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I Questions about gradient and scalar product

  1. Aug 15, 2016 #1
    I recently learned that the general formula for the dot product between two vectors A and B is:

    gμνAμBν

    Well, I now have a few questions:

    1. We know how in Cartesian coordinates, the dot product between a vector and itself (in other words A ⋅ A) is equal to the square of the magnitude |A|2. Well does this hold true in the general case as well for any coordinate system? If I do gμνAμAν, will that give me the square of the magnitude of the vector A?

    2. We know that in Cartesian coordinates, you can find a unit vector in the direction of any vector by dividing the vector by its magnitude ( in other words by doing this: A / |A| ). Can you use this same method to find said unit vector in any coordinate system (whether it be a set of orthogonal coordinates or curvilinear coordinates)?

    3. This next question has to do with the gradient operation. We know that in cylindrical coordinates, the gradient of a function is as follows:

    ∇U = [r-hat * (∂U/∂r)] + [ θ- hat * (1/r) * (∂U/∂θ)] + [z-hat * (∂U/∂z)]

    Note: r-hat, θ-hat and z-hat are all just unit basis vectors.

    Now my question is: Why is it (1/r) that is multiplied by θ-hat and (∂U/∂θ) instead of just being r? I know that the unit basis vector θ-hat is equal to eθ/r (where eθ is the original not normalized basis vector). It is for that reason that I would think that r would be in the place of (1/r). If that were the case, then the gradient vector would be expressed as a linear combination of all of the original basis vectors. However, (1/r) * θ-hat simply equals eθ/r2 , which doesn't seem like any special quantity to me. Why then is it (1/r) instead of just r? Thank you.
     
  2. jcsd
  3. Aug 15, 2016 #2

    Orodruin

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    This is the definition of the magnitude of a vector.

    Yes, this will always give you a unit vector. (You can easily show this by insertion.)

    If it would be r, your result would be dimensionally inconsistent. I suggest you first write down the gradient in Cartesian coordinates and then transform the derivatives to polar coordinates using the chain rule
    $$
    \frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}
    $$
    and then write the Cartesian basis vectors in terms of the polar ones.

    If you want to be more fancy, the partial derivatives (e.g., ##\partial_\theta U## etc) are not the components related to the tangent vector basis (given by ##\vec e_\theta = \partial \vec r/\partial \theta## etc). They are the components related to the covector basis given by ##\vec e^\theta = \nabla\theta## etc.

    Edit: Note that the tangent vector and covector bases coincide for a Cartesian coordinate system. For this reason there is no need to distinguish them in a Cartesian system.

    It is a special quantity, it is the covector basis and is equal to ##\nabla \theta##.
     
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