Covariant derivative and general relativity

[cristo]

I'm not really sure where to put this, so I thought it post it here!

I'm reading through my GR lecture notes, and have come across a comment that has confused me. I quote

Covariant differentiation does not change the character of the object being differentiated; viz, the covariant derivative of a vector is a vector, the covariant derivative of a scalar is a scalar

Now, I don't really see how this is true. For example, consider a scalar field f. The covariant derivative of this is
$$\nabla_af\equiv\partial_af=\frac{\partial f}{\partial x^a}$$

But, aren't $\partial_af$ the components of a covector?

 

[Dick]

I can only assume that they mean an isolated component of a covariant derivative eg $$\nabla_0f$$ is a scalar if f is scalar, vector if f is vector etc. Which would not be true for a partial derivative.

 

[cristo]

Thanks. So, I would be correct in saying that the covariant derivative of a type (r,s) tensor field is a type (r,s+1) tensor field, right?

 

[Dick]

You betcha! And each covariant component of the (r,s+1) field is an (r,s) field - as is always the case.

 

[cristo]

Cheers, Dick!

 

[robphy]

Cristo, what is the definition of "covariant differentiation" [in your notes]?
Note, e.g., http://en.wikipedia.org/wiki/Covariant_derivative
and http://www.utsc.utoronto.ca/~harper/Redten/redten/node36.html

 

[cristo]

Cristo, what is the definition of "covariant differentiation" [in your notes]?
Note, e.g., http://en.wikipedia.org/wiki/Covariant_derivative
and http://www.utsc.utoronto.ca/~harper/Redten/redten/node36.html

He defines the covariant derivative by setting up the situation of two vector fields u and v ona manifold. Then, picking some point P, u determines a flow line passing through P; parametrised by some path parameter s, such that $$\bold{u}_P=\left.\frac{d}{ds}\right|_{s=s_P}$$
Suppose Q$\neq$P is another point on the flowline corresponding to $s_Q\neq s_P$. Then we write
$$\bold{v}(s_Q)\equiv\bold{v}_Q, \hspace{2cm} \tilde{\bold{v}}(s_Q)\equiv \tilde{\bold{v}}_{QP}$$ [he actually uses a symbol with two parallel lines over v, like one side of a "norm" sign-- anyway, it means the vector field parallely transported from P, evaluated at Q]
and define the covariant derivative as
$$\nabla_{\bold{u}}\bold{v}_P=\lim_{s_Q\rightarrow s_P}\frac{\tilde{\bold{v}}(s_Q)-\bold{v}(s_Q)}{s_Q-s_P}$$

 

[robphy]

in that case, you are evaluating

$$u^a \nabla_a v^b$$, which is a vector.
Thus, with the above definition,

the covariant derivative of a vector is a vector

 

[cristo]

Ok, so would I be correct in thinking that whilst $$\nabla_{\bold{u}}v^b=u^a\nabla_av^b$$ is a vector, $$\nabla_bv^a$$ is a not a vector, but is a type (1,1) tensor?

I think this makes sense, since clearly u^a is a vector; if $$\nabla_bv^a$$ is a type (1,1) tensor, then the product of the two will be a vector.

 

[robphy]

Ok, so would I be correct in thinking that whilst $$\nabla_{\bold{u}}v^b=u^a\nabla_av^b$$ is a vector, $$\nabla_bv^a$$ is a not a vector, but is a type (1,1) tensor?

I think this makes sense, since clearly u^a is a vector; if $$\nabla_bv^a$$ is a type (1,1) tensor, then the product of the two will be a vector.

Yes.
Of course, to do the "product", you'd do $$(u^b) (\nabla_b v^a)= u^b\nabla_b v^a$$.

 

[cristo]

Yes.
Of course, to do the "product", you'd do $$(u^b) (\nabla_b v^a)= u^b\nabla_b v^a$$.

Yes. Thanks for your help!

 

[Dick]

Yeah, of course they meant $$\nabla_{\bold{u}}\bold{T}$$ in which case the extra index is contracted. Duh.