The size of a vector

  • #1
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Summary:

How to calculate the size of a vector, confusion with basis vectors.

Main Question or Discussion Point

I'm stumbling on something rather basic here, will explain with an example. (Pardon the LaTeX problems, trying to fix..)

Suppose I have a plane, and in the plane I put the familiar (x,y) Cartesian coordinate system, and the metric is the usual Euclidean metric with [itex] ds^2 = dx ^2 + dy^2 [/itex].

Now suppose I add into this another coordinate system defined by

u=x+2y
v=x-y

and so it follows that

x=(1/3)(u + 2v)
y=(1/3)(u-v).

The basis vectors for u and v are
[itex] \vec{e}_u = < \frac {\partial x}{\partial u},\frac {\partial y}{\partial u} > ,= < 1/3 , 1/3 > [/itex]
[itex] \vec{e}_v = < \frac {\partial x}{\partial v},\frac {\partial y}{\partial v} > ,= < 2/3 , -1/3 > [/itex]
and the corresponding covectors are

[itex] \bar{e}^u = < \frac {\partial u}{\partial x},\frac {\partial u}{\partial y} > ,= < 1 , 2 > [/itex]
[itex] \bar{e}^v = < \frac {\partial v}{\partial x},\frac {\partial v}{\partial y} > ,= < 1 , -1 > [/itex].

The inner products of the basis vectors and covectors are
[itex] \vec{e}_u \cdot \bar{e}^u = < 1/3 , 1/3 > \cdot < 1 , 2 > =1 [/itex]
[itex] \vec{e}_v \cdot \bar{e}^v = < 2/3 , -1/3 > \cdot < 1 , -1 > =1 [/itex]

. . as one would expect.

My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

[itex]
\vec {v} \cdot \vec{v} = v^a v^b g_{ab} =
v^a (v^b g_{ab}) = v^a v_a
[/itex]

This would imply that [itex] \vec{e}_u [/itex], for example, has a magnitude of 1, when clearly it's the square root of [itex] (1/3)^2 + (1/3)^2 = 2/9 [/itex].
 
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Answers and Replies

  • #2
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The inner product changes, too, or better: its calculation changes! You cannot turn to ##(u,v)## coordinates and still compute angles and lengths in ##(x,y)## coordinates. Angles, lengths and the quadratic form might be invariant, its matrix is not.
 
  • #3
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I should add that I thought this rule was good for a tensor of any rank — just use index gymnastics to create the "opposite" tensor and stick them together.

[tex] R^{a}_{bcd} R ^{bcd}_{a} =
[/tex]
the square of the size, or magnitude of the fourth rank tensor [itex] \bf{R} [/itex], although perhaps in cases like this there is no clear meaning to this quantity.
 
  • #4
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So would you say that the size of [itex] \vec {e}_u [/itex] is 1 or the square root of 2/9? I'm guessing the latter since, as you say, vector lengths must be invariant.
 
  • #5
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Yes, I'm mixing up (x,y) and (u,v) here - since I guess in (u,v) [itex] \vec{e}_u [/itex] is simply (1,0). Perhaps that's the root of my confusion.
 
  • #6
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Don't ask me about coordinates, and even less about this index juggling by physicists.
What you have is ##(x,y) \longmapsto x^\tau Q y##. So we have for ##v=Sx, w=Sy##
$$
x^\tau Q y= v^\tau (S^{-1})^\tau Q S^{-1}w = v^\tau \left( \underbrace{(S^{-1})^\tau Q S^{-1}}_{=Q'} \right)w
$$
and the matrix changed from ##Q## to ##Q'##.
 
  • #7
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Thanks fresh_42, that's fascinating. Will see if I can compute the inner products properly in (u,v)..
 
  • #8
DrClaude
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Don't ask me about coordinates, and even less about this index juggling by physicists.
@Orodruin can probably help here.
 
  • #9
Orodruin
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My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

[itex]
\vec {v} \cdot \vec{v} = v^a v^b g_{ab} =
v^a (v^b g_{ab}) = v^a v_a
[/itex]

This would imply that [itex] \vec{e}_u [/itex], for example, has a magnitude of 1, when clearly it's the square root of [itex] (1/3)^2 + (1/3)^2 = 2/9 [/itex].
It is ##\sqrt{2/9}##. ##\vec e_u \cdot \vec e^u## is not the inner product of ##\vec e_u## with itself, it is the inner product between ##\vec e_u## and ##\vec e^u##, which are different vectors.
 
  • #10
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I should add, what led me here was a YouTube video of a man deriving the Christoffel symbols for the surface of a sphere. He took the inner products of the basis vector derivatives with the basis covectors, and I've been trying to understand why he bothered to make sure they were covectors. Will post a link to the video when I find it..

(In that case there was again the complication of using two coordinate systems simultaneously - the phi and theta for the sphere and the (x,y,z) of the Cartesian 3-space.)
 
  • #11
Orodruin
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If you want to treat this in a proper differential geometry fashion, you should not be using an embedding 3D space at all. All you need is the 2-dimensional description of the sphere (although you typically assume that the metric is that induced by the embedding in ##\mathbb R^3##).
 
  • #12
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I agree. My interest was to show by example that the covariant derivative of a metric tensor is zero, even in a non-flat continuum. So I wanted the Christoffel symbols for the surface of a sphere. I know one can produce them using that formula with the partial derivatives of the components of the metric, but of course that assumes [itex] \nabla g = 0 [/itex] in the first place.
cs.jpg
 
  • #13
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Follow up: I realize that one of the sources of my confusion has been that I've been trying to transform basis vectors in the contravariant way instead of covariantly. That they transform covariantly is puzzling to me: Since a vector is a linear combination of basis vectors, aren't basis vectors themselves vectors? Why would they be exceptional in this regard?
 

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