Posted on Categories:Thermodynamics, 热力学, 物理代写

# 物理代写|热力学代写Thermodynamics代考|PHY360 Back to Ohm

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## 物理代写|热力学代写Thermodynamics代考|Back to Ohm

We may obtain further information from Kirchhoff’s principle. Let $\mathbf{j}{e l}$ and $-2 \phi{e l}$ be the Lagrangian coordinate and the Lagrange multiplier, respectively (As for the Lagrange multipliers, see Sect. A.3). After integration by parts, therefore, Kirchhoff’s principle and Gauss’ theorem of divergence give:
\begin{aligned} 0=\delta \int(&\left.\frac{\left|\mathbf{j}{e l}\right|^2}{\sigma{\Omega}}-2 \phi_{e l} \nabla \cdot \mathbf{j}{e l}\right) d \mathbf{x}=2 \int\left(\frac{\mathbf{j}{e l} \cdot \delta \mathbf{j}{e l}}{\sigma{\Omega}}–\phi_{e l} \nabla \cdot \delta \mathbf{j}{e l}\right) d \mathbf{x}=\ =2 \int & {\left[\frac{\mathbf{j}{e l} \cdot \delta \mathbf{j}{e l}}{\sigma{\Omega}}-\nabla \cdot\left(\phi_{e l} \delta \mathbf{j}{e l}\right)+\delta \mathbf{j}{e l} \cdot \nabla \phi_{e l}\right] d \mathbf{x}=-2 \oint \phi_{e l} \delta \mathbf{j}{e l} d \mathbf{a}+} \ &+2 \int \delta \mathbf{j}{e l} \cdot\left(\frac{\mathbf{j}{e l}}{\sigma{\Omega}}+\nabla \phi_{e l}\right)=2 \int \delta \mathbf{j}{e l} \cdot\left(\frac{\mathbf{j}{e l}}{\sigma_{\Omega}}+\nabla \phi_{e l}\right) \end{aligned}
as the surface integral vanishes because $\delta \mathbf{j}{e l}=0$ on the boundary. The relationship above holds for arbitrary $\delta \mathbf{j}{e l}$, hence $\frac{\mathbf{j}{e l}}{\sigma{\Omega}}=-\nabla \phi_{e l}=\mathbf{E}+\mathbf{v} \wedge \mathbf{B}$, i.e. we retrieve Ohm’s law. We have tacitly assumed that $\delta \sigma_{\Omega}=0$ in the computation above. This assumption can e.g. justified under the assumption of uniform temperature, as $\sigma_{\Omega}=$ $0=\sigma_{\Omega}=O(T)$ in most cases. For $\phi_{e l} \rightarrow \phi_{e l}+\delta \phi_{e l}$ we retrieve the constraint $\nabla \cdot \mathbf{j}{e l}=0$. MinEP agrees with Kirchhoff’s principle for $T_1=T_2$ only. If $T_1 \neq T_2$, then Kirchhoff may still hold (it holds in each resistor separately as far as Ohm’s law holds and $\sigma{\Omega}$ is fixed in each resistor), in contrast with LNET. We discuss the stability of the state described by Kirchhoff’s principle below.

## 物理代写|热力学代写Thermodynamics代考|An Auxiliary Relationship

We may rewrite Kirchhoff’s principle in different forms. Let us replace $\mathbf{j}{e l}$ with $\mathbf{B}$ as Lagrangian coordinate. When performing such a change, the constraint too must involve the new Lagrangian coordinate. Maxwell’s equations of electromagnetism include $\nabla \cdot \mathbf{B}=0$ and lead to $\nabla \wedge \mathbf{B}=\mu_0 \mathbf{j}{e l}$ in the nonrelativistic limit $c \rightarrow \infty$. We define a Lagrange multiplier $\xi$ and write $\int\left[\frac{|\nabla \wedge \mathbf{B}|^2}{\mu_0^2 \sigma_{\Omega}}+\xi \nabla \cdot \mathbf{B}\right] d \mathbf{x}=\min$, i.e. ${ }^{16}$ :
$$\begin{gathered} 0=\delta \int\left[\frac{|\nabla \wedge \mathbf{B}|^2}{\mu_0^2 \sigma_{\Omega}}+\xi \nabla \cdot \mathbf{B}\right] d \mathbf{x}=\int\left[\frac{2(\nabla \wedge \mathbf{B}) \cdot(\nabla \wedge \delta \mathbf{B})}{\mu_0^2 \sigma_{\Omega}}+\xi \nabla \cdot \delta \mathbf{B}\right] d \mathbf{x}= \ =\frac{2}{\mu_0^2 \sigma_{\Omega}} \int \delta \mathbf{B} \cdot\left(\nabla \wedge \nabla \wedge \mathbf{B}+\nabla \xi^{\prime}\right) \quad ; \quad\left(\xi^{\prime} \equiv-\frac{\mu_0^2 \sigma_{\Omega} \xi}{2}\right) \end{gathered}$$

for arbitrary $\delta \mathbf{B}$, hence $\nabla \wedge \nabla \wedge \mathbf{B}+\nabla \xi^{\prime}=0$. We get rid of the unknown quantity $\xi^{\prime}$ by taking the curl of both sides ${ }^{17}$ and obtain a relationship involving $\mathbf{B}$ only:
$$\nabla \wedge \Delta \mathbf{B}=0$$
The usefulness of this result will be clear below. For the moment, we may say that we have found a condition satisfied by the generic magnetic field which is produced by a distribution of electric currents which satisfies Kirchhoff’s principle. Finally, we remark that the same relationship could also be obtained by taking the curl of Ohm’s law in the form $\frac{\mathbf{j}{s l}}{\sigma{\Omega}}+\nabla \phi_{e l}=0$ and in the nonrelativistic limit as far as $\nabla \sigma_{\Omega}=0$.

## 物理代写|热力学代写Thermodynamics代考|Back to Ohm

$$0=\delta \int\left(\frac{|\mathbf{j} e l|^2}{\sigma \Omega}-2 \phi_{e l} \nabla \cdot \mathbf{j} e l\right) d \mathbf{x}=2 \int\left(\frac{\mathbf{j} e l \cdot \delta \mathbf{j} e l}{\sigma \Omega}-\phi_{e l} \nabla \cdot \delta \mathbf{j} e l\right) d \mathbf{x}=2 \int\left[\frac{\mathbf{j} e l \cdot \delta \mathbf{j} e l}{\sigma \Omega}-\nabla \cdot\left(\phi_{e l} \delta \mathbf{j} e l\right)+\delta \mathbf{j} e l \cdot \nabla \phi_{e l}\right] d \mathbf{x}=-2 \oint \phi_{e l} \delta \mathbf{j} e l d \mathbf{a}++$$

## 物理代写热力学代写Thermodynamics代考|An Auxiliary Relationship

$$\nabla \wedge \Delta \mathbf{B}=0$$

## MATLAB代写

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