Consider a Lorentz covariant expression that is not a Lorentz scalar, \(C^{\lambda}=\) \(K^{\lambda} h\left(A^{\mu} \eta_{\mu u} B^{u}ight)\), where \(h\) is any function of the quantity in parentheses. Here quantities with a single superscript are four-vectors. Under a Lorentz transformation, \(A^{\mu} \eta_{\mu u} B^{u}\) is Lorentz covariant and is also a Lorentz scalar. Hence, its form and value are unchanged, which means that the function \(h\left(A^{\mu} \eta_{\mu u} B^{u}ight)\) is unchanged in form or value as well. The quantity \(K^{\mu}\) on the other hand is a four-vector; this means that it transforms as \(K^{\mu}=\) \(\Lambda_{\mu^{\prime}}^{\mu} K^{\mu^{\prime}}\). The right-hand side of the equation for \(C^{\lambda}\) transforms as a four-vector as whole, which implies that \(C^{\lambda}\) also transforms as a four-vector and observer \(\mathcal{O}^{\prime}\) can write \(C^{\lambda^{\prime}}=\) \(K^{\lambda^{\prime}} h\left(A^{\mu^{\prime}} \eta_{\mu^{\prime} u^{\prime}} B^{u^{\prime}}ight)\). This quantity is said to be a Lorentz vector (instead of a scalar), since it transforms as a four-vector: That is, its components change, but through the well-defined prescription for a four-vector. Which of the following quantities are Lorentz vectors, given that \(K\) is a Lorentz scalar and any quantity with a single subscript or superscript is a Lorentz vector?

(a) \(K \eta_{\mu u}\)

(b) \(C^{\lambda}=D^{\mu} A^{\lambda} \eta_{\mu u} B_{u}\)

(c) \(K A^{\mu} \eta_{\mu u} B^{\lambda} \eta_{\lambda \sigma} D^{u} F^{\sigma}\)