Magic Tensor Algebra: Part 3 - Curvilinear Coordinates

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Introduction

Reading the reviews for my articles, I realized that I overloaded the reader with theoretical introductory ones. I apologize for this, to be honest, I myself am far from formal mathematics.

However, tensor calculus is replete with concepts, many of which need to be introduced formally. Therefore, the third cycle will also be devoted to the dry theory. Nevertheless, I promise that in the next work I will proceed to what I have long wanted - the description of the practical value of the tensor approach. There is an interesting task in mind, most of which is already dismantled in my head. Tensor calculus is not an idle interest for me, but a way to process some of my theoretical and practical considerations in the field of mechanics. So practice for the full program remains to be.
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For now we will consider some theoretical bases. Welcome under cat.

1. Jacobi matrix and local metric. "Juggling" indices

Those coordinate systems that we considered so far have been oblique. But their axes were straight lines. However, it is extremely often necessary to work in space, the coordinate lines of which are curves. This coordinate system is called curvilinear.

The simplest vital example of a curvilinear coordinate system is geographic coordinates. $(\ varphi, \ lambda, h)$ - latitude, longitude and height above sea level, by which the position of objects near the surface of the Earth is determined. Curvilinear coordinates are widely used in astronomy. In mechanics, the generalized coordinates of a mechanical system that uniquely determine its position in space, taking into account the geometry imposed on a system of relations, can serve as an example of such coordinates. This is the basis of analytical mechanics.

Fig. 1. Curvilinear coordinates in three-dimensional space

Consider the curvilinear coordinates defined in three-dimensional Euclidean space (Figure 1). Let the position of the point is set in these coordinates by the vector

$\ mathbf {q} = \ begin {bmatrix} q ^ 1 \\ q ^ 2 \\ q ^ 3 \ end {bmatrix} \ quad (1)$

and Cartesian coordinates of a point are related to (1) by

$\ vec {r} = \ vec {r} \ left (q ^ 1, q ^ 2, q ^ 3 \ right) \ quad (2)$

or, in component form

$x ^ i \ left (q ^ 1, q ^ 2, q ^ 3 \ right) \ quad (3)$

Consider the partial derivative $\ cfrac {\ partial \ vec {r}} {\ partial q ^ 1}$ . The result of this differentiation is a vector directed tangentially to the coordinate line. $q ^ 1$ . Differentiating (2) over all curvilinear coordinates we get the three vectors

$\ vec {e} _1 = \ cfrac {\ partial \ vec {r}} {\ partial q ^ 1}, \ quad \ vec {e} _2 = \ cfrac {\ partial \ vec {r}} {\ partial q ^ 2}, \ quad \ vec {e} _3 = \ cfrac {\ partial \ vec {r}} {\ partial q ^ 3} \ quad (4)$

These vectors define the basis of the so-called tangent space . And unlike the basis in an oblique coordinate system, the modulus and direction of these vectors will change when moving from one point to another. We obtain a variable basis depending on the position in the space defined by the vector (1). Such a basis is called local.

Vectors (4) are assembled into a matrix

$\ mathbf {J} = \ begin {bmatrix} \ cfrac {\ partial \ vec {r}} {\ partial q ^ 1} & & & amp; \ cfrac {\ partial \ vec {r}} {\ partial q ^ 2} & amp; & amp; \ cfrac {\ partial \ vec {r}} {\ partial q ^ 3} \ end {bmatrix} = \ begin {bmatrix} \ cfrac {\ partial x ^ 1} {\ partial q ^ 1} & &; \ cfrac {\ partial x ^ 1} {\ partial q ^ 2} & amp; & amp; \ cfrac {\ partial x ^ 1} {\ partial q ^ 3} \\ \ cfrac {\ partial x ^ 2} {\ partial q ^ 1} & & & amp; \ cfrac {\ partial x ^ 2} {\ partial q ^ 2} & amp; & amp; \ cfrac {\ partial x ^ 2} {\ partial q ^ 3} \\ \ cfrac {\ partial x ^ 3} {\ partial q ^ 1} & & & amp; \ cfrac {\ partial x ^ 3} {\ partial q ^ 2} & amp; & amp; \ cfrac {\ partial x ^ 3} {\ partial q ^ 3} \ end {bmatrix} \ quad (5)$

which is called the Jacobi matrix, and is essentially defined as a derivative of one vector with respect to another vector. In our case

$\ mathbf {J} = \ cfrac {\ partial \ vec {r}} {\ partial \ vec {q}}$

It is easy to guess that if function (2) is linear with respect to the components of the vector $\ vec {q}$ then it can be expressed by the matrix ratio

$\ vec {r} = \ mathbf {A} _ {01} \ vec {q}$

then we consider the oblique coordinate system, and the Jacobi matrix will be equal to the transformation matrix from oblique coordinates to Cartesian

$\ mathbf {J} = \ mathbf {A} _ {01}$

Now, any vector defined in space (a tensor of rank (1, 0)) can be represented through its contravariant components in a curvilinear coordinate system

$\ vec {a} = a ^ i \, \ vec {e} _i = a ^ i \, \ cfrac {\ partial \ vec {r}} {\ partial q ^ i} \ quad (6)$

However, the components of the vector, due to the variable basis, will depend on the position in space of the point of application of the vector. In addition, in order for the representation (6) to exist, it is necessary that the vectors that make up the basis are not coplanar. From the course of vector algebra, we know that vectors are non-coplanar if their mixed product is non-zero. Hence the condition that the determinant of the Jacobi matrix must satisfy.

$\ det (\ mathbf {J}) \ ne 0 \ quad (7)$

This determinant just determines the mixed product of the basis vectors.

Now we calculate the covariant components of the vector $\ vec {a}$ . For this, in the very first article of the cycle, we multiplied the vector scalarly by the corresponding basis vector

$\ vec {a} = a_i = \ vec {a} \ cdot \ vec {e} _1 = a ^ i \, (\ vec {e} _i \ cdot \ vec {e} _1) = a ^ i \, \ cfrac {\ partial \ vec {r}} {\ partial q ^ i} \ cdot \ cfrac {\ partial \ vec {r}} {\ partial q ^ i}$

In the same, first article, we determined that the covariant components of the vector are contravariant through the metric tensor $g_ {ij}$

$a_j = a ^ i g_ {ij}$

Comparing the last two expressions, we get the definition of the metric tensor in curvilinear coordinates

$g_ {ij} = \ cfrac {\ partial \ vec {r}} {\ partial q ^ i} \ cdot \ cfrac {\ partial \ vec {r}} {\ partial q ^ j}$

which can be represented in a matrix form

$\ mathbf {g} = \ mathbf {J} ^ T \, \ mathbf {J}$

This connection can also be represented in tensor form, but for this it is necessary to introduce explicitly the metric for Cartesian coordinates $\ hat {g} _ {kl}$

$\ hat {g} _ {kl} = \ begin {bmatrix} 1 & amp & amp; 0 & amp; & amp; 0 \\ 0 & amp; & amp; 1 & amp; & amp; 0 \\ 0 & amp; & amp; 0 & amp; & amp; 1 \ end {bmatrix}$

Then, the transformation of the Cartesian metric into curvilinear will look like this

$g_ {ij} = J_i ^ k \, J_j ^ p \, g_ {kl} \ quad (8)$

Expression (8) introduces the metric tensor for curvilinear coordinates. This tensor depends on the position of a point in space, so they say that it is defined locally or determines a local metric

Having defined the metric, we can write the rules for converting contravariant coordinates into covariant

$a_j = g_ {ij} \, a ^ i \ quad (9)$

and covariant coordinates in contravariant

$a ^ k = a_i g ^ {ik} \ quad (10)$

In the tensor calculus, lowering operations (9) and raising (10) indices are called “juggling” indices.

Writing relations (9) and (10), we implied that the matrices $g_ {ij}$ and $g_ {ij}$ mutually reversible. This is only possible if

$\ det | g_ {ij} | \ ne 0$

This condition is satisfied for curvilinear coordinates if the Jacobi matrix is not degenerate, and this directly follows from (8), since

$\ det | g_ {ij} | = \ det | J_i ^ k J_j ^ l g_ {kl} | = \ det | J_i ^ k | \, \ det | J_j ^ l | = J ^ 2 \ ne 0$

that is, condition (7) is satisfied for all points in space — a sufficient condition for the local metric to be nondegenerate.

Consideration of degenerate metrics is a separate and complex issue, so we confine ourselves to metrics in which the matrix of the metric tensor is reversible, that is, the condition

$g_ {ik} \, g ^ {kj} = \ delta_i ^ j$

Where

$\ delta_i ^ j = \ begin {cases} 1, \ quad i = j \\ 0, \ quad i \ ne j \ end {cases}$

unit tensor, called the Kronecker delta. It can be seen that its components are represented by a unit matrix with a dimension corresponding to the dimension of space.

2. Mutual basis

We introduce vectors $\ vec {e} ^ {\, 1}, \, \ vec {e} ^ {\, 2}, \, \ vec {e} ^ {\, 3}$ derived from the vectors of the original base by raising the index

$\ vec {e} ^ {\, j} = g ^ {ji} \, \ vec {e} _i \ quad (11)$

Now we take and multiply (11) by the scalar vector $\ vec {e} _k$

$\ vec {e} ^ {\, j} \ cdot \ vec {e} _k = g ^ {ji} \, \ vec {e} _i \ cdot \ vec {e} _k$

but, we know that $\ vec {e} _i \ cdot \ vec {e} _k = g_ {ik}$ - metric tensor, therefore, we arrive at the equation

$\ vec {e} ^ {\, j} \ cdot \ vec {e} _k = g ^ {ji} \, g_ {ik}$

$\ vec {e} ^ {\, j} \ cdot \ vec {e} _k = \ delta_k ^ j \ quad (12)$

If we take, for example, the vector $\ vec {e} ^ {\, 1}$ , then by virtue of (12) it is perpendicular to the vectors $\ vec {e} _2$ and $\ vec {e} _3$ (its scalar product with them is zero), and the scalar product of this vector on $\ vec {e} _1$ - equal to one

Next we take and multiply (11) scalarly by $\ vec {e} ^ {\, k}$

$\ vec {e} ^ {\, j} \ cdot \ vec {e} ^ {\, k} = g ^ {ji} \, \ vec {e} _i \ cdot \ vec {e} ^ {\, k }$

and by virtue of (12) this gives a contravariant metric tensor

$\ vec {e} ^ {\, j} \ cdot \ vec {e} ^ {\, k} = g ^ {jk} \ quad (13)$

Vector system $\ vec {e} ^ {\, 1}, \, \ vec {e} ^ {\, 2}, \, \ vec {e} ^ {\, 3}$ also forms a basis, which is called reciprocal or conjugate to the basis $\ vec {e} _ {\, 1}, \, \ vec {e} _ {\, 2}, \, \ vec {e} _ {\, 3}$ .

Consider the vector again. $\ vec {a}$ . From the relations (10) and (11) follows a chain of transformations

$\ vec {a} = a ^ i \, \ vec {e} _ {\, i} = a_j \, g ^ {ji} \, \ vec {e} _ {\, i} = a_j \, \ vec {e} ^ {\, j} \ quad (13)$

Multiply (13) scalarly by $\ vec {e} ^ {\, k}$

$\ vec {a} \ cdot \ vec {e} ^ {\, k} = a_j \, \ vec {e} ^ {\, j} \ cdot \ vec {e} ^ {\, k} = a_j \, g ^ {jk} = a ^ k$

we conclude that any vector can be decomposed as a basis $\ vec {e} _ {\, i}$ - then its components will be contravariant, and on the basis $\ vec {e} ^ {\, i}$ - components will be covariant

$\ vec {a} = a ^ i \, \ vec {e} _ {\, i} = a_i \, \ vec {e} ^ {\, i}$

At the same time, the covariant components are the scalar product of the vector and the basis vectors $\ vec {e} _ {\, i}$ and contravariant components are scalar products of a vector and a basis vector $\ vec {e} ^ {\, i}$

$a_i = \ vec {a} \ cdot \ vec {e} _ {\, i}, \ quad a ^ i = \ vec {a} \ cdot \ vec {e} ^ {\, i} \ quad (15)$

which once again illustrates the reciprocity of these bases.

It should be noted here that the basis vectors $\ vec {e} _ {\, i}$ are obtained in a natural way - they are tangent to the corresponding coordinate lines and they can be assigned a geometric meaning. As for the basis $\ vec {e} ^ {\, i}$ , its vectors are not directed along tangent coordinate lines, but perpendicular to pairs of tangent basis vectors. Such a basis is sometimes called nonholonomic.

3. Transform curvilinear coordinates. Formal definition of covariant and contravariant components

Suppose we work in a curvilinear coordinate system defined by a vector $\ vec {q}$ . Let's move to another coordinate system, the position of the points of which is determined by the vector $\ vec {p}$ such that the transformation from the old coordinate system to the new one is determined by the equations

$p ^ i = p ^ i (q ^ 1, \, q ^ 2, \, q ^ 3) \ quad (16)$

We assume that the transformation (16) is reversible, that is, let's say the existence of

$q ^ i = q ^ i (p ^ 1, \, p ^ 2, \, p ^ 3) \ quad (17)$

This requires that the determinant of the Jacobi matrix

$A_j ^ i = \ frac {\ partial p ^ i} {\ partial q ^ j} \ quad (18)$

was non-zero

$\ det | A_j ^ i | \ ne 0$

Then there is a matrix $B_j ^ i$ , inverse of the matrix (18), such that

$A_k ^ j \, B_i ^ k = \ delta_i ^ j$

Matrix $B_j ^ i$ is the Jacobi matrix for the transformation (17). Then you can calculate the vectors of the new basis

$\ vec {e} _ {\, i} ^ {\, '} = \ frac {\ partial \ vec {r}} {\ partial p ^ i} = \ frac {\ partial \ vec {r}} {\ partial q ^ j} \, \ frac {\ partial \ vec {q ^ j}} {\ partial p ^ i} = \ frac {\ partial \ vec {r}} {\ partial q ^ j} \, B_i ^ j$

We get the connection between the old basis and the new

$\ vec {e} _i ^ {\, '} = \ vec {e} _k \, B_i ^ k \ quad (19)$

$\ vec {e} _i = \ vec {e} _k ^ {\, '} \, A_i ^ k \ quad (20)$

Expand the vector $\ vec {a}$ in the new basis

$\ vec {a} = a ^ {'i} \, \ vec {e} _ {\, i} ^ {\,'}$

and using the relation (19), we write

$\ vec {a} = a ^ {'i} \, B_i ^ k \ vec {e} _ {\, k} = a ^ k \, \ vec {e} _ {\, k} \ quad (21)$

Given the fact that the basis vectors are linearly independent, we equate the coefficients with them in (21)

$a ^ k = a ^ {'i} \, B_i ^ k$

Now multiply both sides (21) by $A_k ^ j$

$a ^ k \, A_k ^ j = a ^ {'i} \, B_i ^ k \, A_k ^ j$

Consider that

$B_i ^ k \, A_k ^ j = \ delta_i ^ j$

$a ^ k \, A_k ^ j = a ^ {'i} \, \ delta_i ^ j = a ^ j$

That is, we obtain the formula for the inverse transformation of contravariant components

$a ^ {'j} = a ^ k \, A_k ^ j \ quad (22)$

From (22) and (19) we can conclude the following

The contravariant components of the vector are transformed by the inverse operator of the basis transformation operator.

Indeed, to get the vectors of the new basis, we used the matrix $B_i ^ k$ according to the formula (19). To get the contravariant components of the vector given in the new basis, we use the matrix $A_k ^ j$

Now let's see how the vector is transformed, given by its covariant components.

$a_j ^ {\, '} = \ vec {a} \ cdot \ vec {e} _j ^ {\,'} = \ vec {a} \ cdot \ vec {e} _j \, B_i ^ j = a_j \, B_i ^ j \ quad (23)$

From (23) it is clear that

The covariant components of the vector are transformed by the same operator that is used to transform the basis

Formulas (19), (22) and (23) and the formulated definitions made in the quotation block give a formal definition of contravariant and covariant coordinates and illustrate the difference between them. We can formulate a statement

The rank tensor (1.0) is transformed by the inverse operator used in the basis transformation, and the rank tensor (0.1) is transformed by the same operator that is used in the basis conversion.

4. Covariant derivative. Symbols of Christoffel 2nd Kind

Suppose we want to differentiate a vector defined by arbitrary coordinates along some coordinate. What should we do? Let's try this operation.

$\ frac {\ partial \ vec {a}} {\ partial q ^ j} = \ frac {\ partial} {\ partial q ^ j} (a ^ i \, \ vec {e} _i) = \ frac {\ partial a ^ i} {\ partial q ^ j} \, \ vec {e} _i + a ^ i \ frac {\ partial \ vec {e} _i} {\ partial q ^ j} \ quad (24)$

On what basis did we write the derivative of the base vector? And on the grounds that the basis in curvilinear coordinates depends on them, and therefore its derivative of the coordinate is nonzero. Well, okay, this derivative will also be a vector, which means it can be expanded on a local basis, for example, like this

$\ frac {\ partial \ vec {e} _i} {\ partial q ^ j} = \ Gamma_ {ij} ^ k \, \ vec {e} _k \ quad (25)$

Find the expansion coefficients in (25). To do this, take the covariant metric tensor and differentiate it by the specified coordinate

$\ frac {\ partial g_ {ij}} {\ partial q ^ k} = \ frac {\ partial \ vec {e} _i} {\ partial q ^ k} \ cdot \ vec {e} _j + \ vec {e } _i \ cdot \ frac {\ partial \ vec {e} _j} {\ partial q ^ k} \ quad (26)$

Substitute (25) into (26)

$\ frac {\ partial g_ {ij}} {\ partial q ^ k} = \ Gamma_ {ik} ^ m \, (\ vec {e} _m \ cdot \ vec {e} _j) + \ Gamma_ {jk} ^ m \, (\ vec {e} _m \ cdot \ vec {e} _i)$

The presence of the components of the metric tensor is obvious here; therefore, we perform the replacement

$\ frac {\ partial g_ {ij}} {\ partial q ^ k} = \ Gamma_ {ik} ^ m \, g_ {mj} + \ Gamma_ {jk} ^ m \, g_ {mi} \ quad (27)$

Before we start working with (27), we say that the desired expansion coefficients are symmetric with respect to the lower indices, since, having carried out a direct differentiation of the base vector, we arrive at the expression

$\ frac {\ partial \ vec {e} _i} {\ partial q ^ j} = \ frac {\ partial ^ 2 \ vec {r}} {\ partial q_i \ partial q_j} = \ Gamma_ {ij} ^ k \ , \ vec {e} _k$

whence, by virtue of the continuity of the considered functions, we conclude that

$\ Gamma_ {ij} ^ k = \ Gamma_ {ji} ^ k \ quad (28)$

Now, in (27) we rearrange the indices i and k

$\ frac {\ partial g_ {kj}} {\ partial q ^ i} = \ Gamma_ {ki} ^ m \, g_ {mj} + \ Gamma_ {ji} ^ m \, g_ {mk} \ quad (28)$

Now, let's rearrange the indices j and k in (27)

$\ frac {\ partial g_ {ik}} {\ partial q ^ j} = \ Gamma_ {ij} ^ m \, g_ {mk} + \ Gamma_ {kj} ^ m \, g_ {mi} \ quad (29)$

Now add (29) and (30) while taking into account the symmetry (28)

$\ frac {\ partial g_ {kj}} {\ partial q ^ i} + \ frac {\ partial g_ {ik}} {\ partial q ^ j} = \ Gamma_ {kj} ^ m \, g_ {mj} + \ Gamma_ {ij} ^ m \, g_ {mk} + \ Gamma_ {ji} ^ m \, g_ {mk} + \ Gamma_ {kj} ^ m \, g_ {mi}$

$\ frac {\ partial g_ {kj}} {\ partial q ^ i} + \ frac {\ partial g_ {ik}} {\ partial q ^ j} = \ Gamma_ {kj} ^ m \, g_ {mj} + 2 \, \ Gamma_ {ij} ^ m \, g_ {mk} + \ Gamma_ {kj} ^ m \, g_ {mi} \ quad (31)$

Subtract (27) from (31), again considering (28)

$\ frac {\ partial g_ {kj}} {\ partial q ^ i} + \ frac {\ partial g_ {ik}} {\ partial q ^ j} - \ frac {\ partial g_ {ij}} {\ partial q ^ k} = \ Gamma_ {ki} ^ m \, g_ {mj} + 2 \, \ Gamma_ {ij} ^ m \, g_ {mk} + \ Gamma_ {kj} ^ m \, g_ {mi} - \ Gamma_ {ik} ^ m \, g_ {mj} - \ Gamma_ {jk} ^ m \, g_ {mi}$

$\ frac {\ partial g_ {kj}} {\ partial q ^ i} + \ frac {\ partial g_ {ik}} {\ partial q ^ j} - \ frac {\ partial g_ {ij}} {\ partial q ^ k} = 2 \, \ Gamma_ {ij} ^ m \, g_ {mk} \ quad (32)$

Multiply (32) by $\ cfrac {1} {2} \, g ^ {mk}$ and finally we get

$\ Gamma_ {ij} ^ m = \ frac {1} {2} \, g ^ {mk} \ left (\ frac {\ partial g_ {kj}} {\ partial q ^ i} + \ frac {\ partial g_ {ik}} {\ partial q ^ j} - \ frac {\ partial g_ {ij}} {\ partial q ^ k} \ right) \ quad (33)$

Expression (33) defines the so-called Christoffel symbol of the 2nd kind. Then

$\ frac {\ partial \ vec {a}} {\ partial q ^ j} = \ frac {\ partial a ^ m} {\ partial q ^ j} \ cdot \ vec {e} _m + a ^ i \, \ Gamma_ {ij} ^ m \, \ vec {e} _m = \ left (\ frac {\ partial a ^ m} {\ partial q ^ j} + a ^ i \, \ Gamma_ {ij} ^ m \ right) \ cdot \ vec {e} _m \ quad (34)$

The expression in brackets in (34) is called the covariant derivative of the contravariant components of the vector

$\ nabla_j \, a ^ m = \ left (\ frac {\ partial a ^ m} {\ partial q ^ j} + a ^ i \, \ Gamma_ {ij} ^ m \ right) \ quad (35)$

Proceeding from (35), we must understand that when trying to differentiate along a curvilinear coordinate, we must take into account the dependence of the basis on the coordinates. If the metric does not depend on the position of the application point of the vector in space, then (35) turns into a partial partial derivative, since all Christoffel symbols are equal to zero, due to the fact that the metric tensor does not depend on coordinates. In any oblique coordinate system, and in their particular case, Cartesian coordinates, the Christoffel symbols, according to (33), are equal to zero. So, according to (35), the covariant derivative of the vector with respect to the coordinate will coincide with its partial derivative with respect to this coordinate, which we have been accustomed to for a long time. But if (33) was a tensor, then, being equal to zero, it would remain zero in any other coordinate system. But in curvilinear coordinates (33) are not equal to zero. So the Christoffel symbols are not a tensor. When converting the coordinate system, the components change, but not the essence of the tensor. The zero tensor must be such in any coordinate system.

Conclusion

Primary theoretical foundations dismantled. From the next article we will go into the practice of using tensor calculus to solve specific problems. Thank you for your attention and trust.

To be continued…

Source: https://habr.com/ru/post/261717/

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