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Understanding the physical meaning of multiplication, etc

  1. Oct 6, 2015 #1
    Is there some kind of intuitive way to understand the physical meaning when mathematical operations are applied to equations in physics?

    What I mean is that, say we start with a 'starting point' equation, in this example Ficks law of diffusion (wikipedia:):

    [tex] J = -D \frac{\delta \phi}{\delta x} [/tex]

    In this case, what is the physical meaning of multiplying a scalar (D) by the derivative?

    Can this be generalized to things like, multiplying the gradient of something by the gradient of something else, taking the gradient of a tensor, then multiplying that by the gradient of something else?

    What I am trying to get at is in a lot of physics derivations, equations become huge and complex, but I want to know how they break down, and what physical meaning the 'propagation' of all of these multiplication and additions means?

    And then when you have things like integrals of the gradient of a function

    Essentially what I would like to know is, given some complex equation such as (eq from the navier stokes page on wikpedia:)


    If I was just looking at the equation without knowing its roots, how can I take an educated guess about what It might be describing?

    I hope this question makes sense.

  2. jcsd
  3. Oct 6, 2015 #2


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    I'll pick a few items; can't really give a sensible answer to all questions.
    It literally scales the proportionality: some materials diffuse more easily than others. So transport is proportional to change in concentration with a proportionality coefficient D that depends on the stuff you are considering.
    Actually it is a vector equation: transport has a magnitude and a direction, so Fick really reads $$\vec J = - D \, \nabla\phi$$and as such completely analog to Fourier's law $$\vec q = -k\, \nabla T$$

    Yes and no. It becomes physically meaningless pretty quickly. And note that a gradient is a vector, so you'll have to tell what you mean when you say "multiplying the gradient of something by the gradient of something else" ! Mathematics is a bit more stoic but even there it can go pointless at some point.

    Sure you do. In principle the author should help you with that by building up his (her) relationships in steps. Some authors with their intended readers are at an abstraction level that a normal interested person can't really follow. Especially then, there are no cheap tricks, unfortunately. Nothing left to do than trying to find a text at a more elementary level. What also helps is not to try asking too much about advanced chapters until you've read all the preceding chapters -- and done the exercises !

    Your question makes sense, but e.g. your Navier Stokes example presented bare like that is meaningless to me. The best I can do is ignore the ##\Pi## and see that it's the curl of the vector that itself is a volume integrated curl. I would really have to revert to some context or some introductory stuff before I could make sense out of that. Doesn't have to take all that long: you see "Helmholtz" so link to that etc. etc.
  4. Oct 8, 2015 #3
    Hi BvU,

    Thanks for your reply. Your answers really shone some light on this.

    I want to be able to understand the 'steps' used in derivations, and what they mean physically, why they are done, and how you 'know' what to do and when to do it... In terms of going back to elementary level, I find I don't learn anything doing maths just for maths sake.

    The thing that keeps confusing me is, say we have the equation for a plane wave. I am all cool with what that represents. But then a few equations down a book will have that equations in an integral with all these other symbols and stuff and I just don't know what it means when they smash it all together like that...

    Anyway my question of "how can one better understand these sorts of things" probably doesn't have an answer. I am sure a lot of it just comes with experience but though I should ask if there are any techniques or tricks on the matter.

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