Partial derivatives as basis vectors?

[Rearden]

Hi,

I'm having trouble understanding how people can make calculations using the partial derivatives as basis vectors on a manifold. Are you allowed to specify a scalar field on which they can operate? eg. in GR, can you define f(x,y,z,t) = x + y + z + t, in order to recover the Cartesian basis vectors of flat spacetime? Or am I missing the point?

Thanks!

 

[Fredrik]

They act on smooth functions $f:M\rightarrow\mathbb R$, where M is the manifold. Note that M isn't (usually) a vector space, but its tangent space $T_pM$ at any point $p\in M$ is. The derivative operators are basis vectors for those vector spaces.

See also this post.

 

[bcrowell]

It's purely a matter of notation. There is no interesting physics involved. You never really use the partial derivative operators to take the partial derivative of any actual function. The reason the notation makes sense is that derivative operators have the transformation properties implied by the notation.

 

[Rearden]

I think I understand. So does a four-velocity vector expressed in flat spacetime with a Cartesian basis have exactly the same components as the corresponding tangent vector on a manifold, despite the nominally different basis vectors?

 

[Ben Niehoff]

It's purely a matter of notation. There is no interesting physics involved. You never really use the partial derivative operators to take the partial derivative of any actual function.

False. I use them to take derivatives all the time: as operators in differential equations, for calculating commutators of Killing vectors, etc. I find the literal interpretation of tangent vectors as derivative operators to be quite productive.

I also use the basis 1-forms as linear functionals acting on vectors.

 

[Fredrik]

I think I understand. So does a four-velocity vector expressed in flat spacetime with a Cartesian basis have exactly the same components as the corresponding tangent vector on a manifold, despite the nominally different basis vectors?

Yes. Suppose that $C:[a,b]\rightarrow M$ is a curve in a manifold, and $x:U\rightarrow\mathbb R^n$ is a coordinate system such that $C([a,b])\subset U$. The union of all the tangent spaces $TM=\bigcup_{p\in M}T_pM$ is called the tangent bundle. A function [tex]V:[a,b]\rightarrow TM[/itex] such that $V(t)\in T_{C(t)}M$ for all $t\in[a,b]$, is called a vector field along the curve C. We're interested in a particular vector field along the curve, called the velocity vector field of C, or the tangent vector field of C. It's often written as $$\dot C$$ and is defined by<br /> <br /> $$\dot C(t)(f)=(f\circ C)'(t)$$<br /> <br /> To find its components, we need to express $$\dot C(t)$$ as a linear combination of the $$\frac{\partial}{\partial x^i}\bigg|_{C(t)}$$. This isn't hard if we know the chain rule.<br /> <br /> $$(f\circ C)'(t)=(f\circ x^{-1}\circ x\circ C)'(t)=(f\circ x^{-1})_{,i}(x(C(t)))(x\circ C)^i'(t)=(x\circ C)^i'(t)\frac{\partial}{\partial x^i}\bigg|_{C(t)}f$$<br /> <br /> so<br /> <br /> $$\dot C(t)=(x\circ C)^i'(t)\frac{\partial}{\partial x^i}\bigg|_{C(t)}$$<br /> <br /> Define $$\tilde x=x\circ C$$. This is a curve in $$\mathbb R^n$$, and the tangent vector at $$\tilde x(t)$$ of such a curve, is defined as $$\tilde x'(t)=(\tilde x^1'(t),\dots,\tilde x^n'(t))$$. So its components can be written as $$\tilde x^i'(t)=(x\circ C)^i'(t)$$, which as we just saw, are also the components of $$\dot C(t)$$ in the basis that consists of the partial derivative operators at C(t), constructed from the coordinate system x.[/tex]

 

[Rearden]

Exactly what I needed,

Thanks a lot!