Irodov 1.382: Chasing a Photon Lowers Its Energy – Amit & Deepali Sinha Foundation

Problem Irodov 1.382

A photon travels along the $x$ axis of a frame $K$ , in which its energy is $\varepsilon$ . A frame $K'$ moves relative to $K$ at the constant velocity $V$ along that same axis, in the direction the photon travels. The energy $E$ and the momentum component $p_x$ of any particle transform between the two frames as

E' = \frac{E-p_xV}{\sqrt{1-\beta^2}}, \qquad p_x' = \frac{p_x-EV/c^2}{\sqrt{1-\beta^2}}, \qquad \beta = \frac{V}{c} .

Find the energy $\varepsilon'$ of the photon in the frame $K'$ , and find the value of $V$ at which $\varepsilon' = \varepsilon/2$ .

The photon on a spacetime chart of K. Its worldline is the dotted 45 degree line through the origin, with a dot at one event. A boost along that direction tilts the primed axes x' and ct' toward the line from either side. — The photon on a spacetime chart of $K$ . Its worldline is the dotted $45$ degree line through the origin, with a dot at one event. A boost along that direction tilts the primed axes $x'$ and $ct'$ toward the line from either side.

Given

photon energy $\varepsilon$ in $K$
boost speed $V$ along the photon’s direction
$\beta=V/c$
the stated transformation of $E$ and $p_x$

Find

the energy

\varepsilon'

K'

, and the

V

for which

\varepsilon'=\varepsilon/2

Solution

Step 1What the photon carries in the frame

K

The transformation formulas need both the energy and the momentum of the particle in $K$ , so the photon’s momentum has to be expressed through its energy before anything can be substituted. A photon has no rest mass, so the energy-momentum relation loses its rest term:

\varepsilon^2 = p_x^2c^2+m_0^2c^4, \qquad m_0 = 0

p_x = \frac{\varepsilon}{c}

(1)

Step 2Substituting the photon into the energy transformation

Only the energy formula is needed. Putting $E = \varepsilon$ and the momentum (1) into it, and using $V = \beta c$ to clear the speeds,

\begin{aligned} \varepsilon' &= \frac{\varepsilon-p_xV}{\sqrt{1-\beta^2}} \\[3pt] &= \frac{\varepsilon-\dfrac{\varepsilon}{c}\,\beta c}{\sqrt{1-\beta^2}} \end{aligned}

\varepsilon' = \frac{\varepsilon\left(1-\beta\right)} {\sqrt{1-\beta^2}}

(2)

Step 3Tidying the result into the Doppler factor

The denominator of (2) is a difference of two squares, so it factors into two roots, one of which cancels against the numerator:

\begin{aligned} \sqrt{1-\beta^2} &= \sqrt{1-\beta}\,\sqrt{1+\beta} \\[3pt] \varepsilon' &= \frac{\varepsilon\left(1-\beta\right)} {\sqrt{1-\beta}\,\sqrt{1+\beta}} \end{aligned}

Canceling one power of $\sqrt{1-\beta}$ leaves the two factors under a single root:

\varepsilon' = \varepsilon\sqrt{\frac{1-\beta}{1+\beta}}, \qquad \beta = \frac{V}{c}

(3)

Step 4The boost that halves the energy

Impose the condition $\varepsilon' = \varepsilon/2$ on (3). Squaring removes the root, and $\varepsilon$ cancels because the condition is a ratio:

\begin{aligned} \frac{1-\beta}{1+\beta} &= \frac{1}{4} \\[3pt] 4\left(1-\beta\right) &= 1+\beta \\[3pt] 3 &= 5\beta \end{aligned}

Multiplying $\beta = 3/5$ by the speed of light gives the velocity asked for:

V = \frac{3}{5}c = 1.8\times10^{8}\ \mathrm{m/s}

(4)

The energy ratio plotted against β over 0 <= β <= 0.95. It equals 1 at β = 0 and falls steadily from there. The marked point is β = 3/5, where the ratio has dropped to one half. — The energy ratio (3) plotted against $\beta$ over $0 \le \beta \le 0.95$ . It equals $1$ at $\beta = 0$ and falls steadily from there. The marked point is $\beta = 3/5$ , where the ratio has dropped to one half.

Check

The transformed momentum is the real test, because the answer has to leave the photon a photon. Feeding (1) into the momentum formula gives $p_x' = \left(\varepsilon/c-\varepsilon\beta/c\right)/\sqrt{1-\beta^2}$ , which is the $\varepsilon'$ of (3) divided by $c$ , so $\varepsilon'^2-p_x'^2c^2 = 0$ in $K'$ as well: the photon is still massless and still moves at $c$ , which is what the chart above asserts. Two limits also behave: $\beta \to 0$ returns $\varepsilon' = \varepsilon$ , and $\beta \to 1$ drives $\varepsilon'$ to zero, never below it. Finally the halving condition can be read backwards, since $\sqrt{(1-0.6)/(1+0.6)} = \sqrt{0.25} = 1/2$ exactly.

Remark

The factor in (3) is the longitudinal Doppler factor, and it acts on the whole of the photon’s energy-momentum pair at once: both $\varepsilon$ and $p_x$ are multiplied by the same $\sqrt{(1-\beta)/(1+\beta)}$ , which is what a boost can do to a null four-vector, since a boost along the photon’s own line cannot change the direction of a vector lying along that line. Written with $\varepsilon = h\nu$ it says $\nu' = \nu\sqrt{(1-\beta)/(1+\beta)}$ , the red shift seen by an observer receding from a source, and reversing the sign of $V$ turns it into the blue shift seen by one approaching. However hard $K'$ chases the photon, the energy only tends to zero: no frame catches the light and no frame sees it at rest, which is the same statement about the light line that the chart makes geometrically.