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December 8th, 2017, 12:47 PM   #1
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Scalar Fields vs Vector Fields

I am trying to figure out which expressions are a scalar field and which ones are a vector field.




For (a) I figured that we are taking a radial component into our function and taking the gradient. Which I believe would result in a scalar value thus making this a scalar field. For (b), it is a dot product, so the result has to be a scalar and thus a scalar field. For (c), I would suspect this to be a scalar field for the same reason as (a) except we are doing a laplacian. And (d) is the only vector field since it is a cross product and the result of a cross product is a vector.

I am shaky on my reasoning and unsure if I am even thinking about these fields correctly. Anyone have any tips?

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December 8th, 2017, 01:57 PM   #2
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These all follow from the basic definitions.

Given a scalar valued function of three variables, $\displaystyle \phi(x, y, z)$, the "gradient" of $\displaystyle \phi$ is a vector valued function, $\displaystyle \nabla \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial x}\vec{k}$. You can think of this as the "scalar product" of the vector operator $\displaystyle \nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{j}$ and the scalar function $\displaystyle \phi(x,y,z)$. The product of a scalar and a vector is a vector.

Given a vector valued function of three variables,$\displaystyle \vec{F}(x, y, z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, the "divergence" of $\displaystyle \vec{F}$ is a scalar valued function, $\displaystyle \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$. You can think of this as the "dot product" of the vector operator $\displaystyle \nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{j}$ and the vector function $\displaystyle \vec{F}(x,y,z)$. The dot product of two vectors is a scalar.

Given a scalar valued function, $\displaystyle \phi(x,y,z)$, the "Laplacian", $\displaystyle \nabla^2\phi$ is the scalar valued function $\displaystyle \nabla^2\phi= \frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial y^2}+ \frac{\partial^2\phi}{\partial z^2}$. You can think of this as the dot product of the vector operator $\displaystyle \nabla$ with the vector valued $\displaystyle \nabla \phi$. The dot product of two vectors is a scalar.

Given a vector valued function, $\displaystyle \vec{F}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, the "curl" of $\displaystyle \vec{F}$ is $\displaystyle \nabla\times\vec{F}= \left(\frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$. You can think of this as the cross product of the vector operator, $\displaystyle \nabla$, and the vector valued function $\displaystyle \vec{F}$. The cross product of two vectors is a vector.
Notice that I have consistently used the variables (x, y, z) rather than $\displaystyle \vec{r}$. You mention, for the first, the "radial component". $\displaystyle \vec{r}$ is simply the position vector, not the "radial component" of the position (that would be just "r", not a vector). $\displaystyle \phi(\vec{r})$ and $\displaystyle \vec{F}(\vec{r})$ is the same as $\displaystyle \phi(x,y,z)$ and $\displaystyle \vec{F}(x,y,z)$ respectively.
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Last edited by skipjack; December 8th, 2017 at 05:58 PM.
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December 8th, 2017, 07:26 PM   #3
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Originally Posted by Country Boy View Post
These all follow from the basic definitions.

Given a scalar valued function of three variables, $\displaystyle \phi(x, y, z)$, the "gradient" of $\displaystyle \phi$ is a vector valued function, $\displaystyle \nabla \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial x}\vec{k}$. You can think of this as the "scalar product" of the vector operator $\displaystyle \nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{j}$ and the scalar function $\displaystyle \phi(x,y,z)$. The product of a scalar and a vector is a vector.

Given a vector valued function of three variables,$\displaystyle \vec{F}(x, y, z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, the "divergence" of $\displaystyle \vec{F}$ is a scalar valued function, $\displaystyle \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$. You can think of this as the "dot product" of the vector operator $\displaystyle \nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{j}$ and the vector function $\displaystyle \vec{F}(x,y,z)$. The dot product of two vectors is a scalar.

Given a scalar valued function, $\displaystyle \phi(x,y,z)$, the "Laplacian", $\displaystyle \nabla^2\phi$ is the scalar valued function $\displaystyle \nabla^2\phi= \frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial y^2}+ \frac{\partial^2\phi}{\partial z^2}$. You can think of this as the dot product of the vector operator $\displaystyle \nabla$ with the vector valued $\displaystyle \nabla \phi$. The dot product of two vectors is a scalar.

Given a vector valued function, $\displaystyle \vec{F}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, the "curl" of $\displaystyle \vec{F}$ is $\displaystyle \nabla\times\vec{F}= \left(\frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$. You can think of this as the cross product of the vector operator, $\displaystyle \nabla$, and the vector valued function $\displaystyle \vec{F}$. The cross product of two vectors is a vector.
Notice that I have consistently used the variables (x, y, z) rather than $\displaystyle \vec{r}$. You mention, for the first, the "radial component". $\displaystyle \vec{r}$ is simply the position vector, not the "radial component" of the position (that would be just "r", not a vector). $\displaystyle \phi(\vec{r})$ and $\displaystyle \vec{F}(\vec{r})$ is the same as $\displaystyle \phi(x,y,z)$ and $\displaystyle \vec{F}(x,y,z)$ respectively.
Thank you so so so much. That was very clearly explained!

Happy holidays!
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