My Math Forum Scalar Fields vs Vector Fields

 Calculus Calculus Math Forum

 December 8th, 2017, 11:47 AM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics 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? Thanks!
 December 8th, 2017, 12:57 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. Thanks from SenatorArmstrong and idontknow Last edited by skipjack; December 8th, 2017 at 04:58 PM.
December 8th, 2017, 06:26 PM   #3
Senior Member

Joined: Nov 2015
From: United States of America

Posts: 198
Thanks: 25

Math Focus: Calculus and Physics
Quote:
 Originally Posted by Country Boy 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!

 Tags fields, scalar, vector

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Real Analysis 1 March 19th, 2014 04:45 AM Tooperoo Calculus 8 January 6th, 2014 06:02 PM arron1990 Calculus 7 August 9th, 2012 07:46 AM playthious Calculus 0 December 2nd, 2010 07:43 PM Altairian_dollar Abstract Algebra 1 August 31st, 2010 09:07 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top