Irodov 1.369: Rest Mass over a Shrinking Box – Amit & Deepali Sinha Foundation

Problem Irodov 1.369

A body has density $\rho_0$ as measured in its own rest frame, where density means the body’s rest mass divided by its (proper) volume. Find the speed $v$ of a reference frame, moving relative to the body, in which the density of the body, computed the same way (rest mass over the volume as measured in that frame), comes out $\eta = 25\%$ greater than $\rho_0$ .

Given

$\rho_0$ , the density measured in the body’s own frame
$\eta = 0.25$
density means rest mass divided by volume

Find

Find

v

, the speed of the frame relative to the body

Solution

Step 1Decide which of the two factors in the density can change

Density is a mass divided by a volume, and only one of the two is frame dependent under the convention this problem uses. The mass in the numerator is the rest mass $m_0$ , which is a property of the body itself and is the same number in every frame. The volume in the denominator is not: a body has three dimensions and a boost shortens exactly one of them.

In its own frame the body occupies its proper volume V_0. — In its own frame the body occupies its proper volume $V_0$ .

Seen from a frame in which it moves at v, the dimension along the motion is shortened and the two across it are not, so V = V_0sqrt(1-β^2). — Seen from a frame in which it moves at $v$ , the dimension along the motion is shortened and the two across it are not, so $V = V_0\sqrt{1-\beta^2}$ .

Step 2Contract the one dimension that lies along the motion

Write $\beta = v/c$ and take the body, in its own frame, as a rectangular block of edges $a_0$ , $b_0$ , $c_0$ with $a_0$ along the direction of the relative motion. In the frame where the body moves, the edges measure

a = a_0\sqrt{1-\beta^2}, \qquad b = b_0, \qquad c = c_0 ,

so the volume itself carries a single factor of the contraction:

V = abc = a_0b_0c_0\sqrt{1-\beta^2} = V_0\sqrt{1-\beta^2} .

Dividing the unchanged rest mass by this shrunken volume, and writing $\rho_0 = m_0/V_0$ for the proper density, gives the density in the moving frame:

\rho = \frac{m_0}{V} = \frac{m_0}{V_0\sqrt{1-\beta^2}} .

\rho = \frac{\rho_0}{\sqrt{1-\beta^2}}

(1)

Step 3Impose the required increase and solve for the speed

The condition is that the moving-frame density exceed the proper density by the fraction $\eta$ , that is $\rho = \left(1+\eta\right)\rho_0$ . Putting that into (1) and inverting:

\sqrt{1-\beta^2} = \frac{1}{1+\eta} .

Squaring and solving for $\beta^2$ :

\begin{aligned} \beta^2 &= 1 - \frac{1}{\left(1+\eta\right)^2} \\[2pt] &= \frac{\left(1+\eta\right)^2-1}{\left(1+\eta\right)^2} \\[2pt] &= \frac{\eta\left(2+\eta\right)}{\left(1+\eta\right)^2} . \end{aligned} \tag{2}

Step 4Take the root and put the number in

Substitute $\eta = 0.25$ into the numerator and denominator of (2).

\eta\left(2+\eta\right) = 0.25 \times 2.25 = 0.5625, \qquad \sqrt{0.5625} = 0.75, \qquad 1+\eta = 1.25 .

Hence

v = c\,\frac{\sqrt{\eta\left(2+\eta\right)}}{1+\eta} = 0.60c

(3)

The density ratio of against v/c. A rise of one quarter is already reached at 0.6c. — The density ratio of (1) against $v/c$ . A rise of one quarter is already reached at $0.6c$ .

Check

Run the result back through (1). At $v = 0.60c$ the contraction factor is $\sqrt{1-0.36} = 0.80$ , so $\rho = \rho_0/0.80 = 1.25\rho_0$ , which is precisely $25$ % greater than $\rho_0$ . The formula also behaves at both ends: $\eta \to 0$ gives $v \to 0$ , and $\eta \to \infty$ gives $v \to c$ , so any density increase whatever can be reached by a fast enough frame, but none of them needs a superluminal one.

Remark

The answer depends on what is meant by “the mass of the body”, and the two readings do not agree. Rest mass over volume, the convention used here, gives one factor of $\sqrt{1-\beta^2}$ and the answer above. Counting instead the relativistic mass $m_0/\sqrt{1-\beta^2}$ would put a second factor in and give $\rho = \rho_0/\left(1-\beta^2\right)$ , hence $v = c\sqrt{\eta/\left(1+\eta\right)} = 0.45c$ for the same $\eta$ . The question is only well posed once the convention is stated.