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Natural Science Forum / Physics / Relativity / March 2005Mixed tensor?
Can anyone explain to me what is meant by the covariant and contravariant indices of a mixed tensor? In particular, I can't understand why it is that the covariant index transforms with the basis and the contravariant index transforms against the basis. Also, I am aware that the contravariant index is in the dual vector space and the covariant index is in the real vector space - but what does this mean geometrically? I don't get the meaning. > Can anyone explain to me what is meant by the covariant and contravariant > indices of a mixed tensor? [quoted text clipped - 6 lines] > and the covariant index is in the real vector space - but what does this > mean geometrically? I don't get the meaning. There are lots of ways of looking at it. Random ramblings follow. The elements of the vector space are ordinary vectors, and the elements of the dual space are linear functionals, i.e. functions from vectors to scalars. You can think of the linear functionals as N-1 dimensional hyperplanes in the N-dimensional vector space which don't intersect the origin. Vectors which end on the hyperplane are mapped to 1 by the linear functional, and everything else follows from that and linearity. The surfaces at which the linear functional adopts integer values are evenly spaced parallel hyperplanes in the vector space. The only linear functional that this picture doesn't work for is the zero functional, which corresponds to the zero vector. You can also turn it around and think of the vectors as linear functions from linear functionals to scalars, with V(f) = f(V). At least in finite-dimensional spaces, there's a one-to-one correspondence between linear functionals and vectors: the linear functional corresponding to V is the function that takes any vector W to the inner product <V,W>. For this to make sense you need a notion of inner product, which is what the metric tensor gives you. This is why the metric tensor lets you raise and lower indices. The coordinates in which the two kinds of vectors are represented are in some sense different, even when the coordinate system is the same. Try drawing skewed coordinate axes on a piece of paper, e.g. / - / - -|--|--|--/--|--|--|- - / - / and notice that there are two natural ways of referring a point in the plane to these axes: you can think of the axes as forming part of a skewed grid, or you can drop perpendiculars to the axes. Those correspond to the two kinds of tensor index. A square matrix is like a tensor of rank (1,1). C=AB in matrix notation is like C^a_c = A^a_b B^b_c in tensor notation (with the Einstein summation convention). You can think of the vectors as column-vectors (single-column matrices) and the dual vectors as row-vectors. They transform in opposite ways because the column-vector has to be left-multiplied by the transformation matrix, while the row-vector has to be right-multiplied (otherwise the inner dimensions don't match). The two transformation matrices have to be inverses of each other because otherwise the scalar product of a row- and column-vector wouldn't be invariant under a coordinate transformation. -- Ben Thanks for your explaination, which was very useful. However, as usual with these things, it raises more questions than it answers!: > Vectors which end on the hyperplane are mapped to 1 by the linear > functional, and everything else follows from that and linearity. Can you give more info about this? I have drawn the situation you describe, with a vector touching one of the planes. So if it meets the plane the result is 1? What is the result if the vector fails to meet the plane or goes right through it, stopping at a point on the other side of the plane? Is this how tensors are used to describe distance/curvature at a point? - or am I barking up the wrong tree here? > You can also turn it around and think of the vectors as linear functions > from linear functionals to scalars, with V(f) = f(V). In what sense is this turning the previous description around? > At least in finite-dimensional spaces, there's a one-to-one correspondence > between linear functionals and vectors: the linear functional > corresponding to V is the function that takes any vector W to the inner > product <V,W>. > For this to make sense you need a notion of inner product, which is what > the metric tensor gives you. As I understand it, the inner product multiplies vectors together (it is a general version of the dot product)? How is the metric tensor related to the inner product? >This is why the metric tensor lets you raise and lower indices. I don't understand that bit either! - I know it does, but I still don't appreciate why. The functional u describe there, I write as: f: W -> <V,W> so this is a function that maps a vector to the inner product of the vector and another vector? (there is another vector V, given as a parameter of the inner product - what does this other vector do?) How does the metric tensor fit in to this? > The coordinates in which the two kinds of vectors are represented are in > some sense different, even when the coordinate system is the same. Try [quoted text clipped - 14 lines] > grid, or you can drop perpendiculars to the axes. Those correspond to the > two kinds of tensor index. Which is which? > A square matrix is like a tensor of rank (1,1). C=AB in matrix notation is > like C^a_c = A^a_b B^b_c in tensor notation (with the Einstein summation > convention). Is that a dot product between A^a_b and B^b_c ? > You can think of the vectors as column-vectors (single-column matrices) > and the dual vectors as row-vectors. They transform in opposite ways [quoted text clipped - 3 lines] > inverses of each other because otherwise the scalar product of a row- and > column-vector wouldn't be invariant under a coordinate transformation. You mean this?: A=B^-1 and B=A^-1 > -- Ben >This is why the metric tensor lets you raise and lower indices. I don't understand that bit either! - I know it does, but I still don't appreciate why. And an even better question would be - why would I want to raise and lower indices? What exactly does it achieve? > >This is why the metric tensor lets you raise and lower indices. > [quoted text clipped - 3 lines] > And an even better question would be - why would I want to raise and lower > indices? What exactly does it achieve? Look up basis vectors. In an orthogonal CS, (Coordinate System), covariant and contravariant basis vectors e_i , e^i are the same, where i = 1,2,3 and 0 if time is included. In the circumstance of a g-field the spacetime field requires a "nonorthogonal" CS, and e_i =/= e^i generally. Some arbituary vector "V" can be expressed (I'll place ">" to show the vectors), V> = e>_i V^i = e>^j V_j with V being a scalar invariant that has no units, V^2 = V>.V> (scalar product), and V^i = g^ij V_j allows a conversion between the components, which is desirable in Nonorthogonal CS's that are used in g-fields. Regards Ken S. Tucker *Shameless butting in* One of these days I'd like to hold a competition to see which linear algebra guru can speak in the most incomprehensibly thick jargon possible and spontaneously start a new religion in doing so.  SignatureDavid Cross dcross1 AT shaw DOT ca >*Shameless butting in* > >One of these days I'd like to hold a competition to see which linear algebra >guru can speak in the most incomprehensibly thick jargon possible and >spontaneously start a new religion in doing so. PS. I meant to put a big ;) at the end of that. --- David Cross dcross1 AT shaw DOT ca Hi Mr. Cross > >*Shameless butting in* > > [quoted text clipped - 7 lines] > David Cross > dcross1 AT shaw DOT ca That's the trick, here's the challenge... 1) Take a blank piece of paper draw BIG Happy face :-). 2) Using ordinary graph paper trace the Happy face. 3) Using polar graph paper trace the Happy face again. The Happy face (1), (2) and (3) are all the same, the lines of the Happy face are invariant :-), but the equations for the lines using graph(2) and graph(3) are different because they use different graph paper aka different CS's (Coordinate Systems). The trick is to take the measurements of trace(1) and convert them to the measuremnts of trace(2) so the Happy face remains the same i.e. invariant. Certainly you've converted 1 inch = 2.54 cm, well the math of tensor analysis seeks to do that in a more general way, it saves buying graph paper and helps the environment ;). Genious mathematicians have managed to boil that down to quite simple procedures, really without jargon. When physicists discover an invariant they get a :), but if they can't they get :( a sad face. Quantum physicists have to many :( and want more :). Regards Ken S. Tucker :) Anthony Smales says... >As I understand it, the inner product multiplies vectors together (it is a >general version of the dot product)? [quoted text clipped - 4 lines] >I don't understand that bit either! - I know it does, but I still don't >appreciate why. As I explained in another post, the most straightforward type of vector is a velocity vector. If you have coordinates x,y,z, and you have an object moving through space (or spacetime, if you are doing things relativistically), then its velocity has components V^x = dx/dt, V^y = dy/dt, V^z = dz/dt. (When doing things relativistically, you typically use proper time tau instead of the time t). The most natural kind of product is *not* the product of two vectors, but the product of a vector with a covector (for example, the gradient of a scalar field, which is a function assigning a real number such as temperature to each point in space). If Phi is a scalar field, and V is a velocity vector for some object travelling through space, then the rate at which Phi changes for that object is dPhi/dt = Grad(Phi) . V. If you don't have a metric, then that is the *only* kind of product you can have. Without a metric, you can't multiply two vectors together, you can only multiply a vector with a covector. Now what does a metric do for you? It converts a vector into a corresponding covector. That allows you to multiply two vectors: To multiply vectors A and B, you first use the metric g to convert A into a covector, g(A), and then you multiply *that* by B. Converting a vector into a covector seems trivial if you are used to using Cartesian coordinates, but it isn't trivial if you use other types of coordinates. For example, look at how you compute speed from velocity: If V is a velocity vector, then you define the speed s to be (in Cartesian coordinates) s = square-root(V^x * V^x + V^y * V^y) (let's just consider two dimensions for simplicity). Since square-roots are messy to work with, let's look instead at s^2: s^2 = V^x * V^x + V^y * V^y The right-hand side looks like the product of vector V with itself. But the correct way to think of it is this: V^x * V^x + V^y * V^y = g(V) . V Rather than directly multiplying V by itself, you first convert V to the corresponding covector, and *then* multiply it. For Cartesian coordinates, that seems like a pointless thing to do, because the metric doesn't do anything in Cartesian coordinates, but now let's look at speed in *polar* coordinates (R, Theta): s^2 = R^2 (dTheta/dt)^2 + (dR/dt)^2 If you consider the "velocity vector" to be the vector with components V^R = dR/dt V^Theta = dTheta/dt then we can express s^2 as follows: s^2 = V_R V^R + V_Theta V^Theta where the covector components V_R and V_Theta are defined by V_R = dR/dt V_Theta = R^2 dTheta/dt That's all that lowering indices amounts to: Given a vector A with components A^i, the corresponding covector with components A_i is defined implicitly by the constraint that sum over i of A_i A^i = |A|^2 where |A|^2 is the squared magnitude of A. (and also by the constraint that A_i is a linear combination of the components A^i). The metric g is that linear function that converts the vector with components A^i to the covector with components A_i. -- Daryl McCullough Ithaca, NY Anthony Smales says... >Can anyone explain to me what is meant by the covariant and contravariant >indices of a mixed tensor? > >In particular, I can't understand why it is that the covariant index >transforms with the basis and the contravariant index transforms against the >basis. First you need to understand why vectors and covectors transform differently. The best example of a vector is a velocity vector, while the best example of a covector is a gradient. Let me illustrate with a simple example: Let TG be the vertical temperature gradient, which is defined to be (change in temperature)/(change in altitude). For definiteness, suppose that TG = .001 degree C/meter Let v be the vertical velocity of a rising balloon, which is defined to be (change in altitude)/(change in time). For definiteness, suppose that v = 2 meters/second. Then using these two values, we can compute the rate at which the temperature at the balloon rises. It's just the product of these two numbers: dT/dt = TG * v = .002 degrees/second Note that this quantity does not mention the coordinates used to measure altitude at all. If I had used feet instead of meters to measure altitude, I would have gotten the same answer dT/dt = .002 degrees/second. However, v and TG would certainly have changed. If v = 2 meters/second, then if I change to using feet, then I would have v = 6.5 feet/second So under the coordinate change x' = 3.25 x, we have v' = 3.25 v. The velocity varies in the same way as the coordinate change. Looking at the temperature gradient, I find something quite different: If TG = .001 degrees/meter, then if I change to using feet, then I would have TG = .0003077 degrees/feet So under the coordinate change x' = 3.25 x, we have TG' = TG/3.25. The temperature gradient varies in the opposite way as the coordinate change. But now when I multiply the two numbers together, I get dT/dt = v * TG = 6.5 * .000377 = .002 degrees/second This example was simplified by the fact that I only considered 1 dimension, namely altitude, and I only considered the absolute simplest type of coordinate transformation, namely scaling by a constant. But it illustrates the main point: although a temperature gradient and a velocity both are considered vectors, they vary in opposite ways under coordinate transformations. But the combination of a gradient acting on a vector is invariant---it does not change under coordinate transformations. -- Daryl McCullough Ithaca, NY |
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