Covariant and Contravariant Tensors


by benk99nenm312
Tags: contravariant, covariant, tensors
benk99nenm312
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#1
Jun15-11, 11:28 AM
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Hey everyone, I am reading a Schaum's Outline on Tensor Calculus and came to something I can't seem to understand. I'm admittedly young to be reading this but so far I've understood everything except this. My question is: what is the difference between a contravariant tensor and a covariant tensor, and what do these terms mean conceptually? I've gone online and searched a variety of articles with no luck. I appreciate it, thanks in advance.
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dx
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Jun15-11, 11:45 AM
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First, there are covariant and contravariant vectors. A multilinear function acting on covariant vectors is a contravariant tensor. A multilinear function acting on contravariant vectors is a covariant tensor. A multilinear function which acts on both is a mixed tensor.
Rasalhague
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Jun15-11, 12:07 PM
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Adding to what dx has said, a particular system/structure of tensors is defined with respect to a particular vector space. Vectors of this vector space are called contravariant vectors. Vectors of its dual space (i.e. scalar-valued linear functions of one vector each) are called covariant vectors (or covectors, dual vectors, linear functionals, etc.). As such, covariant vectors are the simplest kind of covariant tensor.

Similarly, contravariant vectors can be thought of as scalar-valued linear functions of one covariant vector each, with the following definition: If w is a covariant vector, and v a contravariant vector, then v(w) is defined as w(v). Thus contravariant vectors (often called simply "vectors") are the simplest kind of contravariant tensor.

benk99nenm312
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#4
Jun15-11, 07:36 PM
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Covariant and Contravariant Tensors


Ok I think I'm starting to understand this, thanks guys. So what would be some typical examples of contravariant and covariant vectors?
WannabeNewton
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#5
Jun15-11, 07:43 PM
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The gradient would be a pretty good example of a covariant vector (although if you know the formalism of differential forms you could just use the more modern definition of a covariant vector as a one - form which actually makes more sense intuitively) and there are countless examples of contravariant vectors (or simply vectors) but in keeping with the previous example, one example of a contravariant vector is the directional derivative along a curve. Evaluated at some point p, this type of vector can form a basis for the tangent vector space that Rasalhague talked about. The gradient then, at that point, can form a basis for the cotangent space (this is all to some manifold).
benk99nenm312
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Jun15-11, 07:49 PM
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Ok I can visualize that without trouble, thanks WannabeNewton :)

Yeah as far as I can see this book doesn't cover the modern definition of the one-form, though I do believe I've heard of that, maybe that'd clear things up a bit too..
atyy
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Jun15-11, 07:49 PM
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I never know which is which, but you can think of contravariant and covariant vectors as row and column vectors. By matrix multiplication, a row vector can be thought of as a function that takes column vectors as input and produces a number as an output.
WannabeNewton
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Jun15-11, 07:57 PM
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Quote Quote by benk99nenm312 View Post
Ok I can visualize that without trouble, thanks WannabeNewton :)

Yeah as far as I can see this book doesn't cover the modern definition of the one-form, though I do believe I've heard of that, maybe that'd clear things up a bit too..
Actually it does with detail at the level of a typical Schaum's Outline book so its a good introduction to it however brief, go to the very last chapter titled Tensor Fields on Manifolds. It pretty much talks about everything I just said and then some (actually a lot more than some =D).
benk99nenm312
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Jun15-11, 09:18 PM
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Quote Quote by atyy View Post
I never know which is which, but you can think of contravariant and covariant vectors as row and column vectors. By matrix multiplication, a row vector can be thought of as a function that takes column vectors as input and produces a number as an output.
In this book it shows the contravariant with upper indices and the covariant with lower indices, so that helps too thanks.
benk99nenm312
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Jun15-11, 09:18 PM
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Quote Quote by WannabeNewton View Post
Actually it does with detail at the level of a typical Schaum's Outline book so its a good introduction to it however brief, go to the very last chapter titled Tensor Fields on Manifolds. It pretty much talks about everything I just said and then some (actually a lot more than some =D).
Oh awesome! Lol I'm only in chapter 3 :P
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Jun15-11, 09:53 PM
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This post might be useful.
PhilDSP
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Jun16-11, 02:35 AM
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The terminology does seem deceptive. The principle of covariance is that 2 different views or perspectives of the same phenomenon should be symmetric or inversely related to one another. Ideally the term covariant tensor would have been used for the pair of tensors which are together covariant concerning a particular phenomenon.
benk99nenm312
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#13
Jun16-11, 11:31 PM
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Thanks Fredrik, yeah so many terms to understand.. that's great to have them all lined up in a single post :)

I see, so if covariance is symmetrical or inversely symmetrical, what would contravariance be?
atyy
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Jun16-11, 11:52 PM
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Quote Quote by benk99nenm312 View Post
I see, so if covariance is symmetrical or inversely symmetrical, what would contravariance be?
That's a very tangentially related use of the term. Just use the definitions in the book you are reading.

Or if we are allowed to change definitions mid-sentence, both covariant and contravariant vectors transform covariantly:P

OK, to be more serious, let's imagine we have a 2D surface covered by coordinates (x,y). Imagine that each point has a different temperature f(x,y). A vehicle, carrying a clock which reads time t moves across the surface making a curve (x(t),y(t)). The variation in temperature versus time that the vehicle experiences is df/dt which is just one dimensional calculus. At a point p, df/dt=df/dx.dx/dt+df/dy.dy.dt, a scalar which we can rewrite as a a row vector (df/dx,df/dy) multiplied by a column vector (dx/dt,dy/dt), with all derivatives evaluated at p. The column vector or contravariant vector is something like the velocity, and the row vector or covariant vector is something like the gradient of the temperature.

We don't conceive of the velocity at that point as belonging to only one curve, since many curves can have the same velocity at that point. We also conceive of the velocity of any particular curve being the same under a change of coordinates from (x,y) to (U(x,y),V(x,y)). The same temperature variation is now described by f(U,V), and the same path is now described by (U(t),V(t)). So the new column vector representing the same velocity will be (dU/dt,dV/dt)=(dU/dx.dx/dt+dU/dy.dy/dt,dV/dx.dx.dt+dV/dy.dy/dt), which is how the coordinate representation of a contravariant vector transforms. Similarly, the new row vector representing the same gradient will be (df/dU,df/dV)=(df/dx.dx/dU+df/dy.dy/dU,df/dx.dx/dV+df/dy.dy/dV), which is how the coordinate representation of a covariant vector transforms. df/dt=df/dU.dU/dt+df/dV.dV/dt remains unchanged.

Something to keep in mind for later, when the metric is introduced: at this stage, we can multiply row and column vectors, but we haven't defined what it means to "multiply" column vectors, which is a job the metric can do.
benk99nenm312
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Jun17-11, 03:49 PM
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Quote Quote by atyy View Post
That's a very tangentially related use of the term. Just use the definitions in the book you are reading.

Or if we are allowed to change definitions mid-sentence, both covariant and contravariant vectors transform covariantly:P

OK, to be more serious, let's imagine we have a 2D surface covered by coordinates (x,y). Imagine that each point has a different temperature f(x,y). A vehicle, carrying a clock which reads time t moves across the surface making a curve (x(t),y(t)). The variation in temperature versus time that the vehicle experiences is df/dt which is just one dimensional calculus. At a point p, df/dt=df/dx.dx/dt+df/dy.dy.dt, a scalar which we can rewrite as a a row vector (df/dx,df/dy) multiplied by a column vector (dx/dt,dy/dt), with all derivatives evaluated at p. The column vector or contravariant vector is something like the velocity, and the row vector or covariant vector is something like the gradient of the temperature.

We don't conceive of the velocity at that point as belonging to only one curve, since many curves can have the same velocity at that point. We also conceive of the velocity of any particular curve being the same under a change of coordinates from (x,y) to (U(x,y),V(x,y)). The same temperature variation is now described by f(U,V), and the same path is now described by (U(t),V(t)). So the new column vector representing the same velocity will be (dU/dt,dV/dt)=(dU/dx.dx/dt+dU/dy.dy/dt,dV/dx.dx.dt+dV/dy.dy/dt), which is how the coordinate representation of a contravariant vector transforms. Similarly, the new row vector representing the same gradient will be (df/dU,df/dV)=(df/dx.dx/dU+df/dy.dy/dU,df/dx.dx/dV+df/dy.dy/dV), which is how the coordinate representation of a covariant vector transforms. df/dt=df/dU.dU/dt+df/dV.dV/dt remains unchanged.

Something to keep in mind for later, when the metric is introduced: at this stage, we can multiply row and column vectors, but we haven't defined what it means to "multiply" column vectors, which is a job the metric can do.
This makes incredible sense! Thanks so much, that really cleared things up :)
Gadhav
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#16
Dec31-11, 11:08 AM
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Quote Quote by WannabeNewton View Post
The gradient would be a pretty good example of a covariant vector (although if you know the formalism of differential forms you could just use the more modern definition of a covariant vector as a one - form which actually makes more sense intuitively) and there are countless examples of contravariant vectors (or simply vectors) but in keeping with the previous example, one example of a contravariant vector is the directional derivative along a curve. Evaluated at some point p, this type of vector can form a basis for the tangent vector space that Rasalhague talked about. The gradient then, at that point, can form a basis for the cotangent space (this is all to some manifold).

1) I understand the diff between co variant and contra variant coordinate system but not sure why velocity is considered contra where as grad is considered "co"variant. What does that have to do with how we lay out coordinate system?

2) Why do we always write covariant basis for contra variant components?
Fredrik
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Dec31-11, 05:29 PM
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Quote Quote by Gadhav View Post
1) I understand the diff between co variant and contra variant coordinate system
A coordinate system is just a function that assigns n-tuples of real numbers ("coordinates") to points in the manifold. They aren't covariant or contravariant.

Quote Quote by Gadhav View Post
but not sure why velocity is considered contra where as grad is considered "co"variant. What does that have to do with how we lay out coordinate system?
It doesn't matter which coordinate system is used. Velocity is defined as a tangent vector to a curve. The gradient of ##\phi## has components ##\partial_i\phi##. Those components transform covariantly (the same way as the basis vectors) because the partial derivative operators ##\partial_i## are the basis vectors for the tangent space at the relevant point. To understand this better, I recommend that you read the post I linked to above, and the three posts I linked to in that one, in particular the first one.

Quote Quote by Gadhav View Post
2) Why do we always write covariant basis for contra variant components?
The convention to write cotangent vectors as ##\omega_i e^i## and tangent vectors as ##v^i e_i## is just that, a convention.
Gadhav
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Jan1-12, 10:45 PM
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Quote Quote by Fredrik View Post
A coordinate system is just a function that assigns n-tuples of real numbers ("coordinates") to points in the manifold. They aren't covariant or contravariant.


It doesn't matter which coordinate system is used. Velocity is defined as a tangent vector to a curve. The gradient of ##\phi## has components ##\partial_i\phi##. Those components transform covariantly (the same way as the basis vectors) because the partial derivative operators ##\partial_i## are the basis vectors for the tangent space at the relevant point. To understand this better, I recommend that you read the post I linked to above, and the three posts I linked to in that one, in particular the first one.


The convention to write cotangent vectors as ##\omega_i e^i## and tangent vectors as ##v^i e_i## is just that, a convention.
Thanks Fredrik.

I thought over it and it now makes sense to me that it does not matter whether they are called covariant or contravariant. I think they should probably be called system(1) and system(2).

These terms are confusing since some books insist that Contravariant component is parallel to axes and Covariant component is perpendicular. In reality vector is simply some of three independent vectors(basis). That makes the whole thing confusing as you wonder why should one of them specifically relate to gradient and other is related to velocity. I think I was missing the point that we can use either. Conversion depends on what is in the denominator when we take differential.

I also happen to come across a tensor book by Sokolnikoff which I think is the best book on tensor that I have read. I strongly suggest that everyone read that book to get the fog of these terms out of your mind. Especially first 150+ pages explain it well. It has a very logical path to go from here to SR/GR but I have yet to read that part. (Search google as "sokolnikoff tensor pdf")

BTW it also mentions on #125 why contravariant components usually have covariant basis. It may be convention but it says that the reason is primarily due to fact that change of coordinate for these basis vary as gradient.


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