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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Let \[\phi(x,y)\] define a scalar field then will \[\frac{\partial^2\phi}{\partial x\partial y}\] and \[\frac{\partial^2 \phi}{\partial y \partial x}\] always be continuous? Also if \[\frac{\partial \phi}{\partial x}=0\] then \[\implies \frac{\partial^2 \phi}{\partial x \partial y}=\frac{\partial}{\partial x}(\frac{\partial \phi}{\partial y})=0?\]

  2. IrishBoy123
    • one year ago
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    .

  3. ganeshie8
    • one year ago
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    Essentially, you're asking if the mixed partials are always equal ?

  4. anonymous
    • one year ago
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    I know they are not always, they must be continuous, but is it true for all cases when we talk about a scalar field?

  5. anonymous
    • one year ago
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    I was thinking about the different ambiguities that arise when we say the following determinant is 0 \[\left[\begin{vmatrix}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}\end{vmatrix}\right]\]

  6. anonymous
    • one year ago
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    and it IS 0 that's an identity

  7. ganeshie8
    • one year ago
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    Curl of gradient is 0, yeah

  8. anonymous
    • one year ago
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    How can it be 0 if we are ambigious about for example the expression \[\frac{\partial^2 \phi}{\partial y \partial z}-\frac{\partial^2 \phi}{\partial z \partial y}\] This is not always 0 yet the curl of gradient is 0 vector

  9. ganeshie8
    • one year ago
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    Clearly \(\phi \) is the potential function of vector field \(F=\nabla \phi\) Since a potential function exists, the field \(F\) is conservative by definition. Equivalently \(\nabla \times F\) is 0.

  10. ganeshie8
    • one year ago
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    wait a second, i think i got your question you're saying the mixed partials are "required" to be equal for the curl to be 0 ?

  11. anonymous
    • one year ago
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    yep that's what im saying

  12. anonymous
    • one year ago
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    at least when we talk about a scalar field, not talking about other multivariable functions

  13. anonymous
    • one year ago
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    The individual components need to be 0 so that curl(grad phi) is a null vector

  14. ganeshie8
    • one year ago
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    Rigt, i get it... I don't seem to have a convincing explanation for myself.. let me think

  15. anonymous
    • one year ago
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    ok

  16. anonymous
    • one year ago
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    I guess I'll doze off for a few hours, really tired. I will check this out later if you have found an explanation

  17. ganeshie8
    • one year ago
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    let me tag few others @eliassaab @zzr0ck3r @SithsAndGiggles @Michele_Laino @Empty @UnkleRhaukus @oldrin.bataku @Astrophysics

  18. ganeshie8
    • one year ago
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    @mukushla @adxpoi @jtvatsim

  19. anonymous
    • one year ago
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    Ok, so I'm back, not much luck huh lol

  20. imqwerty
    • one year ago
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    O-O i jst know that we can use curl method to check if a force is conservative or not.

  21. anonymous
    • one year ago
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    @ganeshie8 Looks like I've the found the solution to this, apparently in physics, we assume our functions to be "nice" http://mathworld.wolfram.com/PartialDerivative.html

  22. Empty
    • one year ago
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    I had some problems when I asked a similar question, you might enjoy because it wasn't given a satisfactory answer I feel. http://math.stackexchange.com/questions/956095/when-does-order-of-partial-derivatives-matter

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