Special Theory of Relativity Questions and Answers

Answer: C

The time difference is \(t_{Q} - t_{P} = \frac{L}{6c}\).

In the ground frame the rod is contracted to \(L/\gamma\), with \(\gamma = 1/\sqrt{1 - 9/25} = 5/4\), so its length there is \(4L/5\). Q is the front end. If P's position is read at \(t_{P}\) and Q's at the later time \(t_{Q}\), Q has moved on by \(v(t_{Q} - t_{P})\) in the meantime:

\[ \begin{aligned} \frac{4L}{5} + \frac{3c}{5}(t_{Q} - t_{P}) &= \frac{9L}{10} \\ t_{Q} - t_{P} &= \frac{L/10}{3c/5} = \frac{L}{6c} \end{aligned} \]

The other values do not satisfy this relation: \(L/3c\), for example, would put the ends \(L\) apart, and \(2L/3c\) even further.

Answer: A

The speed of T relative to S is \(\dfrac{4}{5}c\).

Velocities do not simply add in special relativity. In the observer's frame S moves at -0.5c and T at +0.5c. Using the relativistic subtraction formula:

\[ \begin{aligned} v_{TS} &= \frac{u - v}{1 - uv/c^{2}} = \frac{0.5c + 0.5c}{1 + 0.25} \\ &= \frac{c}{1.25} = 0.8c \end{aligned} \]

c is the Galilean sum 0.5c + 0.5c, and 4/3 c exceeds the speed of light, which no relative speed of material bodies can. 2/3 c comes from dividing by 1.5 instead of 1.25.

Answer: B

The correct relation is \(\vec{E}' = \vec{E} + \vec{v}_0 \times \vec{B}\) and \(\vec{B}' = \vec{B}\).

Under a Galilean transformation a charge with velocity \(\vec{v}\) in the rest frame has velocity \(\vec{v} - \vec{v}_0\) in the moving frame, and the force on it is the same in both frames:

\[ \begin{aligned} & q(\vec{E} + \vec{v} \times \vec{B}) \\ &= q(\vec{E}' + (\vec{v} - \vec{v}_0) \times \vec{B}') \end{aligned} \]

This must hold for every \(\vec{v}\). Matching the terms that depend on \(\vec{v}\) gives \(\vec{B}' = \vec{B}\); the remaining terms then give \(\vec{E} = \vec{E}' - \vec{v}_0 \times \vec{B}\), that is, \(\vec{E}' = \vec{E} + \vec{v}_0 \times \vec{B}\).

Leaving both fields unchanged ignores the change of velocity. The other two options either mix the electric field into the magnetic term or have \(\vec{B}'\) depend on the particle's own velocity, which a field cannot do.

Answer: D

The correct diagram is (D).

In the frame of \(\bar{O}\), P and R lie on the \(\bar{t}\) axis (\(\bar{x} = 0\)), before and after the reflection, and Q lies on the \(\bar{x}\) axis (\(\bar{t} = 0\)) at positive \(\bar{x}\). These facts hold in every frame, because they describe which events share a position or a time for \(\bar{O}\).

Since \(\bar{O}\) moves along \(+x\), its worldline, the \(\bar{t}\) axis, tilts towards \(+x\) in the diagram of \(O\), and its \(\bar{x}\) axis tilts upwards by the same angle, as a mirror image about the light line. So P and R must sit on the tilted \(\bar{t}\) line and Q on the tilted \(\bar{x}\) line, with Q on the right.

(A) puts P on the vertical \(t\) axis, not on the \(\bar{t}\) axis. (B) draws the \(\bar{x}\) axis on top of the \(x\) axis, which is the Galilean picture and ignores the relativity of simultaneity. (C) places Q on the negative side, which would need \(\bar{O}\) to move the other way. Only (D) has P, R on \(\bar{t}\), Q on \(\bar{x}\) and Q to the right.

Answer: D

The correct statement is that events \((9, -1, 1, -1)\) and \((-10, 1, 0, -1)\) are time-like separated.

The invariant interval is \(\Delta s^2 = (c\Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2\): positive for time-like, zero for light-like and negative for space-like separation. The differences for each pair are:

\[ \begin{aligned} \text{A:}\ (-2, 1, 1, 0) &\Rightarrow 4 - 1 - 1 = 2 \\ \text{B:}\ (2, 1, -1, 0) &\Rightarrow 4 - 1 - 1 = 2 \\ \text{C:}\ (1, 1, -2, 0) &\Rightarrow 1 - 1 - 4 = -4 \\ \text{D:}\ (-19, 2, -1, 0) &\Rightarrow 361 - 4 - 1 = 356 \end{aligned} \]

Pair D has a positive interval, so it is time-like, as stated. Pairs A and B are also time-like, not space-like, and pair C is space-like, not light-like. So only D is correct.

Answer: B

The time order can change only for events in the space-like region.

The invariant interval between two events is \(s^{2} = c^{2}\Delta t^{2} - \Delta x^{2}\). If the separation is time-like (\(s^{2} \gt 0\)), one event lies inside the other's light cone; a signal could connect them, and every inertial observer agrees on which came first. If it is light-like (\(s^{2} = 0\)), the events can be joined by a light ray, and again the order is fixed.

Only when the separation is space-like (\(s^{2} \lt 0\)) can a Lorentz boost with speed below c make \(\Delta t\) change sign. Such events cannot be causally connected, so no paradox arises. Exceeding the speed of light is not possible, which rules out the last option.

Answer: D

A single photon can be produced only when at least one particle is in a bound state.

For a free electron and positron, energy and momentum cannot both be conserved if only one photon is emitted. In the centre-of-momentum frame the total momentum is zero but the energy is at least \(2m_{e}c^{2}\); a single photon with that energy would carry momentum \(E/c \neq 0\). So free-pair annihilation gives at least two photons.

If one of the particles is bound, for example an electron bound in an atom, the nucleus or the rest of the atom can take up the recoil momentum, and one-photon annihilation becomes possible. Having one or both particles at rest does not help, since the centre-of-momentum argument still applies.

Answer: A

The rocket's velocity is 0.75 c.

Velocities must be combined with the relativistic addition formula, not simply added:

\[ \begin{aligned} u &= \frac{v + u'}{1 + vu'/c^{2}} = \frac{0.4c + 0.5c}{1 + 0.2} \\ &= \frac{0.9c}{1.2} = 0.75c \end{aligned} \]

The option 0.90 c is the Galilean sum, which ignores the denominator. The other options do not follow from either rule.

Answer: A

The rest mass is 12 GeV/c\(^{2}\).

The relativistic energy-momentum relation gives

\[ \begin{aligned} (mc^{2})^{2} &= E^{2} - (pc)^{2} \\ &= 13^{2} - 5^{2} = 144\ \text{GeV}^{2} \\ mc^{2} &= 12\ \text{GeV} \end{aligned} \]

The numbers form the 5-12-13 right triangle. The MeV options are a thousand times too small, and 1.2 GeV/c\(^{2}\) misplaces the decimal point; the rest energy must be of the same order as the 13 GeV total energy because the momentum is comparatively small.

Answer: B

A is 40.2 years old and B is 22.8 years old.

In A's frame the trip takes

\[ t_{A} = \frac{20\ \text{ly}}{0.99c} = 20.2\ \text{years} \]

so A ages to 40.2 years. B's clock runs slow by the factor \(\sqrt{1 - v^{2}/c^{2}}\):

\[ \begin{aligned} \sqrt{1 - 0.99^{2}} &= \sqrt{0.0199} = 0.141 \\ t_{B} &= 20.2 \times 0.141 = 2.85\ \text{years} \end{aligned} \]

so B is about 22.8 years old. This is the twin paradox: B, who accelerates and turns round, is the younger. Equal ages ignore time dilation, and the other options give the dilation to the wrong twin or not at all.