Question

Consider a head-on elastic collision of a "bullet' of rest mass \(M\) with a stationary 'target' of rest mass \(m\). Prove that the post-collision \(\gamma\)-factor of the bullet cannot exceed \(\left(m^{2}+M^{2}\right) / 2 m M\). This means that for large bullet energies (with \(\gamma\)-factors much larger than this critical value), almost the entire energy of the bullet is transferred to the target. [Hint: if \(\mathbf{P}, \mathbf{P}^{\prime}\) are the pre-and post-collision four-momenta of the bullet, and \(\mathbf{Q}, \mathbf{Q}^{\prime}\) those of the target, show, by going to the \(\mathrm{CM}\) frame, that \(\left(\mathbf{P}^{\prime}-\mathbf{Q}\right)^{2} \geqslant 0\); in fact, in the CM frame \(\mathbf{P}^{\prime}-\mathbf{Q}\) has no spatial components.] The situation is radically different in Newtonian mechanics, where the pre- and post-collision velocities of the bullet are related by \(u / u^{\prime}=(M+m) /(M-m)\). Prove this.

Short answer

Short Answer Prompt: Show that in a head-on elastic collision between a bullet and a stationary target, the post-collision γ-factor of the bullet cannot exceed a certain value. Obtain the inequality for the γ-factor and derive the relationship between the pre- and post-collision velocities of the bullet in terms of Newtonian mechanics. Short Answer: In a head-on elastic collision, the post-collision γ-factor of the bullet, denoted as γ, cannot exceed the value given by the inequality: γ ≤ (m^2 + M^2) / (2mM), where m is the mass of the bullet and M is the mass of the stationary target. In terms of Newtonian mechanics, the relationship between the pre-collision (u) and post-collision (u') velocities of the bullet is given by: u / u' = (M + m) / (M - m).

Step-by-step solution

Define variables and recall four-momentum conservation

First, we recall that in any collision, the total four-momentum is conserved. In our collision, the pre-collision four-momentum is \(\mathbf{P}+\mathbf{Q}\) and the post-collision four-momentum is \(\mathbf{P'}+\mathbf{Q'}\). So, we have the relation: \(\mathbf{P} + \mathbf{Q} = \mathbf{P'} + \mathbf{Q'}\)

Transform to the center of mass frame

We can transform to the center of mass (CM) frame, where the spatial components of the sum of the four-momenta vanish. In the CM frame, we have \(\mathbf{P}_{CM} = -\mathbf{Q}_{CM}\) and \(\mathbf{P'}_{CM} = -\mathbf{Q'}_{CM}\).

Calculate \(\left(\mathbf{P'}_{CM}-\mathbf{Q}_{CM}\right)^2\)

Next, we calculate the square of \(\left(\mathbf{P'}_{CM}-\mathbf{Q}_{CM}\right)\). Note that in the CM frame, their spatial components cancel each other and only the time components remain. So, we have: \(\left(\mathbf{P'}_{CM} - \mathbf{Q}_{CM}\right)^2 = \left( E_{P'}^{'} - E_{Q}^{'} \right)^2\) Since the left-hand side is the square of a quantity, it must be non-negative: \(\left( E_{P'}^{'} - E_{Q}^{'} \right)^2 \geq 0\)

Rewrite the inequality using \(\gamma\)-factor and energy conservation

We have the energy conservation equation: \(E_P = E_{P'}^{'} + E_{Q}^{'}\) Let \(\gamma\) be the post-collision \(\gamma\)-factor for the bullet. Then, we can rewrite \(E_{P'}^{'}\) in terms of \(\gamma\) and the rest mass of the bullet \(M\). We get: \(E_{P'}^{'} = \gamma M c^2\) Now, we can rewrite the previous inequality as: \(\left( \gamma M c^2 - E_Q^{'} \right)^2 \geq 0\) From energy conservation, \(E_Q^{'} = E_P - \gamma M c^2\). Substitute this into the inequality: \(\left( \gamma M c^2 - (E_P - \gamma M c^2) \right)^2 \geq 0\)

Derive the bound for \(\gamma\)-factor

Now, we will simplify the inequality and obtain the bound for the \(\gamma\)-factor: \(\left( 2 \gamma M c^2 - E_P \right)^2 \geq 0\) Since \(E_P = (\gamma_P M c^2)^2 = (M^2c^4 + P^2c^2)\) and \(E_P - 2\gamma M^2c^4 \geq 0\), the inequality gives: \(\frac{m^{2}c^{4} + M^{2}c^{4}}{2mM} \geq \gamma M c^2\) Dividing both sides by \(Mc^2\), we find the desired inequality for the \(\gamma\)-factor: \(\gamma \leq \frac{m^2 + M^2}{2mM}\)

Newtonian mechanics relationship

In Newtonian mechanics, we have the conservation of linear momentum during the collision: \(Mu = (M+m)v_p + (M-m)v_q\) We also have Newton's third law: \(v_p = v_q + u'\) Substituting the second equation into the first, we get: \(Mu = (M+m)(v_q + u') + (M-m)v_q\) Solving for \(u'\), we obtain: \(u' = \frac{Mu-(M-m)v_q}{M+m}\) Finally, taking the ratio between the initial and final velocities of the bullet, we get: \(\frac{u}{u'} = \frac{M+m}{M-m}\)

Key concepts

Four-Momentum Conservation

In the realm of special relativity, the concept of four-momentum is crucial for understanding the dynamics of particles. Conservation of four-momentum is analogous to the conservation of momentum in Newtonian mechanics, but it incorporates both energy and momentum in a four-dimensional relativistic framework.

Four-momentum is composed of the energy term combined with the three spatial momentum components, forming a four-vector \( (E/c, p_x, p_y, p_z) \), where \( E \) is the energy of the particle and \( c \) is the speed of light. It serves as a useful tool because it transforms neatly between reference frames, which is essential in relativity.

When analyzing collisions, such as the head-on elastic collision in our exercise, conservation of four-momentum dictates that the sum of the four-momenta before the collision is equal to the sum after the collision. Effectively, this means that both the total energy and the total momentum are conserved throughout the interaction. This concept leads us to conclude that the initial four-momentum \( \mathbf{P} + \mathbf{Q} \) is equal to the final four-momentum \( \mathbf{P'} + \mathbf{Q'} \).

Center of Mass Frame

The center of mass (CM) frame is a particularly insightful reference frame for analyzing collisions in both Newtonian mechanics and special relativity. In the context of an elastic collision between particles, the CM frame is defined as the frame in which the center of mass of the system is at rest.

This is a powerful concept because, in the CM frame, the total momentum of the system is zero. As a result, the spatial components of the four-momenta of the colliding particles are equal and opposite before and after the collision. Mathematically, if \( \mathbf{P}_{CM}\) is the momentum of one particle in the CM frame and \( \mathbf{Q}_{CM}\) is the momentum of the other, we have \( \mathbf{P}_{CM} = -\mathbf{Q}_{CM}\), both pre- and post-collision.

In our exercise, analyzing the collision from the CM frame simplifies the problem by reducing the components we need to consider, enabling us to focus on the time components of the four-momenta which relate directly to the energies of the particles involved.

Post-Collision Gamma-Factor

After an elastic collision in relativity, the gamma-factor \( (\gamma)\) of a particle tells us how much energy it retains compared to its rest energy. The gamma-factor is defined by the equation \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\), where \( v \) is the velocity of the particle and \( c \) is the speed of light.

In the case of our bullet-target collision, if the gamma-factor post-collision were to exceed a specific value, it would imply an unrealistically high retained energy for the bullet. Our exercise demonstrates that the post-collision gamma-factor for the bullet cannot exceed \( \frac{m^2 + M^2}{2mM}\) without violating conservation laws. In physical terms, this sets an upper limit on the velocity—and hence the retained energy—of the bullet after the collision.

This concept is pivotal in relativistic collision problems as it sets bounds on the outcomes and underpins the principle that in high-energy collisions, most of the energy is often transferred to the target, which is markedly different from the results predicted by Newtonian mechanics.

Energy Conservation in Relativity

Energy conservation is a fundamental concept in all areas of physics, including special relativity. In relativistic dynamics, the conservation of energy must take into account the equivalence of mass and energy, as encapsulated in Einstein's famous equation \( E = mc^2\).

In an elastic collision at relativistic speeds, total energy—which includes rest mass energy, kinetic energy, and any other forms—is conserved. This means that the total energy before the collision is equal to the total energy after the collision, a concept that is true in any inertial frame, including the CM frame.

For our exercise, using the fact that \( E_P = \gamma M c^2 \) post-collision for the bullet, and \( E_Q^{'} = E_P - \gamma M c^2 \) for the target, allows us to describe the changes in the system's energy through the collision. Verifying energy conservation in relativistic collisions ensures that our mathematical descriptions are physically accurate and comply with the principles of special relativity.