Irodov 1.383: Every Frame Agrees on Rest Mass – Amit & Deepali Sinha Foundation

Problem Irodov 1.383

A single particle of rest mass $m_0$ moves freely. An observer in some inertial reference frame measures its total energy $E$ and the magnitude $p$ of its momentum, and forms the combination $E^2 - p^2c^2$ , where $c$ is the speed of light. Show that every inertial observer, whatever the particle’s velocity in that observer’s frame, arrives at the same value for this combination, and find what that value is.

Given

rest mass $m_0$
total energy $E$
momentum magnitude $p$
an arbitrary inertial frame

Find

demonstrate that

E^2-p^2c^2

is the same in every inertial frame, and find its value

Solution

Step 1Energy and momentum from one Lorentz factor

Both quantities in the combination are built from the same two ingredients: the particle’s rest mass, which belongs to the particle, and its velocity, which belongs to the frame. Let $v$ be the particle’s speed in the frame doing the measuring, and collect the velocity dependence in the Lorentz factor $\gamma$ :

\gamma = \frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}, \qquad E = \gamma m_0c^2, \qquad p = \gamma m_0 v . \tag{1}

Step 2Eliminate the velocity from the combination

Square each of the two and subtract $c^2$ times the momentum square. The common factor $\gamma^2m_0^2c^2$ comes out of both terms at once.

\begin{aligned} E^2-p^2c^2 &= \left(\gamma m_0c^2\right)^2-\left(\gamma m_0v\right)^2c^2 \\[2pt] &= \gamma^2m_0^2c^4-\gamma^2m_0^2v^2c^2 \\[2pt] &= \gamma^2m_0^2c^4\left(1-\frac{v^2}{c^2}\right) \end{aligned}

The bracket is precisely the quantity $\gamma$ was defined to undo, so squaring the definition of $\gamma$ and rearranging gives the identity that finishes the calculation:

\gamma^2\left(1-\frac{v^2}{c^2}\right) = 1 .

Applying it to the last line leaves no trace of $v$ at all:

E^2-p^2c^2 = m_0^2c^4 = \left(m_0c^2\right)^2

(2)

E against pc. The dashed upper branch is E^2-(pc)^2 = (m_0c^2)^2; changing frame slides the point along it, from the rest frame R through K and K'. The dotted lines E = pc are the massless limit. — $E$ against $pc$ . The dashed upper branch is $E^2-\left(pc\right)^2 = \left(m_0c^2\right)^2$ ; changing frame slides the point along it, from the rest frame $R$ through $K$ and $K'$ . The dotted lines $E = \pm\,pc$ are the massless limit.

Step 3The value names the particle, not the frame

The right-hand side just obtained contains no velocity, no direction and no frame: it is built from $m_0$ , which is by definition the mass measured in the particle’s own rest frame and so is one number attached to the particle, and from $c$ , which is the same in every inertial frame. The calculation above was carried out in an arbitrary frame, so it holds equally in a second frame $K'$ in which the particle’s speed, energy and momentum are $v'$ , $E'$ and $p'$ :

E'^2-p'^2c^2 = m_0^2c^4 .

One particle, two inertial frames, drawn to one scale. In each panel E is the hypotenuse of a right triangle whose legs are pc and the rest energy m_0c^2; the panels differ in pc and in E but share the vertical leg. — One particle, two inertial frames, drawn to one scale. In each panel $E$ is the hypotenuse of a right triangle whose legs are $pc$ and the rest energy $m_0c^2$ ; the panels differ in $pc$ and in $E$ but share the vertical leg.

Both frames therefore return the same number, and the two right triangles of the figure have the same vertical leg:

E'^2-p'^2c^2 = E^2-p^2c^2

(3)

The combination has one value in every inertial frame.

Step 4The same conclusion straight from the boost

The argument so far assumed only that each frame applies (1) to its own $v$ . It is worth seeing the cancellation happen inside the transformation itself. For a boost of speed $V$ along the $x$ axis, with $\beta = V/c$ , energy and the momentum component along the boost transform as

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

Form the combination from these two and expand the numerators. The cross terms $2Ep_xV$ appear once with each sign and cancel:

\begin{aligned} 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} \\[4pt] &= \frac{E^2-2Ep_xV+p_x^2V^2 -p_x^2c^2+2Ep_xV-\dfrac{E^2V^2}{c^2}}{1-\beta^2} \\[4pt] &= \frac{E^2\left(1-\beta^2\right)-p_x^2c^2\left(1-\beta^2\right)} {1-\beta^2} \end{aligned}

The factor $1-\beta^2$ divides out of numerator and denominator, which settles the boost component alone. A boost along $x$ leaves the two transverse components of the momentum untouched, so the same amount is taken from either side when the full momentum magnitude is put back in place of $p_x$ :

\begin{aligned} p_y' &= p_y, \qquad p_z' = p_z , \\[2pt] p'^2 &= p_x'^2+p_y'^2+p_z'^2, \qquad p^2 = p_x^2+p_y^2+p_z^2 , \\[2pt] E'^2-p'^2c^2 &= E'^2-p_x'^2c^2-\left(p_y^2+p_z^2\right)c^2 \\[2pt] &= E^2-p_x^2c^2-\left(p_y^2+p_z^2\right)c^2 \end{aligned}

The last line is $E^2-p^2c^2$ , so this is (3) again, reached without any appeal to the particle’s velocity.

Check

Evaluate the invariant in the frame that makes it easiest, the particle’s own rest frame, marked $R$ on the hyperbola: there $p = 0$ and $E = m_0c^2$ , so the combination is $\left(m_0c^2\right)^2-0$ , the value in (2). The dimensions agree as well, since $pc$ has the dimensions of an energy, so both terms of $E^2-p^2c^2$ are an energy squared and so is $m_0^2c^4$ . The massless limit is the other useful test: setting $m_0 = 0$ makes the invariant vanish and forces $E = pc$ in every frame, which is the dotted line the hyperbola branch approaches.

Remark

The combination $E^2-p^2c^2$ stands to the pair $\left(E/c,\,\vect p\right)$ exactly as $c^2t^2-r^2$ stands to the pair $\left(ct,\,\vect r\right)$ : same boost, same cancellation, same invariance. That is what makes energy and momentum one four-vector whose “length” is the rest energy, and it is why the hyperbola figure can borrow the axes of a spacetime diagram wholesale. The practical use is that an invariant may be computed in whichever frame is convenient and then read in any other, which is what turns awkward collision problems into one line of algebra. One caution about the extension to several particles: forming the same combination from a system’s total energy and total momentum gives a number that is invariant for the same reason, but it is not $\left(\sum m_i c^2\right)^2$ . The system’s invariant mass is at least the sum of the parts’ rest masses, and equals it only when all the particles share one common velocity, so a system’s rest mass is not additive even though each particle’s own is fixed.