)|_(x(t),y(t),t) + (δ/δx u)|_(x(t),y(t),t) dx(t)/dt + (δ/δy u)|_(x(t),y(t),t) dy(t)/dt. Note that the |_ means that you plug in the underscored variables after calculating the partial derivative.
    This is an important idea for something like Liouville's theorem, which says that d/dt ρ = 0, where ρ is the phase-space distribution function. It seems strange at first that something that clearly has time dependence like ρ would have zero time derivative! But what it's really saying is that for any trajectory (q(t),p(t)) that satisfies Hamilton's equations, the partial time derivative of ρ will satisfy (δ/δt ρ)|_(q(t),p(t),t) = -(δ/δq ρ)|_(q(t),p(t),t) dq(t)/dt - (δ/δp ρ)|_(q(t),p(t),t) dp(t)/dt. So the d/dt ρ = 0 is really talking about the possible "valid" trajectories through time, not ρ by itself. In fact, the underlying idea is very simple: the weighting of a trajectory to the probability distribution shouldn't change through time, thus d/dt ρ = 0.

  • @landsgevaer
    @landsgevaer 6 месяцев назад +1

    The way U understood is, the partial derivative is the derivative of a function with respect to one argument, *while fixing all other arguments constant*. I see no problem there.
    Partial derivative of u is undefined if it isn't defined as a function.

  • @kevinboyd2011
    @kevinboyd2011 7 месяцев назад +2

    Engineer here.
    This notation of using the partial derivative symbol for the "complete [partial] derivative" with respect to t (or any other variable, but let's stick to t) seems obfuscatory.
    For the example function u=f(t,x(t),y(t);z) (with dz/dt=0), the quantity [ ∂f/∂t+(∂f/∂x)(dz/dt)+(∂f/∂y)(dy/dt) ] *is* the total derivative of f wrt t --- that f coincidentally has other arguments (in this case, z) which are independent of t doesn't change that. (In fact, one might as well include the term (∂f/∂z)(dz/dt), except óila!, dz/dt=0 and the term vanishes.)
    Using ∂ for anything other than the explicit partial derivative just seems like a pedagogical nightmare.

  • @thallesaraujo7814
    @thallesaraujo7814 7 месяцев назад +1

    I use ∂ for "single-layer" derivatives (do not go into dependencies of dependencies) and d for "every-layer" derivatives (go into every possible sub-dependency, i.e. a total derivative) - regardless of the quantity of variables of a function (one or more). This solves the ambiguity presented in this video still in a third way (which seems cleaner, in my opinion). A consequence is that it no longer holds that d is for when there is only one variable and ∂ is for when there are more (which has always added more confusion than usefulness, in my opinion). Another consequence is that, for a function with a single variable, f(t), we have that ∂f/∂t (its derivative - no word "partial" needed) equals df/dt (its total derivative). Of course, mathematics is kept intact and this is just an alternative notation that changes the meaning of the symbols ∂ and d - also, one should always be cautious of what is a function and what is a variable.

  • @davidawakim5473
    @davidawakim5473 7 месяцев назад +5

    What a great channel!

  • @الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ

    4:45
    So THAT'S what it means... I've been seeing that notation in my thermodynamics course all over and was wondering what the point of it was. It's why whenever I do an exercise I'm always confused whether, when differentiation with respect to T for example, I should take P to be constant, or replace it with its expression as a function of T.
    This really was very helpful. We're not taking multi-variable calculus as a module in maths until next year, so we've had to navigate our way through it using butchered physics math. Thank you for this video.

  • @bowfinger26
    @bowfinger26 7 месяцев назад +3

    Thanks for the video. This has always confused me *a lot*. Now, at least, I learned that I wasn't totally stupid.

  • @jukmifggugghposer
    @jukmifggugghposer 7 месяцев назад +2

    oh hey now i kinda see why we were using the parentheses and subscripts for partials in thermodynamics and nowhere else. that was always weird to me. cool!

  • @mikeflowerdew7877
    @mikeflowerdew7877 5 месяцев назад

    Nice video, this broke down a few things I'd just taken for granted. As someone with a physics background, it was also great to see things from another perspective.
    One small correction for the triangular graph: the red arrow's head should be in the bottom left corner, where you have 100% silver. As it is, you end up with Ag:Au:Cu = 60:40:0, which doesn't satisfy the original premise. Those graphs can be pretty tricky to read, angled tick marks would definitely have helped!

  • @whatitmeans
    @whatitmeans 6 месяцев назад +1

    I hope I would had this video before learning thermodynamics a decade ago

  • @curtiswfranks
    @curtiswfranks 7 месяцев назад +8

    Please, no, do not use "Д" or "д" (or, while we are at it, "Ð" or "ð")! I am running out if symbols that I can use when I was to use "d" or something similar.

  • @joluju2375
    @joluju2375 7 месяцев назад +4

    Very interesting, tough a little hard for my level. However, I can understand the ambiguity in notation, thanks to the dependency graphs, and despite the vocabulary. But my most complete incomprehension is of a logical nature and is as follows: said very simply, how is possible that someone decides that x(t) is fixed when t varies ? My question may sound silly to specialists, but this is what is really blocking me. To me, fixing something that is bound to vary cannot be decided, because there is no choice. I hope I'm clear.

  • @Mr.Nichan
    @Mr.Nichan 5 месяцев назад

    This video was much easier for me to understand than your later "They Use ∂ Differently in Math and Physics. Which is Better?" video. Also, all the examples here are things I've kind of seen before.

  • @MathsSciencePhilosophy
    @MathsSciencePhilosophy 7 месяцев назад +2

    Maybe I am not getting the main point you are trying to convey, but it cleared some of my other doubts and confusions ❤

  • @ldc0322
    @ldc0322 7 месяцев назад

    I think you explained the difference between partial and total derivatives better than my rational mechanics professor. I think now I actually visualise the difference correctly, so really cool video!

  • @KirkWaiblinger
    @KirkWaiblinger 6 месяцев назад +2

    Sooooo much of mathematics and physics is context dependent. Yes, the way "functions" and their derivatives work in physics is one. Or the Einstein summation notation. Keep your eyes open and it's all over though, right down to things you'd think would be set in stone like order of operations

  • @StCharlos
    @StCharlos 7 месяцев назад +2

    I bet you don’t know how much this video has saved me just in time 😎
    🕶️👌🏻😭

  • @Arkunter
    @Arkunter 5 месяцев назад +1

    Thank you for this video I've always wondered about this!

  • @davidherrera4837
    @davidherrera4837 6 месяцев назад

    Great video.
    In Spivak's Calculus on Manifolds, he uses the notations \partial_1 f, \partial_2 f, etc to indicate the partial derivative of f with respect to the 1st argument, 2nd argument, etc.
    This sort of notation is similar to how when writing computer code you can call a function with the first argument and second argument specified however you like but what really matters is the first argument is the first argument regardless of name.

  • @scapegoatoftheuniverse7302
    @scapegoatoftheuniverse7302 7 месяцев назад +2

    we need a type system for math that you can "import" from to give you some set of expressions or functions but its not an actual program its just like virtual. But the type checker could still work

  • @stokedfool
    @stokedfool 6 месяцев назад +1

    Your visuals are top notch

  • @SVVV97
    @SVVV97 6 месяцев назад +1

    As a mathematician this annoyed me so much in the physics and engineering lectures I had. People overcomplicate things so much for no good reason - just write your shit down properly and suddenly all the problems resolve themselves

  • @woodreauxwoodreaux6298
    @woodreauxwoodreaux6298 7 месяцев назад +1

    Good choice of notation; giving Cyrillic some love. If I ever need to crunch or reference a PDE with such a dependency arrangements, I'm gunna use this notation. And when my colleagues are like WTF? I'm going to send them to this video, and they'll likely adopt this, too.

  • @AlphaDestroyer-pw8on
    @AlphaDestroyer-pw8on 7 месяцев назад +1

    Ahh I knew there was something about partials that was tripping me up and this video hit it right on the mark!
    Also I genuinely wish my p-chem prof explained the equational approach to me. I had no idea what the difference the subscripts made before watching this video.

  • @alengm
    @alengm 7 месяцев назад

    These are the most useful yet easy to understand math explainers I have seen. So straightforward. 3b1b videos are too clever for me 😅

  • @shipisleaving
    @shipisleaving 7 месяцев назад +5

    great video, always happy for a new upload

  • @robfielding8566
    @robfielding8566 7 месяцев назад +1

    See Johnathan Bartlett's minor change to differential notation. He is a computer graphics programmer, rather than a mathematician. There is a bug in standard calculation that causes people to say "dy/dx is not really a fraction". But they can be if you are careful. You can even solve for "dx/dy" without too much trouble after you fix the notation. It basically means to ONLY use implicit differentiation for everything. That causes dividing by the var we are respect to happens later, as does holding things constant. The second derivative is kind of surprising. It's a bug to assume that d^2[x]=0, though it usually is. Explicitly nest second derivative, and it's richer than what you normally get:
    d[ dy/dx ]/dx
    = d[ dy * dx^(-1) ]/dx
    = (dy * d[dx^(-1) + d[dy] * dx^(-1)])/dx
    = (dy * -dx^(-2)*d[dx] + d^2[y]/dx)/dx
    = (-dy * d^2[x]/(dx^2) + d^2[y]/dx)/dx
    = d^2[y]/(dx^2) - (dy/dx)*(d^2[x]/(dx^2))
    wow that subtracted term looks strange. these get huge for higher derivatives too.
    "acceleration is d^2[y]/(dx^2)" is only true when d^2[x]=0. And when x is a line, it's true:
    d[c]=0 "c is constant"
    d[d[t]] = "t is a line"
    d[] is an operator with binary operations, and it is recursively evaluated. we treat everything as variable. declaring "a is constant" is done exlicitly like "d[a]=0". To simplify with infinitesimals you could say that "b is positive infinitesimal", we could use dual numbers like "d[b] > 0 and d[b]^2=0" ... though maybe "d[b]^2 -> 0" is a one-way relation (not sure).
    Anyways, the point is to make Calculus fit on one page by explicitly defining the implicit differentiation operator d[]:
    d[a+b] = d[a] + d[b]
    d[a-b] = d[a + -b]
    d[a*b] = d[a]*b + a*d[b]
    d[a/b] = d[a * b^(-1)]
    d[a^b] = b*a^(b-1)*d[a] + log_e[a] * a^b * d[b]
    d[log_a[b]] = -log_e[b] /(a * log_e[a]^2) * d[a] + 1/(b * log_e[a]) * d[b]
    S[d[f]] = f - f_0; d[f_0]=0
    d[d[x]] = d^2[x]
    d[x] = dx
    The reason for doing this is that when you implicitly differentiate, you don't lose the information about what var it was respect to:
    f = x^2 + y^2
    df = 2x dx + 2y dy
    assuming dy=0, df/dx = 2x + 2y(dy/dx) = 2x
    assuming dx=0, df/dy = 2y
    Doing this, the derivative notation can be tedious, but it is legitimately algebraic now.

    • @thallesaraujo7814
      @thallesaraujo7814 7 месяцев назад

      I am glad you wrote this! Jonathan's paper on this seems really interesting (although I just skimmed through it yet). I just found it for free here: . In my opinion, there are many ways in which mathematics could be more pedagogical. Thanks!

  • @curtiswfranks
    @curtiswfranks 7 месяцев назад +2

    9:12 :
    What if the graph is not three layers deep, but is instead four or more layers deep?

  • @66sbjaygoti80
    @66sbjaygoti80 7 месяцев назад +1

    Another fantastic video!!👍👍

  • @koktszfung
    @koktszfung 7 месяцев назад +2

    Thank you for using examples in physics!
    I learnt about the material derivative in the context of fluid dynamics, maybe it would be easier to explain the operator if we imagine a particle flowing in a velocity field

  • @TimTeatro
    @TimTeatro 7 месяцев назад

    This reminds me of the introductory chapter of Sussman and Wisdom's “Functional Differential Geometry”. Nicely done.

  • @JonnyMath
    @JonnyMath 7 месяцев назад +2

    Thanks a lot!!! Nice video!!! I would also like to watch something about the differentials in thermodynamics because I don't want get why using differentials instead of simply derivatives... Thanks!!!🤩🤗

  • @baptistesirvente2697
    @baptistesirvente2697 7 месяцев назад +1

    Is this why we use the big D derivative for the navier stokes equation?

  • @jonathan3372
    @jonathan3372 7 месяцев назад

    This would also be very enlightening for students learning the Lagrangian formalism for the first time, with its quirky functional derivative symbols.

  • @Gailon1000
    @Gailon1000 7 месяцев назад +2

    This seems to be more about the imprecise notation rather than a flaw rooted in mathematics. Mathematics cant to anything about physicist abusing various notations.

  • @trogdorbu
    @trogdorbu 7 месяцев назад +1

    Something about this video makes me want to do multivariate calculus with my hands at my side and my feet a-flailing.

  • @TheIllerX
    @TheIllerX 7 месяцев назад +1

    Some remarks on the equation at 0.58.
    This is not necessarily true. It depends on how you represent z.
    If you write z as the function z(x,y) then it has only two possible partial derivatives: the one with respect to x and the one with respect to y. No t partial derivative exists in this setting.
    The partial derivatives refers to positions in the function signature.
    If you then let x and y depend on s and t, you get a function f(t,s) = z(x(t,s), y(t,s)). The full derivative of f with respect to t is then the partial derivative with respect to t of the second expression using the chain rule, as in your expression at 0.58.
    My point is that partial derivatives should be thought of as derivatives of the argument positions in the function signature.
    Let us take an example:
    Take the function f(t, x) = t * x, where x = t^2. If we replace x by t^2 so that we get the function g(t) = t^3, the full (and partial) derivative of this with respect to t is 3t^2.
    But the partial derivative of f with respect to t is x, which is t^2.

  • @stevendouglas4437
    @stevendouglas4437 7 месяцев назад +3

    Can't we just make z depend on t in a trivial way so that dz/dt = 0?

  • @paysongough
    @paysongough 7 месяцев назад +2

    0:19s terrified me. I thought my computer had finally had it with me.

  • @JCisHere778
    @JCisHere778 7 месяцев назад +1

    It’s all very clear once you learn differential geometry.

  • @etienneparcollet727
    @etienneparcollet727 7 месяцев назад +2

    Well you could always use the total derivative. Two unrelated variables are functions of each other, as constants.

    • @98danielray
      @98danielray 7 месяцев назад

      that id very nonrigorous

  • @Cyrusislikeawsome
    @Cyrusislikeawsome 7 месяцев назад +2

    How is equation vs function being defined here?

  • @donaldtimpson4320
    @donaldtimpson4320 7 месяцев назад +3

    Lol! Solid ending.

  • @dustinsoodak8954
    @dustinsoodak8954 6 месяцев назад

    I learned how to solve a bunch of different types of dif eqs in various math and physics courses, but whenever I tried to deeply understand the notation (like I would for a new programming language) it felt like my brain was returning "incomplete and inconsistent notation" errors.
    No one told me that there are technically several extra types of "d"'s or that physics & math sometimes used the symbols in slightly different ways.
    Before I started finding videos like this I thought I'd have to either dedicate a year of my life to reverse engineering the subject or wait till AI got advanced enough to explain it to me.

  • @rv706
    @rv706 7 месяцев назад

    Explaining the whole thing (for mathematicians) in few paragraphs:
    There are two setups for partial derivative notations.
    *1) Setup number one.*
    Have a function of several (say, two) variables f: (x,y)-->f(x,y).
    The notation ∂f/∂x means the usual partial derivative of f w.r.t. its first argument (the one named x) and evaluated at (x,y).
    *2) Setup number two.*
    Let's make the case of two variables. Have a constraint F(x,y)=0, and you want to define a rate of change ∂y/∂x between two variables.
    Under usual assumptions, F(x,y)=0 represents a 1-submanifold of the plane.
    The notation ∂y/∂x means the following:
    - utilize the Implicit Function Theorem (where you can) to locally write y=g(x) such that F(x,g(x))=0,
    - then compute the usual derivative dg(x)/dx.
    By the way, the IFT gives you an expression of dg(x)/dx in terms of F and its partial derivatives (in the sense of Setup n.1).
    Let's now define *Setup n.2* in the case of constraints of more than two variables. In this case, you need to specify further constraints. For example, suppose you want to compute the rate of change ∂y/∂x of the variable y w.r.t. to the variable x under the constraint F(x,y,z)=0. You need a curve to project into the (x,y) plane to which to apply the IFT and compute ∂y/∂x according to Setup n.2.
    But F(x,y,z)=0 only gives you a surface in 3-space. So you need to specify an _additional_ equation, say h(x,y,z)=0, and then the notation
    (∂y/∂x)_(h(x,y,z)=0)
    makes sense (under usual transversality/genericity assumptions).

  • @markykid8760
    @markykid8760 7 месяцев назад +2

    sweet rant dude. I totally agree 👍

  • @Reddles37
    @Reddles37 7 месяцев назад +2

    At 2:30, I don't really see the problem with just writing the total derivative du/dt. You already have the t there indicating that you should ignore Z and focus on the time-dependent variables, and I'm not really sure what the 'real' total derivative including Z is supposed to be. Obviously there are still cases where the notation gets ambiguous, but I don't think this is one of them.

    • @98danielray
      @98danielray 7 месяцев назад

      the problem is g(t)= u(x(t),y(t),t,z) does not depend only on t.

  • @EmissaryOfSmeagol
    @EmissaryOfSmeagol 7 месяцев назад +2

    Clear and engaging, well done.

  • @Ertplays
    @Ertplays 7 месяцев назад +1

    Inverse notation, powers, and derivative notation video next?

  • @curtiswfranks
    @curtiswfranks 7 месяцев назад +6

    "Complete partial" is a funny name.
    Is there really a difference between it and a total though? Just take every direct or nth-order-implicit parameter which does not depend on, say, t to be a constant wrt t. Those terms are reduced to 0.

    • @HaramGuys
      @HaramGuys 7 месяцев назад +3

      nonissue for mathematics where explicit function is usually provided.
      but in science and engineering you never know how some of these variables are interrelated, maybe even in a different equation

    • @angeldude101
      @angeldude101 7 месяцев назад +2

      ​@@HaramGuys Then just leave those terms explicitly as du/dx or what have you and not simplify it. If you know they're completely independent, then you can simplify it to zero. If you know they're dependent, then it can simplify it with the chain rule. If you don't know which it is, then it's already as simplified as it can be.

    • @98danielray
      @98danielray 7 месяцев назад

      ​@@angeldude101the "simplification to zero" will be implicit and nom-standard. it needs to be explained

    • @markusklyver6277
      @markusklyver6277 7 месяцев назад

      @@HaramGuys Just denote which variables depend on what when you write down your equation. If you don't know any dependencies, assume they do so that the additional terms just vanish when there is no dependency.

  • @chrislubs1341
    @chrislubs1341 4 месяца назад

    The ambiguity is historic, with a solution to be explicit, but the notation expands. Always question what it is that makes sense.

  • @jasonleong5811
    @jasonleong5811 6 месяцев назад

    Thanks for the informative video. Quick question, at 4:46 shouldn't the direct partial be written with 3 subscripts, x, y, z? as the direct partial assumes x,y,z are constant

  • @4thalt
    @4thalt 7 месяцев назад

    I fully support using дu/дt as the notation for an implicit derivative. Math only uses latin and greek letters, when there's all of cyrillic, hebrew, arabic, hindi, chinese, korean, japanese, and ethiopic scripts all for free.

  • @88coolv
    @88coolv 7 месяцев назад +1

    z is z(t) = const, so total derivative is ok here

  • @liamturman
    @liamturman 7 месяцев назад +2

    Great video man!

  • @kisaragi-hiu
    @kisaragi-hiu 6 месяцев назад +1

    "Spaghetti code of mathematics" exactly lol

  • @valentinlishkov9540
    @valentinlishkov9540 6 месяцев назад

    Issue:
    What is a differential of an irrational argument?
    Let a= some rational approximation, and A be the irrational number itself (if that makes sense).
    Then A - a > dA and there is no way a + dA > A
    can there be a length commensurate with all lengths (differential of length)?
    then all numbers would be rational

  • @active285
    @active285 5 месяцев назад

    I (as a mathematician) am not sure about any "Ambiguity" here, maybe it's a problem for physicists? The problem is just sloppy notation. If you suppress the dependency of the functions (not variables!) x = x(t) and y = y(t) on the time t for convenience, it might be helpful to state it explicitly before any calculations are done. But of course there is difference in the partial derivative of a function with three variables u = u(x, y, t) and a function depending on time and two parametrised curves u = u(x(t), y(t), t). For the former taking the derivative in t is just taking a partial derivative of the function u in one variable, for the latter you actually try to differentiate a function with indirect dependencies u = u(x(t), y(t), t) in the variable t (hence t, x(t), y(t)) so that the chain rule leads to the total derivative of u(x(t), y(t), t).

  • @gabrielsantiago7318
    @gabrielsantiago7318 7 месяцев назад +1

    Would it kill mathematicians to just write out something instead of using 83 different notations? I just finished calculus 3 and my biggest problem was honestly just trying to interpret e ridiculous and confusing notation

    • @mariusj.2192
      @mariusj.2192 6 месяцев назад

      Something would die inside of them probably if they had to do that, so yes it would :D
      but agreed, for learning that would be helpful

  • @tens0r884
    @tens0r884 6 месяцев назад

    its my opinion that f(x,y,z,t) is itself flawed notation as im pretty sure its supposed to be taken that the independent variables are, well ... *independent* (i know its technically not flawed conventionally speaking since its used everywhere, but i personally like writing f(x(t), y(t), z, s(t)) instead, where s(t) = t, removing the dependence)

  • @Adam123a
    @Adam123a 7 месяцев назад

    The issue is not with the derivative, it’s with whether you are taking a derivative of a function OR a function composition. If u depends on x,y,z. Then you can ONLY differentiate with respect to x, y, or z. That’s it. Plain and simple. If you want to differentiate with respect to t, it makes no sense and I mean that. However, if you make a composite function where x,y,z all or some depend on t, then you now have a *completely different function*. You have an outside function and and an inside(s) function. It is *not* possible to differentiate u with respect to t. But the composite function (emphasis on composite function) can be differentiated with respect to t. And if all 3 x,y,z depend on t, then the composite function is now a **single variable** function, while the outside function was a **multi variable** function. Do you see how an outside function and the full composite function are 2 completely different functions? **Always and forever** a function can **only** be differentiated with respect to its variables. But if you **overload** a function to mean different things, that’s where the confusion comes in. (A function is just a name given to a mapping. If I call a certain mapping h, and some other mapping h, do you see how confusion comes? There nothing wrong with the derivative, it’s your notation of the function you’re differentiating. Writing u(x,y,z) is **completely different** than u(x,y(t), z). These are 2 different functions. The first u can be differentiated with respect to x, y, or z. Not b or t or s or k. The second is a composite function. It’s domain is not xyz, but xtz. The outside function can be differentiated with respect to y, but not the composite function. That’s all)
    There is no “total” derivative. Just derivatives and partial derivatives. Single or multivariable calculus. And if you form a composite function, now you need the chain rule for the outside and inside functions.

  • @Blackrobe
    @Blackrobe 7 месяцев назад

    7:44 where can I read more about this diagram? I'm interested of its use

  • @ashes2ashes3333
    @ashes2ashes3333 7 месяцев назад

    I want to defend the “material derivative” you mention at the end in a joke. It actually is really important to distinguish between the material derivative and the total derivative, and here is why.
    Suppose you are standing above a fluid, looking down at it (formally, I mean an Eulerian description of a fluid). You are interested in the change in some property, say temperature or fluid velocity or something, in the fluid. The fluid has some velocity function u(x,y,z,t), (u should be a three vector but I can’t draw arrows lol) which depends on the position in the fluid in your coordinate system, and the time.
    Now I ask, what is the time dependence of my property, e.g. fluid velocity?
    The problem is, I could mean two different things here. If I’m looking from above at a fluid that is in some steady flow state, there might be no change in the velocity field at any time: the fluid is just moving in the same pattern it always has been, and u does not change with time. Sure, u might be different at different spatial coordinates, but from my perspective, asking about the change of the fluid properties at a particular x, y, z with time, I would only get \partial u/\partial t.
    Now I ask: how does each fluid parcel (each fluid element that moves with the fluid) experience a change in time? Well now you want to account for HOW THE PARCEL MOVES IN x, y, z, as it is moving to new points. Ok, so simple, you want something that looks like
    \partial u/\partial t + \partial u/\partial x dx/dt + same for y and z.
    Which LOOKS LIKE the total derivative. BUT IT ISNT! If we actually compute this, well we are unfortunately stuck because x,y,z is MY FIXED COORDINATE SYSTEM! None of the coordinates depend on time! The x that enters the thing I want is not the same as the x in my coordinate system. So in fact, the expression I wrote above is just \partial u/\partial t, and I still haven’t computed the change of the fluid from the perspective of the parcel.
    So the thing you actually compute is not the total derivative, because it doesn’t really make sense here, x,y,z don’t depend on t. Instead, you compute the MATERIAL DERIVATIVE, which is
    D/Dt = \partial/\partial t + velocity_vector\cdot gradient_vector
    It’s just that the velocity vector is not the three vector of dx/dr, because again, x,y,z are fixed coordinates.
    I think it’s unfortunate that you didn’t mention this, because I think the notation of the material derivative is one of the best examples of GOOD notation that is clear about what you’re actually computing. It’s just when you write it like the total derivative, you’re missing that the x that appears in the right hand side is a different x to the coordinate system x (namely, that x is the parcel position in the coordinates).
    The material derivative is one of the nice things we have.

  • @legendariersgaming
    @legendariersgaming 7 месяцев назад

    I think part of the ambiguity is that the same glyph is often used a placeholder (to distinguish input position) *and* as an actual variable used as an input. This naming conflict arises even in one dimension, it just typically does not cause any issues. To illustrate, suppose we have a function f that takes a single variable as input. If I write df/dx(x^2), do I mean to evaluate the derivative of f (a map R -> R) at the point x^2, or do I mean the derivative of the function x ↦ f(x^2)?
    For this reason, I tend to use notation like D^α f. For a function f taking n inputs, the superscript α is a tuple of length n whose entries are nonnegative integers. The ith entry of α tells you how many times the ith input position to f is differentiated. The drawback here is that Clairaut's theorem does not always hold, but I don't encounter these too often in my work

  • @markusklyver6277
    @markusklyver6277 7 месяцев назад

    Just write down the limit definition before using a notation - no confusion arises. For example, what you call "total derivative" is just
    lim_{h to 0} 1/h[u(x(t+h), y(t+h), t+h) - u(x(t), y(t), t)],
    and what you call the "direct partial" or "the explicit partial" is just
    lim_{h to 0} 1/h[u(x(t), y(t), t+h) - u(x(t), y(t), t)].

  • @resistancefm5133
    @resistancefm5133 7 месяцев назад +1

    personally, I would just say that the partial z with respect to t is zero, then the notation holds.

  • @yuGesreveR
    @yuGesreveR 7 месяцев назад +1

    Hm.... I thought that in the case of one extra variable z, we should not invent any additional notation, because if we do not suppose that z = z(t), then both partial or full derivative of z by t is zero. So, basically we just add zero when use du/dt. But I have never thought about the case, when we want just to omit dependency of z on t even if we know that this dependency exists. Interesting. Although I don't like the proposed notation. Maybe something like Du/D_xy t would be better. Imho

  • @otterlyso
    @otterlyso 7 месяцев назад +1

    At (6:08) is that actually a PVT diagram of a general substance that isn't water? Water expands on freezing so there should be some region of solid with volume greater (less dense) than some region of liquid. But excellent video either way.

    • @HaramGuys
      @HaramGuys 7 месяцев назад

      nice catch

    • @EpsilonDeltaMain
      @EpsilonDeltaMain  7 месяцев назад

      You are right, I added that to the list of corrections

  • @christophergame7977
    @christophergame7977 7 месяцев назад +1

    Very helpful.

  • @JustPassingBy_
    @JustPassingBy_ 7 месяцев назад

    One way i kinda managed to by-pass the ambiguity is by considering two different things, let's give an example let u(x(t), y(t), t, z), if i want to know how u changes with respect the t input i would write it as ∂u/∂t, but if i want the total partial i would think it as an operator (∂/∂t)(u). With this i feel like i am asking the total partial derivative, since it feels like i am asking all the changes u when i move t. But i recognize that this solution is kinda jank

  • @mahiainti678
    @mahiainti678 7 месяцев назад +2

    I hoped there'd be a new, "better" solution at the end..

    • @LemoUtan
      @LemoUtan 7 месяцев назад +1

      I do hope there's a place for ð.

  • @trabek123
    @trabek123 6 месяцев назад

    please do a video on cantor schroder theorem.

  • @KakoriGames
    @KakoriGames 7 месяцев назад +1

    2:44 why can't you just do du/dt = pu/pt + pu/px * dx/dt + pu/py * dy/dt + pu/pz * dz/dt? Since z doesn''t depend on t, dz/dt = 0 and that's it, I don't see the problem.

    • @KakoriGames
      @KakoriGames 7 месяцев назад

      @@koktszfung When you define the Total Derivative as I described, with dz/dt = 0 included, then the Total Derivative and the "Complete Partial Derivative" describe the same thing, at least in the case shown, where that's supposedly a problem. If there's a actual reason why this notation doesn't work, then the video didn't properly showcase it.

    • @koktszfung
      @koktszfung 7 месяцев назад

      I was wrong, forget about what I said

    • @98danielray
      @98danielray 7 месяцев назад

      the problem is the notation can be confusing and imply u has no dependence on z, or that the writer confused d/dt with partial/partial t since u depends on z.

  • @HelloWorlds__JTS
    @HelloWorlds__JTS 7 месяцев назад +9

    [
    Updated response after @HaramGuys' first reply:
    ***In the context of what is presented in this video*** there is no ambiguity as claimed... @HaramGuys gives an example that implies the following: There might be an ambiguity for non-invertible representations, e.g. z(t) isn't invertible as t(z). But those sorts of ambiguities are handled differently than what was shown here, and they are out of scope for this video and the ambiguity implied in it. So when I said "...but is never necessary" below, I should've said in any relevant context implied by this video, and for most problems of interest to people watching it.
    ]
    Original response:
    There is no ambiguity as you claim. Making these distinctions between, e.g., ∂u/∂t and (∂u/∂t)x,y, is useful sometimes but never necessary.
    And denoting du/dt as ∂u/∂t is misleading, unnecessary, and never useful. The dependence of u on z doesn't change just because the dependence of z on t is different than the dependence of {x,y} on t; the full derivative du/dt is still valid and complete, it's just that ∂u/∂z * ∂z/∂t = 0 when ∂z/∂t = 0.
    The only reason people have introduced all of the supposed ambiguities is because they have prior knowledge of the functional dependencies on all variables, so it helps to keep things in order.

    • @HaramGuys
      @HaramGuys 7 месяцев назад +6

      you are vastly underestimating sloppy math done in science and engineering. things aren't so formally defined with explicitly defined functions, and frequently the relationship between any tuple of variables are in a completely different equation or equations.
      say you have 2 equations u(x,y,z,t) = x^2+y^2+z^2-t^2 and t^2+z^2 = const,
      du/dt would implicitly imply that you are treating second equality to mean z as a (*single branch of multivalued) function of t, but second equality could just as well mean t as a function of z say when taking du/dz.

    • @HelloWorlds__JTS
      @HelloWorlds__JTS 7 месяцев назад

      ​@@HaramGuys great example! I'll edit my response to account for what you are implying.

    • @98danielray
      @98danielray 7 месяцев назад

      you are advocating for pragmatic notation, but the ramificatioj of using du/dt for when there is another variable z is that du/dt is not defined unless you use currying.

    • @HaramGuys
      @HaramGuys 7 месяцев назад

      ​@@98danielray I read the pinned comment and I just found out that mathematicians and physicists use the symbol ∂ and the very word "partial" differently.
      mathematicians use the word partial as opposed to ordinary, physicists use the word partial as opposed to total.
      using du/dt is considered wrong in calculus 3, but perfectly fine in fluid dynamics

  • @kamilrichert8446
    @kamilrichert8446 7 месяцев назад

    1:20 I don't understand how does x and y depend on t in the example. Are coordinates not constant?

    • @HaramGuys
      @HaramGuys 7 месяцев назад

      Walk around, position changes over time