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Gradient in other coordinates

  1. Jun 7, 2015 #1
    Sorry again for all these ongoing question as I try to fix my math deficiencies. (Back to working on differential forms.)

    So...

    I understand that the equation of steepest ascent/descent in Cartesian coordinates is written:

    dxi/dt = ∂f/∂xi

    And I can follow the "physical interpretations" of how the right side is in the path of the greatest change IN Cartesian coordinates.

    But we know this cannot hold in cylindrical coordinates: the left is contravariant and the right is covariant.

    So we can convert the covariant term on the right, with the metric tensor...

    dui/dt = gij∂f/∂uj

    OK, I get all that...

    But it seems a contrived way to get the form of the gradient used in this equation.

    Is there some more objective way that one can state that...

    ∇fi = gij∂f/∂uj...

    while providing an interpretation of this nabla?

    Or should I just content myself with knowing.... "all things flow from the Cartesian... get it there first and then use the metric tensor for all other coordinates"

    It seems really contrived to base the whole thing on Cartesian and not be able to give an interpretation of
    gij∂f/∂uj
    as a direction of greatest change.
     
  2. jcsd
  3. Jun 7, 2015 #2

    Orodruin

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    What you have stumbled across is nothing else than the fact that the gradient is an inherently covariant vector.

    In order to realise how
    comes about, I suggest that you start from
    $$
    \frac{df}{dt} = \dot x^\mu \partial_\mu f
    $$
    and maximise it with respect to ##\dot x^\mu## under the condition ##g_{\mu\nu}\dot x^\mu \dot x^\nu = 1## (Hint: Lagrange multipliers). This should give you something proportional to the formula you quote.
     
  4. Jun 7, 2015 #3

    Would you be so kind as to help me by filling in those obvious steps (that are not obvious to me).
     
  5. Jun 7, 2015 #4

    Orodruin

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    So you want to find the direction in which the function grows the fastest. The way to do this is to maximise the function
    $$
    \frac{df}{dt} = \dot x^\mu \partial_\mu f.
    $$
    But you can always make this bigger by increasing all ##\dot x^\mu## by the same factor so for this to have any meaning, you want to do it for velocities of unit length, i.e., for ##\dot x^\mu## such that ##g_{\mu\nu}\dot x^\mu \dot x^\nu = 1##.

    Are you familiar with how to find the extremal values of a function? (In this case it is a function of ##\dot x^\mu##.) Do you know how to use Lagrange multipliers?
     
  6. Jun 7, 2015 #5
    1. Yes
    2. I have used them in the most simplest of cases, many years ago. So I will say no with the caveat that it will likely come back fast.
     
  7. Jun 7, 2015 #6

    Orodruin

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    So if I say that if you want to find the extreme values of a function ##f(x_i)## under the constraint ##g(x_i) = c_0##, then you:
    1. Create the function ##h = f - \lambda g##.
    2. Find the extreme values of ##h##.
    3. Fix ##\lambda## such that the constraint is satisfied.
    Does this ring a bell?
     
  8. Jun 7, 2015 #7
    So I will work this backwards and spend time reviewing. I do remember.
    But could you just tell me what lamda will be in this problem
     
  9. Jun 7, 2015 #8

    Orodruin

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    I suggest you work through steps 1 and 2. The constant ##\lambda## will only serve to normalise the vector ##\dot x^\mu##.
     
  10. Jun 7, 2015 #9

    WWGD

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    What do you mean by inherently covariant in spaces where you have a metric ( or any non-degenerate form) that gives you a natural isomorphism between a space and its dual? Or are you referring to the cases where you do not have such form?
     
  11. Jun 8, 2015 #10
    WWGD: you have helped me before, thank you.

    Would you be willing to fill in the missing steps?

    I make no excuses: my physics is bad -- I am a 56 year old engineer (I am very good at that, but my physics is bad).
    I am not here for HW help. I am here to rapidly fill in gaps in my knowlege.

    And Orodruin would rather treat me like a studen and that is OK (but it is not what I am looking for).

    In all likelyhood, once I see it done, the light will shine and I will realize what a fool I have been. This has happened many times in the last few weeks.

    But right now, I cannot finish those steps.

    Again, this is NOT HW and a I am not a student.
     
  12. Jun 8, 2015 #11

    Orodruin

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    If this is your attitude, Physics Forums might not be the best fit for you. Physics Forums is not a place to go to be handed the answer on a silver platter, regardless whether you are doing homework or not.

    The "rapidly fill gaps" is a warning bell to me. You are not going to do this by taking shortcuts, the only way of doing it is spending time on doing it properly.

    Yes, I intended it for spaces where there is no metric, since df/dt does not require one. I agree that it might be confusing that there then is a metric necessary to define the speed.
     
  13. Jun 8, 2015 #12
    No, it just means YOU cannot help me. And that is OK.

    I have learned a great deal from others here who took the time to step me through and not dispense information slowly.

    I understand you probably teach this way and it probably works well for you and that is very good.

    You are interpreting "rapidly filling in gaps" as if I was a student cutting corners. I have my own agenda to understand steps toward a specific goal (and it really makes no sense to elaborate on my personal goals other than to say I just want to learn a straight path to what I want and once that path is cleared THEN I can go back and fill things in.) And the examples on the web or in book on Lagrange multipliers are simple; and this problem has a directional aspect: and there are other things I want to learn in the process of this done correctly (and I do not want to belabor anyone with those other issues.)
     
    Last edited: Jun 8, 2015
  14. Jun 8, 2015 #13

    Orodruin

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    If you had spent half the time you spent on complaining that I did not yet hand you the answer on a silver platter on actually following the very simple steps I outlined, you would have already solved your problem and learnt something in the process.

    The fact is that the problem outlined is very simple in terms of Lagrange multipliers.
     
  15. Jun 8, 2015 #14
    Here... it's like this...

    Person comes to me to ask for help on their transmission: I spoon feed. I give it them. I explain all steps. And do it for them becasue I respect their fundamental intelligence and KNOW they are not manipulating me. I know they just are stuck on putting pieces together. That is the kind of person I am. Please stop now. I have not insulted you. Please stop.
     
  16. Jun 8, 2015 #15

    Drakkith

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    Thread locked, pending moderation.
     
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