Question

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.

Short answer

In summary, we have proven 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. We first found the time derivative of the angular momentum and related it to the velocity and force. Then we differentiated the four-angular momentum with respect to proper time and correlated it with the fourcouple. Finally, we confirmed that the space-space components of this equation reflected the three-vector result we previously derived.

Step-by-step solution

Define the variables

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.

Relate angular momentum to applied force

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.

Show that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\)

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 \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\).

Find the space-space part of the equation

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.

Key concepts

Angular Momentum

Angular momentum in mechanics is the rotational equivalent of linear momentum. It's a measure of the rotational motion of an object relative to a point, often an origin. For a particle, angular momentum is defined as \( h = \mathbf{r} \times p \), where \( \mathbf{r} \) is the position vector and \( p \) is the linear momentum.
The cross product in this definition indicates a vector perpendicular to both \( \mathbf{r} \) and \( p \), signifying rotational behavior.

• In Newtonian mechanics, it represents how difficult it is to change the rotation of an object—akin to inertia for linear motion.
• In relativistic mechanics, the definition stays similar, but it becomes vital to factor in the time dilation effects at significant velocities.

Understanding angular momentum allows us to compare and relate rotational effects in various frames of reference as done in the exercise above.

Four-Vectors

Four-vectors extend regular vectors into the realm of relativity by adding a time component. In relativity, physical quantities like position and momentum are bundled into four-vectors to ensure equations remain consistent at high speeds, near the speed of light. A four-vector is denoted typically by \( x^{\mu} = (ct, \mathbf{x}) \), where \( ct \) represents the time component scaled by the speed of light, and \( \mathbf{x} \) is the spatial part.
Four-vectors maintain the form of the laws of physics under Lorentz transformations, obeying the structure of spacetime.

• Position vectors become four-vectors to incorporate time:
• Momentum vectors do similarly, including energy as the time-like component.

This concept ensures that quantities like position and momentum correctly transform between different observers moving relative to each other.

Proper Time

Proper time \( \tau \) is the time measured by a clock moving with the particle, representing the time interval between events as experienced by the particle itself. It's pivotal in relativity because it remains invariant for an observer moving along with the particle.
Unlike regular time, proper time does not dilate; it’s akin to the particle’s internal clock showing the passage of time.
For a particle's trajectory, the proper time \( \tau \) is used as the parameter for describing its motion in four-dimensional spacetime.

• This makes it useful for deriving quantities like the four-velocity \( V^{\mu} = \frac{dx^{\mu}}{d\tau} \), which incorporates both world and particle perspectives.

Proper time serves as a fundamental bridge in calculations where relativistic effects are significant.

Four-Force

Four-force is the relativistic extension of the classical force concept, converted into four-vector form. Defined as \( F^{\mu} = (\gamma \mathbf{f} \cdot \mathbf{v}/c, \gamma \mathbf{f}) \), where \( \gamma \) is the Lorentz factor, \( \mathbf{f} \) is the three-force, and \( \mathbf{v} \) the velocity.
The four-force relates directly to the four-momentum \( p^{\mu} \) of the object via proper time intervals:
\[ F^{\mu} = \frac{dp^{\mu}}{d\tau} \]
This expression underscores the force’s role in influencing the four-momentum's rate of change over proper time.

• Encompassing both acceleration (three-force) and its effects into four-momentum changes, four-force ensures consistency with Einstein's relativity.

Understanding four-force is crucial for solving problems that involve forces acting on particles moving at relativistic speeds, as exemplified in the exercise with the equation \( (\mathrm{d} / \mathrm{d} \tau) L^{\mu u}=G^{\mu u} \).