Does the Laplace operator equal the Del operator squared?

[Safinaz]

TL;DR Summary

Dose Laplace operator $\Delta$ equal nabla operator squared $\bigtriangledown^2$ ?

Hello ,

The Laplace operator equals

$\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$

so does it equal as well nable or Del operator squared $\bigtriangledown^2$ ?

where

$\bigtriangledown =\frac{\partial}{\partial x} { \bf x} + \frac{\partial}{\partial y} { \bf y} + \frac{\partial}{\partial z } { \bf z}$

Edit:

Now about this equation$\Delta \sigma + \frac{1}{2} \partial_\mu \sigma \partial^\mu \sigma = \frac{3}{2} e^{-2\sigma} \partial_\mu h \partial^\mu h$

where $\sigma$ and $h$ are fields, and m and n are constants. I wonder how this equation to be solved to give $\sigma = m ~ln h$ as given by equations : (21), (23) in Paper

 

[Dr Transport]

$\Delta = \vec{\nabla}\cdot \vec{\nabla}$ is a notational convention

 

[vanhees71]

Indeed $\Delta=\vec{\nabla} \cdot \vec{\nabla}$. It's not a convention but can be derived from the meaning of the operators, which is independent on the chosen basis and coordinate system, because it is a vector operator. The nice thing of this fact is that you can always use the most convenient basis for your problem. For general properties of the various differential operators Cartesian coordinates are the most simple ones. So let $\vec{e}_k$ ($k \in \{1,2,3 \}$) be a right-handed Cartesian basis, for which $\vec{e}_k=\text{const}$ and $\vec{e}_1 \times \vec{e}_2=\vec{e}_3$.

Then the gradient of a scalar field is defined by
$$\vec{\nabla} \Phi=\vec{e}_k \partial_k \Phi,$$
where you have to some over $k=1 \ldots 3$ (Einstein summation convention). This is a vector field expressed with help of the Cartesian basis and its components wrt. this basis.

Now it is immediately clear that by taking the formal scalar product of $\vec{\nabla}=\vec{e}_k \partial_k$ with this vector field you get another scalar field,
$$\vec{\nabla} \cdot \vec{\nabla} \Phi = (\vec{e}_j \partial_j) \cdot (\vec{e}_k \partial_k \Phi) = (\vec{e}_j \cdot \vec{e}_k) \partial_j \partial_k \Phi = \delta_{jk} \partial_j \partial_k \Phi,$$
where
$$\delta_{jk}=\begin{cases} 1 & \text{for} \quad j=k, \\ 0 & \text{for} \quad j \neq k. \end{cases}$$
So finally you get
$$\vec{\nabla} \cdot \vec{\nabla} \Phi = \partial_j \partial_j \Phi = (\partial_1^2 + \partial_2^2 + \partial_3^2) \Phi=\Delta \Phi.$$

 

[Safinaz]

Indeed $\Delta=\vec{\nabla} \cdot \vec{\nabla}$. It's not a convention but can be derived from the meaning of the operators, which is independent on the chosen basis and coordinate system, because it is a vector operator. The nice thing of this fact is that you can always use the most convenient basis for your problem. For general properties of the various differential operators Cartesian coordinates are the most simple ones. So let $\vec{e}_k$ ($k \in \{1,2,3 \}$) be a right-handed Cartesian basis, for which $\vec{e}_k=\text{const}$ and $\vec{e}_1 \times \vec{e}_2=\vec{e}_3$.

Then the gradient of a scalar field is defined by
$$\vec{\nabla} \Phi=\vec{e}_k \partial_k \Phi,$$
where you have to some over $k=1 \ldots 3$ (Einstein summation convention). This is a vector field expressed with help of the Cartesian basis and its components wrt. this basis.

Now it is immediately clear that by taking the formal scalar product of $\vec{\nabla}=\vec{e}_k \partial_k$ with this vector field you get another scalar field,
$$\vec{\nabla} \cdot \vec{\nabla} \Phi = (\vec{e}_j \partial_j) \cdot (\vec{e}_k \partial_k \Phi) = (\vec{e}_j \cdot \vec{e}_k) \partial_j \partial_k \Phi = \delta_{jk} \partial_j \partial_k \Phi,$$
where
$$\delta_{jk}=\begin{cases} 1 & \text{for} \quad j=k, \\ 0 & \text{for} \quad j \neq k. \end{cases}$$
So finally you get
$$\vec{\nabla} \cdot \vec{\nabla} \Phi = \partial_j \partial_j \Phi = (\partial_1^2 + \partial_2^2 + \partial_3^2) \Phi=\Delta \Phi.$$

Hello, thanks for the answer. May you please see the edit to the question.