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Mathematical Methods for Physicists: A concise introduction - Site Map

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VECTOR AND TENSOR ANALYSIS<br />

which shows that @A=@x are not the components of a tensor because of the<br />

second term on the right hand side. The same also applies to the di€erential of<br />

a contravariant vector. But we can construct a tensor by the following device.<br />

From Eq. (1.111) we have<br />

ˆ @x @x @ x <br />

<br />

@ x @ x @x ‡<br />

@2 x @ x <br />

@ x @ x <br />

@x :<br />

Multiplying (1.114) by A and subtracting from (1.113), we obtain<br />

<br />

@ A <br />

@ x A <br />

ˆ @A <br />

<br />

@x<br />

@x A <br />

@x <br />

<br />

@ x @ x :<br />

…1:114†<br />

…1:115†<br />

If we de®ne<br />

then (1.115) can be rewritten as<br />

A ; ˆ @A <br />

@x A ;<br />

…1:116†<br />

A ; ˆ A ;<br />

@x <br />

@ x @x <br />

@ x ;<br />

which shows that A ; is a covariant tensor of rank 2. This tensor is called the<br />

covariant derivative of A with respect to x . The semicolon denotes covariant<br />

di€erentiation. In a Cartesian coordinate system, the Christo€el symbols vanish,<br />

and so covariant di€erentiation reduces to ordinary di€erentiation.<br />

The contravariant derivative is found by raising the index which denotes di€erentiation:<br />

A ; ˆ g A ;:<br />

…1:117†<br />

We can similarly determine the covariant derivative of a tensor of arbitrary<br />

rank. In doing so we ®nd the following simple rule helps greatly:<br />

Thus,<br />

To obtain the covariant derivative of the tensor T with respect to<br />

x , we add to the ordinary derivative @T =@x <strong>for</strong> each covariant<br />

index …T:† a term T:, and <strong>for</strong> each contravariant index<br />

† a term ‡ T <br />

…T <br />

<br />

... .<br />

T ; ˆ @T <br />

@x T T ;<br />

T ; ˆ @T <br />

@x T ‡ T :<br />

The covariant derivatives of both the metric tensor and the Kronnecker delta<br />

are identically zero (Problem 1.38).<br />

56

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