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Invariance of del^2 operator under rotation of axes

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A scalar function can be represented as a position on the x-y plane, or on the u-v plane, where u and v are axes rotated by θ from the x and y axes.

    Prove that the 2-dimensional [tex]\nabla^2[/tex] operator is invariant under a rotation of axes.

    ie,

    [tex]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}[/tex]

    2. Relevant equations

    u = x cos θ + y sin θ
    v = - x sin θ + y cos θ

    I think I got the answer, *but* I wonder if there is an easier and more elegant way to do it. My method was to write out (this is just for [tex]\partial^2 f / \partial x[/tex]

    [tex]\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) = \frac{\partial}{\partial u}(\frac{\partial f}{\partial x})\frac{\partial u}{\partial x} + \frac{\partial}{\partial v}(\frac{\partial f}{\partial x})\frac{\partial v}{\partial x} [/tex]

    And then substitute in

    [tex]\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x } + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} cos \theta } + \frac{\partial f}{\partial v}(-sin \theta)[/tex]

    Eventually I get

    [tex]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({sin}^2 \theta + {cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({sin}^2 \theta + {cos}^2 \theta)[/tex]

    to get the result needed.

    But this becomes incredibly long and tedious, like one to two pages of working and it's pretty easy to make a careless mistake somewhere. Is there an easier/shorter way to do this? For instance, is there a way that I could do it without the rotation of axes formula?
     
  2. jcsd
  3. May 12, 2009 #2

    CompuChip

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    I think what you did is the standard way to prove it.
    If you have the expression for [itex]\nabla^2[/itex] in polar coordinates you could have used that one, and note that the kind of rotation you are looking at is simply a transformation [itex]\theta \to \theta + \theta_0[/itex] (theta0 a constant) which changes neither r nor theta-derivatives.
     
  4. May 12, 2009 #3
    a) You can prove in one line that that [itex]\nabla f[/itex] is a vector.

    b) You can prove in one line that [itex]\nabla\cdot \vec{F}[/itex] is a scalar.

    So, you can prove it in two lines.
     
  5. May 13, 2009 #4
    Thanks Chip.

    Count Iblis, ok, I understand a), but what's the vector F? (Do you mean nabla f?) After all I only have scalar function f. How do I proceed from there? Thanks.
     
  6. May 13, 2009 #5
    You can do b) for the general case. So F is some arbitrary vector field and then you prove that the divergence is a scalar. Then, since in a) you have proven that the gradient of a scalar is a vector, you can take F to be the gradient of the scalar f and then you find that the Laplacian of f is a scalar.

    a) is a trivial application of the chain rule, it doesn't depend on the fact thatthe new coordinates are related to the old coordinates via an orthogonal transformation. Indeed a) is also true in case of manifolds (curved space) were one defines vectors as transforming covariantly under general coordinate transformations.

    In b) you do need to use that the derivative of the old coordinates w.r.t. the new coordinates does not itself depend on the position, so you can bring it inside the derivative and then use that to transform the vector. In case of general coordinate transformations this is not true. This means that in case of manifolds one needs to define a different kind of derivative that when applied to vectors does lead to scalars or tensors that transform covariantly under coordinate transformations. This derivative is called the "covariant derivative".
     
  7. May 14, 2009 #6
    Count Iblis, if I am getting you right, to sum up your method: the Laplacian of f is a scalar. Then, since scalars are invariant under rotation of axes, the Laplacian is invariant under rotation of axes.

    Would it be sufficient to leave it as that, or would I need to prove it a bit more rigorously?
     
  8. May 14, 2009 #7
    Well, you would already have proven it. Suppose the coordinates are denoted by (x_1, x_2, x_3) and we introduce a new coordinate system (y_1, y_2, y_3), such that each y_{i} is a function of x_1, x_2 and x_3 and vice versa. Then in case of rotations the y_i have a special form, but we don't need to use that if it isn't necessary. The transformation matrix to transform a vector from the old basis to the new basis is then:

    [tex]M_{i,j} =\frac{\partial y_{i}}{\partial x_{j}}[/tex] (1)

    M is an orthogonal matrix. The transpose of M is its inverse, so we have:

    [tex]\frac{\partial y_{i}}{\partial x_{j}} = \frac{\partial x_{j}}{\partial y_{i}}[/tex] (2)



    If [itex]\vec{F} =\nabla f[/itex], we have in the old coordinates

    [tex]F_{i} = \frac{\partial f}{\partial x_{i}}[/tex] (3)

    Now, we define the components of [itex]\vec{F}[/tex] in any coordinate system according to (3). To prove that [itex]\vec{F}[/itex]is indeed a vector, we must check that trasforming from one coordinate system to another can also be done by applying the transformation matrix. If we denote the components of [itex]/vec{F}[/itex] in the coordinate system (y_1, y_2, y_3) by F'_i we have on the one hand:

    [tex]F'_{i} = \frac{\partial f}{\partial y_{i}}[/tex]

    because this is simply how we define the components of F in any coordinate system. But using the chain rule we can write:

    [tex]F'_{i} = \frac{\partial f}{\partial y_{i}} =
    \sum_{j}\frac{\partial f}{\partial x_{j}}\frac{x_{j}}{y_{i}}[/tex]

    Using (2) and (1), we can write this as:

    [tex]F'_{i} = \sum_{j}M_{i,j}F_{j}[/tex]

    So, we see that [itex]\vec{F}[/itex] indeed transforms as a vector.

    To prove that div F is a scalar for any arbitrary vector field is just an easy appication of the chain rule. We have in the old coordinates:

    [tex]\sum_{i}\frac{\partial F_{i}}{\partial x_{i}} =\sum_{i,j}\frac{\partial F_{i}}{\partial y_{j}}\frac{\partial y_{j}}{\partial x_{i}} =
    \sum_{i,j}\frac{\partial F_{i}}{\partial y_{j}}M_{j,i}
    [/tex]


    M is a constant matrix that can be brought inside the derivative. Then we can use that:

    [tex]\sum_{i}M_{j,i}F_{i} = F'_{j}[/tex]

    So, we get:

    [tex]\sum_{i}\frac{\partial F_{i}}{\partial x_{i}} =\sum_{j}\frac{\partial F'_{j}}{\partial y_{j}}[/tex]

    and we see that div F is the same in any two coordinate systems related to each other by orthogonal transformations, proving that it is a scalar.

    Now, the only reason I needed more than two lines is because I added some extra explanations. The mathematical manipulations themselves were nothing more than just applying the chain rule. In particular, you did not need to use the precise way how the old coordinates are related to the new coordinates in terms of the angles. You only used the fact that the transformation is orthogonal and that the components transformation matrix are constant and thus do not depend on the position.
     
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