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Michael Fowler
UVa Physics
As you can see from the previous lecture, although Einstein's Theory of Special Relativity solves the problem posed by the Michelson-Morley experiment -- the nonexistence of an ether -- it is at a price. The simple assertion that the speed of a flash of light is always c in any inertial frame leads to consequences that defy common sense. When this was pointed out somewhat forcefully to Einstein, his response was that common sense is the layer of prejudices put down before the age of eighteen. All our intuition about space, time and motion is based on childhood observation of a world in which no objects move at speeds comparable to that of light. Perhaps if we had been raised in a civilization zipping around the universe in spaceships moving at relativistic speeds, Einstein's assertions about space and time would just seem to be common sense. The real question, from a scientific point of view, is not whether Special Relativity defies common sense, but whether it can be shown to lead to a contradiction. If that is so, common sense wins. Ever since the theory was published, people have been writing papers claiming it does lead to contradictions.
One of the first paradoxes to be aired was based on the Fitzgerald contraction. Recall that any object moving relative to an observer will be seen by that observer to be contracted, foreshortened in the direction of motion by the ubiquitous factor sqrt(1 - v²/c²). Einstein lived in Switzerland, a very mountainous country where the railroads between towns often go through tunnels deep in the mountains.
Suppose a train of length L is moving along a straight track at a relativistic speed and enters a tunnel, also of length L. There are bandits inhabiting the mountain above the tunnel. They observe a short train, one of length Lsqrt (1 - v²/c²), so they wait until this short train is completely inside the tunnel of length L, then they close doors at the two ends, and the train is trapped fully inside the mountain. Now look at this same scenario from the point of view of someone on the train. He sees a train of length L, approaching a tunnel of length Lsqrt (1 - v²/c²), so the tunnel is not as long as the train from his viewpoint! What does he think happens when the bandits close both the doors?
The key to understanding what is happening here is that we said the bandits closed the two doors at the ends of the tunnel at the same time. How could they arrange to do that, since the doors are far apart? They could use walkie-talkies, which transmit radio waves, or just flash a light down the tunnel, since it's long and straight. Remember, though, that the train is itself going at a speed close to that of light, so they have to be quite precise about this timing! The simplest way to imaging them synchronizing the closings of the two doors is to assume they know the train's timetable, and at a prearranged appropriate time, a light is flashed halfway down the tunnel, and the end doors are closed when the flash of light reaches the ends of the tunnel. Assuming the light was positioned correctly in the middle of the tunnel, that should ensure that the two doors close simultaneously.
Now consider this door-closing operation from the point of view of someone on the train. Assume he's in an observation car and has incredible eyesight, and there's a little mist, so he actually sees the light flash, and the two flashes traveling down the tunnels towards the two end doors. Of course, the train is a perfectly good inertial frame, so he sees these two flashes to be traveling in opposite directions, but both at c, relative to the train. Meanwhile, he sees the tunnel itself to be moving rapidly relative to the train. Let us say the train enters the mountain through the "front" door. The observer will see the door at the other end of the tunnel, the "back" door, to be rushing towards him, and rushing to meet the flash of light. Meanwhile, once he's in the tunnel, the front door is receding rapidly behind him, so the flash of light making its way to that door has to travel further to catch it. So the two flashes of light going down the tunnel in opposite directions do not reach the two doors simultaneously as seen from the train.
The concept of simultaneity, events happening at the same time, is not invariant as we move from one inertial frame to another. The man on the train sees the back door close first, and, if it is not quickly reopened, the front of the train will pile into it before the front door is closed behind the train.
The above discussion is based on Einstein's prediction that objects moving at relativistic speed appear shrunken in their direction of motion. How do we know that they're not shrunken in all three directions, i.e. moving objects maybe keep the same shape, but just get smaller? This can be seen not to be the case through a symmetry argument, also due to Einstein. Suppose two trains traveling at equal and opposite relativistic speeds, one north, one south, pass on parallel tracks. Suppose two passengers of equal height, one on each train, are standing leaning slightly out of open windows so that their noses should very lightly touch as they pass each other. Now, if N (the northbound passenger) sees S as shrunken in height, N's nose will brush against S's forehead, say, and N will feel S's nose brush his chin. Afterwards, then, N will have a bruised chin (plus nose), S a bruised forehead (plus nose). But this is a perfectly symmetric problem, so S would say N had the bruised forehead, etc. They can both get off their trains at the next stations and get together to check out bruises. They must certainly be symmetrical! The only consistent symmetrical solution is given by asserting that neither sees the other to shrink in height (i.e. in the direction perpendicular to their relative motion), so that their noses touch each other. Therefore, the Lorentz contraction only operates in the direction of motion, objects get squashed but not shrunken.
Perhaps the most famous of the paradoxes of special relativity, which was still being hotly debated in national journals in the fifties, is the twin paradox. The scenario is as follows. One of two identical twins is an astronaut. He sets off in a relativistic spaceship to alpha-centauri, four light-years away, at a speed close to that of light. When he gets there, he immediately turns around and comes back. As seen by his brother on earth, the astronauts clocks ran very slowly, so although the trip took over eight years by earth time, the astronaut has only aged by, say, one year. So as he steps down out of the spaceship, he is seven years younger than his twin brother. But wait a minute-how does this look from the astronaut's point of view? He sees the earth to be moving close to the speed of light, first away from him then towards him. So he must see the clock of his brother on earth to be running slowly. So doesn't he expect his brother on earth to be the younger one after this trip?
The key to this paradox is that this situation is not as symmetrical as it looks. The two brothers have quite different experiences. The one on the spaceship is not in an inertial frame during the initial acceleration and the turnaround and braking periods. (To get an idea of the speeds involved, to get close to the speed of light at the acceleration of a falling stone would take about a year.) Suppose the two kept in touch with each other by flashing a light once a month, using their own calendars, so the other could keep tabs on his brother's age by counting flashes. After the initial acceleration, the two are parting at a constant rate, and each will see the other's flashes at some steady rate which will be less frequent than monthly, because the other guy's clock appears to be running slowly, and also the distance between them is increasing, so each succeeding flash has further to travel. When the astronaut turns around, however, halfway through his trip, he will be traveling towards the light flashes, each succeeding one has less far to travel, so he will see the flashes coming in at a faster rate, in other words, he will see his brother on earth to be aging rapidly. From the earthbound brother's point of view, after he sees the astronaut to turn around, he sees the astronaut to age at the rapid rate. But since the astronaut is traveling close to the speed of light, he arrives back on earth very shortly after his brother on earth sees him turn around! Thus for the brother on earth, watching the signals coming in from the spaceship, he sees them coming in at the slower rate for almost the entire trip, whereas the astronaut looking at the signals from earth sees them to come in at the slow rate for the first half of the trip, until he turns around, and after that, they come in at the fast rate-so the brother on earth ages more.
Consider now the following. As seen from earth, two spaceships, A and B, are traveling towards each other at equal and opposite relativistic velocities, along, let us call it, the x-direction. They collide with a glancing blow, which does no damage but gives them equal and opposite small velocities in the y-direction, which is just some direction perpendicular to the original x-direction. (Of course, they both still have tremendous (equal and opposite) velocities in the x-direction.) Now recall Newton's Second and Third Laws of motion. The Second states that the rate of change of momentum is proportional to the external force. Remember, too, that momentum, being mass x velocity, is a vector. Let's concentrate on the rate of change of momentum in the y-direction. The Second Law tells us that during the actual collision, the rate at which spaceship A picks up velocity in the y-direction is proportional to the force in the y-direction it is experiencing from colliding with the spaceship B. The Third Law tells us that the pushing between the two spaceships is equal and opposite throughout the collision. We conclude that the total change in momentum for spaceship A during the collision is equal but opposite to that for spaceship B. In other words, as seen from the earth, this is a symmetrical collision and afterwards the two ships A and B have equal and opposite velocities.
Now let us consider this same collision in the inertial frame of reference spaceship A is in at the beginning. We call this "A's Initial Inertial Frame", and before the collision, spaceship A is of course at rest in this frame. (Note that a frame attached to the spaceship A itself is not an inertial frame throughout, because it is accelerated briefly by the jolt of the collision.) After the collision, spaceship A is moving slowly in "A's Initial Inertial Frame". To be specific, let us say that after the collision, A is moving at 15 meters per second in the y-direction in A's Initial Inertial Frame.
Let us now consider the motion of spaceship B after the collision, as seen by the A-crew. Of course, B is retreating rapidly at the relativistic speed v, say, in the x-direction. But it also has a small velocity in the y-direction corresponding to the 15 meters per second the A-crew measure their y-direction velocity to be. In fact, by the complete symmetry of the situation the B-crew must also measure their own y-direction velocity to be 15 meters per second.
Imagine now the A-crew eavesdropping on the B-crew's measurement of spaceship B's y-direction velocity. Don't forget that B is moving at a relativistic speed compared with A. We established above that distances measured in the y-direction (perpendicular to relativistic motion) are not Fitzgerald contracted. Does this mean that the A-crew and the B-crew will agree that spaceship B is moving at 15 meters per second in the y-direction? The answer is no, because although they agree on what 15 meters in the y-direction is, they don't agree on what one second is! As the A-crew, equipped with powerful binoculars, watch the B-crew making their y-direction speed measurements, they will note that the B-crew's clocks are running slow by the famous time dilation factor sqrt(1 - v²/c²). In fact, everything they see on the B spaceship will look like slow motion by this same factor, and in particular they will see b to be moving sideways not at 15 meters per second, but at 15sqrt(1 - v²/c²) meters per second.
The bottom line is that, as seen from A's Initial Inertial Frame, after the collision the two spaceships have different speeds in the y-direction, so apparently momentum in the y-direction is not conserved. But that would imply that Newton's Laws cannot be true in A's Initial Inertial Frame, because the equal and opposite forces (from the Third Law) acting between the two spaceships during the collision must, from the Second Law, generate equal and opposite y-direction momenta!
There is only one way out. We have established that in A's Initial Inertial Frame, the two spaceships after the collision do not have equal but opposite y-direction velocities. Newton's Laws require them to have equal but opposite y-direction momenta. The loophole is that momentum is mass x velocity, so Newton's Laws could be rescued if we assume the masses are not the same. Of course, we took the two spaceships to be identical, so what we are saying is that a moving object gains weight. That is to say, spaceship B, as observed by the A-crew, is heavier than their spaceship A. In fact, we can be more precise. Notice from above that spaceship B was moving sideways at speed 15 sqrt(1 - v²/c²) meters per second as observed by the A-crew. For A and B to have equal but opposite sideways momenta, as seen by the A-crew, the observed mass of spaceship B would have to be up by a compensating factor 1/ sqrt(1 - v²/c²).
So, to save Newton's Laws in all inertial frames, we are forced to assume that if we try to measure the mass of a moving object, we will find it increase with speed, so if an object when at rest has a mass M, moving at a speed v it will have a mass M/sqrt(1 - v²/c²). Note that this is an undetectably small effect at ordinary speeds, but as an object approaches the speed of light, the mass increases without limit!
Deciding that masses of objects must depend on speed like this seems a heavy price to pay to rescue conservation of momentum! However, it is a prediction that is not difficult to check by experiment. The first confirmation came in 1908, measuring the mass of fast electrons in a vacuum tube. In fact, the electrons in a color TV tube are about half a percent heavier than electrons at rest, and this must be allowed for in calculating the magnetic fields used to guide them to the screen.
Much more dramatically, in modern particle accelerators very powerful electric fields are used to accelerate electrons, protons and other particles. It is found in practice that these particles become heavier and heavier as the speed of light is approached, and hence need greater and greater forces for further acceleration. Consequently, the speed of light is a natural absolute speed limit. Particles are accelerated to speeds where their mass is thousands of times greater than their mass measured at rest, usually called the "rest mass".
Let's think about the kinetic energy of one of these particles traveling close to the speed of light. Recall that in an earlier lecture we found the kinetic energy of an ordinary non-relativistic (i.e. slow moving) mass m was ½mv². The way we did that was by accelerating it with a constant force F, and finding the work done by the force (force x distance) to get it to speed v from a standing start. The kinetic energy of the mass, E = ½mv², is exactly equal to the work done by the force in bringing the mass up to that speed. (It can be shown in a similar way that if a force is applied to a particle already moving at speed u, say, and it is accelerated to speed v, the work necessary is ½mv² - ½mu².)
It is interesting to try to repeat the exercise for a particle moving very close to the speed of light, like the particles in the accelerators mentioned in the previous paragraph. Newton's Second Law, in the form
is still true, but close to the speed of light the speed changes negligibly as the force continues to work - instead, the mass increases. Therefore, we can write to an excellent approximation,
where as usual c is the speed of light. To get more specific, suppose we have a constant force F pushing a particle. At some instant, the particle has mass M, and speed extremely close to c. One second later, since the force is continuing to work on the particle, and thus increase its momentum from Newton's Second Law, the particle will have mass M + m say, where m is the increase in mass as a result of the work done by the force.
What is the increase in the kinetic energy E of the particle during that one second period? By exact analogy with the non-relativistic case reviewed above, it is just the work done by the force during that period. Now, since the mass of the particle changes by m in one second, m is also the rate of change of mass. Therefore, from Newton's Second Law in the form Force = (rate of change of mass) x c, we can write Force = mc. The increase in Kinetic energy E over the one second period is just the work done by the force, force x distance. Since the particle is moving essentially at the speed of light, the distance the force acts over in the one-second period is just c meters, c = 3.10^8. So the total work the force does in that second is force x distance = mc.c = mc².
Hence the relationship between the increase in mass of the relativistic particle and its increase in kinetic energy is:
Recall that to get Newton's Laws to be true in all inertial frames, we had to assume an increase of mass with speed by the factor 1/sqrt(1 - v²/c²). This implies that even a slow-moving mass has a tiny increase when it moves. How does that tiny increase relate to the kinetic energy? Consider a mass M, moving at speed v, much less than the speed of light. Its kinetic energy E =½mv², as discussed above. Its mass is M/sqrt(1 - v²/c²), which we can write as M + m. What is m? For small v, we can approximate sqrt(1 - v²/c²) as (1 - ½v²/c²) , and 1/(1 - ½v²/c²) as (1 + ½v²/c²). (These approximations become more and more exact as v/c gets smaller, check them out). This means the total mass at speed v is M(1 + ½v²/c²), and writing this as M + m, we see the mass increase m equals ½ Mv²/c². This means that again, the mass increase m is related to the kinetic energy E by E = mc².
In fact, it is not difficult to show, using a little calculus, that over the whole range of speed from zero to as close as you like to the speed of light, a moving particle experiences a mass increase related to its kinetic energy by E = mc². To understand why this isn't noticed in everyday life, try an example, such as a jet airplane weighing 100 tons moving at 2,000mph. 100 tons is 100,000 kilograms, 2,000mph is about 1,000 meters per second. That's a kinetic energy ½Mv² of ½.10^11joules, but the corresponding mass change of the airplane down by the factor c², 9.10^16, giving an actual mass increase of about half a milligram, not too easy to detect!
We have seen above that when a force does work accelerating a body to give it kinetic energy, the mass of the body increases by an amount equal to the total work done by the force, the energy E transferred, divided by c². What about when a force does work on a body that is not speeding it up, so there is no increase in kinetic energy? For example, if I just lift something at a steady rate, giving it potential energy? It turns out that in this case, too, there is a mass increase given by E = mc², of course unmeasurably small for everyday objects. However, there is a measurable and important effect in nuclear physics. For example, the helium atom has a nucleus which has two protons and two neutrons bound together very tightly by a strong nuclear attraction force. This can be separated if sufficient outside force is applied into two separate "heavy hydrogen" nuclei, each of which has one proton and one neutron. A lot of outside energy has to be spent to achieve this separation, and it is found that the total mass of the two heavy hydrogen nuclei is measurably (about half a percent) heavier than the original helium nucleus. This extra mass, multiplied by c², is just equal to the energy needed to split the helium nucleus into two. Even more important, this energy can be recovered by letting the two heavy hydrogen nuclei collide and join to form a helium nucleus again. (They are both electrically charged positive, so they repel each other, and must come together fairly fast to overcome this repulsion and get to the closeness where the much stronger nuclear attraction kicks in.) This is the basic power source of the hydrogen bomb, and of the sun.
It turns out that all forms of energy, kinetic and different kinds of potential energy, have associated mass given by E = mc². For nuclear reactions, the mass change is typically of order one thousandth of the total mass, and readily measurable. For chemical reactions, the change is of order a billionth of the total mass, and I think not currently measurable.
Copyright © Michael Fowler, 1996