# Finding the equation a potential must satisfy in order to minimize E

• Wavefunction
In summary, the equation for the electric potential that minimizes the total energy is given by the Poisson equation, which can be obtained by minimizing the average value of the square of the gradient of the potential over a given volume. This equation is intuitive as it tells us that in a region with no charge distribution, the EM field will hold the least amount of energy while still respecting the boundary conditions.
Wavefunction

## Homework Statement

Find an expression involving the function $ϕ(x_1, x_2, x_3)$ that has a minimum average value of the
square of its gradient within a certain volume V of space.

If ϕ is the electric potential, $\vec{E} = -\nabla ϕ$ is the electric field, and $ρ = \frac{1}{2} ϵ_0|\vec{E}\cdot\vec{E}|$ is the energy density of
the electric field, this result tells us what equation the electric potential must satisfy to minimize
the total energy.

## Homework Equations

(1)$\frac{∂f}{∂ϕ}+\sum_{i=1}^3\frac{∂}{∂x_i}\frac{∂f}{∂ϕ'_i} = 0$ where $ϕ'_i$ are the partial derivatives of $ϕ$

(2)min$(\frac{1}{V}\iiint ϕ dV) = \nabla ϕ \cdot \nabla ϕ$

## The Attempt at a Solution

First, since minimizing the average value of $ϕ$ yields a function $\nabla ϕ \cdot \nabla ϕ$ I will use this as my functional since it must satisfy (1). Doing so yields:

$\frac{∂f}{∂ϕ}= 0$ and $\nabla \cdot [\frac{∂(\nabla ϕ \cdot \nabla ϕ)}{∂(\nabla ϕ)}] = 0$

The second part of (1) yields:

$2\nabla \cdot \nabla ϕ =0$ or alternatively, $\nabla^2 ϕ = 0$ (Laplace's Equation)

Now this makes sense to me ( or at least tells me that I'm on the right track) since I know that the electrostatic potential does satisfy this equation; however, I know that Laplace's equation is a simpler case of Poisson's equation and when I plug my result in using $\vec{E} = -\nabla ϕ$ I get $\nabla \cdot \vec{E} = 0$ which is only true if the charge density inside of the object is zero. So here's my question should I have gotten the more general Poisson equation, or am I overthinking this whole thing?

Wavefunction said:

## Homework Statement

Find an expression involving the function $ϕ(x_1, x_2, x_3)$ that has a minimum average value of the
square of its gradient within a certain volume V of space.

If ϕ is the electric potential, $\vec{E} = -\nabla ϕ$ is the electric field, and $ρ = \frac{1}{2} ϵ_0|\vec{E}\cdot\vec{E}|$ is the energy density of
the electric field, this result tells us what equation the electric potential must satisfy to minimize
the total energy.

## Homework Equations

(1)$\frac{∂f}{∂ϕ}+\sum_{i=1}^3\frac{∂}{∂x_i}\frac{∂f}{∂ϕ'_i} = 0$ where $ϕ'_i$ are the partial derivatives of $ϕ$

(2)min$(\frac{1}{V}\iiint ϕ dV) = \iiint \nabla ϕ \cdot \nabla ϕ dV$

## The Attempt at a Solution

First, since minimizing the average value of $ϕ$ yields a function $\nabla ϕ \cdot \nabla ϕ$ I will use this as my functional since it must satisfy (1). Doing so yields:

$\frac{∂f}{∂ϕ}= 0$ and $\nabla \cdot [\frac{∂(\nabla ϕ \cdot \nabla ϕ)}{∂(\nabla ϕ)}] = 0$

The second part of (1) yields:

$2\nabla \cdot \nabla ϕ =0$ or alternatively, $\nabla^2 ϕ = 0$ (Laplace's Equation)

Now this makes sense to me ( or at least tells me that I'm on the right track) since I know that the electrostatic potential does satisfy this equation; however, I know that Laplace's equation is a simpler case of Poisson's equation and when I plug my result in using $\vec{E} = -\nabla ϕ$ I get $\nabla \cdot \vec{E} = 0$ which is only true if the charge density inside of the object is zero. So here's my question should I have gotten the more general Poisson equation, or am I overthinking this whole thing?

I think I have figured it out I actually need to use (2) as a constraint on (1):

$f+μg = \frac{\phi}{V} + μ\nabla \phi \cdot \nabla \phi$ and (1) becomes $\frac{∂(f+μg)}{∂\phi}+\frac{∂}{∂x_i}\frac{∂(f+μg)}{∂\phi'_i}$

The first term of (1) is just $\frac{1}{V}$ and the second term of (1) is the same as before except multiplied by μ.

Now I have: $\frac{1}{V}+μ\nabla^2\phi = 0 → \nabla^2\phi = \frac{-1}{μV}$

Because $μ$ is an arbitrary constant I'll define it to be: $μ \equiv \frac{ε_0}{Vρ}$

Then I get the Poisson equation: $\nabla^2\phi = \frac{-ρ}{ε_0}$

Now if I make the substitution $\vec{E} = -\nabla \phi$ I get the desired result:

$\nabla \cdot \vec{E} = \frac{ρ}{ε_0}$

I think your first answer was most likely what they were looking for. you minimised the average value of the square of the vector norm of the gradient of the potential, over a given volume. And it gave you Laplace's equation. Which is good. intuitively, it means that in a region with zero charge distribution, the EM field will hold the least amount of energy possible, while still respecting the boundary conditions.

## 1. What is the equation that a potential must satisfy in order to minimize E?

The equation that a potential must satisfy in order to minimize E is known as the Euler-Lagrange equation. It is a second-order partial differential equation that relates the potential to the energy of the system.

## 2. How is the Euler-Lagrange equation derived?

The Euler-Lagrange equation is derived from the principle of least action, which states that the path taken by a system between two points in time is the one that minimizes the action integral. The action integral is the integral of the Lagrangian, which is a function of the system's position and velocity.

## 3. What is the significance of minimizing E in a system?

Minimizing E in a system is important because it represents the most stable and energetically favorable state of the system. This is known as the equilibrium state, where the system has reached a balance between its potential and kinetic energy.

## 4. Can the Euler-Lagrange equation be applied to any system?

Yes, the Euler-Lagrange equation can be applied to any system that can be described by a Lagrangian function. This includes classical mechanics, quantum mechanics, and field theory.

## 5. Are there any limitations to the Euler-Lagrange equation?

While the Euler-Lagrange equation is a powerful tool for finding the equilibrium state of a system, it does have some limitations. It may not be applicable to systems with constraints or systems that are in a non-equilibrium state. Additionally, the equation may not have a unique solution in some cases.

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