Change of Coordinate for V.Field in Mfld.

Tags: coordinate, mfld, vfield
WWGD is offline
Feb27-09, 09:57 PM
P: 393
Hi, again:
Just a quick question; I have "notation indigestion", i.e., I have been trying

to figure way too many technicalities recently; I would appreciate a quick yes/no:

Say X_p is a V.Field defined at p in a C^k manifold; k>0 . Say (U,Phi) and (U',Phi')

are both charts containing p . Just wondering if the rule for coordinate change

of X from chart-to-chart is; is it given by

d(Phi o Phi'^-1) and d(Phi' o Phi^-1) ?

I mean, I know it involves the chain rule, but I wonder if it is the chain rule

applied to the two formulas above.

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CompuChip is offline
Feb28-09, 03:46 AM
Sci Advisor
HW Helper
P: 4,301
Think about the geometrical picture. You have patches U and U' on the manifold, which are mapped to some open subsets V and V' of Euclidean space.
Suppose you want the coordinate change from V to V'. Suppose you have a coordinate [itex]\vec v \in V[/itex]. Then you can go to [itex]\Phi^{-1}(\vec v) \in U[/tex]. Assuming that this point is also in U', you can go to V': [itex]\Phi'( \Phi^{-1}(\vec v) ) \in V'[/itex]. You can write this as [itex]\kappa(\vec v)[/itex], where depending on your convention for composition of functions,
[tex]\kappa = \Phi^{-1} \circ \Phi'[/tex]
[tex]\kappa = \Phi' \circ \Phi^{-1}[/tex]

You can also go from V' to V by a similar reasoning (and in fact you will find that the coordinate change you get is [itex]\kappa^{-1}[/itex]).
WWGD is offline
Mar2-09, 12:13 AM
P: 393
Thanks, Compuchip; I don't know if I misunderstood ( or misunderestimated )
your reply:

But I think that both k, k' as you described them ( I am sorry, I can't make the

'quote' function work well in here ) give me a coordinate change between the patches

V,V' =Phi(U) and Phi'(U') respectively , as you described, but I don't see that this

gives me a way of changing the coordinate representation of the V.Field X_p , which

lives in U/\U' (Sorry, I am still learning Tex.).

I know that we get the coordinate rep. ( in terms of the basis for T_pM , in

each of the charts U,U' ) by pulling back (Thru Phi, Phi' respectively), the basis

of T_Phi(p)R^n and T_Phi'(p) R^n respectively, to get different bases for T_pM.

Does your k , k' give me a way of going from one basis representation of T_pR^n

to another basis rep. of T_pR^n ?

Thanks, and sorry for writing in ASCII. Hopefully this summer I will have time to

learn Latex.

zhentil is offline
Mar2-09, 01:24 AM
P: 491

Change of Coordinate for V.Field in Mfld.

It's just the chain rule from good old calculus.
WWGD is offline
Mar2-09, 11:39 AM
P: 393
Yes, Zhentil,thanks, I understand that. I was looking for what specific map
we apply the chain rule to:

Is it to the composition of chart maps (Phi o Phi' ^-1) ;with (U,Phi) and (U',Phi')

overlapping charts for p ?(or, of course, the inverse of the map above, if we want to

change in the opposite direction).
WWGD is offline
Mar2-09, 10:01 PM
P: 393
This is my issue: If M,N, are open subsets of R^n, this is easy. Now, if M,N are not
open subsets of R^n, then , we get a basis representation for X_p as above in U,
by pulling back the basis vectors in T_Phi(p)R^n , by the chart map Phi^-1 , and
we get the basis rep. in the chart U' , by pulling back (using Phi'^-1) ,the basis
vectors in T_Phi'p R^n.

Then, to change , I think we need to push forward the vector field X_p ( in whichever
basis) to R^n , then use the fuch forward to map this image into the other basis in R^n,
and then pull back again. I get the idea, I think, but this is becoming a "tress v. forest"
thing , here, and I lost track of just how to figure out the change of expression, and I
was hoping that if someone was familiar with it, they could give me the "forest" --
to push the analogy to its limit.

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