In a given frame, a particle A moves hyperbolically with proper acceleration \(\alpha\) from rest at \(t=0\). At \(t=0\) a photon B is emitted in the same direction, a distance \(c^{2} / \alpha\) behind A. Prove that in A's instantaneous rest frames the distance \(A B\) is always \(c^{2} / \alpha\).

To solve this problem, we first need to understand the motion of particle A. Since A moves hyperbolically with proper acceleration \(\alpha\), we can use the equation of hyperbolic motion to describe its position as a function of time in the lab frame: \(x_A(t) = \dfrac{c^2}{\alpha}\sqrt{1+ \left(\dfrac{\alpha t}{c}\right)^2}\) We will need this equation later to calculate the distance between A and B.

At time \(t=0\), the photon B is emitted behind A by a distance \(\dfrac{c^2}{\alpha}\). The photon moves with the speed of light, \(c\), so its position in the lab frame at time \(t\) is: \(x_B(t) = -\dfrac{c^2}{\alpha} + ct\)

The distance between particle A and photon B in the lab frame at time \(t\) is the difference between their positions: \(D_{lab}(t) = x_A(t) - x_B(t) = \dfrac{c^2}{\alpha}\left(\sqrt{1+ \left(\dfrac{\alpha t}{c}\right)^2} - 1 + \dfrac{\alpha t}{c}\right)\)

Now we need to convert the distance and time to A's instantaneous rest frame using the Lorentz transformation. The velocity of A at time \(t\) in the lab frame is: \(v_A(t) = \dfrac{\alpha t}{\sqrt{1 + \left(\dfrac{\alpha t}{c}\right)^2}}\) Using this velocity, we can find the Lorentz factor \(\gamma\): \(\gamma(t) = \dfrac{1}{\sqrt{1 - \left(\dfrac{v_A(t)}{c}\right)^2}} = \dfrac{1}{\sqrt{1 + \left(\dfrac{\alpha t}{c}\right)^2}}\) Now, we can use the Lorentz transformation to convert the distance \(D_{lab}\) to the distance \(D_A\) in A's instantaneous rest frame: \(D_A(t) = \gamma(t)D_{lab}(t)\) Substituting the expressions for \(\gamma(t)\) and \(D_{lab}(t)\), we get: \(D_A(t) = \gamma(t)D_{lab}(t) = \dfrac{c^2}{\alpha}\left(1 + \dfrac{\alpha t}{c}\right)\)

Finally, we need to show that the distance \(D_A(t)\) remains constant in A's instantaneous rest frame. Simplifying the expression obtained in the previous step, we get: \(D_A(t) = \dfrac{c^2}{\alpha}\left(1 + \dfrac{\alpha t}{c}\right) = \dfrac{c^2}{\alpha}\left(1 + \dfrac{\alpha t}{c}\right) = \dfrac{c^2}{\alpha}\) Thus, the distance between particle A and photon B in A's instantaneous rest frame remains constant and equal to \(\dfrac{c^2}{\alpha}\), proving the statement.