# Intro # Gradient and the Laplacian Sometimes, you'll be asked to take a double-gradient! This operator is known as the **Laplacian** and is given as: $\nabla^2 f = \frac{\delta^2 f}{\delta x^2} + \frac{\delta^2 f}{\delta y^2} + \frac{\delta^2 f}{\delta z^2}$ Which can also be written as: $\nabla^2 \cdot f = \frac{\delta^2 f}{\delta x^2} + \frac{\delta^2 f}{\delta y^2} + \frac{\delta^2 f}{\delta z^2}$ Where the dot product puts the 'vector' in vector calculus! # The Divergence and Curl Operators Remember from [[Surface and Line Integrals]] that the divergence operator $\mathrm{div} f$ is written as: $\mathrm{div} f = \frac{\delta f_{1}}{\delta x} \imath + \frac{\delta f_{2}}{\delta y} \jmath + \frac{\delta f_{3}}{\delta z}k $ Now, think about it. Doesn't this look a lot like a dot product between two vectors? Well, that's because it is. Cue it! $\mathrm{div} f = \nabla \cdot f$ If the Divergence Operator is the dot product between the Gradient of the vector field $f = P(x,y,z) \imath + Q(x,y,z) \jmath + R(x,y,z)k$, the *Curl* Operator is the **cross product** between the gradient and the function itself! $\nabla \times f = \begin{pmatrix} i & j & k \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z} \\ P(x,y,z) & Q(x,y,z) & R(x,y,z) \end{pmatrix}$ Now, we take the [[Matrices#The Determinant|determinant]] of this matrix to find the expression for the Curl operator: $\det\begin{pmatrix} i & j & k \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z} \\ P(x,y,z) & Q(x,y,z) & R(x,y,z) \end{pmatrix} = (\frac{\delta R}{\delta y} - \frac{\delta Q}{\delta z}) \imath + (\frac{\delta P}{\delta x} - \frac{\delta R}{\delta x}) \jmath + ( \frac{\delta Q}{\delta x} - \frac{\delta P}{\delta y}) k$ $\mathrm{curl} f = (\frac{\delta R}{\delta y} - \frac{\delta Q}{\delta z}) \imath + (\frac{\delta P}{\delta x} - \frac{\delta R}{\delta x}) \jmath + ( \frac{\delta Q}{\delta x} - \frac{\delta P}{\delta y}) k$ ## Geometrically Interpreting the Operators Now, let's first draw an unassuming vector field to act as our base. ![[divfield1.png]] Take this one, of the gravitational forces around the moon (for this field, the system includes the Earth, which is far away. NO, STOP ASKING QUESTIONS ABOUT THAT!!! PLEAS!!) That notwithstanding, let's try to take a point, $P$, and find the *divergence* and *curl* of the field at that point. ![[pointp.png]] Let's zoom in on the general 'field' around this point. ![[pointzoom.png]] The gradient of the field at point P is equivalent to the instantaneous change between the blue and green arrows. If we take the dot product of this gradient with that of the vector field at P, we get another vector, with magnitude that of how *perpendicular* the two vectors are. In other words, **if the direction of the vector changes a lot, the divergence of the field at that point P will be larger.** Conversely, if we choose to take the cross product between the gradient and the vector, we note that the smaller the angle between the gradient and vector field functions, the smaller the curl! Now, notice how the angle between the vectors gets larger the closer we get to our object, meaning that the curl of the field is higher the closer we are to its centre. ![[physicalanglecurl.png]] So, what can we get from these two statements? 1. **Divergence is a measure of how much a field 'diverges' from each other, or in other words, how fast each of the points within the field move!** 2. **Curl is a measure of how much a field 'encloses' itself, or how fast the points go into the field's center.** Got it? ## BONUS: Maxwell's Equations *Vector fields will call. Cut me some slack for now! Already got a lot to worry about.* $$