Relating basis vectors at different points in a neighborhood

  • B
  • Thread starter Shirish
  • Start date
  • #1
165
21
I'm reading a section on the derivative of a vector in a manifold. Quoting (the notation ##A^{\alpha}_{\beta'}## means ##\partial x^{\alpha}/\partial x^{\beta'}## - instead of using primed and unprimed variables, we use primed/unprimed indices to distinguish different bases):
Consider the total differential of a vector field ##\mathbf{t}=t^{\alpha}(x^1,\ldots,x^n)\mathbf{e}_{\alpha}## (without specifying what "##\text{d}##" means):
$$\text{d}\mathbf{t}=\text{d}t^{\alpha}\mathbf{e}_{\alpha}+t^{\alpha}\text{d}\mathbf{e}_{\alpha}=\bigg(\frac{\partial t^{\alpha}}{\partial x^{\beta}}\text{d}x^{\beta}\bigg)\mathbf{e}_{\alpha}+t^{\alpha}\bigg(\frac{\partial\mathbf{e}_{\alpha}}{\partial x^{\beta}}\text{d}x^{\beta}\bigg)$$
The second term involves derivatives of vectors - the very quantiuty we're trying to formulate with the covariant derivative. We _expect_ a change ##\mathbf{e}_{\beta}\to\mathbf{e}_{\beta}+\text{d}\mathbf{e}_{\beta}## under ##x^{\alpha}\to x^{\alpha}+\text{d}x^{\alpha}## because coordinate basis vectors are tangent to coordinate curves. We know that the manifold ##M## is locally flat. Thus in a sufficiently small neighborhood ##p\in M## there is a local inertial frame, the basis vectors of which are _constants_; call them ##\{\mathbf{e}^0_{\beta}\}##. The coordinate basis ##\{\mathbf{e}_{\alpha}\}## in a neighborhood of ##p\in M## can be expressed in the basis ##\{\mathbf{e}^0_{\beta}\}##, ##\mathbf{e}_{\alpha}(x)=A^{\beta'}_{\alpha}(x)\mathbf{e}^0_{\beta'}##. Differentiating this formula (the ##\mathbf{e}^0_{\beta'}## are constants), $$\partial_{\mu}\mathbf{e}_{\alpha}=(\partial_{\mu}A^{\beta'}_{\alpha})\mathbf{e}^0_{\beta'}=(\partial_{\mu}A^{\beta'}_{\alpha})A^{\rho}_{\beta'}\mathbf{e}_{\rho}$$
where we've inverted the basis transformation, ##\mathbf{e}^0_{\beta'}=A^{\rho}_{\beta'}\mathbf{e}_{\rho}##
Now this "we know that the manifold ##M## is locally flat" doesn't seem like a good enough explanation for why basis vectors at two different points were related by the transformation law. The book covers the explicit, mathematical derivation for why basis vectors in the same tangent space but under different charts ##(U,x)## and ##(U',x')## can be related by ##\mathbf{e}_{\beta'}=A^{\rho}_{\beta'}\mathbf{e}_{\rho}##.

But in the above quote, basis vectors at different points, and hence belonging to different tangent spaces have been related by the exact same formula. Can anyone help me with a proper mathematical proof on how the same transformation law comes about in this context? Without that, "##M## is locally flat" is a handwavy argument.
 

Answers and Replies

  • #2
strangerep
Science Advisor
3,166
1,011
I'm reading a section [...]
Which textbook?? Or, if you're reading lecture notes, give a weblink.

[I'll never understand why questioners think it's a good idea not to post their sources (sigh).]
 
  • #3
165
21
Which textbook?? Or, if you're reading lecture notes, give a weblink.

[I'll never understand why questioners think it's a good idea not to post their sources (sigh).]
Core Principles of Special and General Relativity, Luscombe
 
  • #4
strangerep
Science Advisor
3,166
1,011
Core Principles of Special and General Relativity, Luscombe
OK, so it's Luscombe sect 14.2 on p250 for those playing along at home.

I, too, don't like the handwavy reasoning. But it really all just falls out from a simple change of coordinate variables, and the chain rule. E.g., for a change of variables $$x^\mu ~\to~ x'^\mu ~=~ x'^\mu(x) ~,$$ we can use the chain rule to figure out the corresponding transformation of derivative operators,. Let ##f(x)## be an arbitrary scalar field. Then $$ \frac{\partial f}{\partial x'^\mu} ~=~ \frac{\partial x^\alpha}{\partial x'^\mu} \; \frac{\partial f}{\partial x^\alpha} ~.$$ The vector basis elements ##{\mathbf e}_\mu## are just the coordinate derivative operators here. You don't even need to think about tangent spaces It's just an elementary change of variable.

HTH.
 
  • #5
165
21
OK, so it's Luscombe sect 14.2 on p250 for those playing along at home.

I, too, don't like the handwavy reasoning. But it really all just falls out from a simple change of coordinate variables, and the chain rule. E.g., for a change of variables $$x^\mu ~\to~ x'^\mu ~=~ x'^\mu(x) ~,$$ we can use the chain rule to figure out the corresponding transformation of derivative operators,. Let ##f(x)## be an arbitrary scalar field. Then $$ \frac{\partial f}{\partial x'^\mu} ~=~ \frac{\partial x^\alpha}{\partial x'^\mu} \; \frac{\partial f}{\partial x^\alpha} ~.$$ The vector basis elements ##{\mathbf e}_\mu## are just the coordinate derivative operators here. You don't even need to think about tangent spaces It's just an elementary change of variable.

HTH.
It helps for sure! As far as I've read (so by my limited understanding), we interpret $$\bigg(\frac{\partial f}{\partial x^i}\bigg)_p=\partial_i(f\circ x^{-1})(x(p))$$ And here's the derivation of chain rule I wrote that is as explicit as I could make it $$\begin{align}
\bigg(\frac{\partial}{\partial x^i}\bigg)_pf&=\partial_i(f\circ x^{-1})(x(p))
=\partial_i\big((f\circ x'^{-1})\circ(x'\circ x^{-1})\big)(x(p))
\\&=\big(\partial_j(f\circ x'^{-1})\big)(x'(p))\cdot \big(\partial_i(x'\circ x^{-1})\big)^j(x(p))
\\&=\big(\partial_j(f\circ x'^{-1})\big)(x'(p))\cdot \big(\partial_i(x'^j\circ x^{-1})\big)(x(p))
\\&=\bigg[\bigg(\frac{\partial}{\partial x'^j}\bigg)_pf\bigg]\cdot\bigg[\bigg(\frac{\partial}{\partial x^i}\bigg)_px'^j\bigg]
\\&=\bigg(\frac{\partial x'^j}{\partial x^i}\bigg)_p\bigg(\frac{\partial}{\partial x'^j}\bigg)_pf
\end{align}
$$ The point is that while employing the chain rule, we're calculating the derivatives at a single point only. It doesn't work if we try calculating derivative w.r.t. ##x## chart map at one point and w.r.t. the ##x'## chart map at another point. So that was the primary source of confusion.

Having said that, I know that the chain rule that you wrote in your answer definitely works in flat spaces where there's a global chart. So just to be clear, your version of chain rule follows because of the assumption that the neighborhood of a point ##p## is locally flat, right? And the precise mathematical description for "locally flat" would be that that neighborhood can be covered by a single chart, correct?
 
  • #6
strangerep
Science Advisor
3,166
1,011
I know that the chain rule that you wrote in your answer definitely works in flat spaces where there's a global chart. So just to be clear, your version of chain rule follows because of the assumption that the neighborhood of a point ##p## is locally flat, right? And the precise mathematical description for "locally flat" would be that that neighborhood can be covered by a single chart, correct?
I don't think the concept of "locally flat" matters for this. I.e., the concepts of connection and curvature are not needed.

Consider the Lie derivative: it needs neither a connection nor curvature. Rather, it relies on the basic properties of a generic manifold: by definition, any manifold's topology is such that every point has a neighbourhood which is homeomorphic to Euclidean space. Usually in physics, we assume differentiable manifolds -- i.e., each point has an infinitesimal neighbourhood isomorphic to a linear space which allows us to differentiate functions defined over the manifold. (That's why Lie derivatives make sense even in the absence of a connection.)
 

Related Threads on Relating basis vectors at different points in a neighborhood

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
19
Views
9K
  • Last Post
Replies
1
Views
3K
Replies
6
Views
1K
Replies
5
Views
4K
Top