Q56P You may have noticed that the fo... [FREE SOLUTION]

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Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 12: Q56P (page 570)

You may have noticed that the four-dimensional gradient operator $鈭� / 鈭� x^{渭}$ functions like a covariant 4-vector鈥攊n fact, it is often written ${鈭�}_{渭}$ , for short. For instance, the continuity equation, ${鈭�}_{渭} J^{渭} = 0$ , has the form of an invariant product of two vectors. The corresponding contravariant gradient would be ${鈭�}^{渭} 鈮� 鈭� / 鈭� x_{渭}$ . Prove that ${鈭�}^{渭} f$ is a (contravariant) 4-vector, if $蠒$ is a scalar function, by working out its transformation law, using the chain rule.

Short Answer

Expert verified

It is proved that ${鈭�}^{渭} 蠒$ is a contravariant 4-vector.

Step by step solution

Expression for the value of ∂0ϕ¯:

Write the expression for the value of $\overset{炉}{{鈭�}^{0} 蠒}$ .

role="math" localid="1655877718000" $\begin{matrix} \overset{炉}{{鈭�}^{0} 蠒} = \frac{鈭�}{鈭� {\overset{炉}{x}}_{0}} 蠒 \\ \overset{炉}{{鈭�}^{0} 蠒} = 鈭� \frac{1}{c} \frac{鈭�}{鈭� t} 蠒 \\ \overset{炉}{{鈭�}^{0} 蠒} = 鈭� \frac{1}{c} (\frac{鈭� 蠒}{鈭� t} \frac{鈭� t}{鈭� t} + \frac{鈭� 蠒}{鈭� x} \frac{鈭� x}{鈭� t} + \frac{鈭� 蠒}{鈭� y} \frac{鈭� y}{鈭� t} + \frac{鈭� 蠒}{鈭� z} \frac{鈭� z}{鈭� t}) \end{matrix}$ 鈥︹€�. (1)

Determine the value of ∂0ϕ¯

It is known that:

$t = 纬 (\overset{炉}{t} + \frac{v}{c} \overset{炉}{x})$

Here, v is the velocity and c is the speed of light.

Using equation 12.19, write the expression for the transformation equations.

$\begin{matrix} \frac{鈭� t}{鈭� t} = 纬 \\ x = 纬 (\overset{炉}{x} + v \overset{炉}{t}) \\ \frac{鈭� x}{鈭� t} = 纬 v \\ y = \overset{炉}{y} \end{matrix}$

It is also known that:

$\begin{matrix} z = \overset{炉}{z} \\ \frac{鈭� y}{鈭� t} = 0 \\ \frac{鈭� z}{鈭� t} = 0 \end{matrix}$

Substitute $\frac{鈭� t}{鈭� t} = 纬$ , $x = 纬 (\overset{炉}{x} + v \overset{炉}{t})$ , $\frac{鈭� x}{鈭� t} = 纬 v$ , $\frac{鈭� y}{鈭� t} = 0$ and $\frac{鈭� z}{鈭� t} = 0$ in equation (1)l.

$\begin{matrix} \overset{炉}{{鈭�}^{0} 蠒} = 鈭� \frac{1}{c} (\frac{鈭� 蠒}{鈭� t} (纬) + \frac{鈭� 蠒}{鈭� x} (纬 v) + \frac{鈭� 蠒}{鈭� y} (0) + \frac{鈭� 蠒}{鈭� z} (0)) \\ \overset{炉}{{鈭�}^{0} 蠒} = 鈭� \frac{1}{c} (\frac{鈭� 蠒}{鈭� t} (纬) + \frac{鈭� 蠒}{鈭� x} (纬 v)) \\ \overset{炉}{{鈭�}^{0} 蠒} = 鈭� \frac{1}{c} 纬 (鈭� \frac{鈭� 蠒}{鈭� t} + v \frac{鈭� 蠒}{鈭� x}) \\ \overset{炉}{{鈭�}^{0} 蠒} = 纬 [\frac{鈭� 蠒}{鈭� x_{0}} 鈭� \frac{v}{c} \frac{鈭� 蠒}{鈭� x^{'}}] \end{matrix}$

Here,

$尾 = \frac{v}{c}$

On further solving,

$\overset{炉}{{鈭�}^{0} 蠒} = 纬 [{鈭�}^{0} 蠒鈭� 尾 ({鈭�}^{1} 蠒)]$

Prove that ∂μϕ is a contravariant 4-vector:

Calculate the value of $\overset{炉}{{鈭�}^{1} 蠒}$ .

$\begin{matrix} \overset{炉}{{鈭�}^{1} 蠒} = \frac{鈭� 蠒}{鈭� x} \\ \overset{炉}{{鈭�}^{1} 蠒} = \frac{鈭� 蠒}{鈭� t} \frac{鈭� t}{鈭� \overset{炉}{x}} + \frac{鈭� 蠒}{鈭� x} \frac{鈭� x}{鈭� \overset{炉}{x}} + \frac{鈭� 蠒}{鈭� y} \frac{鈭� y}{鈭� \overset{炉}{x}} + \frac{鈭� 蠒}{鈭� z} \frac{鈭� z}{鈭� \overset{炉}{x}} \end{matrix}$ 鈥︹€� (2)

It is known that:

$\begin{matrix} \overset{炉}{x} = 纬 (\overset{炉}{x} + v \overset{炉}{t}) \\ t = 纬 (\overset{炉}{t} + \frac{v}{c^{2}} \overset{炉}{x}) \\ \frac{鈭� x}{鈭� \overset{炉}{x}} = 纬, \frac{鈭� y}{鈭� \overset{炉}{x}} = 0 \\ \frac{鈭� t}{鈭� \overset{炉}{x}} = \frac{v}{c^{2}} 纬, \frac{鈭� z}{鈭� \overset{炉}{x}} = 0 \end{matrix}$

Substitute $\frac{鈭� x}{鈭� \overset{炉}{x}} = 纬$ , $\frac{鈭� y}{鈭� \overset{炉}{x}} = 0$ , $\frac{鈭� t}{鈭� \overset{炉}{x}} = \frac{v}{c^{2}} 纬$ and $\frac{鈭� z}{鈭� \overset{炉}{x}} = 0$ in equation (2).

$\begin{matrix} \overset{炉}{{鈭�}^{1} 蠒} = \frac{鈭� 蠒}{鈭� t} (\frac{v}{c^{2}} 纬) + \frac{鈭� 蠒}{鈭� x} (纬) + \frac{鈭� 蠒}{鈭� y} (0) + \frac{鈭� 蠒}{鈭� z} (0) \\ \overset{炉}{{鈭�}^{1} 蠒} = \frac{鈭� 蠒}{鈭� t} (\frac{v}{c^{2}} 纬) + \frac{鈭� 蠒}{鈭� x} (纬) \\ \overset{炉}{{鈭�}^{1} 蠒} = 纬 [\frac{v}{c^{2}} \frac{鈭� 蠒}{鈭� t} + \frac{鈭� 蠒}{鈭� x}] \\ \overset{炉}{{鈭�}^{1} 蠒} = 纬 [({鈭�}^{'} 蠒) 鈭� 尾 ({鈭�}^{0} 蠒)] \end{matrix}$

Calculate the value of $\overset{炉}{{鈭�}^{2} 蠒}$ .

$\begin{matrix} \overset{炉}{{鈭�}^{2} 蠒} = \frac{鈭� 蠒}{鈭� \overset{炉}{y}} \\ \overset{炉}{{鈭�}^{2} 蠒} = \frac{鈭� 蠒}{鈭� t} \frac{鈭� t}{鈭� \overset{炉}{y}} + \frac{鈭� 蠒}{鈭� x} \frac{鈭� x}{鈭� \overset{炉}{y}} + \frac{鈭� 蠒}{鈭� y} \frac{鈭� y}{鈭� \overset{炉}{y}} + \frac{鈭� 蠒}{鈭� z} \frac{鈭� z}{鈭� \overset{炉}{y}} \\ \overset{炉}{{鈭�}^{2} 蠒} = {鈭�}^{2} 蠒 \end{matrix}$

Calculate the value of $\overset{炉}{{鈭�}^{3} 蠒}$ .

$\begin{matrix} \overset{炉}{{鈭�}^{3} 蠒} = \frac{鈭� 蠒}{鈭� \overset{炉}{z}} \\ \overset{炉}{{鈭�}^{3} 蠒} = \frac{鈭� 蠒}{鈭� t} \frac{鈭� t}{鈭� \overset{炉}{z}} + \frac{鈭� 蠒}{鈭� x} \frac{鈭� x}{鈭� \overset{炉}{z}} + \frac{鈭� 蠒}{鈭� y} \frac{鈭� y}{鈭� \overset{炉}{z}} + \frac{鈭� 蠒}{鈭� z} \frac{鈭� z}{鈭� \overset{炉}{z}} \\ \overset{炉}{{鈭�}^{3} 蠒} = \frac{鈭� 蠒}{鈭� z} \end{matrix}$

Therefore, ${鈭�}^{渭} 蠒$ is a contravariant 4-vector.

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Short Answer

Step by step solution

Expression for the value of ∂0ϕ¯:

Determine the value of ∂0ϕ¯

Prove that ∂μϕ is a contravariant 4-vector:

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