Covariant and Contravariant Tensors

In summary: So the idea is that if you have two contravariant tensors $T_1$ and $T_2$, then $T_1$ is related to $T_2$ in a way that is not symmetrical with respect to the vector $x$.
  • #1
benk99nenm312
302
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
Ok I think I'm starting to understand this, thanks guys. So what would be some typical examples of contravariant and covariant vectors?
 
  • #5
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).
 
  • #6
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..
 
  • #7
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.
 
  • #8
benk99nenm312 said:
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).
 
  • #9
atyy said:
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.
 
  • #10
WannabeNewton said:
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
 
  • #11
  • #12
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.
 
  • #13
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?
 
  • #14
benk99nenm312 said:
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.
 
Last edited:
  • #15
atyy said:
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 :)
 
  • #16
WannabeNewton said:
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?
 
  • #17
Gadhav said:
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.

Gadhav said:
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.

Gadhav said:
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.
 
  • #18
Fredrik said:
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.
 
Last edited:
  • #19
Gadhav said:
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).
I don't understand what you have in mind here. Are you talking about the two coordinate systems that are involved in the "transformation" of a tensor? You could call them something like x and y, or x and x'.

There are infinitely many coordinate systems on all the interesting manifolds, but there aren't two kinds of coordinate systems associated with the terms "covariant" and "contravariant". I'm not saying that it doesn't matter if you call a coordinate system covariant or contravariant. I'm saying that those terms don't apply to coordinate systems. It doesn't make sense to call a coordinate system covariant or contravariant.

Gadhav said:
These terms are confusing since some books insist that Contravariant component is parallel to axes and Covariant component is perpendicular.
I don't think I've seen this claim, and I don't understand it.

Gadhav said:
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 don't understand what you're saying here. Algebra defines "vector" as "member of a vector space". In differential geometry, "vector" is sometimes a lazy term for "tangent vector" or "tangent vector field", meaning "member of the tangent space at a point of the manifold" and "function that takes points in the manifold to tangent vectors at that point" respectively. "Vector" can also refer to an association of an n-tuple of real numbers with each coordinate system, such that the relationship between the tuples associated with any two coordinate systems is described by the tensor transformation law.
 
  • #20
Gadhav said:
These terms are confusing since some books insist that Contravariant component is parallel to axes and Covariant component is perpendicular.

Fredrik said:
I don't think I've seen this claim, and I don't understand it.

He's probably talking about the ability to identify forms and tangent vectors in the presence of a metric (lowering and raising indices). Something like http://www.mathpages.com/rr/s5-02/5-02.htm "As can be seen, the jth contravariant component consists of the projection of P onto the jth axis parallel to the other axis, whereas the jth covariant component consists of the projection of P into the jth axis perpendicular to that axis."

That's a different approach from what you've taken where you define forms and vectors first without a metric. Then only with a metric are we allowed to identify forms and vectors. In this approach, the metric always exists, which is an assumption of classical GR.
 
Last edited:

1. What is the difference between covariant and contravariant tensors?

Covariant and contravariant tensors are two types of tensors used in differential geometry and tensor calculus. The main difference between them lies in how they transform under a change of coordinates. Covariant tensors transform in the same way as the coordinate system, while contravariant tensors transform in the opposite way.

2. How are covariant and contravariant tensors related?

Covariant and contravariant tensors are related through the metric tensor. The metric tensor is used to raise and lower indices on tensors, converting them from one type to the other. This allows for the manipulation and use of both types of tensors in mathematical equations.

3. What is the significance of covariant and contravariant tensors in physics?

Covariant and contravariant tensors play a crucial role in the field of physics, particularly in the theory of relativity. They are used to describe the geometric properties of spacetime and the laws of physics in a coordinate-independent manner. They also aid in the formulation of equations that are invariant under coordinate transformations.

4. How are covariant and contravariant tensors used in practical applications?

Covariant and contravariant tensors are used in various fields, such as engineering, computer graphics, and image processing. They are used to represent physical quantities, such as stress and strain in solid mechanics, or color and intensity in image processing. They also have applications in machine learning and data analysis.

5. Are there any limitations to the use of covariant and contravariant tensors?

While covariant and contravariant tensors have numerous applications and are widely used in physics and engineering, they are limited to working in a manifold with a fixed number of dimensions. This means that they may not be applicable in some mathematical models or systems with varying dimensions. Additionally, they may become cumbersome to work with in higher dimensions due to the need for multiple indices.

Similar threads

  • Special and General Relativity
Replies
5
Views
1K
Replies
5
Views
1K
  • Special and General Relativity
Replies
28
Views
3K
Replies
1
Views
873
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
7
Views
1K
  • Special and General Relativity
Replies
15
Views
1K
  • Special and General Relativity
5
Replies
151
Views
22K
  • Differential Geometry
Replies
13
Views
2K
  • Differential Geometry
Replies
21
Views
16K
Back
Top