Irodov 1.381: Energy and Momentum Are One Vector – Amit & Deepali Sinha Foundation

Problem Irodov 1.381

A particle moves along the $x$ axis of a frame $K$ , carrying momentum $p$ (directed along that axis, with component $p_x$ ) and total energy $E$ . A second frame $K'$ moves at a constant velocity $V$ relative to $K$ , along the positive direction of the shared $x$ axis. Writing $\beta = V/c$ , show that the particle’s momentum component $p_x'$ and total energy $E'$ , as measured in $K'$ , are given by

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

The standard configuration: K' slides along the shared x axis at V, and the particle P runs along that same axis at u as measured in K. — The standard configuration: $K'$ slides along the shared $x$ axis at $V$ , and the particle $P$ runs along that same axis at $u$ as measured in $K$ .

Given

particle speed $u$ along $x$ in $K$
$p_x = m_0u/\sqrt{1-u^2/c^2}$
$E = m_0c^2/\sqrt{1-u^2/c^2}$
boost speed $V$ , $\beta = V/c$

Find

Demonstrate the stated transformation of

p_x

and

E

Solution

Step 1Write both quantities in terms of the one variable that transforms

Energy and momentum are not independent unknowns here: both are built from the particle’s velocity $u$ and its rest mass, and only $u$ changes under the boost. With $\gamma_u = 1/\sqrt{1-u^2/c^2}$ ,

p_x = \gamma_u m_0 u, \qquad E = \gamma_u m_0 c^2 .

Dividing one by the other removes both $m_0$ and $\gamma_u$ and leaves a relation we shall use twice:

p_x = \frac{Eu}{c^2}

(1)

Step 2Transform the velocity, which is the only thing that moves

For motion along the boost direction the velocity transformation is

u' = \frac{u-V}{1-\dfrac{uV}{c^2}} .

The primed energy and momentum are built from $u'$ by exactly the same two definitions, with the same invariant $m_0$ , so the whole demonstration reduces to evaluating $\gamma_{u'}$ .

Step 3Evaluate the primed Lorentz factor

This is the one piece of algebra in the problem. Form $1-u'^2/c^2$ over a common denominator:

\begin{aligned} 1-\frac{u'^2}{c^2} &= \frac{\left(1-\dfrac{uV}{c^2}\right)^{2} -\dfrac{\left(u-V\right)^{2}}{c^2}} {\left(1-\dfrac{uV}{c^2}\right)^{2}} . \end{aligned}

Expand the numerator on its own. The two cross terms in $uV/c^2$ cancel:

\begin{aligned} 1-\frac{2uV}{c^2}+\frac{u^2V^2}{c^4} -\frac{u^2}{c^2}+\frac{2uV}{c^2}-\frac{V^2}{c^2} &= 1+\frac{u^2V^2}{c^4}-\frac{u^2}{c^2}-\frac{V^2}{c^2} \\[2pt] &= \left(1-\frac{u^2}{c^2}\right)\left(1-\frac{V^2}{c^2}\right) . \end{aligned}

Putting the factored numerator back over its denominator, with $V^2/c^2 = \beta^2$ ,

1-\frac{u'^2}{c^2} = \frac{\left(1-\dfrac{u^2}{c^2}\right)\left(1-\beta^2\right)} {\left(1-\dfrac{uV}{c^2}\right)^{2}} ,

and taking the reciprocal square root of the whole quotient gives the primed factor as a product of the two unprimed ones:

\gamma_{u'} = \gamma_u\,\frac{1}{\sqrt{1-\beta^2}} \left(1-\frac{uV}{c^2}\right)

(2)

Step 4Read off the transformed energy

Multiply (2) by $m_0c^2$ , which is the same number in both frames:

\begin{aligned} E' = \gamma_{u'}m_0c^2 &= \frac{\gamma_u m_0c^2}{\sqrt{1-\beta^2}}\left(1-\frac{uV}{c^2}\right)\\[2pt] &= \frac{E-\dfrac{Eu}{c^2}V}{\sqrt{1-\beta^2}} \\[2pt] &= \frac{E-p_xV}{\sqrt{1-\beta^2}} . \end{aligned}

The last line used $Eu/c^2 = p_x$ from (1). That is the first of the two formulas asked for; it is collected with the second in the box below.

Step 5Read off the transformed momentum

Apply relation (1) in the primed frame, where it reads $p_x' = E'u'/c^2$ , and substitute both the primed energy just obtained and the transformed velocity:

\begin{aligned} p_x' &= \frac{E'u'}{c^2} = \frac{1}{c^2}\cdot \frac{E\left(1-\dfrac{uV}{c^2}\right)}{\sqrt{1-\beta^2}}\cdot \frac{u-V}{1-\dfrac{uV}{c^2}} \\[4pt] &= \frac{E\left(u-V\right)}{c^2\sqrt{1-\beta^2}} \\[4pt] &= \frac{\dfrac{Eu}{c^2}-\dfrac{EV}{c^2}}{\sqrt{1-\beta^2}} . \end{aligned}

The bracket $\left(1-uV/c^2\right)$ has canceled between the two factors, and $Eu/c^2 = p_x$ once more, which completes the demonstration:

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

(3)

E against p_xc: the particle sits at P on the hyperbola E^2-p_x^2c^2 = (m_0c^2)^2, and the boost tilts the axes to E' and p_x'c by equal angles in opposite senses. The dotted lines are the massless limit E = p_xc. — $E$ against $p_xc$ : the particle sits at $P$ on the hyperbola $E^2-p_x^2c^2 = \left(m_0c^2\right)^2$ , and the boost tilts the axes to $E'$ and $p_x'c$ by equal angles in opposite senses. The dotted lines are the massless limit $E = \pm p_xc$ .

Check

The transformation must leave the invariant $E^2-p_x^2c^2$ alone, and that is a real test because it uses both formulas at once. Substituting,

E'^2-p_x'^2c^2 = \frac{\left(E-p_xV\right)^2 -\left(p_x-\dfrac{EV}{c^2}\right)^2c^2}{1-\beta^2} .

Expanding the two squares, the cross terms $2Ep_xV$ cancel and the numerator collapses to

E^2\left(1-\beta^2\right)-p_x^2c^2\left(1-\beta^2\right) ,

so the denominator divides out and $E'^2-p_x'^2c^2 = E^2-p_x^2c^2$ , which is the invariant restored. Two limits also check out: $V \to 0$ returns $E' = E$ and $p_x' = p_x$ , and putting $u = V$ , so that the particle is at rest in $K'$ , gives $p_x = EV/c^2$ and hence $p_x' = 0$ , as it must.

Remark

Compare the pair with the coordinate transformation $x' = \left(x-Vt\right)/\sqrt{1-\beta^2}$ and $ct' = \left(ct-\beta x\right)/\sqrt{1-\beta^2}$ : they are the same equations with $E/c$ playing the part of $ct$ and $p_x$ the part of $x$ . That is what is meant by calling $\left(E/c,\,\vect p\right)$ a four-vector, and it is why the figure above can borrow a spacetime diagram wholesale. The derivation divided by $m_0$ , yet the result needs no mass: hold $E$ fixed and let $m_0 \to 0$ , so that $u \to c$ and (1) becomes the massless relation $p_x = E/c$ . Both formulas are linear in $E$ and $p_x$ , so they pass to that limit unchanged, and for a photon the energy one reads $E' = E\sqrt{\left(1-\beta\right)/\left(1+\beta\right)}$ , the relativistic Doppler shift.