Theory of relativity/Special relativity/energy

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This article presumes that the reader has read Special relativity/momentum.

This article will deduce, on theoretical grounds, the relativistically correct formula for kinetic energy of a moving object. As in the previous article, we will perform a "gedanken experiment" on a collision. As before, the collision will be set up to be perfectly elastic, with no gain or loss of total kinetic energy.

Because of the formula for momentum that was worked out previously, this experiment is simpler than in the previous article. The collision doesn't need to be analyzed in two dimensions—one dimension will suffice.

The derivation[edit]

The collision in the CM frame. Particles A and B both approach and depart with the same speed.

First, set up the experiment in a reference frame such that the total momentum, before and after, is zero. This means that this is the "center of mass frame" or "CM frame". (Physicists commonly use the CM frame when analyzing collisions.)

The two colliding particles have greatly different masses. Particle A, of small mass , approaches from the left at speed . Particle B, of large mass , approaches from the right at speed . They have exactly equal and opposite momenta when approaching. After the collision, each particle moves away at exactly the speed it had before, in the opposite direction. Clearly, momentum and total kinetic energy are both conserved.

We assume that particle A is very much lighter than B, so K is enormous—so much so that particle B's motion is non-relativistic. In fact, we will ultimately take the limit as K goes to infinity and (B's speed) goes to zero.

Particles A's motion, with speed , is presumed to be relativistic.

The conservation of momentum puts constraints on the speeds and . From the derivation of the preceding article, the momentum of A before the collision was

and that of B was

Since these must exactly cancel each other, we have


Some messy high-school algebra gives

            (1)
The collision in the frame in which B is initially at rest. Particle A's departure speed is slightly less than its approach speed; it transmits energy to B during the collision.

Now we change to a reference frame in which B is at rest prior to the collision. In this frame, A approaches the stationary B from the left at high speed, gives it a nudge, and bounces back to the left at slightly lower speed. In response to the nudge, B moves slowly off to the right. A has given up some of its energy to B. But observers in this frame still insist that total kinetic energy is conserved.

Because this is in only one dimension, we don't need to use the full Lorentz transform to calculate speeds. We can just use the formula for addition of relativistic velocities. The speed is added to everything.

B's speed before the collision was zero.

B's speed after the collision is

A's speed before the collision was

A's velocity after the collision is which is negative, but we are only interested in the speed, which is

The change in A's speed before and after the collision is

Now is extremely small compared to c, so the denominator is effectively 1, so the change in A's speed is essentially

            (2)

Now B's speed after the collision is

Because is very small, this is essentially . B's mass is , so, by the classical formula (E = 1/2 mv2), B's kinetic energy after the collision is essentially

Now, by equation (1), we have

Since is huge, is effectively equal to , so this is effectively

Therefore

which is the kinetic energy gained by B and lost by A.

So the kinetic energy lost by A is

and, by equation (2),

If we let go to infinity, so goes to zero, we get

giving the integral:

Solving this, we get:

where C is the usual constant of integration. Since the energy is zero when the speed is zero, we set .

So the formula for kinetic energy is

It follows that an object's kinetic energy grows unboundedly large as its speed approaches c.

A little calculation will show that, in the non-relativistic limit, this reduces to the classical formula:

Intrinsic energy and total energy[edit]

It is common to define a particle's intrinsic energy or rest energy as:

and its total energy as:

Or, using the common abbreviation ,

Then we have:

Controversy about "relativistic mass"[edit]

There has been controversy about the term "mass" and its symbol . Many older textbooks use the term intrinsic mass or rest mass or invariant mass for what we call mass, and use the symbol to denote it. They use to denote what they call relativistic mass or effective mass, with the equation:

Doing this preserves the formula , but it is wrong. It makes the meaning of "mass" dependent on the observer.

While the notion of relativistic mass is convenient, it is frame-dependent. The intrinsic mass is the "true" mass. Everyone, in all reference frames, agrees on the intrinsic mass of the proton—it is 1.6726217×10-27 kg, and any observer can look it up in a textbook. See [1] for a discussion of this point by a physics educator.

Rather than thinking that a particle's mass increases when it moves, it is better to think of the momentum as increasing with the extra factor of .

we have:

and

The energy/momentum formula[edit]

A Pythagoras-like representation of the relationship among intrinsic energy (on the bottom), momentum (on the right), and total energy (on the hypotenuse). The author of this diagram used the "m0" notation. The segment labeled "K" is the kinetic energy.

A little algebra will show that

This means that

which gives a Pythagoras-like formula relating the momentum and the energy.

The next article in this series is Special relativity/E = mc².

See also[edit]

Footnotes[edit]

  1. See this article (from the web archive) from www.worldscientific.com.