From 'The Physics Library': http://www.orkut.co.in/Main#CommMsgs?cmm=5545399&tid=2494726829739691225
25/10/2006
RengaW
What are tensors?
I've written a short primer on tensors, here:
http://www.orkut.com/CommMsgs.aspx?cmm=
http://www.orkut.com/CommMsgs.aspx?cmm=
04/01/2007
RengaW
Follow-up in this thread:
Doubt about co-variant and contra-variant tensors
http://www.orkut.com/CommMsgs.aspx?cmm=
Doubt about co-variant and contra-variant tensors
http://www.orkut.com/CommMsgs.aspx?cmm=
15/07/2007
RengaW
Quite a bit more technical than the above, but a good read:
http://www.grc.nasa.gov/WWW/K-12/Number
(Link courtesy of srinivasa <http://www.orkut.com/Profile.aspx?uid=
http://www.grc.nasa.gov/WWW/K-12/Number
(Link courtesy of srinivasa <http://www.orkut.com/Profile.aspx?uid=
24/10/2006
RengaW
Glad you asked... tensors are cool.
Tensors are a generalization of vectors so, before I explain what they are, let me explain what scalars and vectors are.
Scalars:
Suppose you have a room and you want to describe the temperature at various points in it. First, you set up a coordinate system (x, y, and z axes) and then you take your thermometer and move it around, recording the temperature at various locations, like this:
T(0, 0, 0) = 20 degrees Celsius
T(0, 0, 1) = 20.2 degrees C
T(0, 1, 0) = 19.9 degrees C
and so on. Note that you only need a single number to specify the temperature at any one location.
Now, imagine that someone else decides to do the same thing but chooses a different set of x, y, and z axes. Obviously, the coordinates of the points will now be different than the coordinates of the same points in your coordinate system. However, the temperature measured at those points will be the same. In other words, the temperature depends on the location of points, but not on the coordinates of those points. A point may be described by different coordinates (because of different coordinate systems) but the point doesn't really care about that and neither does the temperature at that point.
When we have a function that depends on points, such as temperature, such that the value of that function is a single number (for each point) that does not depend on the coordinates of the points, that is, a single number for each point that does not depend on how the coordinates change when you change the coordinate system, we call that function a scalar.
Note that the function does not have to have the same value for all points. Each point can have its own function value. The important thing is that when you change the coordinate system in some way, thereby changing the coordinates of the points, the function values at each point remain the same.
Tensors are a generalization of vectors so, before I explain what they are, let me explain what scalars and vectors are.
Scalars:
Suppose you have a room and you want to describe the temperature at various points in it. First, you set up a coordinate system (x, y, and z axes) and then you take your thermometer and move it around, recording the temperature at various locations, like this:
T(0, 0, 0) = 20 degrees Celsius
T(0, 0, 1) = 20.2 degrees C
T(0, 1, 0) = 19.9 degrees C
and so on. Note that you only need a single number to specify the temperature at any one location.
Now, imagine that someone else decides to do the same thing but chooses a different set of x, y, and z axes. Obviously, the coordinates of the points will now be different than the coordinates of the same points in your coordinate system. However, the temperature measured at those points will be the same. In other words, the temperature depends on the location of points, but not on the coordinates of those points. A point may be described by different coordinates (because of different coordinate systems) but the point doesn't really care about that and neither does the temperature at that point.
When we have a function that depends on points, such as temperature, such that the value of that function is a single number (for each point) that does not depend on the coordinates of the points, that is, a single number for each point that does not depend on how the coordinates change when you change the coordinate system, we call that function a scalar.
Note that the function does not have to have the same value for all points. Each point can have its own function value. The important thing is that when you change the coordinate system in some way, thereby changing the coordinates of the points, the function values at each point remain the same.
24/10/2006
RengaW
Note also that the definition of a scalar depends on how you change the coordinate system. A function which is a scalar under, say, rotations of the coordinate system may not be a scalar under, say, an inversion of the coordinate system, or under a stretch of the coordinate system, and so on.
A scalar is also called an invariant.
Vectors:
Now imagine that you have a fluid inside a pipe, such as oil in an oil pipe, and you want to measure the velocity of the oil at each point inside the pipe. Once again, you build a coordinate system and, to each point, you assign a velocity. But, wait... velocity is a quantity that has both direction and magnitude. You can't simply say that the velocity of the oil at point (0, 0, 0) is 5 meters per second. You have to specify also in which direction the oil is flowing at that point.
It turns out (as you can convince yourself easily) that you need 3 numbers to specify completely the velocity of the oil at any given point, namely, the speed of the oil in its motion along each of the three coordinate axes. Thus, you can write:
vx(0, 0, 0) = 0 m/s
vy(0, 0, 0) = 3 m/s
vz(0, 0, 0) = 4 m/s
vx(0, 0, 1) = 2 m/s
vy(0, 0, 1) = 1 m/s
vz(0, 0, 1) = 0 m/s
and so on. Note that now you need 3 numbers for every point, as I said already. Compare this to the scalar case above, where you only needed 1 number per point.
Once again, imagine now that someone else does the same thing, but chooses a different coordinate system than the one you chose. The question arises as to whether or not the values of those 3 numbers per point change.
It turns out that, for velocity fields (which is what the distribution of velocities in a fluid is called), those numbers DO change, but in a very specific way. The numbers in one coordinate system are linearly related to the numbers in the other coordinate system.
A scalar is also called an invariant.
Vectors:
Now imagine that you have a fluid inside a pipe, such as oil in an oil pipe, and you want to measure the velocity of the oil at each point inside the pipe. Once again, you build a coordinate system and, to each point, you assign a velocity. But, wait... velocity is a quantity that has both direction and magnitude. You can't simply say that the velocity of the oil at point (0, 0, 0) is 5 meters per second. You have to specify also in which direction the oil is flowing at that point.
It turns out (as you can convince yourself easily) that you need 3 numbers to specify completely the velocity of the oil at any given point, namely, the speed of the oil in its motion along each of the three coordinate axes. Thus, you can write:
vx(0, 0, 0) = 0 m/s
vy(0, 0, 0) = 3 m/s
vz(0, 0, 0) = 4 m/s
vx(0, 0, 1) = 2 m/s
vy(0, 0, 1) = 1 m/s
vz(0, 0, 1) = 0 m/s
and so on. Note that now you need 3 numbers for every point, as I said already. Compare this to the scalar case above, where you only needed 1 number per point.
Once again, imagine now that someone else does the same thing, but chooses a different coordinate system than the one you chose. The question arises as to whether or not the values of those 3 numbers per point change.
It turns out that, for velocity fields (which is what the distribution of velocities in a fluid is called), those numbers DO change, but in a very specific way. The numbers in one coordinate system are linearly related to the numbers in the other coordinate system.
24/10/2006
RengaW
And that's true for each point:
vx'(P) = axx vx(P) + axy vy(P) + axz vz(P)
vy'(P) = ayx vx(P) + ayy vy(P) + ayz vz(P)
vz'(P) = azx vx(P) + azy vy(P) + azz vz(P)
In the above, P is an arbitrary point, (vx, vy, vz) are the components of the velocity at P, measured in one coordinate system, and (vx', vy', vz') are the components of the velocity at the same point P, measured in the other coordinate system.
Now, any set of fuctions of points which behave in that way, that is, whose function values at any given point, measured in one coordinate system, depend linearly on the corresponding values at the same point, but measured in another coordinate system, is called a vector function or vector field.
Note that, just as with scalars, a quantity is not a vector all by itself. A quantity is a vector only with respect to specific kinds of coordinate transformations (and, for all of those transformations which the quantity is a vector, the change in the quantity's values is linear). Thus, for example, velocity and the electric field are vectors under rotations of the coordinate system, but not under inversions of the coordinate system. The magnetic field and angular momentum, on the other hand, are vectors under both of those kinds of changes of coordinate system.
Note also that vectors need not be 3-dimensional quantities. You can define vectors in any number of dimensions. The same is true of scalar functions and of tensors. For example, in relativity, we define scalars such as the "interval between events", rest-mass, proper-time, and others, as well as 4-vectors. The coordinate systems in relativity have 4 axes, x, y, z, and one for the time, t. As a result, vectors in relativity have 4 components (hence the name 4-vector). For example, the spatial components of 4-momentum are the regular 3-dim components of the momentum vector and the time-component of the 4-momentum is energy.
vx'(P) = axx vx(P) + axy vy(P) + axz vz(P)
vy'(P) = ayx vx(P) + ayy vy(P) + ayz vz(P)
vz'(P) = azx vx(P) + azy vy(P) + azz vz(P)
In the above, P is an arbitrary point, (vx, vy, vz) are the components of the velocity at P, measured in one coordinate system, and (vx', vy', vz') are the components of the velocity at the same point P, measured in the other coordinate system.
Now, any set of fuctions of points which behave in that way, that is, whose function values at any given point, measured in one coordinate system, depend linearly on the corresponding values at the same point, but measured in another coordinate system, is called a vector function or vector field.
Note that, just as with scalars, a quantity is not a vector all by itself. A quantity is a vector only with respect to specific kinds of coordinate transformations (and, for all of those transformations which the quantity is a vector, the change in the quantity's values is linear). Thus, for example, velocity and the electric field are vectors under rotations of the coordinate system, but not under inversions of the coordinate system. The magnetic field and angular momentum, on the other hand, are vectors under both of those kinds of changes of coordinate system.
Note also that vectors need not be 3-dimensional quantities. You can define vectors in any number of dimensions. The same is true of scalar functions and of tensors. For example, in relativity, we define scalars such as the "interval between events", rest-mass, proper-time, and others, as well as 4-vectors. The coordinate systems in relativity have 4 axes, x, y, z, and one for the time, t. As a result, vectors in relativity have 4 components (hence the name 4-vector). For example, the spatial components of 4-momentum are the regular 3-dim components of the momentum vector and the time-component of the 4-momentum is energy.
24/10/2006
RengaW
Now, if you've heard of the Lorentz transformations (LTs), they are the equations that relate the coordinates of points in spacetime, measured in two different reference frames. In other words, the scalars, vectors, and tensors of relativity are scalars, vectors, and tensors under LTs.
One more comment about vectors. The linear equations for how the vector components change can be written more succintly in terms of a matrix equation:
v'(P) = A v(P)
where A is the matrix of those axx, axy, axz, etc coefficients.
Tensors:
Now, scalars have only 1 number per point and vectors have as many numbers per point as there are dimensions in the space you're considering (3-dim vectors have 3 components, 4-vectors have 4, and so on). Rather than refer to those vector components as vx, vy, and vz, and so on, we could refer to them using an index: vi, where i goes from 1 to 3 in 3 dimensions (1 to 4, or 0 to 3, in 4 dimensions), and so on for higher dimensions.
Thus, scalars have no indices and vectors have only 1 index. Can you imagine now a set of quantities that has 2 indices? Something like this: Tij, where i goes from 1 to 3 and j goes from 1 to 3 (in 3 dimensions). There would be 9 quantities in 3 dims rather than 3: T11 = Txx, T12 = Txy, and so on, all the way to Tzz = T33.
Is there such a quantity? Yes, pressure is an example. Pressure is force over area, but both force and area are vectors in 3 dimensions. The area vector may be a strange notion to you, but it works like this: imagine a little flat patch of area in space. It has a certain size and its perpendicular points in a certain direction. Thus, you can associate a vector to that little patch of area. Now, I'm claiming it's a vector, but I'd have to prove that it does behave like a vector. I won't do it here, so you'll have to trust me on that.
One more comment about vectors. The linear equations for how the vector components change can be written more succintly in terms of a matrix equation:
v'(P) = A v(P)
where A is the matrix of those axx, axy, axz, etc coefficients.
Tensors:
Now, scalars have only 1 number per point and vectors have as many numbers per point as there are dimensions in the space you're considering (3-dim vectors have 3 components, 4-vectors have 4, and so on). Rather than refer to those vector components as vx, vy, and vz, and so on, we could refer to them using an index: vi, where i goes from 1 to 3 in 3 dimensions (1 to 4, or 0 to 3, in 4 dimensions), and so on for higher dimensions.
Thus, scalars have no indices and vectors have only 1 index. Can you imagine now a set of quantities that has 2 indices? Something like this: Tij, where i goes from 1 to 3 and j goes from 1 to 3 (in 3 dimensions). There would be 9 quantities in 3 dims rather than 3: T11 = Txx, T12 = Txy, and so on, all the way to Tzz = T33.
Is there such a quantity? Yes, pressure is an example. Pressure is force over area, but both force and area are vectors in 3 dimensions. The area vector may be a strange notion to you, but it works like this: imagine a little flat patch of area in space. It has a certain size and its perpendicular points in a certain direction. Thus, you can associate a vector to that little patch of area. Now, I'm claiming it's a vector, but I'd have to prove that it does behave like a vector. I won't do it here, so you'll have to trust me on that.
24/10/2006
RengaW
Anyway, pressure is force divided by area. But how do you divide a vector (force) by another vector (area)? You don't! Instead, you write:
fx = pxx Ax + pxy Ay + pxz Az
fy = pyx Ax + pyy Ay + pyz Az
fz = pzx Ax + pzy Ay + pzz Az
And there we have a set of 9 quantities, the pxx, pxy, etc, which need 2 indices.
Now, how do these p's change when one changes the coordinate system? They change the same way vectors do, linearly. EACH index behaves as if it was the index of a vector (in fact, it is).
In matrix form, we then write
pij' = sum over k and m of (aik ajm pkm)
where, as before, the primed quantities are measured in one system while the unprimed are measured in another. Do not confuse the a's here with the A's above. The A's above are the 3-dim components of the area vector. The a's here and before are parameters describing how the coordinate system has changed (for example, they're the parameters of a Lorentz Transformation).
Quantities like the pressure (and moment of inertia, as you mentioned), which are sets of quantities which transform among themselves in a linear manner when the coordinate system changes, are called tensors or tensor fields.
You can have tensors with any number of indices and in any number of dimensions. For example, in general relativity there is a tensor called the Riemann tensor. It has 4 indices, each in 4 dimensions. There is also another, called the Ricci tensor, with 2 indices instead, but still in 4 dimensions.
And since scalars and vectors have similar properties as tensors but have no index and 1 index, respectively, we call scalars and vectors, respectively, tensors of rank 0 and tensors of rank 1. The rank of a tensor is the number of indices it carries.
And that's a short primer on tensors. There's more, but I realize that this is already a lot for most people here.
fx = pxx Ax + pxy Ay + pxz Az
fy = pyx Ax + pyy Ay + pyz Az
fz = pzx Ax + pzy Ay + pzz Az
And there we have a set of 9 quantities, the pxx, pxy, etc, which need 2 indices.
Now, how do these p's change when one changes the coordinate system? They change the same way vectors do, linearly. EACH index behaves as if it was the index of a vector (in fact, it is).
In matrix form, we then write
pij' = sum over k and m of (aik ajm pkm)
where, as before, the primed quantities are measured in one system while the unprimed are measured in another. Do not confuse the a's here with the A's above. The A's above are the 3-dim components of the area vector. The a's here and before are parameters describing how the coordinate system has changed (for example, they're the parameters of a Lorentz Transformation).
Quantities like the pressure (and moment of inertia, as you mentioned), which are sets of quantities which transform among themselves in a linear manner when the coordinate system changes, are called tensors or tensor fields.
You can have tensors with any number of indices and in any number of dimensions. For example, in general relativity there is a tensor called the Riemann tensor. It has 4 indices, each in 4 dimensions. There is also another, called the Ricci tensor, with 2 indices instead, but still in 4 dimensions.
And since scalars and vectors have similar properties as tensors but have no index and 1 index, respectively, we call scalars and vectors, respectively, tensors of rank 0 and tensors of rank 1. The rank of a tensor is the number of indices it carries.
And that's a short primer on tensors. There's more, but I realize that this is already a lot for most people here.
03/01/2007
Varun
Doubt about co-variant and contra-variant tensors
Hi,
Can anyone please explain the basic definition of and the difference between: co-variant and contra-variant tensors?
Regards
Varun Sakalkar
Can anyone please explain the basic definition of and the difference between: co-variant and contra-variant tensors?
Regards
Varun Sakalkar
03/01/2007
RengaW
First, I'd recommend reading these two threads, which explain what tensors are and what the metric tensor is, respectively:
http://www.orkut.com/CommMsgs.aspx?cmm=
http://www.orkut.com/CommMsgs.aspx?cmm=
It's easier to understand the difference if you look at the simplest non-invariant tensors, namely, vectors.
Consider a vector v. Under a transformation of the coordinate system, that is, when the components of the position vector r are replaced with r' = Ar, where A is a matrix, the components of the vector v are transformed the same way as those of r, that is, v' = Av. That's what characterizes v as a vector under the coordinate transformation in question.
Now, a scalar is, by definition, an invariant under the coordinate transformation in question. Consider the contraction of two vectors uand v, which in more elementary applications is more commonly known as the dot product: u⋅v. The contraction of two vectors is a scalar, so it must be invariant under the coordinate transformation in question, that is, we must have:
u'⋅v' = u⋅v
But that doesn't work, unless the components of v' do not transform according to v' = Av but according to v' = Bv with some other matrix B. To find out what this matrix must be, we plug in u' and v' into the contraction above:
u'⋅v' = (Au)⋅(Bv) = ∑i (∑j Aij uj) (∑k Bik vk) = ∑ijk (Aij Bik uj vk)
which equals u⋅v = ∑i ui vi only if ∑i Aij Bik = ∑i Aij B*ki = ∑i B*ki Aij = δkj, that is, only if B*A is the identity matrix. In the above, I'm representing the transpose of B by an asterisk (*).
http://www.orkut.com/CommMsgs.aspx?cmm=
http://www.orkut.com/CommMsgs.aspx?cmm=
It's easier to understand the difference if you look at the simplest non-invariant tensors, namely, vectors.
Consider a vector v. Under a transformation of the coordinate system, that is, when the components of the position vector r are replaced with r' = Ar, where A is a matrix, the components of the vector v are transformed the same way as those of r, that is, v' = Av. That's what characterizes v as a vector under the coordinate transformation in question.
Now, a scalar is, by definition, an invariant under the coordinate transformation in question. Consider the contraction of two vectors uand v, which in more elementary applications is more commonly known as the dot product: u⋅v. The contraction of two vectors is a scalar, so it must be invariant under the coordinate transformation in question, that is, we must have:
u'⋅v' = u⋅v
But that doesn't work, unless the components of v' do not transform according to v' = Av but according to v' = Bv with some other matrix B. To find out what this matrix must be, we plug in u' and v' into the contraction above:
u'⋅v' = (Au)⋅(Bv) = ∑i (∑j Aij uj) (∑k Bik vk) = ∑ijk (Aij Bik uj vk)
which equals u⋅v = ∑i ui vi only if ∑i Aij Bik = ∑i Aij B*ki = ∑i B*ki Aij = δkj, that is, only if B*A is the identity matrix. In the above, I'm representing the transpose of B by an asterisk (*).
03/01/2007
RengaW
So, it seems that in order to construct invariants by contraction, we need to define two kinds of components for a given vector. The contravariant components are those which transform according to u' = A u and the covariant components are those which transform according to the inverse of the transpose of A, that is, u' = (A*)^(-1) u.
Note that the vector is one and the same, because the vector itself does not change when the coordinate system changes; it's only the components that change, and they do so differently depending on whether you're talking about contravariant or covariant components.
Now, every vector has both kinds of components but some vectors are more naturally represented by one kind than the other. The position vector, for instance, is more naturally represented by its contravariant components. On the other hand, the components of the gradient operator are more naturally represented by its covariant components. Here's why.
Consider a scalar function Φ(r) of the coordinates. An infinitesimal change dΦ in its value due to changing its argument by an amount dr is also a scalar. But we know what dΦ is: dΦ = dr⋅∇Φ.
Since dr is more naturally represented by contravariant components, we see that ∇Φ is more naturally represented by covariant ones.
Now, all of this is irrelevant for spaces with Euclidean geometries and Cartesian coordinates, that is, in those cases the contravariant and covariant components are identical. In the case of special relativity, spacetime is a non-Euclidean 'space' where the metric (in Cartesian coordinates) has the simple form g00 = 1, gxx = gyy = gzz = -1. Because of the extra negative sign and because of the way the dot product of four vectors is defined, it follows that the temporal contravariant and covariant components are equal but corresponding spatial contravariant and covariant components differ by a negative sign.
Note that the vector is one and the same, because the vector itself does not change when the coordinate system changes; it's only the components that change, and they do so differently depending on whether you're talking about contravariant or covariant components.
Now, every vector has both kinds of components but some vectors are more naturally represented by one kind than the other. The position vector, for instance, is more naturally represented by its contravariant components. On the other hand, the components of the gradient operator are more naturally represented by its covariant components. Here's why.
Consider a scalar function Φ(r) of the coordinates. An infinitesimal change dΦ in its value due to changing its argument by an amount dr is also a scalar. But we know what dΦ is: dΦ = dr⋅∇Φ.
Since dr is more naturally represented by contravariant components, we see that ∇Φ is more naturally represented by covariant ones.
Now, all of this is irrelevant for spaces with Euclidean geometries and Cartesian coordinates, that is, in those cases the contravariant and covariant components are identical. In the case of special relativity, spacetime is a non-Euclidean 'space' where the metric (in Cartesian coordinates) has the simple form g00 = 1, gxx = gyy = gzz = -1. Because of the extra negative sign and because of the way the dot product of four vectors is defined, it follows that the temporal contravariant and covariant components are equal but corresponding spatial contravariant and covariant components differ by a negative sign.
03/01/2007
RengaW
In general, however, that is, in non-Euclidean spaces, or even in Euclidean spaces with non-Cartesian coordinate systems, the contravariant and covariant components are not respectively equal to one another.
Once you understand how the covariant and the contravariant components of a vector are defined, it's easy to extend that to tensors of higher ranks.
I recommend reading the first few sections of Landau's The Classical Theory of Fields, as well as its first few sections on general relativity, for an excellent explanation of these matters.
Once you understand how the covariant and the contravariant components of a vector are defined, it's easy to extend that to tensors of higher ranks.
I recommend reading the first few sections of Landau's The Classical Theory of Fields, as well as its first few sections on general relativity, for an excellent explanation of these matters.
04/01/2007
Varun
Thanks for the reply Mr. Wagner, and I understand it now, also will I be correct to conclude that all metric tensors can are covariant tensors? (did some math with my own nomenclature, and I am getting this result).
04/01/2007
RengaW
will I be correct to conclude that all metric tensors can are covariant tensors?
All tensors have contravariant and covariant components. It's up to you to decide which components to use at any particular moment. For example, when using the metric tensor, you can use its covariant components g_ij, its contravariant components g^ij, or even its mixed components, g^i_j or g_i^j.
As it happens, the mixed components of the metric tensor are nothing but the components of the Kronecker delta, that is, g^i_j = δ^i_j and also g_i^j = δ_i^j. But that's true only for the metric tensor. The mixed components of other tensors are not necessarily equal to those of the Kronecker delta.
Two other famous tensors, both of which appear in general relativity, are the Ricci tensor (with contravariant components R^ij, covariant components R_ij, and mixed components R^i_j and R_i^j) and the Riemman tensor (a tensor of rank 4 with contravariant components R^ijrs, covariant components R_ijrs, and many mixed components).
All tensors have contravariant and covariant components. It's up to you to decide which components to use at any particular moment. For example, when using the metric tensor, you can use its covariant components g_ij, its contravariant components g^ij, or even its mixed components, g^i_j or g_i^j.
As it happens, the mixed components of the metric tensor are nothing but the components of the Kronecker delta, that is, g^i_j = δ^i_j and also g_i^j = δ_i^j. But that's true only for the metric tensor. The mixed components of other tensors are not necessarily equal to those of the Kronecker delta.
Two other famous tensors, both of which appear in general relativity, are the Ricci tensor (with contravariant components R^ij, covariant components R_ij, and mixed components R^i_j and R_i^j) and the Riemman tensor (a tensor of rank 4 with contravariant components R^ijrs, covariant components R_ijrs, and many mixed components).
04/01/2007
Johann
While I (mostly) agree with everything Wagner said, I find it easier to think of covariant and contravariant tensors in terms of the underlying geometrical objects. Here, my thinking has been shaped by Schutz's Geometrical Methods of Mathematical Physicsand Misner, Thorne, and Wheeler's Gravitation.
For example, the gradient of a scalar field is naturally a one-form, which is a covariant vector, given by d phi = phi_i dx^i, using the normal Einstein sum convention. The velocity of a path through space is naturally contravariant, v = v^i d/dx^i. The combination of the two lets you evaluate a path integral.
Hm. I'm afraid I'm rambling a little. If you can find a copy of Schutz's book, I like it a lot. My basic point is that some things are in fact naturally covariant or contravariant. It's not just a matter of moving indices up and down. Gradients and densities are N-forms and hence covariant, as they are maps from displacements and volumes to numbers. Velocities are N-vectors and hence contravariant. Sure, once you have a metric, you can transform from one to the other, but it's good to know what you can do without invoking it.
Things get a little more interesting once you start thinking about Hamiltonian mechanics and conjugate momenta, but it's been a while since I thought my way through that, and I don't really want to get into it now.
For example, the gradient of a scalar field is naturally a one-form, which is a covariant vector, given by d phi = phi_i dx^i, using the normal Einstein sum convention. The velocity of a path through space is naturally contravariant, v = v^i d/dx^i. The combination of the two lets you evaluate a path integral.
Hm. I'm afraid I'm rambling a little. If you can find a copy of Schutz's book, I like it a lot. My basic point is that some things are in fact naturally covariant or contravariant. It's not just a matter of moving indices up and down. Gradients and densities are N-forms and hence covariant, as they are maps from displacements and volumes to numbers. Velocities are N-vectors and hence contravariant. Sure, once you have a metric, you can transform from one to the other, but it's good to know what you can do without invoking it.
Things get a little more interesting once you start thinking about Hamiltonian mechanics and conjugate momenta, but it's been a while since I thought my way through that, and I don't really want to get into it now.
04/01/2007
RengaW
As it happens, I have both Geometrical Methods of Mathematical Physics and Gravitation, and I like them both.
In the end, it's a personal preference, really, at least in my case. I prefer the "component view" I offered rather than the more "geometrical view", for the following reasons:
(a) when it's all said and done, at the end of the day, one does always revert to using components for (almost) all calculations involving tensors in modern physics. The only example I'm aware of where there is a clear advantage in using the more geometric view is in computing the components of the Riemann tensor by means of Cartan's formulation.
(b) although it's true that tensors and forms can be defined independently in more general spaces, without the need for a metric, the fact is that our universe does seem to have a metric nature, which then defines a natural duality between tensors and forms, which is how the contravariant and covariant components get paired up.
(c) in my opinion, it's easier for people starting in the field to learn the component approach before they learn the geometrical approach.
As for Hamiltonian mechanics, it's true that its symplectic nature lends itself to a natural and elegant geometric formulation but, once again, any computational work is done using its more traditional formulation.
In summary, it seems to me that all the machinery of differential geometry, although very elegant, is really not so fundamental from aphysical point of view to justify "turning physics into math".
Then, again, I could be wrong. Maybe it is a more fundamental description of nature, in which case I'll have to read those books again.
In the end, it's a personal preference, really, at least in my case. I prefer the "component view" I offered rather than the more "geometrical view", for the following reasons:
(a) when it's all said and done, at the end of the day, one does always revert to using components for (almost) all calculations involving tensors in modern physics. The only example I'm aware of where there is a clear advantage in using the more geometric view is in computing the components of the Riemann tensor by means of Cartan's formulation.
(b) although it's true that tensors and forms can be defined independently in more general spaces, without the need for a metric, the fact is that our universe does seem to have a metric nature, which then defines a natural duality between tensors and forms, which is how the contravariant and covariant components get paired up.
(c) in my opinion, it's easier for people starting in the field to learn the component approach before they learn the geometrical approach.
As for Hamiltonian mechanics, it's true that its symplectic nature lends itself to a natural and elegant geometric formulation but, once again, any computational work is done using its more traditional formulation.
In summary, it seems to me that all the machinery of differential geometry, although very elegant, is really not so fundamental from aphysical point of view to justify "turning physics into math".
Then, again, I could be wrong. Maybe it is a more fundamental description of nature, in which case I'll have to read those books again.
04/01/2007
Bilwaj
Hi
Well hey i am getting a bit confused here but weren't covariant and contravariant vectors classified directly based on how they transform under any transformation of a co-ordinate system.
So I could not really understand what u meant by saying the same vector has contravariant and covariant components??
A Tensor might have both but isn't a vector supposed to be either of the two?
So I could not really understand what u meant by saying the same vector has contravariant and covariant components??
A Tensor might have both but isn't a vector supposed to be either of the two
?04/01/2007
RengaW
No, the vectors aren't contravariant or covariant; it's their components which are one or the other. In other words, it's possible to define two sets of components for each vector. The same is true of higher-ranked tensors.
The contravariant components transform according to (u')^i = ∑j A^i_j u^j while the covariant ones transform according to (u')_i = ∑j A_i^j u_j. Note that A^i_j is not, in general, equal to A_i^j. In fact, the two matrices are the inverse transpose of each other.
And since there's a metric (defined by the contraction), there's a natural relation connecting the contravariant components to the covariant components: if the metric has covariant components g_ij, then u_i = ∑j g_ij u^j.
The contravariant components transform according to (u')^i = ∑j A^i_j u^j while the covariant ones transform according to (u')_i = ∑j A_i^j u_j. Note that A^i_j is not, in general, equal to A_i^j. In fact, the two matrices are the inverse transpose of each other.
And since there's a metric (defined by the contraction), there's a natural relation connecting the contravariant components to the covariant components: if the metric has covariant components g_ij, then u_i = ∑j g_ij u^j.
04/01/2007
Bilwaj
OK I got it now
Well yes so the same vector has two kind of components
and these two kinds of components transform differently under the
transformations but essentially represent the same object.
and these two kinds of components transform differently under the
transformations but essentially represent the same object.
Some replies on this page have been deleted or are under review.
I do not fear for the third link (one with 'link coursey of Srinivasa'). It is from NASA's educational resources.
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