*Michael Fowler, UVa Physics, 12/1/07*

As we discussed in the last lecture, even before *motion*—always
stayed the same. The momentum of a moving object is defined as the product of
the mass and the velocity, and so is a *vector*: it has magnitude *and
direction*. If you’re
standing on frictionless skates and you throw a ball, you move backwards: you
have momentum equal in magnitude, but *opposite*
in direction, to that of the ball, so the total momentum (yours plus the
ball’s) remains zero. Rockets
work the same way, by throwing material out at high speed. They do *not* work by
“pushing against the air”, they work by pushing against the stuff
they’re pushing out, just as you push against a ball you’re
throwing, and it pushes you back, causing your acceleration.

If you still suspect that really rockets push against the air, remember they
work just as well in space! In
fact, it was widely believed when Goddard, an early American rocketeer (the *New York Times*
editorial written in 1921: “*Professor Goddard does not know the
relation between action and reaction and the need to have something better than
a vacuum against which to react. He
seems to lack the basic knowledge ladled out daily in our high schools.*” Obviously, *the New York Times*
editorial writers of the time lacked the basic knowledge being ladled out in
this course!

In fact, as we discussed, the conservation of momentum in a collision
follows from *conservation of momentum should still hold true in any
inertial frame*. This led him to
some surprising conclusions, as we shall see.

As a warm-up exercise, let us consider conservation of momentum for a
collision of two balls on a pool table.
We draw a chalk line down the middle of the pool table, and shoot the
balls close to, but on opposite sides of, the chalk line from either end, at
the same speed, so they will hit in the middle with a glancing blow, which will
turn their velocities through a small angle. In other words, if initially we
say their (equal magnitude, opposite direction) velocities were parallel to the
*x*-direction—the chalk line—then after the collision they
will also have equal and opposite small velocities in the *y*-direction.
(The *x*-direction velocities will have decreased very slightly).

Now let us repeat the exercise on a grand scale. Suppose somewhere in space, far from any
gravitational fields, we set out a string one million miles long. (It could be between our two clocks in
the time dilation experiment). This
string corresponds to the chalk line on the pool table. Suppose now we have two identical
spaceships approaching each other with equal and opposite velocities parallel
to the string from the two ends of the string, aimed so that they suffer a
slight glancing collision when they meet in the middle. It is evident from the symmetry of the
situation that momentum is conserved in both directions. In particular, the rate at which one
spaceship moves away from the string after the collision—its* y*-velocity—is
equal and opposite to the rate at which the other one moves away from the
string.

But now consider this collision as observed by someone in one of the
spaceships, call it *A*.
Before the collision, he sees the string moving very fast by the window,
say a few meters away. After the
collision, he sees the string to be moving away, at, say, 15 meters per
second. This is because spaceship *A*
has picked up a velocity perpendicular to the string of 15 meters per
second. Meanwhile, since this is a
completely symmetrical situation, an observer on spaceship *B* would
certainly deduce that her spaceship was moving away from the string at 15
meters per second as well.

The crucial question is:* how fast does an observer in spaceship A see
spaceship B to be moving away from the string? * Let us suppose that relative to spaceship
*A*, spaceship *B* is moving away (in the *x*-direction) at 0.6*c*. First, recall that distances
perpendicular to the direction of motion are not Lorentz contracted. Therefore, when the observer in
spaceship *B* says she has moved 15 meters further away from the string in
a one second interval, the observer watching this movement from spaceship *A*
will agree on the 15 meters—but disagree on the one second! He will say her clocks run slow, so as
measured by his clocks 1.25 seconds will have elapsed as she moves 15 meters in
the *y*-direction.

It follows that, as a result of time dilation, this collision as viewed from
spaceship *A* does *not* cause equal and opposite velocities for the
two spaceships in the *y*-direction.
Initially, both spaceships were moving parallel to the *x*-axis,
there was *zero* momentum in the *y*-direction. So how can we argue there is zero total
momentum in the *y*-direction *after* the collision, when the
identical spaceships do *not* have equal and opposite velocities?

Einstein was so sure that momentum conservation must always hold that he
rescued it with a bold hypothesis: the mass of an object must depend on its
speed! In fact, the mass must
increase with speed in just such a way as to cancel out the lower *y*-direction
velocity resulting from time dilation.
That is to say, if an object at rest has a mass *M*, moving at a
speed *v* it will have a mass _{}. 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 an old style 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 ½*mv*². The way we did that was by considering
how much work we had to do to raise it through a certain height: we had to
exert a force equal to its weight *W* to lift it through height *h*,
the total work done, or energy expended, being force x distance, *Wh*. As it fell back down, the force of
gravity, *W*, did an exactly equal amount of work *Wh* on the falling
object, but this time the work went into accelerating the object, to give it
kinetic energy. Since we know how
fast falling objects pick up speed, we were able to conclude that the kinetic
energy was ½*mv*². (For details, see the previous lecture.)

More generally, we could have accelerated the mass with any constant force *F*,
and found 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* = ½*mv*², 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 ½*mv*²
- ½*mu*².)

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.

Force = rate of change of momentum

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,

Force = (rate of
change of mass) x *c*

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 _{} 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

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*
= *mc*².

Hence the relationship between the increase in mass of the relativistic particle and its increase in kinetic energy is:

*E* = *mc*²

Recall that to get _{} This
implies that even a slow-moving object has a tiny increase in mass 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*
=½*Mv*², as discussed above. Its mass is _{}, which we can write as *M* + *m*. What is *m*?

Since we’re talking about speeds we are familiar with, like a jet
plane, where *v*/*c* is really small, we can use some simple mathematical tricks to
make things easier.

The first one is a good approximation for the square root of 1 – *x* when *x* is a lot less than one:

_{}

You can easily check this with your calculator: try _{}, you find _{} which is
extremely close to _{}.

The next approximation is

_{}

This is also easy to check: again take
_{}:

_{}

and

_{}

Using these approximations with _{} we can approximate
_{} as _{}, and then _{} as _{}.

This means the total mass at speed *v*

_{}

and writing this as *M* + *m*, we see the mass increase *m*
equals ½ *Mv*²/*c*².

This means that—again—the mass increase *m* is related to
the kinetic energy *E *by _{}

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*
= *mc*². 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 ½*Mv*² of
½×10^{11}joules, but the corresponding mass change of the
airplane down by the factor *c*², 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 *c*². 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, what 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* = *mc*², of course
unmeasurably small for everyday objects.

However, this *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. If sufficient outside force is applied,
this can be separated into two “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 *c*²,
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* = *mc*². 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 not currently
measurable.

Copyright © Michael Fowler, 1996, 2007