Prove the 'zero component' lemma for an antisymmetric tensor \(T^{\mu v}\) : if any one of its off-diagonal components is zero in all inertial frames, then the entire tensor is zero.

An antisymmetric tensor \(T^{\mu v}\) has the property that when the indices are swapped, the tensor changes its sign, i.e. \(T^{\mu v} = -T^{v\mu}\).

Let's assume that one off-diagonal component of the tensor is zero in all inertial frames. Without loss of generality, let's assume that this component is when \(\mu=1\) and \(v=2\), or \(T^{12}=0\) in all inertial frames. We will use this assumption to show that all the other components must also be zero.

If the off-diagonal component \(T^{12}\) is zero, then using the antisymmetric property, we get: \(T^{21} = -T^{12} = -0 = 0\) So, both \(T^{12}\) and \(T^{21}\) are zero. Now, let's consider the other possible pairs of indices, \((1,0)\), \((1,3)\), \((2,0)\), \((2,3)\), and \((3,0)\). For each pair, we use the antisymmetric property to show that the corresponding component is zero: - For \(T^{10}\), we have \(T^{01} = -T^{10}\). Since \(T^{12}=0\), we know that the off-diagonal component \(T^{01}\) must also be zero in all inertial frames. Therefore, \(T^{10} = -T^{01} = 0\). - Similarly, for \(T^{13}\), we have \(T^{31} = -T^{13}\) and as \(T^{12}=0\), the off-diagonal component \(T^{31}\) must also be zero in all inertial frames. Therefore, \(T^{13} = -T^{31} = 0\). - Continuing this logic for the other components, we get: - \(T^{20} =-T^{02} = 0\) - \(T^{23} = -T^{32} = 0\) - \(T^{30} = -T^{03} = 0\)

Since all the off-diagonal components are zero, the entire tensor \(T^{\mu v}\) is antisymmetric with zero off-diagonal components. This means that \(T^{\mu v}\) is zero for all indices \(\mu\) and \(v\), proving the 'zero component' lemma for an antisymmetric tensor.