Monday, 12 November 2018

Question on co- / contra-variant coordinates

I am studying co- and contra- variant vectors and I found the video at very useful. It discusses the slanted coordinate system where the X, Y axes are at an angle of α. One can get the components of v either by dropping perpendiculars to the axes (vi) or by dropping a line parallel to the other axis (vi). These give correct results for the norm vi vi and the dot product vi wi. (I have not shown w). So the vi are called contravariant and the vi are called covariant. According to the video Dirac thought this was a great example.

But both vi and vi  contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Submitted as comment on video and
on Physics Forums at

Orodruin answered on PF seven hours later with "The covariant and contravariant components belong to different basis vectors." Cryo came up with something a bit more complicated and I followed Orodruin. So here's how it goes:

To get v with the covariant coordinates we must have
v = vxex + vyey      (1)
I have drawn them in on the diagram and ex is clearly smaller than ex, which I have conveniently drawn so that vx = 1. We must also have
 vxex = vxex       (2)
and similarly
vyey = vyey       (3)
which give us
v = vxex + vyey       (4)
which is as it should be.
(3) gives us
vy = ey(ey)-1vy       (5)
where (ey)-1 is the inverse of ey: or ey(ey)-1 = 1.

(5) shows us that vy does indeed vary with ey with a strange constant of proportionality (ey)-1vy.

Back to the drawing board

Orodruin did not like that at his #5 and then gave me a tip at #7 that the dual (or covariant) bases are given by êa = ∇xa. I will use ê for basis vectors in future, following Carroll. From the formula I could calculate the covariant basis vectors in terms of the Cartesian basis vectors because I already had the inverse metric of the covariant system. This was calculated from the transformations of the systems to and from Cartesian.

The answer came:

and draw them properly (x = 1, y = 2)
From (NF12) it is clear that  êx is π/2 - α below the Cartesian X axis. Therefore the angle between êy and êx is a right angle, and the projected line from vy  is parallel to êx as we should have guessed and Orodruin intimated. Likewise êy is parallel to the projected line from vx and always on the Cartesian Y axis. In addition |êi | > |êi |.

Getting back to the original question, what happens to vx if we double the contravariant basis êx to ê'x? We could leave  vx, êx, vy, êy alone and they would still give v. We could double êx and halve vx. Neither option would co-vary. Both would give incorrect values for |v|. We need to find the metric of the primed system so we can just calculate the primed covariant coordinates and bases. We also cannot use the previous technique (calculating from the transformations of the systems to and from Cartesian), because we don't know how to transform primed covariant coordinates to Cartesian - we don't know what the primed coordinates are.

Therefore we use the pullback operator which we met two posts ago in Commentary on Appendix A: Mapping S2 and R3.

If we have our contravariant primed bases, with scale factors a,b so
ê'x = aêx  ,  ê'y = bêy 
which give
v'x  = vx  ⁄ a   ,  v'y =vy  ⁄ b 

Turning all the handles we end up with

v'x = a2v'x  + abv'ycos⁡α
v'y =b2v'y + abv'xcos⁡α

Since v' i  decrease as a, b and v'i  increase as a2  ,b2 ,ab, we can safely say that v'i  increase as êi . They co-vary. It also looks as if they will 'compensate' for v' i , giving correct |v| and v.w. We would just have to follow the trail.

The full details of all the above are contained in sections 5b and Note F of  Commentary 1.1 Tensors matrices and indexes.pdf

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