13.07.2015 Views

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

TENSORS<strong>and</strong> sov i ; i = ∂vρ∂ρ + ∂vφ∂φ + ∂vz∂z + 1 ρ vρ= 1 ∂ρ ∂ρ (ρvρ )+ ∂vφ∂φ + ∂vz∂z .This result is identical to the expression <strong>for</strong> the divergence of a vector field in cylindricalpolar coordinates given in section 10.9. This is discussed further in section 26.20. ◭So far we have considered only the covariant derivative of the contravariantcomponents v i of a vector. The corresponding result <strong>for</strong> the covariant componentsv i may be found in a similar way, by considering the derivative of v = v i e i <strong>and</strong>using (26.77) to obtainv i; j = ∂v i∂u j − Γk ijv k . (26.88)Comparing the expressions (26.87) <strong>and</strong> (26.88) <strong>for</strong> the covariant derivative ofthe contravariant <strong>and</strong> covariant components of a vector respectively, we see thatthere are some similarities <strong>and</strong> some differences. It may help to remember thatthe index with respect to which the covariant derivative is taken (j in this case), isalso the last subscript on the Christoffel symbol; the remaining indices can thenbe arranged in only one way without raising or lowering them. It only remains tonote that <strong>for</strong> a covariant index (subscript) the Christoffel symbol carries a minussign, whereas <strong>for</strong> a contravariant index (superscript) the sign is positive.Following a similar procedure to that which led to equation (26.87), we mayobtain expressions <strong>for</strong> the covariant derivatives of higher-order tensors.◮By considering the derivative of the second-order tensor T with respect to the coordinateu k , find an expression <strong>for</strong> the covariant derivative T ij ; kof its contravariant components.Expressing T in terms of its contravariant components, we have∂T∂u = ∂k ∂u (T ij e k i ⊗ e j )ij∂T=∂u e k i ⊗ e j + T ij ∂e i∂u ⊗ e k j + T ij e i ⊗ ∂e j∂u . kUsing (26.75), we can rewrite the derivatives of the basis vectors in terms of Christoffelsymbols to obtainij∂T ∂T=∂uk ∂u e k i ⊗ e j + T ij Γ l ike l ⊗ e j + T ij e i ⊗ Γ l jke l .Interchanging the dummy indices i <strong>and</strong> l in the second term <strong>and</strong> j <strong>and</strong> l in the third termon the right-h<strong>and</strong> side, this becomes( )∂T ∂Tij∂u = +Γ i k ∂ulkT lj +Γ j k lk T il e i ⊗ e j ,970

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!