Researchers often upload problem sets under “Educational Resources.” Look for contributions by authors like Ivan Avramidi (New Mexico Tech) or D.C. Kay.
Compute covariant derivative ( \nabla_j V^i ) for ( V^i = (r,0,0) ) in cylindrical coords.
Solution:
( \nabla_j V^i = \partial_j V^i + \Gamma^i_jk V^k )
Nonzero:
( \nabla_1 V^1 = \partial_1 r + \Gamma^1_11r + ... = 1 + 0 = 1 )
( \nabla_2 V^1 = 0 + \Gamma^1_22V^2 = (-r)(0) = 0 )
etc.
Check ( \nabla_2 V^2 = \partial_2 0 + \Gamma^2_21V^1 = 0 + (1/r)(r) = 1 ).
Here is an example of the type of solution you should look for in a high-quality PDF resource.
Problem: Prove the "Quotient Rule" for tensors: If $A_i$ is known to be a covariant vector and the relation $B^ijA_j = C^i$ holds for arbitrary $A_j$, and $C^i$ transforms as a contravariant vector, prove that $B^ij$ is a contravariant tensor of rank 2. tensor analysis problems and solutions pdf free
Solution: Step 1: Write the transformation laws. Since $A_j$ is covariant and $C^i$ is contravariant: $$ A'j = \frac\partial x^k\partial x'^j Ak $$ $$ C'^i = \frac\partial x'^i\partial x^m C^m $$
Step 2: Transform the relation in the primed system. The relation holds in the new coordinate system: $$ B'^ij A'_j = C'^i $$
Step 3: Substitute the transformation laws. Substitute the expressions for $A'j$ and $C'^i$ into the relation: $$ B'^ij \left( \frac\partial x^k\partial x'^j Ak \right) = \left( \frac\partial x'^i\partial x^m C^m \right) $$
Step 4: Substitute the original relation. We know from the original system that $C^m = B^mkA_k$. Substitute this into the right side: $$ B'^ij \frac\partial x^k\partial x'^j A_k = \frac\partial x'^i\partial x^m (B^mkA_k) $$ Write in index notation: (a) Dot product (
Step 5: Rearrange terms. Bring all terms to one side. Since $A_k$ is arbitrary, its coefficient must be zero. $$ \left[ B'^ij \frac\partial x^k\partial x'^j - \frac\partial x'^i\partial x^m B^mk \right] A_k = 0 $$
Since this holds for any $A_k$, the bracket must vanish: $$ B'^ij \frac\partial x^k\partial x'^j = \frac\partial x'^i\partial x^m B^mk $$
Step 6: Solve for $B'^ij$. Multiply both sides by $\frac\partial x'^j\partial x^k$ (summing over $j$) to isolate $B'^ij$: $$ B'^ij = \frac\partial x'^i\partial x^m \frac\partial x'^j\partial x^k B^mk $$
Conclusion: This is the transformation law for a contravariant tensor of rank 2. Thus, $B^ij$ is a tensor. After solving in flat space, try the same problem for:
Write in index notation:
(a) Dot product ( \mathbfa \cdot \mathbfb )
(b) Matrix product ( (AB)_ij )
(c) Trace of matrix ( C )
Solution:
(a) ( a_i b_i )
(b) ( A_ik B_kj )
(c) ( C_ii )
A quality problem set will include 50–200 exercises ranging from basic index manipulation to advanced curvature calculations.
After solving in flat space, try the same problem for:
Some advanced free PDFs include metric-specific exercises.
Covers index notation, tensor algebra, tensor calculus, metrics, curvature, applications to continuum mechanics and GR. Each week includes a short theory summary, 8–12 problems (graded easy→hard), and fully worked solutions.
Researchers often upload problem sets under “Educational Resources.” Look for contributions by authors like Ivan Avramidi (New Mexico Tech) or D.C. Kay.
Compute covariant derivative ( \nabla_j V^i ) for ( V^i = (r,0,0) ) in cylindrical coords.
Solution:
( \nabla_j V^i = \partial_j V^i + \Gamma^i_jk V^k )
Nonzero:
( \nabla_1 V^1 = \partial_1 r + \Gamma^1_11r + ... = 1 + 0 = 1 )
( \nabla_2 V^1 = 0 + \Gamma^1_22V^2 = (-r)(0) = 0 )
etc.
Check ( \nabla_2 V^2 = \partial_2 0 + \Gamma^2_21V^1 = 0 + (1/r)(r) = 1 ).
Here is an example of the type of solution you should look for in a high-quality PDF resource.
Problem: Prove the "Quotient Rule" for tensors: If $A_i$ is known to be a covariant vector and the relation $B^ijA_j = C^i$ holds for arbitrary $A_j$, and $C^i$ transforms as a contravariant vector, prove that $B^ij$ is a contravariant tensor of rank 2.
Solution: Step 1: Write the transformation laws. Since $A_j$ is covariant and $C^i$ is contravariant: $$ A'j = \frac\partial x^k\partial x'^j Ak $$ $$ C'^i = \frac\partial x'^i\partial x^m C^m $$
Step 2: Transform the relation in the primed system. The relation holds in the new coordinate system: $$ B'^ij A'_j = C'^i $$
Step 3: Substitute the transformation laws. Substitute the expressions for $A'j$ and $C'^i$ into the relation: $$ B'^ij \left( \frac\partial x^k\partial x'^j Ak \right) = \left( \frac\partial x'^i\partial x^m C^m \right) $$
Step 4: Substitute the original relation. We know from the original system that $C^m = B^mkA_k$. Substitute this into the right side: $$ B'^ij \frac\partial x^k\partial x'^j A_k = \frac\partial x'^i\partial x^m (B^mkA_k) $$
Step 5: Rearrange terms. Bring all terms to one side. Since $A_k$ is arbitrary, its coefficient must be zero. $$ \left[ B'^ij \frac\partial x^k\partial x'^j - \frac\partial x'^i\partial x^m B^mk \right] A_k = 0 $$
Since this holds for any $A_k$, the bracket must vanish: $$ B'^ij \frac\partial x^k\partial x'^j = \frac\partial x'^i\partial x^m B^mk $$
Step 6: Solve for $B'^ij$. Multiply both sides by $\frac\partial x'^j\partial x^k$ (summing over $j$) to isolate $B'^ij$: $$ B'^ij = \frac\partial x'^i\partial x^m \frac\partial x'^j\partial x^k B^mk $$
Conclusion: This is the transformation law for a contravariant tensor of rank 2. Thus, $B^ij$ is a tensor.
Write in index notation:
(a) Dot product ( \mathbfa \cdot \mathbfb )
(b) Matrix product ( (AB)_ij )
(c) Trace of matrix ( C )
Solution:
(a) ( a_i b_i )
(b) ( A_ik B_kj )
(c) ( C_ii )
A quality problem set will include 50–200 exercises ranging from basic index manipulation to advanced curvature calculations.
After solving in flat space, try the same problem for:
Some advanced free PDFs include metric-specific exercises.
Covers index notation, tensor algebra, tensor calculus, metrics, curvature, applications to continuum mechanics and GR. Each week includes a short theory summary, 8–12 problems (graded easy→hard), and fully worked solutions.
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