laplacian of a scalar field
At first sight, it looks like we have a very big problem here. hence yz\,\text{d}S =0 \end{gather*}\], \[\begin{align*} \iint_{S_\varepsilon}\vecs{v} (x,y,z)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}S &= \frac{1}{\varepsilon} \iint_{S_\varepsilon}\big( \textbf{B}\cdot\hat{\pmb{\imath}}\,x^2 +\textbf{C}\cdot\hat{\pmb{\jmath}}\,y^2 +\textbf{D}\cdot\hat{\mathbf{k}}\, z^2 \big)\,\text{d}S\\ &\hskip1in +\frac{1}{\varepsilon} \iint_{S_\varepsilon}\textbf{R}(x,y,z)\cdot (x,y,z)\,\text{d}S \end{align*}\], As well \(S_\varepsilon\) is invariant15under the interchange of \(x\) and \(y\) and also under the interchange of \(x\) and \(z\text{. By precisely the same argument, we could come up with another vector potential whose second component is zero, and with a third vector potential whose first component is zero. In fact, here are a very large number of them. }\) Compute and simplify the following quantities. Oof! We have, \[\begin{gather*} \vecs{ \nabla} \cdot\vecs{v} (\vecs{0}) = 3 \end{gather*}\]. The vector identity Theorem 4.1.7.a is indeed another screening test. Note that the operator is commonly written as by mathematicians (Krantz 1999, p. 16). }\), If you are at \(\vecs{r} \text{,}\) the component of the fluid velocity in your direction of motion, i.e. The divergence, \[\begin{gather*} \vecs{ \nabla} \cdot\vecs{v} (\vecs{0}) = 0 \end{gather*}\], Here is a sketch of the vector field \(\vecs{v} (x,y,z) = \hat{\pmb{\imath}}\) and a sphere centered on the origin, like \(S_\varepsilon\text{. 0. }\) So that term is \((\vecs{ \nabla} f)\cdot\vecs{F} \text{. and so passes the screening test of Theorem 4.1.12. We have put in the minus sign just to provide compatibility with the usual physics terminology. dU/dx, second derivative of U(x) at the point M(x) is the slope of the derivative U'(x)=dU/dx at this point. }\) Let, \[ \vecs{r} (t)=tx\,\hat{\pmb{\imath}}+ty\,\hat{\pmb{\jmath}}+tz\,\hat{\mathbf{k}},\qquad 0\le t\le 1 \nonumber \], be a parametrization of the line segment from the origin to \((x,y,z)\text{. Taking the divergence of both sides of Eq. }\) Experiments also show that \(\vecs{F} \) must be of the form, \[ \vecs{F} (x,y,z)=(xz+xy)\hat{\pmb{\imath}}+\alpha(yz-xy)\hat{\pmb{\jmath}}+\beta(yz+xz)\hat{\mathbf{k}} \nonumber \]. 3,681 7 39 68 The Laplacian takes a scalar valued function and gives back a scalar valued function. When \(\vecs{F} = \vecs{ \nabla} \times\textbf{A}\text{,}\) \(\textbf{A}\) is called a vector potential for \(\vecs{F} \text{. Now, use the relation (A B) = (A) B + (B) A ( A B) = ( A) B + ( B) A and get To end up with a scalar, rather than a vector, we must take the dot product of \(\vecs{ \nabla} f\) and \(\vecs{F} \text{. We'll now develop the interpretation of the curl, or more precisely, of \(\vecs{ \nabla} \times\vecs{v} (\vecs{r} _0)\cdot\hat{\textbf{n}}\) for any unit vector \(\hat{\textbf{n}}\text{. Two vector fields \(\textbf{A}\) and \(\tilde{\textbf{A}}\) are both vector potentials for the same vector field if and only if, \[ \vecs{ \nabla} \times\textbf{A}=\vecs{ \nabla} \times\tilde{\textbf{A}} \iff \vecs{ \nabla} \times(\textbf{A}-\tilde{\textbf{A}})=\vecs{0} \nonumber \], That is, if and only if the difference \(\textbf{A}-\tilde{\textbf{A}}\) passes the conservative field screening test of Theorems 2.3.9 and 2.4.8. }\), We can use exactly the strategy of the last examples to prove, Let \(\textbf{B}\) be a vector field that is defined and has all of its first order partial derivatives continuous on all of \(\mathbb{R}^3\text{. }\), \(\displaystyle \vecs{ \nabla} \cdot\vecs{r} \), \(\displaystyle \vecs{ \nabla} \times(\vecs{r} \times\textbf{a})\), \(\displaystyle \vecs{ \nabla} \cdot\big(\vecs{ \nabla} (r)\big)\), Compute \(a\) where \(\vecs{ \nabla} \big(\frac{1}{r}\big) =- r^a\,\vecs{r} \text{. This page titled 4.1: Gradient, Divergence and Curl is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. }\) That arc has length \(\Omega\varepsilon\text{. What does this value tell us about the field or it's behaviour in the given spot? The du/dx and du/dy field are what I expect. In a poll of prominent physicists, Maxwell was voted the third greatest physicist of all time. Example 4.17 Let r(x, y, z) = xi + yj + zk be the position vector field on R3. All paddles have to move at the same speed. }\) There should be no \(x\)'s, \(y\)'s, or \(z\)'s in your answers. }\) This is the rate of change per unit time. Its name pays homage to the work done by the mathematician, astronomer and French physicist Pierre-Simon Laplace (1749-1827). So the derivative must be a scalar, i.e. So the paddlewheel should not rotate at all. That is, it should have positive angular velocity. Compute the Laplacian of this function with respect to the vector ( x, y, z). For the second proof, we again write out both sides, but this time we express them in terms of determinants. We write the Laplacian of a scalar field U As an example of an application in which both the divergence and curl appear, we have Maxwell's equations345, which form the foundation of classical electromagnetism. \[\begin{align*} \dfrac{d\ }{dx}\big(af(x)+bg(x)\big) &=a\dfrac{df}{dx}(x)+b\dfrac{dg}{dx}(x)\\ \dfrac{d\ }{dx}\big(f(x)\,g(x)\big) &=g(x)\,\dfrac{df}{dx}(x)+f(x)\,\dfrac{dg}{dx}(x) \end{align*}\]. Exchanging two rows in a determinant changes the sign of the determinant. That is, \(\textbf{A}+\vecs{ \nabla} \psi\) is another vector potential for \(\textbf{B}\text{. We havea }\) If that is not the case, for example because the vector field is not defined on all of \(\mathbb{R}^3\text{,}\) then \(\textbf{B}\) can fail to have a vector potential. }\) In particular there is a constant \(K\) so that, on \(S_\varepsilon\text{,}\), \[ |\textbf{R}(x,y,z)|\le K\varepsilon^2 \nonumber \], \[\begin{align*} &\iint_{S_\varepsilon}\vecs{v} (x,y,z)\cdot\hat{\textbf{n}}(x,y,z)\,\text{d}S\\ &\hskip1in= \frac{1}{\varepsilon} \iint_{S_\varepsilon}\big( \textbf{A} + \textbf{B}\,x +\textbf{C}\,y +\textbf{D}\, z +\textbf{R}(x,y,z) \big)\cdot (x,y,z)\,\text{d}S \end{align*}\], Multiply out the dot product so that the integrand becomes, \[\begin{alignat*}{3} &\phantom{+}\ \textbf{A}\cdot\hat{\pmb{\imath}}\,x &&+\ \textbf{A}\cdot\hat{\pmb{\jmath}}\,y &&+\textbf{A}\cdot\hat{\mathbf{k}}\,z\\ &+\textbf{B}\cdot\hat{\pmb{\imath}}\,x^2 &&+\ \textbf{B}\cdot\hat{\pmb{\jmath}}\,xy &&+\textbf{B}\cdot\hat{\mathbf{k}}\,xz\\ &+\textbf{C}\cdot\hat{\pmb{\imath}}\,xy &&+\ \textbf{C}\cdot\hat{\pmb{\jmath}}\,y^2 &&+\textbf{C}\cdot\hat{\mathbf{k}}\,yz\\ &+\textbf{D}\cdot\hat{\pmb{\imath}}\,xz\ &&+\ \textbf{D}\cdot\hat{\pmb{\jmath}}\,yz &&+\textbf{D}\cdot\hat{\mathbf{k}}\,z^2\\ &+\textbf{R}(x,y,z)\cdot (x,y,z) \end{alignat*}\], That's a lot of terms. We will later see that each has a physical significance. tangential to \(C_\varepsilon\text{,}\) is \(\vecs{v} (\vecs{r} )\cdot\dfrac{d\vecs{r} }{ds}\text{,}\) because \(\hat{\textbf{t}}=\dfrac{d\vecs{r} }{ds}\text{,}\) with \(s\) denoting arc length along the circle, is a unit vector tangential to \(C_\varepsilon\text{.}\). He was Napoleon's examiner when Napoleon attended the Ecole Militaire in Paris. James Clerk Maxwell (18311879) was a Scottish mathematical physicist. An important application of the . Let \(\vecs{F} = P\,\hat{\pmb{\imath}} + Q\,\hat{\pmb{\jmath}}\) be the two dimensional vector field shown below. Stack Exchange Network. First, here are the statements of a bunch of them. So we are free to eliminate one of the unknowns by setting, for example, \(M=0\text{,}\) and then choose any \(N\) that obeys, \[ \frac{\partial N}{\partial x}(x,y) = xy \nonumber \], Integrating with respect to \(x\) gives, as one possible choice, \(N(x,y) = \frac{x^2y}{2}\text{. }\), This velocity field has fluid going around in circles, counterclockwise. }\) Typically, we can easily solve one linear equation in one unknown. \(\displaystyle \textbf{a}\cdot(\textbf{b}\times\textbf{c})=(\textbf{a}\times\textbf{b})\cdot\textbf{c}\), \(\displaystyle \textbf{a}\times(\textbf{b}\times\textbf{c})=(\textbf{c}\cdot\textbf{a})\textbf{b}-(\textbf{b}\cdot\textbf{a})\textbf{c}\). Definition of Laplacian In Euclidean space, the Laplacian is defined as the divergence of the gradient of a scalar function f f that can be twice differentiable. (b) The proof is not exceptionally difficult just write out both sides and grind. It represents the infinitesimal variation of U(x) relative to an infinitesimal change in x at this point. Rename the coordinates \(x,y,z\) to \(x_1,x_2,x_3\) and the standard unit basis vectors \(\hat{\pmb{\imath}}\text{,}\) \(\hat{\pmb{\jmath}}\text{,}\) \(\hat{\mathbf{k}}\) to \(\hat{\pmb{\imath}}_1\text{,}\) \(\hat{\pmb{\imath}}_2\text{,}\) \(\hat{\pmb{\imath}}_3\text{. You can't apply a vector laplacian to a scalar field. Two computationally extremely important properties of the derivative \(\dfrac{d\ }{dx}\) are linearity and the product rule. Let \(\vecs{r} (x,y,z) = x\,\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}+z\,\hat{\mathbf{k}}\) and let \(\psi(x,y,z)\) be an arbitrary function. xy\,\text{d}S =\iint_{S_\varepsilon}\!\! and is called del or nabla. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, . Substituting in, \[ \textbf{b}\times\textbf{c} \ =\ (b_2c_3-b_3c_2)\hat{\pmb{\imath}}-(b_1c_3-b_3c_1)\hat{\pmb{\jmath}} + (b_1c_2-b_2c_1)\hat{\mathbf{k}} \nonumber \], \[\begin{align*} \textbf{a}\times(\textbf{b}\times\textbf{c}) =\phantom{-}&\!\!\!\det\left[\begin{matrix}\hat{\pmb{\imath}}&\hat{\pmb{\jmath}} &\hat{\mathbf{k}} \\ a_1&a_2&a_3 \\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1 \end{matrix}\right]\\ =\phantom{-}&\hat{\pmb{\imath}}\big[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)\big]\\ -&\hat{\pmb{\jmath}}\big[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)\big]\\ +&\hat{\mathbf{k}}\big[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)\big] \end{align*}\], \[\begin{align*} (\textbf{a}\cdot\textbf{c})\textbf{b}-(\textbf{a}\cdot\textbf{b})\textbf{c}\ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hat{\pmb{\imath}}+b_2\hat{\pmb{\jmath}}+b_3\hat{\mathbf{k}})\\ &\hskip0.5in-(a_1b_1+a_2b_2+a_3b_3)(c_1\hat{\pmb{\imath}}+c_2\hat{\pmb{\jmath}}+c_3\hat{\mathbf{k}})\\ =\ & \hat{\pmb{\imath}}\ \big[{\color{blue}{a_1b_1c_1}} +a_2b_1c_2+a_3b_1c_3- {\color{blue}{a_1b_1c_1}} -a_2b_2c_1-a_3b_3c_1\big]\\ {+}&\hat{\pmb{\jmath}}\ \big[a_1b_2c_1 +{\color{blue}{a_2b_2c_2}} +a_3b_2c_3-a_1b_1c_2 -{\color{blue}{a_2b_2c_2}} -a_3b_3c_2\big]\\ {+}&\hat{\mathbf{k}}\ \big[a_1b_3c_1+a_2b_3c_2 +{\color{blue}{a_3b_3c_3}} -a_1b_1c_3-a_2b_2c_3 -{\color{blue}{a_3b_3c_3}}\big]\\ {=}\ & \hat{\pmb{\imath}}\ [a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1]\\ {+}&\hat{\pmb{\jmath}}\ [a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2]\\ {+}&\hat{\mathbf{k}}\ [a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] \end{align*}\]. }\) In particular there is a constant \(K\) so that, on \(C_\varepsilon\text{,}\), \[\begin{align*} &\oint_{C_\varepsilon} \vecs{v} (\vecs{r} )\cdot \text{d}\vecs{r} \\ &\hskip0.25in=\int_0^{2\pi} \big(\textbf{A} + \textbf{B}\,\varepsilon\,\cos t +\textbf{C}\,\varepsilon\,\sin t +\textbf{R}(\vecs{r} (t))\big) \cdot \big(-\varepsilon\sin t\,\hat{\pmb{\imath}} +\varepsilon\cos t\,\hat{\pmb{\jmath}}\big)\ \text{d}t \end{align*}\], Again, multiply out the dot product so that the integrand becomes, \[\begin{alignat*}{2} & -\varepsilon\textbf{A}\cdot\hat{\pmb{\imath}}\,\sin t &&+\ \varepsilon\textbf{A}\cdot\hat{\pmb{\jmath}}\,\cos t\\ &-\varepsilon^2\textbf{B}\cdot\hat{\pmb{\imath}}\,\sin t\cos t &&+\varepsilon^2\textbf{B}\cdot\hat{\pmb{\jmath}}\,\cos^2t\\ &-\varepsilon^2\textbf{C}\cdot\hat{\pmb{\imath}}\,\sin^2 t &&+\ \varepsilon^2\textbf{C}\cdot\hat{\pmb{\jmath}}\,\sin t\cos t\\ &+\textbf{R}(\vecs{r} (t))\cdot \big(-\varepsilon\sin t\,\hat{\pmb{\imath}} +\varepsilon\cos t\,\hat{\pmb{\jmath}}\big) \end{alignat*}\], Again most of these terms integrate to zero, because, \[\begin{alignat*}{2} \int_0^{2\pi}\sin t\ \text{d}t &=\hskip10pt\int_0^{2\pi}\cos t\ \text{d}t &&=0\\ \int_0^{2\pi}\sin t\cos t\ \text{d}t &= \frac{1}{2}\int_0^{2\pi}\sin(2t)\ \text{d}t &&=0 \end{alignat*}\], and the \(\sin^2t\) and \(\cos^2 t\) terms are easily integrated using (see Example 2.4.4), \[ \int_0^{2\pi}\sin^2 t\ \text{d}t=\int_0^{2\pi}\cos^2 t\ \text{d}t =\frac{1}{2}\int_0^{2\pi}\big[\sin^2t+\cos^2 t\big]\ \text{d}t=\pi \nonumber \], \[\begin{align*} \oint_{C_\varepsilon} \vecs{v} (\vecs{r} )\cdot \text{d}\vecs{r} &= \pi\varepsilon^2\textbf{B}\cdot\hat{\pmb{\jmath}} - \pi\varepsilon^2\textbf{C}\cdot\hat{\pmb{\imath}} +\int_0^{2\pi} \textbf{R}(\vecs{r} (t)) \cdot \big(-\varepsilon\sin t\,\hat{\pmb{\imath}} +\varepsilon\cos t\,\hat{\pmb{\jmath}}\big)\ \text{d}t \end{align*}\], \[\begin{align*} \lim_{\varepsilon\rightarrow 0}\frac{1}{\pi\varepsilon^2} \oint_{C_\varepsilon} \vecs{v} (\vecs{r} )\cdot \text{d}\vecs{r} &= \frac{\partial \vecs{v} _2}{\partial x}(0,0,0) - \frac{\partial \vecs{v} _1}{\partial y}(0,0,0)\\ &\hskip0.5in +\lim_{\varepsilon\rightarrow 0}\! }\), All of the proofs (except for those of Theorem 4.1.7.c,d, which we will return to later) consist of, For Theorem 4.1.3.a,b, Theorem 4.1.4.a,b, Theorem 4.1.5.a,b and Theorem 4.1.6.a,b, the computation is trivial one line per identity, if one uses some efficient notation. }\) We shall see in Theorem 4.1.16, below, that, conversely, if \(\vecs{F} (\textbf{x})\) is defined and has continuous first order partial derivatives on all of \(\mathbb{R}^3\text{,}\) and if \(\vecs{ \nabla} \cdot\vecs{F} =0\text{,}\) then \(\vecs{F} \) has a vector potential8. }\) Imagine standing on the circle with your feet on the plane through \(\vecs{r} _0\) perpendicular to \(\hat{\textbf{n}}\text{,}\) with the vector from your feet to your head in the same direction as \(\hat{\textbf{n}}\) and with your left arm point towards \(\vecs{r} _0\text{. }\) We shall do so in two steps. }\), Further experiments show that \(\textbf{G}=xyz\hat{\pmb{\imath}}-xyz\hat{\pmb{\jmath}}+g(x,y,z)\hat{\mathbf{k}}\text{. }\) But it appears like the right hand side depends on \(z\) too. }\), This velocity field just has fluid moving uniformly to the right. }\) Express your answer in terms of \(r\text{. We can extend this reasoning to a two-dimesnional scalar field U(x, y). Because \(S_\varepsilon\) is invariant under \(x\rightarrow -x\) and under \(y\rightarrow -y\) and under \(z\rightarrow -z\) we have, \[\begin{gather*} \iint_{S_\varepsilon}\!\! In contrast, the gradient operator acts on a scalar field to produce a vector field. The vector field \(\textbf{A}\) is said to be a vector potential for the vector field \(\textbf{B}\) if, \[ \textbf{B}=\vecs{ \nabla} \times\textbf{A} \nonumber \]. 2A = x2Ax + y2Ay + z2Az. writing out the definition of the left hand side and, writing out the definition of the right hand side and. When evaluating Laplacian of some scalar field at a given point one can get a value. CLP-4 Vector Calculus (Feldman, Rechnitzer, and Yeager), { "4.01:_Gradient_Divergence_and_Curl" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.
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