36.10.3 Conservation of Energy

The creation and annihilation of particles with mass, processes strictly forbidden in Newtonian mechanics, are vivid proof that neither mass nor the Newtonian definition of energy is conserved. Even so, the total energy—the kinetic energy and the energy equivalent of mass—remains a conserved quantity.

Mass and energy are not the same thing, but, as the last few examples have shown, they are equivalent in the sense that mass can be transformed into energy and energy can be transformed into mass as long as the total energy is conserved.

Probably the most well-known application of the conservation of total energy is nuclear fission. The uranium isotope 236U, containing 236 protons and neutrons, does not exist in nature. It can be created when a 235U nucleus absorbs a neutron, increasing its atomic mass from 235 to 236. The 236U nucleus quickly fragments into two smaller nuclei and several extra neutrons, a process known as nuclear fission. The nucleus can fragment in several ways, but one is

n+ 235U→ 236U→ 144Ba+ 89Kr+3n

Ba and Kr are the atomic symbols for barium and krypton.

This reaction seems like an ordinary chemical reaction—until you check the masses. The masses of atomic isotopes are known with great precision from many decades of measurement in instruments called mass spectrometers. If you add up the masses on both sides, you find that the mass of the products is 0.185 u smaller than the mass of the initial neutron and 235U, where, you will recall, 1 u=1.66 × 10−27 kg is the atomic mass unit. In kilograms the mass loss is 3.07×10−28 kg.

Mass has been lost, but the energy equivalent of the mass has not. As FIGURE 36.41 shows, the mass has been converted to kinetic energy, causing the two product nuclei and three neutrons to be ejected at very high speeds. The kinetic energy is easily calculated: ΔK=mlostc2=2.8×10−11 J.

FIGURE 36.41 In nuclear fission, the energy equivalent of lost mass is converted into kinetic energy.

The figure illustrates the conversion of mass into kinetic energy in a nuclear fission reaction.
FIGURE 36.41 Full Alternative Text

This is a very tiny amount of energy, but it is the energy released from one fission. The number of nuclei in a macroscopic sample of uranium is on the order of NA, Avogadro’s number. Hence the energy available if all the nuclei fission is enormous. This energy, of course, is the basis for both nuclear power reactors and nuclear weapons.

We started this chapter with an expectation that relativity would challenge our basic notions of space and time. We end by finding that relativity changes our understanding of mass and energy. Most remarkable of all is that each and every one of these new ideas flows from one simple statement: The laws of physics are the same in all inertial reference frames.

CHAPTER 36 CHALLENGE EXAMPLE Goths and Huns

The rockets of the Goths and the Huns are each 1000 m long in their rest frame. The rockets pass each other, virtually touching, at a relative speed of 0.8c. The Huns have a laser cannon at the rear of their rocket that fires a deadly laser beam perpendicular to the rocket’s motion. The captain of the Huns wants to send a threatening message to the Goths by “firing a shot across their bow.” He tells his first mate, “The Goths’ rocket is length contracted to 600 m. Fire the laser cannon at the instant the tail of their rocket passes the nose of ours. The laser beam will cross 400 m in front of them.”

But things are different in the Goths’ reference frame. The Goth captain muses, “The Huns’ rocket is length contracted to 600 m, 400 m shorter than our rocket. If they fire as the nose of their ship passes the tail of ours, the lethal laser beam will pass right through our side.”

The first mate on the Huns’ rocket fires as ordered. Does the laser beam blast the Goths or not?

MODEL Both rockets are inertial reference frames. Let the Huns’ rocket be frame S and the Goths’ rocket be frame S′. S′ moves with velocity v=0.8c relative to S. We need to describe the situation in terms of events.

VISUALIZE Begin by considering the situation from the Huns’ reference frame, as shown in FIGURE 36.42.

FIGURE 36.42 The situation seen by the Huns.

The figure illustrates a rocket firing a laser beam as it passes by another rocket as seen in Frame S perspective.

SOLVE The key to resolving the paradox is that two events simultaneous in one reference frame are not simultaneous in a different reference frame. The Huns do, indeed, see the Goths’ rocket length contracted to LGoths=(1−(0.8)2)1/2(1000 m)=600 m. Let event 1 be the tail of the Goths’ rocket passing the nose of the Huns’ rocket. Since we’re free to define the origin of our coordinate system, we define this event to be at time t1=0 μs and at position x1=0 m. Then, in the Huns’ reference frame, the spacetime coordinates of event 2, the firing of the laser cannon, are (x2, t2)=(1000 m, 0 μs). The nose of the Goths’ rocket is at x=600 m at t=0 μs; thus the laser cannon misses the Goths by 400 m.

Now we can use the Lorentz transformations to find the space-time coordinates of the events in the Goths’ reference frame. The nose of the Huns’ rocket passes the tail of the Goths’ rocket at (x1′, t1′)=(0 m, 0 μs). The Huns fire their laser cannon at

x2′=γ(x2−vt2)=53(1000 m−0 m)=1667 mt2′=γ(t2−vx2c2)=53(0 μs−(0.8)1000 m300 m/μs)=−4.444 μs

where we calculated γ=5/3 for v=0.8c. Events 1 and 2 are not simultaneous in S′. The Huns fire the laser cannon 4.444 μs before the nose of their rocket reaches the tail of the Goths’ rocket. The laser is fired at x2′=1667 m, missing the nose of the Goths’ rocket by 667 m. FIGURE 36.43 shows how the Goths see things.

FIGURE 36.43 The situation seen by the Goths.

The figure illustrates a rocket firing a laser beam as it passes by another rocket as seen in Frame S prime perspective.

In fact, since the Huns’ rocket is length contracted to 600 m, the nose of the Huns’ rocket is at x′=1667 m−600 m= 1067 m at the instant they fire the laser cannon. At a speed of v=0.8c =240 m/μs, in 4.444 μs the nose of the Huns’ rocket travels Δx′ =(240 m/μs)(4.444 μs)=1067 m—exactly the right distance to be at the tail of the Goths’ rocket at t1′=0 μs. We could also note that the 667 m “miss distance” in the Goths’ frame is length contracted to (1−(0.8)2)1/2(667 m)=400 m in the Huns’ frame—exactly the amount by which the Huns think they miss the Goths’ rocket.

REVIEW Thus we end up with a consistent explanation. The Huns miss the Goths’ rocket because, to them, the Goths’ rocket is length contracted. The Goths find that the Huns miss because event 2 (the firing of the laser cannon) occurs before event 1 (the nose of one rocket passing the tail of the other). The 400 m distance of the miss in the Huns’ reference frame is the length-contracted miss distance of 667 m in the Goths’ reference frame.