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Trying to understand contravariance

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cin-bura
#1
May9-12, 03:17 PM
P: 7
hello!

i am having a hard time understanding this:
contravariance is defined in the textbooks as some entity that transforms like
[tex]\tilde A^{\mu}(u)= \frac{\partial u^{\mu}}{\partial x^{\nu}} A^{\nu}(x)[/tex].
du/dx is not constant in space because the relations between 2 coordinate systems don't have to be linear, but is rather a function of the position. so how can this hold not only infinitesimally but in general? to phrase it in an another way:
why is the first relation sufficient and it is not necessary to write
[tex]\tilde A^{\mu}(u)= u^{\mu}( A^{\nu}(x))[/tex] ?

does anybody have an example of a contravariant field under the transformation between plane polars and caresian coordinates, say? it would help me very much to picture it!

thank you!
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Naty1
#2
May9-12, 04:22 PM
P: 5,632
Have you read here:

http://en.wikipedia.org/wiki/Covaria...nce_of_vectors

I haven't studied it but it seems to have some explanations, diagrams, and examples.
Matterwave
#3
May9-12, 04:37 PM
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A (contravariant) vector is always defined at a point on the manifold, and it lives in the tangent space to that point on the manifold. Vector spaces must be linear simply because of the definition of a vector space.

So, even though the manifold itself might be curved, or the coordinates you use might be curvilinear, the vector space in which the vector resides is a linear space.

The relation that you put is valid AT EACH POINT in the manifold.

Whenever you see equations like that you should always remember that there is an implied "AT POINT P in the manifold" always in statements like this.

cin-bura
#4
May10-12, 02:15 AM
P: 7
Trying to understand contravariance

thank you for your replies!
i think that it is valid at each point is a very valuable statement. still, i am not completely clear about that: doesn't the definition of a derivative require an infinitesimal displacement away from this point?

to show more clearly what i mean, here is an example that shows my problem:

take plane polars u<-(r,Φ) and cartesian coordinates (x,y). Define a (linear) function A(x,y)=(2x,2y).
then, (r,phi) is (2r cos Φ, 2r sinΦ).
the jacobian is
[tex]\frac{\partial x^{1}}{\partial u^{1}} = cos \Phi , \frac{\partial x^{2}}{\partial x^{1}} = sin \Phi[/tex] and so on.
note that the entries depend non-linearily on the polar angle.

then, claiming that it would transform covariantly, i could write
[tex]A^{1}(x)= \frac{\partial x^{1}}{\partial u^{1}}\tilde A^{1}(u)+\frac{\partial x^{1}}{\partial u^{2}}\tilde A^{2}(u)[/tex]
and insert to get the obviously wrong relation
[tex] 2x = cos(\Phi) 2r cos(\Phi)+sin(\Phi) 2r sin(\Phi)[/tex]

where is the mistake?
the_emi_guy
#5
May10-12, 10:40 PM
P: 589
cin-bura,
A example of a contravariant vector is ordinary velocity.
Contravariant simply means that the components of the vector must move in the opposite direction (so to speak) of a change in coordinates. The idea being that the vector represents something absolute, like wind velocity, and that the physical reality will not depend on which set of coordinates we choose to use to quantify it. Consider 2D cartesian space with 1meter unit vectors. At a certain point wind is blowing east at 5m/s. Our velocity vector is (5,0) at this point in space. Now if we inflate our coordinate system by a factor of 1000, so our unit vectors are now 1Km, our velocity vector at this point is now (.005,0). Coordinates expanded, our velocity vector magnitude shrank, that's the "contra" in contravariant.
The partial derivative computes the relationship between the unit vectors *at this point in space* (basis vectors) in each coordinate system. Consider wiggling each unit vector in the x coordinates by some tiny amount and seeing how much change that induces in each unit vector in the u coordinates at the point in space of interest. In our simple example if I wiggle the 1meter East unit vector by .0001m, the 1Km unit vector East will wiggle by .000000Km, so their ratio will be 1/1000. I also note that wiggling the East unit vector in x coordinates does not change the North unit vector in u coordinates.
Same idea in polar coordinates. Consider the point in x coordinates (cartesian) 1 meters east, 1 meter north. The cartesian basis vectors at this point are directed east and north. Easterly wind has velocity vector (5,0) at this point. Now lets change to polar. The polar basis vectors at this point (r, theta) will be rotated counterclockwise 45 degrees (r is pointing NE, theta pointing NW). If we wiggle the cartesian easterly basis vector and see what change that induces in r and theta we will find that the velocity vector in polar is rotated 45 degrees clockwise. Basis vectors rotated counterclockwise, vector components rotated clockwise, again the "contra".


Hope this helps.
Cheers.
cin-bura
#6
May11-12, 02:18 AM
P: 7
thank you for your vivid explanation! i really do see clearer now.
i think the point i missed in the example was that (u) not only depends on u but is also expressed in the u coordinates.
cheers
Chestermiller
#7
May11-12, 07:03 AM
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Quote Quote by cin-bura View Post

take plane polars u<-(r,Φ) and cartesian coordinates (x,y). Define a (linear) function A(x,y)=(2x,2y).
then, (r,phi) is (2r cos Φ, 2r sinΦ).
the jacobian is
[tex]\frac{\partial x^{1}}{\partial u^{1}} = cos \Phi , \frac{\partial x^{2}}{\partial x^{1}} = sin \Phi[/tex] and so on.
note that the entries depend non-linearily on the polar angle.

where is the mistake?
Where you made your mistake was in your guess of the cylindrical coordinate components. If the components of A in cartesian coordinates are 2x and 2y, then the contravariant components of A in cylindrical coordinates are 2r and 0. Try these in the transformation law and see what you get. Incidentally, these are also the covariant components of A. In this example, the magnitude of A is 2r, and, at all locations in the plane, it is directed parallel to a radius vector from the origin.

Chet
cin-bura
#8
May14-12, 01:19 PM
P: 7
you are perfectly right. i did not write (u) in terms of the polar basis.
also, it should be

[tex] \frac{\partial x}{\partial r} = cos(\phi) [/tex]
[tex] \frac{\partial x}{\partial \phi} = -r sin(\phi) [/tex]

but what about the non symmetrical case?

consider a map that assigns a constant vector to every point in space, say:
[tex]
\vec{A}(\vec{x})=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}
[/tex]
then, geometrically, i am guessing:
[tex] A^r(\vec{u})=1[/tex] and [tex] A^{\phi}(\vec{u}) =\frac{\pi}{4} [/tex]
inserting:
[tex]
A^x=\frac{\partial x}{\partial r}\tilde{A}^r + \frac{\partial x}{\partial \phi}\tilde{A}^{\phi}
=cos(\phi)\cdot 1 -r sin(\phi)\cdot \frac{\pi}{4}
[/tex]
what is wrong this time?
Matterwave
#9
May14-12, 02:17 PM
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Where do you get the "guessed" components? Because r and phi basis vectors change, and you have a constant vector field, then the vector field components expressed in r and phi must also change and should not be constant.
Chestermiller
#10
May14-12, 08:08 PM
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Matterwave is right. You have to think of A as a vector point function, which can change in magnitude and direction with position within the plane. In the particular case that you described, A is a constant vector, independent of position. Why don't you use the transformation relation from cartesian to polar to see what you get for the vector A as you defined it in cartesian coordinates? You will find that your guessed components in cylindrical coordinates are wrong, and you will get to see what the correct components are. You should also obtain a better understanding of what is happening.

r = [itex]\sqrt{x2+y2}[/itex]
θ = arctan(y/x)

You can also get the same result from

A = (1/[itex]\sqrt{}2[/itex])ix+(1/[itex]\sqrt{}2[/itex])iy = Ar ir + Aθ r iθ

Determine ix and iy in terms of ir and iθ, and then substitute into the above equation.

chet
cin-bura
#11
May15-12, 02:46 PM
P: 7
well, as you said:
[tex]r =\sqrt{x^2+y^2} = \sqrt{1/2+1/2} = 1[/tex]
[tex]θ = arctan(y/x) = arctan(\frac{\sqrt{2}}{\sqrt{2}})=\frac{\pi}{4}[/tex]
it represents a vector with magnitude 1 pointing in 45 direction
DrGreg
#12
May15-12, 07:33 PM
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In non-cartesian coordinate systems, there is a difference between the (contravariant) components of a vector and coordinates of a point in space. Think of the vector as an arrow attached to a point in space. In the case of polar coordinates, you have a point with coordinates [itex](r,\theta)[/itex], and a vector A which you resolve into two components "in the r direction" and "in the θ direction", i.e. radially and tangentially, i.e. in Chestermiller's notation, along [itex]\textbf{i}_r[/itex] and [itex]\textbf{i}_\theta[/itex].

So, the vector given by [itex](2x,2y)[/itex] in [itex](x,y)[/itex] coordinates means
[tex]\textbf{A} = 2x\cdot\textbf{i}_x + 2y\cdot\textbf{i}_y[/tex]
The vector given by [itex](2r,0)[/itex] in [itex](r,\theta)[/itex] coordinates means
[tex]\textbf{A} = 2r\cdot\textbf{i}_r + 0\cdot r\textbf{i}_\theta[/tex]
They are both the same. ir and iθ are not constant vectors; they depend on r and θ at the point in question.

I don't know how far you have studied this, so you may not have come across the terminology yet, but each (contravariant) vector resides in the "tangent space" associated with a specific point on the manifold (=event in spacetime).

Here's one way of thinking about this. If you have a curve on the manifold, parameterised by arclength s as, say, ([itex]x^\alpha(s)[/itex]), then [itex](dx^a/ds)[/itex] represents the unit tangent vector to the curve and is a contravariant vector. (In spacetime terminology, [itex](dx^a/d\tau)[/itex] is the 4-velocity vector for the worldline ([itex]x^\alpha(\tau)[/itex].)

Components of contravariant vectors transform the same way as tangent vectors (or 4-velocities) do:
[tex]\frac{du^\mu}{d\tau} = \frac{\partial u^{\mu}}{\partial x^{\nu}} \frac{dx^\nu}{d\tau}[/tex][tex]\tilde A^{\mu}(u)= \frac{\partial u^{\mu}}{\partial x^{\nu}} A^{\nu}(x)[/tex]
Chestermiller
#13
May15-12, 08:29 PM
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Quote Quote by cin-bura View Post
well, as you said:
[tex]r =\sqrt{x^2+y^2} = \sqrt{1/2+1/2} = 1[/tex]
[tex]θ = arctan(y/x) = arctan(\frac{\sqrt{2}}{\sqrt{2}})=\frac{\pi}{4}[/tex]
it represents a vector with magnitude 1 pointing in 45 direction
These are not the contravariant components (or any other components for that matter) of the vector A discussed in my recent reply (#10). For polar coordinates, the contravariant components of A in that reply are:

Ar = (1/√2)(cos(θ) + sin(θ))

Aθ = (1/√2)(cos(θ) - sin(θ)) / r

This actually does represent a vector of magnitude 1 pointing in the 45 degree direction at all locations within the plane. Substitute these expressions into your transformation formula and see what you get for Ax and Ay!
Matterwave
#14
May15-12, 08:30 PM
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The basic gist is that you are transforming one point as expressed in one coordinate system (the point (1,1)) into the same point as expressed in another coordinate system. You are not transforming the components of a vector.
the_emi_guy
#15
May16-12, 03:16 AM
P: 589
I have uploaded three images that I have created that show the cartesian to polar coordinate transformation for a contravariant vector. Specifically it shows step by step transformation for the easterly blowing wind example I discussed earlier. I am showing *all* of the steps, so it may be to verbose for many, but if anyone is having trouble digesting how this really works feel free to view.
Attached Thumbnails
contravariant.jpg   contravariant2.jpg   contravariant3.jpg  
Matterwave
#16
May16-12, 03:28 AM
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I believe for your vector based at (1,1), the theta component of the vector is off. You have v_theta=-5*r*sin(theta)=-5*sqrt(2)*(1/sqrt(2))=-5 instead of -5sqrt(2)/2.

The polar basis vectors are not of unit length if one uses the regular coordinate basis. They are, however, at least orthogonal.

Most calculus books will use a normalized set of basis vectors in polar coordinates.
the_emi_guy
#17
May16-12, 06:11 AM
P: 589
Matterwave,
Thanks so much for checking this. I actually had the r upstairs instead of downstairs in the partial derivatives for d(theta)/dx, d(theta)/dy. I have corrected the second and third images accordingly.
Attached Thumbnails
contravariant.jpg   contravariant2a.jpg   contravariant3a.jpg  
Chestermiller
#18
May16-12, 06:26 AM
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Here is another example for you to consider.

Assume that you have a flat compact disc (CD), and that the CD lies within the horizontal x-y plane of a rectangular cartesian coordinate system x-y-z. The axis of the CD coincides with the z-axis of the cartesian coordinate system, and the CD is rotating as a rigid body about its axis (relative to the x-y-z coordinate system) with a constant angular velocity of ω. Each material particle of the CD travels in a perfect circle about the z-axis. Assume that there is also a cylindrical polar coordinate system (r-θ-z) present, that coincides with the x-y-z coordinate system.

As reckoned from the r-θ-z coordinate system, the velocities of the various material particles comprising the CD are given by:

V = (ωr) iθ = ω (r iθ) = ω eθ

where eθ is the coordinate basis vector in the θ direction:

eθ = r iθ

According to the equations above, even though the velocity vector V varies with r and θ, its contravariant component in the r-direction is zero, and its contravariant component in the θ-direction is a constant, and equal to ω:

Vr = 0

Vθ = ω

Note that the velocity V is a vector tangent to the circles (trajectories) around which the material particles are traveling.

Now use your transformation formula to show that, in cartesian coordinates, the contravariant components of the velocity V are given by:

Vx = -ωy

Vy = +ωx

so that

V = -ωy ix + ωx iy


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