HW7 Answer

1 Relativistic Collision

1.1

Refer to the total momentum conservation, one yield

\frac{Q}{c} = γ M v

Similarly, energy's conserved

Q + m c^{2} = γ M c^{2}

Thus

\begin{matrix} (1) & v = \frac{Q c}{Q + m c^{2}} \end{matrix}

Using the relativistic invariant $E^{2} = M^{2} c^{4} + p^{2} c^{2}$ , one can obtain that

(Q + m c^{2})^{2} = M^{2} c^{4} + Q^{2}

So the rest mass of the complex is

\begin{matrix} (2) & M = m \sqrt{1 + \frac{2 Q}{m c^{2}}} \end{matrix}

1.2

The atom recoils with momentum equal to the photon's momentum, $p_{a t o m} = p_{p h o t o n} = \frac{Q}{c}$ , So energy's conservation yield that

M c^{2} = Q + \sqrt{(m c^{2})^{2} + (p_{a t o m} c)^{2}} = Q + \sqrt{(m c^{2})^{2} + Q^{2}}

Since $M c^{2} - m c^{2} = Q_{0}$ ,

M c^{2} = Q + \sqrt{(M c^{2} - Q_{0})^{2} + Q^{2}}

Thus

Q = Q_{0} (1 - \frac{Q_{0}}{2 M c^{2}})

The photon energy $Q$ is slightly less than the lost rest energy $Q_{0}$ because a small portion of the energy must go into the kinetic energy of the recoiling atom.

2 Photon Rocket

2.1

Use subscript ' $p$ ' to indicate the photons emitted, refer tor the energy and momentum conservation,

m c^{2} = γ f m c^{2} + E_{p}, P_{p} = γ f m v, with E_{p} = P_{p} c

Thus

f = [γ (1 + β)]^{- 1} = \sqrt{\frac{1 - β}{1 + β}}

2.2

γ = 10 = \sqrt{\frac{1}{1 - β^{2}}} ⟹ f = \sqrt{\frac{1 - \sqrt{0.99}}{1 + \sqrt{0.99}}} \approx 0.0501

Namely $β = \sqrt{0.99} \approx 1 \to f = [γ (1 + β)]^{- 1} \approx \frac{1}{2 γ}$ .

2.3

Because the rocket is effectively "consuming" its own mass to change velocity at each step, one can find the symmetry relation

f_{t o t a l} = f^{4} \approx 6 \times 10^{- 6}

3 Energy-Momentum Conservation

For a time interval in an arbitrary inertial frame, The problem say that we have the energy conservation as follows,

Δ P^{' 0} = Δ P^{0} = \frac{1}{c} [E (t + Δ t) - E (t)] = 0

So according to the Lorentz transformation in x-axis (x-boost):

Δ P^{0^{'}} = γ (Δ P^{0} - \frac{v}{c} Δ P^{1}) ⟹ Δ P^{1} = 0

By choosing frames moving along the $y$ and $z$ axes, one can similarly prove:

Δ P^{2} = 0 and Δ P^{3} = 0

4 Energy-Momentum Tensor of Pure Radiation

For free E.D. field in the problem, the energy-momentum tensor is

T^{μ ν} = [\begin{matrix} \frac{1}{2} ϵ_{0} (E^{2} + c^{2} B^{2}) & \frac{1}{μ_{0} c} E \times B \\ \frac{1}{μ_{0} c} E \times B & \frac{1}{2} ϵ_{0} (E^{2} + c^{2} B^{2}) δ_{i j} - ϵ_{0} (E E + c^{2} B B) \end{matrix}]

where one find obviously

T^{μ ν} = (\begin{matrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{matrix}) ϵ_{0} E_{0}^{2} \cos^{2} k (z - c t)

For a wave traveling in the $z$ -direction, the momentum flux is also only in the $z$ -direction, so only $T^{z z}$ is none-zero in the maxwell stress tensor.

5 Particle in the E.D. Field

5.1

Since for $H = \int L_{t} d t = \int L_{τ} d τ$ , where

d τ = \frac{d t}{γ}

One obtain that

L_{t} = \frac{L_{τ}}{γ} = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}} + e A \cdot v - e φ

substitute to the Euler-Lagrange equation $\frac{d}{d t} \frac{\partial L}{\partial v} - \frac{\partial L}{\partial r} = 0$ , one find

\frac{d}{d t} γ m v = e (- \nabla φ - \frac{\partial A}{\partial t}) + e v \times B = e E + e v \times B

5.2

Use the Legendre Transformation and $p = γ m v = P - e A, P = \frac{\partial L}{\partial v}$ , the Hamiltonian can be calculated as

H = p \cdot v - L = \sqrt{c^{2} (P - e A)^{2} + (m c^{2})^{2}} + e φ

5.3

For Non-relativistic approximation ( $β ≪ 1$ ), we also have $γ β ≪ 1$ . Thus

L_{t} = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}} + e A \cdot v - e φ \approx - m c^{2} + \frac{1}{2} m v^{2} + e A \cdot v - e φ

H = \sqrt{c^{2} (P - e A)^{2} + (m c^{2})^{2}} + e φ \approx m c^{2} + \frac{(P - e A)^{2}}{2 m} + e φ

where the constant rest energy $m c^{2}$ can be ignored.