Chapter 10: Q1P (page 438)

Show that the differential equations for V and A (Eqs. 10.4 and 10.5) can be written in the more symmetrical form
$\begin{array}{r} \begin{matrix} 𝆏^{2} V + \begin{array}{r} \partial L \\ \partial t \end{array} = - \frac{1}{ε_{\circ}} p \\ 𝆏^{2} A - \nabla L = - μ_{\circ} J \end{matrix} \end{array}}$
Where
$𝆏^{2} \equiv \nabla^{2} - μ_{\circ} ε_{\circ} \frac{\partial^{2}}{\partial t^{2}} a n d L \equiv \nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}$

Short Answer

Expert verified

The differential equations for V and Ain the symmetrical form are derived as

$\nabla^{2} V + \frac{\partial}{\partial t} (\nabla . A) = - \frac{1}{ε_{\circ}} p a n d (\nabla^{2} A - μ_{\circ} ε_{\circ} \frac{\partial^{2} A}{\partial t}) = - μ_{\circ} J$

Step by step solution

Expression for the differential equations for V and A:

Using equation 10.6, write the differential equation for V.

$𝆏 \nabla^{2} V + \frac{\partial}{\partial t} = - \frac{1}{ε_{\circ}} p$ …… (1)

Here. $𝆏$ is d’ Alembertian.

Similarly, write the differential equation for A.

$𝆏^{2} A - \nabla L = - μ_{\circ} J$

Determine the differential equations for V and A in the symmetric form:

Substitute $𝆏^{2} = \nabla^{2} - μ_{\circ} ε_{\circ} \frac{\partial^{2}}{\partial t^{2}} and L = \nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}$ and in equation (1).

$(𝆏^{2} - μ_{\circ} ε_{\circ} \frac{\partial^{2}}{\partial t^{2}}) V + \frac{\partial}{\partial t} (\nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}) = - \frac{1}{ε_{\circ}} p \nabla^{2} V - μ_{\circ} ε_{\circ} \frac{\partial^{2} V}{\partial t^{2}} + \frac{\partial}{\partial t} (\nabla . A) + μ_{\circ} ε_{\circ} \frac{\partial^{2} V}{\partial t^{2}} = - \frac{1}{ε_{\circ}} p \nabla^{2} V + \frac{\partial}{\partial t} (\nabla . A) = - \frac{1}{ε_{\circ}} p$

Which is equal to the equation 10.4 as $\nabla^{2} V + \frac{\partial}{\partial t} (\nabla . A) = - \frac{1}{ε_{\circ}} p$ .

Substitute $𝆏^{2} = \nabla^{2} - μ_{\circ} ε_{\circ} \frac{\partial^{2}}{\partial t^{2}} and L = \nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}$ in equation (2).

$(\nabla^{2} - μ_{\circ} ε_{\circ} \frac{\partial^{2}}{\partial t^{2}}) A - \nabla (\nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}) = - μ_{\circ} J (\nabla^{2} A - μ_{\circ} ε_{\circ} \frac{\partial^{2} A}{\partial t^{2}}) - \nabla (\nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}) = - μ_{\circ} J$

Which is equal to the equation 10.5 as .

$(\nabla^{2} A - μ_{\circ} ε_{\circ} \frac{\partial^{2} A}{\partial t^{2}}) - \nabla (\nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}) = - μ_{\circ} J$

Therefore, the differential equations for V and Ain the symmetrical form are derived as $\nabla^{2} V + \frac{\partial}{\partial t} (\nabla . A) = - \frac{1}{ε_{\circ}} p$ and .

$(\nabla^{2} A - μ_{\circ} ε_{\circ} \frac{\partial^{2} A}{\partial t^{2}}) - \nabla (\nabla . A + μ_{\circ} ε_{\circ} \frac{\partial V}{\partial t}) = - μ_{\circ} J$

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Short Answer

Step by step solution

Expression for the differential equations for V and A:

Determine the differential equations for V and A in the symmetric form:

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Show that the differential equations for V and A (Eqs. 10.4 and 10.5) can be written in the more symmetrical form𝆏2V+∂L∂t=-1ε∘p𝆏2A-∇L=-μ∘J}Where𝆏2≡∇2-μ∘ε∘∂2∂t2andL≡∇.A+μ∘ε∘∂V∂t

Short Answer

Step by step solution

Expression for the differential equations for V and A:

Determine the differential equations for V and A in the symmetric form:

One App. One Place for Learning.

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