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I am trying to derive the $D$ field (or electric displacement field):

$$D=\varepsilon _{0}E+P$$

I am using the following relationships:

$$\rho=\rho_{f} + \rho_{b}$$

$$ \rho_{b} =-\nabla \cdot P$$

$$\rho_{f}= \nabla \cdot D$$

Here is what I understand starting with Gauss Law:

$$\nabla \cdot E=\frac{\rho }{\varepsilon _{0}}$$

$$\nabla \cdot E=\frac{\rho_{f} + \rho_{b} }{\varepsilon _{0}}$$

$$\nabla \cdot E=\frac{\rho_{f} -\nabla \cdot P}{\varepsilon _{0}}$$

$$\varepsilon _{0}\nabla \cdot E = \rho_{f} -\nabla \cdot P $$

$$ \varepsilon _{0}\nabla \cdot E +\nabla \cdot P = \rho_{f} $$

$$ \nabla \cdot(\varepsilon _{0}E+P) = \nabla \cdot D$$

I am confused on how to get from my final step to the equation for the D field. I know that the D field relies only on the free charge but there appears to be a gap in my of understanding of the math to arrive at the original equation. Any assistance is appreciated.

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3 Answers 3

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To summarize the above, you have proven that the divergence of two vector fields is the same, and you want to use that to prove that the vector fields themselves are equal.

A theorem by Helmholtz states the following: given a certain divergence and curl, there is a unique vector field that has that divergence and curl and also vanishes at infinity.* In other words, a well-behaved* vector field is uniquely described by its divergence and curl, so if you can prove that the divergence and curl of two well-behaved vector fields are the same, you know that the vector fields themselves are the same.

What you would need to do this is an expression for $\nabla\times\mathbf{E}$ and $\nabla\times\mathbf{D}$. Faraday's Law gives you the expression for $\nabla\times\mathbf{E}$, but it is almost never stated in terms of the electric displacement without already using the definition that you set out to prove. If you want to take $\nabla\times\mathbf{D}=-\varepsilon_0\frac{d\mathbf{B}}{dt}+\nabla\times\mathbf{P}$ as a given, then this will complete your argument, but this is not a particularly common assumption, and smacks of circular reasoning because of this.

In the electrostatic case, there actually is a commonly-used expression for the curl of both $\mathbf{E}$ and $\mathbf{D}$, namely, $\nabla\times\mathbf{E}=0$ and $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$ (source for the second: https://en.wikipedia.org/wiki/Electric_displacement_field). Therefore, in electrostatic situations,

$$\nabla\times(\varepsilon_0\mathbf{E}+\mathbf{P})=\varepsilon_0\cdot 0 + \nabla\times\mathbf{P}=\nabla\times\mathbf{D}$$

which completes the argument; since $\mathbf{D}$ and $\varepsilon_0\mathbf{E}+\mathbf{P}$ have the same divergence and curl, they are the same field by the above theorem by Helmholtz.

In any case, what you're trying to prove is usually taken as a definition, so it shouldn't be surprising that attempting to "derive" it using other assumptions is a bit tricky.

*If the vector field doesn't vanish at infinity, then it has to be at least twice differentiable over a given bounded domain. Either of these two conditions is what is meant by "well-behaved" above.

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Your last equation is $$ \nabla \cdot (\epsilon_0 \bf{E} + \bf{P}) = \nabla \cdot \bf{D} \tag{1} $$

Now, two objects $a$ and $b$ with the same divergence are the same thing up to an additive function which is a rotor of something:

$$ \nabla \cdot a = \nabla \cdot b \Rightarrow a=b + \nabla \times f \tag{2} $$ I think $\bf{D}$ is simply defined to be the thing between pharentesis in the left hand side of equation $(1)$, but as you can see, it follows from $(2)$ that $(1)$ holds even if you replace $\bf{D}$ with ${\bf{D'}} = {\bf{D}} + \nabla \times f$

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You don’t really “derive” the $\vec{D}$ field. It is, by definition, $\epsilon_0\vec{E}+\vec{P}$ which is motivated by the series of arguments you present above I.e. a field whose divergence is only free charge, usually the only information given in a problem. The permittivity constant is simply determined by experiment and defined as exactly this constant between charges and electric fluxes.

If you really wanted to “derive” it, you would have to, by Helmholtz theorem, also specific it’s curl to get a unique answer I.e. you must require $\vec{\nabla}\times\vec{B}=\vec{\nabla}\times\vec{P}$. Your current defining relationships don’t yield a unique answer.

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