(i) Prove the zero-component lemma for four-vectors: if a fourvector \(V^{\mu}\) has a particular one of its four components zero in \(a l l\) inertial frames then the entire vector must be zero. [Hint: suppose first \(V^{i} \equiv 0\); if there is a frame in which \(V^{2} \neq 0\) or \(V^{3} \neq 0\), we can rotate the spatial axes to make \(V^{t} \neq 0\); if there is a frame in which \(V^{0} \neq 0\), we can apply a Lorentz transformation to make \(V^{1} \neq 0\); and so on.] (ii) If \(A_{\mu v}\) is a symmetric tensor and \(A_{00} \equiv 0\), prove \(A_{\mu \nu}=0\). Is this result true for the vanishing of any one component of \(A_{\mu v}\) ?

To prove the zero-component lemma for four-vectors, let's assume that one of the components of the four-vector \(V^{\mu}\) is zero in all inertial frames. We will consider different cases based on which component is zero and use rotations and Lorentz transformations accordingly. Case 1: Let \(V^{i} \equiv 0\), where \(i = 1, 2, 3\) represents the spatial components. If there is a frame in which \(V^{2} \neq 0\) or \(V^{3} \neq 0\), we can rotate the spatial axes to make \(V^{t} \neq 0\). But this contradicts the given condition that one component must be zero in all inertial frames. Therefore, \(V^{0} \equiv 0\) as well, which implies that \(V^{\mu} = 0\) for all components. Case 2: Let \(V^{0} \equiv 0\), where \(0\) represents the time component. If there is a frame in which \(V^{1} \neq 0\), we can apply a Lorentz transformation to make \(V^{0} \neq 0\). But this contradicts the given condition that one component must be zero in all inertial frames. Thus, all components \(V^{i}\) must also be zero. Therefore, \(V^{\mu} = 0\) for all components. In both cases, we have shown that if one component of the four-vector is zero in all inertial frames, the entire vector must be zero. This concludes the proof for the zero-component lemma for four-vectors.

Now, let's consider a symmetric tensor \(A_{\mu \nu}\) with the condition \(A_{00} \equiv 0\). As the tensor is symmetric, we have \(A_{\mu \nu} = A_{\nu \mu}\). Since \(A_{00} \equiv 0\), we can write \(A_{\mu 0} = A_{0 \mu} = 0\). Now, to prove \(A_{\mu \nu} = 0\), we can analyze the other components. a) For \(\mu = 0\) and \(\nu = 0, 1, 2, 3\), we have \(A_{\mu \nu} = A_{0 \nu} = 0\). b) For \(\mu = 1, 2, 3\) and \(\nu = 0, 1, 2, 3\), we have \(A_{\mu \nu} = A_{\nu \mu} = A_{\nu 0} = 0\). In both cases, we have shown that \(A_{\mu \nu} = 0\). Thus, if \(A_{00} \equiv 0\), then \(A_{\mu \nu} = 0\). Now, let's consider the case where any one component of \(A_{\mu \nu}\) is zero, and analyze if the result is still true. Let's take the general case where \(A_{12} = 0\). Then, due to the symmetric property, we have \(A_{21} = 0\). However, we cannot prove that all other components of \(A_{\mu \nu}\) are also zero. For example, there may exist other components such as \(A_{01}\) or \(A_{02}\) which may or may not be zero. Thus, the result is not true for the vanishing of any one component of \(A_{\mu \nu}\).