Prove that, in relativistic as in Newtonian mechanics, the time rate of change of the angular momentum \(h=\mathbf{r} \times p\) of a particle about an origin \(\mathrm{O}\) is equal to the couple \(\mathrm{r} \times \mathrm{f}\) of the applied force about \(\mathrm{O}\). If 1 \(^{\mu v}\) is the particle's four- angular momentum, and if we define the fourcouple \(G^{\mu v}=x^{\mu} F^{v}-x^{v} F^{\mu}\), where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), prove that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) and that the spacc-space part of this equation corresponds to the above threevector result.

Begin by defining the given variables in the problem statement: - \(h = \mathbf{r} \times p\) is the angular momentum, where \(\mathbf{r}\) is the position vector of the particle and \(p\) is its linear momentum in both relativistic and Newtonian mechanics. - \(\mathbf{f}\) is the applied force on the particle. - \(L^{\mu \nu}\) is the particle's four-angular momentum. - \(G^{\mu \nu} = x^{\mu} F^{v} - x^{v} F^{\mu}\) is the fourcouple, where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), respectively.

Let's first establish that, in relativistic mechanics, the time rate of change of angular momentum is equal to the couple of the applied force: \(h = \mathbf{r} \times p\) Take the time derivative of the angular momentum: \(\frac{dh}{dt} = \frac{d}{dt} (\mathbf{r} \times p)\) Now, using the product rule for vector derivatives: \(\frac{dh}{dt} = \frac{d\mathbf{r}}{dt} \times p + \mathbf{r} \times \frac{dp}{dt}\) The time derivative of the position vector (\(\frac{d\mathbf{r}}{dt}\)) is the velocity vector (\(\mathbf{v}\)), and the time derivative of the momentum (\(\frac{dp}{dt}\)) is the force (\(\mathbf{f}\)): \(\frac{dh}{dt} = \mathbf{v} \times p + \mathbf{r} \times \mathbf{f}\) Since velocity and momentum are parallel in both relativistic and Newtonian mechanics, their cross product is zero: \(\frac{dh}{dt} = \mathbf{r} \times \mathbf{f}\) This proves that the time rate of change of angular momentum is equal to the couple of the applied force in both relativistic and Newtonian mechanics.

First, let's write the four-angular momentum: \(L^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}\) Now, differentiate with respect to the proper time \(\tau\): \(\frac{d}{d\tau} L^{\mu \nu} = \frac{d}{d\tau} (x^{\mu} p^{\nu} - x^{\nu} p^{\mu})\) Using the product rule: \(\frac{d}{d\tau} L^{\mu \nu} = (\frac{dx^{\mu}}{d\tau} p^{\nu} + x^{\mu} \frac{dp^{\nu}}{d\tau}) - (\frac{dx^{\nu}}{d\tau} p^{\mu} + x^{\nu} \frac{dp^{\mu}}{d\tau})\) Now, we know that \(\frac{dx^{\mu}}{d\tau} = V^{\mu}\) (four-velocity) and \(\frac{dp^{\mu}}{d\tau} = F^{\mu}\) (four-force). Replacing these in the equation, we get: \(\frac{d}{d\tau} L^{\mu \nu} = (V^{\mu} p^{\nu} + x^{\mu} F^{\nu}) - (V^{\nu} p^{\mu} + x^{\nu} F^{\mu})\) Comparing this with the definition of fourcouple: \(G^{\mu \nu} = x^{\mu} F^{\nu} - x^{\nu} F^{\mu}\) We see that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\).

To show that the space-space part of the equation \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) corresponds to the three-vector result, i.e., \(\frac{dh}{dt} = \mathbf{r} \times \mathbf{f}\), we only need to consider the space components (\(\mu,\nu=1,2,3\)) of the equation: \((\mathrm{d} / \mathrm{d} \tau) L^{ij} = G^{ij}\) Since \(L^{ij} = x^i p^j - x^j p^i\) and \(G^{ij} = x^i F^j - x^j F^i\), this equation represents the same three-vector relation we have already found: \(\frac{dh}{dt} = \mathbf{r} \times \mathbf{f}\) In conclusion, we have proved that in relativistic mechanics, the time rate of change of the angular momentum of a particle about an origin is equal to the couple of the applied force about the origin, and the space-space part of the equation \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) corresponds to this three-vector result.