Question

In \(E^{2}\) let \(\left(\xi^{1}, \xi^{2}\right)\) be standard coordinates and let \(\left(x^{1}, x^{2}\right)\) be new coordinates given by \(x^{1}=3 \xi^{1}+2 \xi^{2}, x^{2}=4 \xi^{1}+3 \xi^{2}\). Find the \(\left(x^{i}\right)\) components of the following tensors, for which the components in standard coordinates are given. a) \(u_{i}\), where \(U_{1}=\xi^{1} \xi^{2}, U_{2}=\xi^{1}-\xi^{2}\). b) \(v^{i}\), where \(V^{1}=\xi^{1} \cos \xi^{2}, V^{2}=\xi^{1} \sin \xi^{2}\). c) \(w_{i j}\), where \(W_{11}=0, W_{12}=\xi^{1} \xi^{2}, W_{21}=-\xi^{1} \xi^{2}, W_{22}=0\). d) \(z_{j}^{i}\), where \(Z_{1}^{1}=\xi^{1}+\xi^{2}, Z_{2}^{1}=Z_{1}^{2}=3 \xi^{1}+2 \xi^{2}, Z_{2}^{2}=\xi^{1}-\xi^{2}\).

Short answer

In summary, we have transformed the tensor components from the standard coordinates \((\xi^1,\xi^2)\) to the new coordinate system \((x^1,x^2)\) using the calculated Jacobian and its inverse. The new components of the tensors are as follows: a) \((u_1)' = \frac{1}{5}(3\xi^1 - 5\xi^2), (u_2)' = \frac{1}{5}(5\xi^1 - 3\xi^2)\) b) \((v^1)' = \xi^1(3\cos \xi^2 + 2\sin \xi^2), (v^2)' = \xi^1(4\cos \xi^2 + 3\sin \xi^2)\) c) \((w_{11})' = 0, (w_{12})' = \frac{2}{25}, (w_{21})' = -\frac{12}{25}, (w_{22})' =-\frac{4}{25}\) d) \((z^1_1)' = 15\xi^1+8\xi^2, (z^1_2)' = 0, (z^2_1)' = 21\xi^1+18\xi^2, (z^2_2)' = 3(\xi^1-\xi^2)\)

Step-by-step solution

Calculating Jacobian of the Coordinate Transformation

To change the tensor components from one coordinate system to another, we need the Jacobian of the coordinate transformation and its inverse. If the transformation is given by \(x^i=f^i(\xi^j)\), the Jacobian is: \(J^i_j = \frac{\partial f^i}{\partial \xi^j}\). Therefore, we can calculate the Jacobian for our given transformation: \(J^1_1 = \frac{\partial (3\xi^1+2\xi^2)}{\partial \xi^1} = 3\) \(J^1_2 = \frac{\partial (3\xi^1+2\xi^2)}{\partial \xi^2} = 2\) \(J^2_1 = \frac{\partial (4\xi^1+3\xi^2)}{\partial \xi^1} = 4\) \(J^2_2 = \frac{\partial (4\xi^1+3\xi^2)}{\partial \xi^2} = 3\) To find the inverse Jacobian \(J^{-1} = \frac{\partial \xi^i}{\partial x^j}\), we have to first find the inverse transformation \(x^j=g^j(\xi^i)\). We can find from the given transformation that \((3\xi^1+2\xi^2 -x^1) = 0\), and \((4\xi^1+3\xi^2 -x^2) = 0\). Using the Cramer’s rule, we can find the inverse transformation: \(\xi^1=\frac{3x^2 - 2x^1}{5}\) \(\xi^2=\frac{-4x^1 + x^2}{-5}\) Now, we can find the inverse Jacobian as: \(J^{-1}_{11} = \frac{\partial (3x^2 - 2x^1)/5}{\partial x^1} = -\frac{2}{5}\) \(J^{-1}_{12} = \frac{\partial (3x^2 - 2x^1)/5}{\partial x^2} = \frac{3}{5}\) \(J^{-1}_{21} = \frac{\partial (-4x^1 + x^2)/(-5)}{\partial x^1} =\frac{4}{5}\) \(J^{-1}_{22} = \frac{\partial (-4x^1 + x^2)/(-5)}{\partial x^2} =\frac{1}{5}\) Now we have both the Jacobian and its inverse.

a) Transforming \(u_i\) Components

To transform the components of a covariant vector \(u_i\), we use the inverse Jacobian: \((u_i)' = J^{-1}_j (u^j)' = J^{-1}_{ij} u^j \) Using the calculated inverse Jacobian and given tensor components, we get: \((u_1)' = -\frac{2}{5}(\xi^1 \xi^2) + \frac{3}{5} (\xi^1 - \xi^2) = \frac{1}{5}(3\xi^1 - 5\xi^2)\) \((u_2)' = \frac{4}{5}(\xi^1 \xi^2) + \frac{1}{5} (\xi^1 - \xi^2) = \frac{1}{5}(5\xi^1 - 3\xi^2)\)

b) Transforming \(v^i\) Components

To transform the components of a contravariant vector \(v^i\), we use the Jacobian: \((v^i)'= J^{i}_j (v^j)' = J^{ij} v_j\) Using the calculated Jacobian and given tensor components, we get: \((v^1)' = 3(\xi^1 \cos \xi^2) + 2(\xi^1 \sin \xi^2) = \xi^1(3\cos \xi^2 + 2\sin \xi^2)\) \((v^2)' = 4(\xi^1 \cos \xi^2) + 3(\xi^1 \sin \xi^2) = \xi^1(4\cos \xi^2 + 3\sin \xi^2)\)

c) Transforming \(w_{ij}\) Components

To transform the components of a covariant rank-2 tensor \(w_{ij}\), we use the inverse Jacobian and the product rule: \((w_{ij})' = J^{-1}_m J^{-1}_n w^{mn}\) Using the calculated inverse Jacobian and given tensor components, we get: $(w_{11})' = (-\frac{2}{5}) (-\frac{2}{5})(0) + (-\frac{2}{5}) (\frac{3}{5})(-1) + (\frac{3}{5}) (-\frac{2}{5})(0) + (\frac{3}{5}) (\frac{3}{5})(0) = 0$ $(w_{12})' = (-\frac{2}{5}) (\frac{4}{5})(0) + (-\frac{2}{5}) (\frac{1}{5})(-1) + (\frac{3}{5}) (\frac{4}{5})(0) + (\frac{3}{5}) (\frac{1}{5})(0) = \frac{2}{25}$ \((w_{21})' = (\frac{4}{5}) (-\frac{2}{5})(0) + (\frac{4}{5}) (\frac{3}{5})(-1) + (\frac{1}{5}) (-\frac{2}{5})(0) +(\frac{1}{5})(\frac{3}{5})(0) = -\frac{12}{25}\) \((w_{22})' = (\frac{4}{5}) (\frac{4}{5})(0) + (\frac{4}{5}) (\frac{1}{5})(-1)+(\frac{1}{5})(\frac{4}{5})(0)+(\frac{1}{5})(\frac{1}{5})(0) =-\frac{4}{25}\)

d) Transforming \(z^i_j\) Components

To transform the components of a mixed rank-2 tensor \(z^i_j\), we use the Jacobian and its inverse and the product rule: \((z^i_j)'= J^{i}_m J^{-1}_n z^m_n\) Using the calculated Jacobian and inverse Jacobian and given tensor components, we get: \((z^1_1)' = 3(\xi^1+\xi^2)+2(3\xi^1+2\xi^2)=15\xi^1+8\xi^2\) \((z^1_2)' = 3\cdot 0 + 2\cdot0=0\) \((z^2_1)' = 4(\xi^1+\xi^2)+3(3\xi^1+2\xi^2)=21\xi^1+18\xi^2\) \((z^2_2)' = 4\cdot 0 + 3(\xi^1-\xi^2)=3(\xi^1-\xi^2)\)

Key concepts

Jacobian Matrix

The Jacobian matrix is a fundamental concept in multivariable calculus utilized extensively to perform coordinate transformations. It comprises of partial derivatives and serves as a transformation matrix between different coordinate systems. When a point's coordinates are transformed from one system to another, the Jacobian matrix represents the linear map which aligns the coordinates to the new system.

Consider a coordinate transformation where each new coordinate is a function of the old coordinates, expressed as \( x^i = f^i(\xi^j) \), the Jacobian matrix's entries are calculated by \( J^i_j = \frac{\partial f^i}{\partial \xi^j} \). It essentially encapsulates how a small change in one set of coordinates affects the change in another set. This is particularly useful in tensor transformation exercises, where converting the components of tensors between different coordinate systems requires both the Jacobian and its inverse to relate the old and new coordinate systems.

Coordinate Transformation

Coordinate transformation is a mathematical operation that transitions points, vectors, and tensors from one coordinate system to another. This is highly relevant in fields like physics and engineering where different coordinate systems are used to simplify problems or match the geometry of the situation. For instance, converting Cartesian coordinates to polar coordinates may simplify an integration problem due to the symmetry of the system.

Through the use of a Jacobian matrix, tensors of various ranks can be converted preserving the appropriate linear transformation. The solution provided employs coordinate transformation to redefine tensor components from standard coordinates \( (\xi^1, \xi^2) \) to new coordinates \( (x^1, x^2) \), applying the calculated Jacobian matrix and its inverse. This process is inherent to the manipulation of geometric and physical quantities over manifold domains, where changes in point of view or reference frame are common.

Covariant and Contravariant Tensors

Covariant and contravariant tensors are types of tensors that differ in how they transform under a change of coordinates. Tensors, in general, are geometric objects that describe linear relations between vectors, scalars, and other tensors. Covariant tensors (denoted with subscripts) scale inversely to the scaling of coordinates, whereas contravariant tensors (denoted with superscripts) scale in the same way as the coordinates do.

In a transformation scenario, covariant components are transformed using the inverse Jacobian matrix, as seen for \( u_i \) in the problem. Conversely, contravariant components, like \( v^i \) in the given problem, are transformed using the original Jacobian matrix. The rank of the tensor determines the number of times the transformation matrices are applied, whether the Jacobian or its inverse. For instance, for rank-2 tensors, you apply the transformation rule twice, once for each index.

Cramer's Rule

Cramer's rule is a classical method used to solve systems of linear equations with as many equations as unknowns. It's particularly convenient when dealing with small systems or when the solution needs to be expressed in terms of the coefficients from the systems' equations. The rule uses determinants to find the values of the unknowns and is particularly handy in finding inverse transformations in tensor conversion exercises.

In the context of the solution provided, Cramer's rule was employed to obtain the inverse coordinates \( \xi^1 \) and \( \xi^2 \) from the given new coordinates \( x^1 \) and \( x^2 \). The rule allowed for a straightforward calculation of the inverse Jacobian, which was then used to transform the tensor components. The application showcases that despite Cramer's rule not being efficient for larger systems due to computational complexity, it remains a powerful tool for analytical solutions in theoretical exercises.