Chapter 2: Tensor Analysis

Tensors are the natural mathematical objects that transform properly under coordinate changes. They form the language of general relativity, expressing physical laws in coordinate-independent form.

What is a Tensor?

A tensor of type (r, s) is a multilinear map that takes r covectors and s vectors as inputs and produces a real number. In component form:

$T^{\mu_1...\mu_r}_{\;\;\;\;\;\;\;\nu_1...\nu_s}$

r contravariant indices (upper), s covariant indices (lower)

Transformation Law

Under coordinate transformation x^μ → x^μ', tensor components transform as:

$T'^{\mu'}_{\;\;\nu'} = \frac{\partial x^{\mu'}}{\partial x^\mu} \frac{\partial x^\nu}{\partial x^{\nu'}} T^{\mu}_{\;\;\nu}$

Important Tensor Types

Scalars (0,0)

Invariant quantities: proper time τ, rest mass m, spacetime interval ds²

Vectors (1,0) and Covectors (0,1)

$Vectors: V^\mu = (V^0, V^1, V^2, V^3) — tangent to curves$

$Covectors: \omega_\mu = (\omega_0, \omega_1, \omega_2, \omega_3) — gradients of functions$

The Metric Tensor (0,2)

$g_{\mu\nu} — defines distances, angles, and raises/lowers indices$

The Riemann Tensor (1,3)

$R^\rho_{\;\sigma\mu\nu} — encodes spacetime curvature$

Tensor Operations

Addition

$(A + B)^{\mu\nu} = A^{\mu\nu} + B^{\mu\nu}$

Same type tensors only

Tensor Product

$(A \otimes B)^{\mu\nu} = A^\mu B^\nu$

Creates higher-rank tensor

Contraction

$A^\mu_{\;\mu} = \sum_\mu A^\mu_{\;\mu}$

Reduces rank by 2

Index Raising/Lowering

$V_\mu = g_{\mu\nu}V^\nu$

Uses metric tensor

Python: Tensor Transformations

Python

tensor_transform.py70 lines

#!/usr/bin/env python3
"""
Tensor Transformations - Coordinate Transformations in SR/GR
Demonstrates Lorentz boosts, vector and tensor transformations,
metric invariance, and 4-velocity transformation.
"""
import numpy as np

def lorentz_boost(v, c=1.0):
    """Create Lorentz boost matrix for velocity v in x-direction"""
    beta = v / c
    gamma = 1.0 / np.sqrt(1 - beta**2)
    Lambda = np.array([
        [gamma, -gamma*beta, 0, 0],
        [-gamma*beta, gamma, 0, 0],
        [0, 0, 1, 0],
        [0, 0, 0, 1]
    ])
    return Lambda

def transform_vector(V, Lambda):
    """Transform contravariant vector: V'^mu = Lambda^mu_nu V^nu"""
    return Lambda @ V

def transform_covector(omega, Lambda):
    """Transform covariant vector: omega'_mu = (Lambda^-1)^nu_mu omega_nu"""
    Lambda_inv = np.linalg.inv(Lambda)
    return omega @ Lambda_inv

def transform_tensor_02(T, Lambda):
    """Transform (0,2) tensor: T'_mu_nu = (L^-1)^a_mu (L^-1)^b_nu T_ab"""
    L_inv = np.linalg.inv(Lambda)
    return L_inv.T @ T @ L_inv

# Example: Minkowski metric under Lorentz boost
eta = np.diag([1, -1, -1, -1])  # Minkowski metric (+---)

v = 0.6  # velocity as fraction of c
Lambda = lorentz_boost(v)

print("Lorentz boost matrix (v = 0.6c):")
print(Lambda)
print()

# Transform the metric
eta_prime = transform_tensor_02(eta, Lambda)
print("Minkowski metric after boost:")
print(np.round(eta_prime, 10))
print()
print("Note: eta_prime = eta (Lorentz invariance!)")
print()

# Transform a 4-velocity
u = np.array([1, 0, 0, 0])  # particle at rest
u_prime = transform_vector(u, Lambda)
gamma = 1/np.sqrt(1 - v**2)
print(f"4-velocity at rest: {u}")
print(f"4-velocity after boost: {u_prime}")
print(f"Expected: [gamma, gamma*v, 0, 0] = [{gamma:.4f}, {gamma*v:.4f}, 0, 0]")
print()

# Verify invariant: u_mu u^mu = 1
u_cov = eta @ u  # lower the index
invariant = u @ u_cov
print(f"u_mu u^mu (rest frame) = {invariant}")

u_prime_cov = eta_prime @ u_prime
invariant_prime = u_prime @ u_prime_cov
print(f"u'_mu u'^mu (boosted frame) = {invariant_prime:.10f}")
print("Invariant is preserved!")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran: Tensor Contraction

Fortran + Python: Tensor Operations

Python

Compiles Fortran tensor code and visualizes components as heatmaps

tensor_plot.py123 lines

import subprocess, sys
import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

fortran_source = r"""
program tensor_contraction
    implicit none
    integer, parameter :: dp = selected_real_kind(15)
    real(dp) :: g(4,4), A(4,4), B(4,4)
    real(dp) :: trace_A, contraction
    integer :: i, j, k

g = 0.0_dp
    g(1,1) = 1.0_dp; g(2,2) = -1.0_dp; g(3,3) = -1.0_dp; g(4,4) = -1.0_dp

A = 0.0_dp
    A(1,1) = 2.0_dp; A(1,2) = 1.0_dp
    A(2,1) = 1.0_dp; A(2,2) = 3.0_dp
    A(3,3) = 1.0_dp; A(4,4) = 1.0_dp

trace_A = 0.0_dp
    do i = 1, 4
        trace_A = trace_A + A(i,i)
    end do

write(*,'(A)') 'Tensor Operations in GR'
    write(*,'(A)') '========================'
    write(*,'(A,F10.4)') 'Trace A^mu_mu = ', trace_A

B = 0.0_dp
    do i = 1, 4
        do j = 1, 4
            do k = 1, 4
                B(i,j) = B(i,j) + g(i,k) * A(k,j)
            end do
        end do
    end do

contraction = 0.0_dp
    do i = 1, 4
        do j = 1, 4
            contraction = contraction + A(i,j) * B(i,j)
        end do
    end do
    write(*,'(A,F10.4)') 'Double contraction = ', contraction
    write(*,'(A)') ''

write(*,'(A)') 'DATA_START'
    write(*,'(A)') 'row,col,A_up,A_down,g'
    do i = 1, 4
        do j = 1, 4
            write(*,'(I2,A,I2,A,F12.6,A,F12.6,A,F12.6)') &
                i, ',', j, ',', A(i,j), ',', B(i,j), ',', g(i,j)
        end do
    end do
    write(*,'(A)') 'DATA_END'
end program tensor_contraction
"""

with open('tensor.f90', 'w') as f:
    f.write(fortran_source)

comp = subprocess.run(['gfortran', '-o', './tensor', 'tensor.f90'],
                      capture_output=True, text=True)
if comp.returncode != 0:
    print("Compilation error:", comp.stderr)
    sys.exit(1)

result = subprocess.run(['./tensor'], capture_output=True, text=True, timeout=30)

lines = result.stdout.strip().split('\n')
data_lines = []
in_data = False
for line in lines:
    if 'DATA_START' in line:
        in_data = True
        continue
    if 'DATA_END' in line:
        in_data = False
        continue
    if in_data:
        data_lines.append(line.strip())
    else:
        print(line)

data = np.array([[float(x) for x in line.split(',')] for line in data_lines[1:]])
A_up = data[:, 2].reshape(4, 4)
A_down = data[:, 3].reshape(4, 4)

fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12, 5))
fig.patch.set_facecolor('#0f172a')

labels = ['t', 'x', 'y', 'z']
im1 = ax1.imshow(A_up, cmap='coolwarm', aspect='equal', vmin=-3, vmax=3)
ax1.set_xticks(range(4)); ax1.set_yticks(range(4))
ax1.set_xticklabels(labels, color='#cbd5e1')
ax1.set_yticklabels(labels, color='#cbd5e1')
for i in range(4):
    for j in range(4):
        ax1.text(j, i, f'{A_up[i,j]:.1f}', ha='center', va='center',
                color='white' if abs(A_up[i,j]) > 1.5 else '#cbd5e1', fontsize=12, fontweight='bold')
ax1.set_title(r'$A^\mu{}_\nu$ (mixed tensor)', color='#e2e8f0', fontsize=14, fontweight='bold')
fig.colorbar(im1, ax=ax1, shrink=0.8)
ax1.set_facecolor('#1e293b')

im2 = ax2.imshow(A_down, cmap='coolwarm', aspect='equal', vmin=-3, vmax=3)
ax2.set_xticks(range(4)); ax2.set_yticks(range(4))
ax2.set_xticklabels(labels, color='#cbd5e1')
ax2.set_yticklabels(labels, color='#cbd5e1')
for i in range(4):
    for j in range(4):
        ax2.text(j, i, f'{A_down[i,j]:.1f}', ha='center', va='center',
                color='white' if abs(A_down[i,j]) > 1.5 else '#cbd5e1', fontsize=12, fontweight='bold')
ax2.set_title(r'$A_{\mu\nu}=g_{\mu\rho}A^\rho{}_\nu$', color='#e2e8f0', fontsize=14, fontweight='bold')
fig.colorbar(im2, ax=ax2, shrink=0.8)
ax2.set_facecolor('#1e293b')

plt.tight_layout()
plt.savefig('output.png', dpi=150, bbox_inches='tight', facecolor='#0f172a')
print("\nVisualization saved to output.png")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Einstein Summation Convention

Repeated indices (one up, one down) are implicitly summed over:

$A^\mu B_\mu \equiv \sum_{\mu=0}^{3} A^\mu B_\mu = A^0 B_0 + A^1 B_1 + A^2 B_2 + A^3 B_3$

This convention eliminates cumbersome summation symbols and makes equations more readable. Free indices must match on both sides of an equation.

Practice Problems

Problem 1:Contract the tensor $T^{\mu\nu}$ with the Minkowski metric $\eta_{\mu\nu} = \text{diag}(1, -1, -1, -1)$ to find the trace $T = \eta_{\mu\nu}T^{\mu\nu}$ for the electromagnetic stress-energy tensor, which is traceless.

Solution:

Step 1: The trace is $T = \eta_{\mu\nu}T^{\mu\nu} = T^{00} - T^{11} - T^{22} - T^{33}$.

Step 2: For the EM field, $T^{00} = \frac{1}{2}(E^2 + B^2)$ (energy density) and $T^{ii} = \frac{1}{2}(E^2 + B^2)\delta^{ii} - E_iE_i - B_iB_i$ (Maxwell stress).

Step 3: Sum the spatial diagonal: $T^{11} + T^{22} + T^{33} = \frac{3}{2}(E^2 + B^2) - E^2 - B^2 = \frac{1}{2}(E^2 + B^2)$.

Step 4: Therefore $T = \frac{1}{2}(E^2 + B^2) - \frac{1}{2}(E^2 + B^2) = 0$.

Answer: $T = \eta_{\mu\nu}T^{\mu\nu} = 0$. The EM stress-energy tensor is traceless, reflecting the conformal invariance of Maxwell's equations in 4D.

Problem 2:Given a 4-velocity $u^\mu = (\gamma, \gamma v, 0, 0)$ in Minkowski space, lower the index to find $u_\mu$ and verify $u_\mu u^\mu = 1$.

Solution:

Step 1: Lower with the metric: $u_\mu = \eta_{\mu\nu}u^\nu$. With signature $(+,-,-,-)$:

$u_0 = \eta_{00}u^0 = \gamma, \quad u_1 = \eta_{11}u^1 = -\gamma v, \quad u_2 = 0, \quad u_3 = 0$

Step 2: So $u_\mu = (\gamma, -\gamma v, 0, 0)$.

Step 3: Compute the invariant: $u_\mu u^\mu = \gamma \cdot \gamma + (-\gamma v)(\gamma v) + 0 + 0 = \gamma^2(1 - v^2)$.

Step 4: Since $\gamma = 1/\sqrt{1-v^2}$: $u_\mu u^\mu = \frac{1-v^2}{1-v^2} = 1$.

Answer: $u_\mu = (\gamma, -\gamma v, 0, 0)$ and $u_\mu u^\mu = 1$. ✓

Problem 3:Compute the Christoffel symbol $\Gamma^r_{tt}$ for the Schwarzschild metric $ds^2 = (1-2M/r)\,dt^2 - (1-2M/r)^{-1}dr^2 - r^2 d\Omega^2$.

Solution:

Step 1: The Christoffel symbol formula is $\Gamma^\rho_{\mu\nu} = \frac{1}{2}g^{\rho\sigma}(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu})$.

Step 2: For $\Gamma^r_{tt}$: $\Gamma^r_{tt} = \frac{1}{2}g^{rr}(\partial_t g_{tr} + \partial_t g_{tr} - \partial_r g_{tt})$. Since the metric is static, $\partial_t g_{\mu\nu} = 0$.

Step 3: This reduces to $\Gamma^r_{tt} = -\frac{1}{2}g^{rr}\partial_r g_{tt}$.

Step 4: With $g_{tt} = 1 - 2M/r$ and $g^{rr} = -(1 - 2M/r)$: $\partial_r g_{tt} = 2M/r^2$.

Step 5: Therefore $\Gamma^r_{tt} = -\frac{1}{2}\left(-(1-2M/r)\right)\frac{2M}{r^2} = \frac{M}{r^2}(1-2M/r)$.

Answer: $\Gamma^r_{tt} = \frac{M}{r^2}\left(1 - \frac{2M}{r}\right)$. This reduces to $M/r^2$ for $r \gg 2M$, recovering Newtonian gravity.

Problem 4:Prove that the covariant derivative of the metric tensor vanishes: $\nabla_\rho g_{\mu\nu} = 0$ (metric compatibility).

Solution:

Step 1: Write out the covariant derivative: $\nabla_\rho g_{\mu\nu} = \partial_\rho g_{\mu\nu} - \Gamma^\sigma_{\rho\mu}g_{\sigma\nu} - \Gamma^\sigma_{\rho\nu}g_{\mu\sigma}$.

Step 2: Lower the Christoffel symbols using $\Gamma_{\sigma\rho\mu} = g_{\sigma\lambda}\Gamma^\lambda_{\rho\mu} = \frac{1}{2}(\partial_\rho g_{\mu\sigma} + \partial_\mu g_{\rho\sigma} - \partial_\sigma g_{\rho\mu})$.

Step 3: The two Christoffel terms become: $\Gamma_{\nu\rho\mu} + \Gamma_{\mu\rho\nu} = \frac{1}{2}(\partial_\rho g_{\mu\nu} + \partial_\mu g_{\rho\nu} - \partial_\nu g_{\rho\mu}) + \frac{1}{2}(\partial_\rho g_{\nu\mu} + \partial_\nu g_{\rho\mu} - \partial_\mu g_{\rho\nu})$.

Step 4: The $\partial_\mu g_{\rho\nu}$ and $\partial_\nu g_{\rho\mu}$ terms cancel pairwise, leaving $\partial_\rho g_{\mu\nu}$.

Answer: $\nabla_\rho g_{\mu\nu} = \partial_\rho g_{\mu\nu} - \partial_\rho g_{\mu\nu} = 0$. This is guaranteed by the definition of the Levi-Civita connection.

Problem 5:Given $A_\mu = (3, -1, 4, 0)$ and $B^\mu = (2, 1, 0, -1)$ in Minkowski space, compute the scalar $A_\mu B^\mu$ and raise $A_\mu$ to get $A^\mu$.

Solution:

Step 1: The contraction is $A_\mu B^\mu = A_0 B^0 + A_1 B^1 + A_2 B^2 + A_3 B^3$.

Step 2: Evaluate: $A_\mu B^\mu = (3)(2) + (-1)(1) + (4)(0) + (0)(-1) = 6 - 1 + 0 + 0 = 5$.

Step 3: To raise $A_\mu$: $A^\mu = \eta^{\mu\nu}A_\nu$. With $\eta^{\mu\nu} = \text{diag}(1,-1,-1,-1)$:

$A^0 = A_0 = 3, \quad A^1 = -A_1 = 1, \quad A^2 = -A_2 = -4, \quad A^3 = -A_3 = 0$

Answer: $A_\mu B^\mu = 5$ and $A^\mu = (3, 1, -4, 0)$.

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