covariant and contravariant ?

by Symmetry
Tags: contravariant, covariant
 P: 3 I have a little question. I hope someone can help me. When we learn the theory of relativity and its formalism, we'll meet concepts : covariant and contravariant, such as covariant vector, covariant tensor... I wonder that why we need to use the concepts ? What are advantages of them ? I think that one of their advantages is to generalize the formalism of the special theory of relativity to the formalism of general theory. Is it true ? Thanks.
 Emeritus Sci Advisor PF Gold P: 5,540 That certainly is one of the advantages. You can do SR without making the distinction between contravariant and covariant, but I do not think you can do GR that way. The reason I prefer to do SR with the distinction is that it does away with that stupid "ict" notation. In my opinion, vectors that correspond to measurable quantities should not have imaginary components.
 Emeritus PF Gold P: 8,147 When you do the tensors in GR they have covariant and contravariant indices, which tell how the tensor transforms when you change coordinates. Contravariant indices are shown superscript and covariant indices are shown subscript. In tensor equations the tensors have to match in number of upper and lower indices. But you can lower an index by doing an inner product with the metric tensor, which is rank two covariant, or raise one by an inner product with the inverse of the metric tensor, which is rank two contravariant.
P: 1,341

covariant and contravariant ?

Do these two terms refer to opposed vectors then ?
And if so, what's a "variant" ? (You mean like
the defined quarters in a 2D graph ?)

P.S. I'm sorry if I totally messed up the meaning. [:)]

Live long and prosper.
Astronomy
PF Gold
P: 22,809
 Originally posted by selfAdjoint When you do the tensors in GR they have covariant and contravariant indices, which tell how the tensor transforms when you change coordinates. Contravariant indices are shown superscript and covariant indices are shown subscript. In tensor equations the tensors have to match in number of upper and lower indices. But you can lower an index by doing an inner product with the metric tensor, which is rank two covariant, or raise one by an inner product with the inverse of the metric tensor, which is rank two contravariant.
SA, two things.
I happen to have just been reading a Lee Smolin paper
http://xxx.lanl.gov/PS_cache/hep-th/...03/0303185.pdf
it is a great paper, recent---"how far are we from the quantum theory of gravity" 18March2003. Baez recommended it in "recent finds" which you mentioned liking to read. At least the introduction and conclusions seemed worth printing out.

The other thing is covariant and contravariant. I think of every point x on the surface or manifold as having a tangent space T
and that tangent space having a dual T* which is the vector space of linear functionals on T.

I think of a contravariant vector field as one with values in T,
and a covariant field as one with values in T*.

The metric is a bilinear functional on T, say written (X, X')
so what you said about "lowering an index" translates to
this: you have a contravariant thing F(x) with values in T
and you form (F(x), .) which is a linear functional for each x, and
so has values in T*.

is this an OK way for someone who has read Halmos "finite dimensional vector spaces" or an equally clear brief vector
text to look at covariant and contravariant?

let me know if I have things confused----this is how I seem to
recall the standard differential geometry of manifolds, always working either with the tangent space or its dual.
Astronomy
PF Gold
P: 22,809
 Originally posted by Symmetry I have a little question. I hope someone can help me. When we learn the theory of relativity and its formalism, we'll meet concepts : covariant and contravariant, such as covariant vector, covariant tensor... I wonder that why we need to use the concepts ? What are advantages of them ? I think that one of their advantages is to generalize the formalism of the special theory of relativity to the formalism of general theory. Is it true ? Thanks.
Hello symmetry, your sig says you are from SE Asia!
I am in n. california. I am about as confused as you about
the words covariant and contravariant. So I am laughing
at myself for trying to answer. But it will help me to try to answer.

Imagine a differential manifold, maybe like the surface of a potato.

Imagine a MAPPING f(x) = y from a little neighborhood around point x to a little neighborhood around point y.

If you have a vectorfield defined around the point y, say F(y') is always a vector in the tangent space at that point y' near y, then
by composing it with the mapping f(x) = y and using the chain rule
one can PULL BACK the vectorfield to make a vectorfield around x.

F(f(x)) is a vectorfield around x.

I am trying to remember lectures on differential geometry, some very nice ones, of some years back. I am not sure this is right.
But anyway both Tom and SA are around here and they know the subject.
So if I am making an error they will catch it.

Anyway that is why i think a vectorfield is CONTRAvariant----because it pulls BACK when you map it between x and y with
the mapping f(x) = y.

the words covariant and contravariant are confusing to me and I am happy to find someone else who calls himself Symmetry who is also confused. The confusion is covariant between us and symmetric. Be well.
 P: 56 Contravariant and covariant makes reference whether you are speaking on vector fields or differential forms. You can relate them by dualization, recall that dual maps are given by transposing of matrices. When you consider transformations these are the things that happen (see volume element, etc). Thus the index of covariancy and contravariancy on a tensor tells you how it is constructed as an element of the tensor product of spaces and dual spaces (e.g. in mixed tensors). This simplifies the presentation of metrics, line elements, usw.
Astronomy
PF Gold
P: 22,809
 Originally posted by rutwig Contravariant and covariant makes reference whether you are speaking on vector fields or differential forms. You can relate them by dualization, recall that dual maps are given by transposing of matrices. When you consider transformations these are the things that happen (see volume element, etc). Thus the index of covariancy and contravariancy on a tensor tells you how it is constructed as an element of the tensor product of spaces and dual spaces (e.g. in mixed tensors). This simplifies the presentation of metrics, line elements, usw.
I see. It seems to agree with what I was saying about the tangent space and its dual.

a vector field has values in the tangent space at each point in its domain
while a differential form has values in the dual or higher analogs of that (multilinear forms on the tangent space)

a mapping from a neighborhood of x to a neighborhood of y pulls BACK vectorfields on its range and makes them defined on its domain-----F(f(x))----so vectorfields transform CONTRAry to the direction of the mapping
but differential forms move with the direction of the mapping, one defined on the domain (the neighborhood of x) will by carried
by the mapping to the range (the neighborhood of y) also by composition----to find its value on a tangent vector at y, pull it back to x and evaluate it. So differential forms move COdirectionally, or along with, the direction of the mapping.

und so weiter

if you don't picture to yourself what a mapping y = f(x) does to these things, how do you remember what is "co" and what is "contra"?
but it pushes forward the other type of object
P: n/a
Symmetry wrote
 When we learn the theory of relativity and its formalism, we'll meet concepts : covariant and contravariant, such as covariant vector, covariant tensor... I wonder that why we need to use the concepts ? What are advantages of them?
Tensors are used in relativity because they allow the equations of physics to be represented in a coordinate independant manner. A tensor has both covariant and contravariant forms.

(For a nice little geometric description see -- www.geocities.com/physics_world/co_vs_contra.htm)

The two forms allow for the definition of the scalar product as you can see in the above link. This is true for both SR and GR.

Pmb
P: 124
 Originally posted by pmb Symmetry wrote Tensors are used in relativity because they allow the equations of physics to be represented in a coordinate independant manner. A tensor has both covariant and contravariant forms. (For a nice little geometric description see -- www.geocities.com/physics_world/co_vs_contra.htm) The two forms allow for the definition of the scalar product as you can see in the above link. This is true for both SR and GR. Pmb
Contra- and co-variant are falling out of favour in physicists language for non-Euclidean spaces - the preferred terms now being (tangent) vector and differential form respectively.

As an engineer used to working in Euclidean 3-space, I still tend to use the 'old' terms.

ron.
P: n/a
 Originally posted by rdt2 Contra- and co-variant are falling out of favour in physicists language for non-Euclidean spaces - the preferred terms now being (tangent) vector and differential form respectively. As an engineer used to working in Euclidean 3-space, I still tend to use the 'old' terms. ron.
I don't think it is a precisely true statement. Indeed covariant and contra-variant is a very generic term in math. If you have a mathematical object live on a set and consider a mapping from the set to another. You can use the mapping to "induce" an image object to another set. It turns out their are two ways of doing this. For some object natural way of doing it is the same direction as the mapping on the basis sets (Co-variant), for the others it's backward (contra-variant.)

In differential geometry the map usually used is diffeomorphism between two manifolds. You can show objects like vectors and forms are "push-forwarded" and "pulled-back" accordingly. One easy way to show this is to realize that 0-form is a function (which is a map from manifold to real) and the vector is a curve (map from real to manifold). This direction difference makes the differences when you combine with diffeo between manifold.

Instanton
P: n/a
 Originally posted by rdt2 Contra- and co-variant are falling out of favour in physicists language for non-Euclidean spaces - the preferred terms now being (tangent) vector and differential form respectively. As an engineer used to working in Euclidean 3-space, I still tend to use the 'old' terms. ron.
I disagree. While the what one might call the "purley geometrical" approach is finding more use, there are still plenty of physicists who used the more analytical approach. And even in the purely geometrical approach one does not find a decrease in the use of both concepts. In fact it's not possible to do much without them. Although there is a tendancy to now refer to a covariant vector as a 1-form and a contravariant vector as simply "vector."

Pete
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,904 An engineer working in "Euclidean 3-space" or any Euclidean space would never have to worry about "covariant" and "contravariant". As long as you use Cartesian coordinates (axes straight lines and perpendicular to each other) there is no difference. Try this: Let x and y be Cartesian coordinate axes, let the x' axis be the same as the x axis and the y' axis the line at angle [theta] with the x (x') axis. If you measure the "components" a point by drawing lines from the point parallel to each axis to the other axis. That will give "covariant" components. If instead you drop perpendiculars from from the point to the axes and measure along the axes to (0,0), you get the "covariant" components. There is no such thing as a "varient". The terms "covariant" and "contravariant" mean that the components "vary" (change) when we change coordinate systems in the same way as the unit vectors along the axes (that's "co" variant) or oppositely (that's "contra" variant).
P: 124
 Originally posted by pmb I disagree....Although there is a tendancy to now refer to a covariant vector as a 1-form and a contravariant vector as simply "vector." Pete
Is that not what I just said?

ron.
P: 124
 Originally posted by HallsofIvy An engineer working in "Euclidean 3-space" or any Euclidean space would never have to worry about "covariant" and "contravariant". As long as you use Cartesian coordinates (axes straight lines and perpendicular to each other) there is no difference.
As I said, used to but not always. There is also the matter of whether a quantity is 'more naturally' a vector or a differential form. This is crucial to moving from E^3 to other spaces - including non-metric spaces - which are commonplace in engineering, although they are seldom identified as such.

ron.
P: n/a
 Originally posted by rdt2 Is that not what I just said? ron.
No. (I thought I responded to this but it seems like it got lost or it didn't get posted).

When I said there is a tendancy to use that terminology I was agreeing with the notion of an increased use. But the purpose of my comment was to disagree with your assertion that its falling out of favour. E.g. two of my favorite new texts on GR use the analytical notation (One is by Francis Low and the other by Wolfgang Rindler).

Pmb

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