Theory of relativity/Special relativity/energy

Physics

Special relativity

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

File:Relativity103fig1.svg

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 $m$, approaches from the left at speed $v$. Particle B, of large mass $Km$, approaches from the right at speed $r$. 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 $r$ (B's speed) goes to zero.

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

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

$p_A=\frac{mv}{\sqrt{1-v^2/c^2}}$

and that of B was

$p_B=\frac{Kmr}{\sqrt{1-r^2/c^2}}$

Since these must exactly cancel each other, we have

$\frac{v}{\sqrt{1-v^2/c^2}}=\frac{Kr}{\sqrt{1-r^2/c^2}}$

Some messy high-school algebra gives

$r=\frac{v}{\sqrt{K^2-[K^2-1]v^2/c^2}}$            (1)

File:Relativity103fig2.svg

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 $r$ is added to everything.

B's speed before the collision was zero.

B's speed after the collision is $\frac{r+r}{1+r^2/c^2}=\frac{2r}{1+r^2/c^2}$

A's speed before the collision was $\frac{v+r}{1+vr/c^2}$

A's velocity after the collision is $\frac{r-v}{1-vr/c^2}$ which is negative, but we are only interested in the speed, which is

$\frac{v-r}{1-vr/c^2}$

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

$\frac{v+r}{1+vr/c^2}-\frac{v-r}{1-vr/c^2}=2r\frac{1-v^2/c^2}{1-v^2{r}^2/c^4}$

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

$\mathrm{\Delta }v=2r(1-v^2/c^2)$            (2)

Now B's speed after the collision is

$\frac{2r}{1+r^2/c^2}$

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

$\frac{1}{2}Km(2r{)}^2=2Km{r}^2$

Now, by equation (1), we have

$r=\frac{v}{\sqrt{K^2-[K^2-1]v^2/c^2}}$

Since $K$ is huge, $K^2-1$ is effectively equal to $K^2$, so this is effectively

$r=\frac{v}{K\sqrt{1-v^2/c^2}}$

Therefore

$2Km{r}^2=\frac{2mrv}{\sqrt{1-v^2/c^2}}$

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

So the kinetic energy lost by A is

$\mathrm{\Delta }E=\frac{2mrv}{\sqrt{1-v^2/c^2}}=\frac{mv}{(1-v^2/c^2{)}^{3/2}}2r(1-v^2/c^2)$

and, by equation (2),

$\mathrm{\Delta }v=2r(1-v^2/c^2)$

If we let $K$ go to infinity, so $r$ goes to zero, we get

$\frac{dE}{dv}=\frac{mv}{(1-v^2/c^2{)}^{3/2}}$

giving the integral:

$E=\int \frac{mv}{(1-v^2/c^2{)}^{3/2}}dv$

Solving this, we get:

$E=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}+C$

where C is the usual constant of integration. Since the energy is zero when the speed is zero, we set $C=-m{c}^2$.

So the formula for kinetic energy is

$E_{\text{kinetic}}=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}-m{c}^2$

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:

$E_{\text{kinetic}}=\frac{1}{2}m{v}^2$

Intrinsic energy and total energy

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

$E_{\text{intrinsic}}=m{c}^2$

and its total energy as:

$E_{\text{total}}=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}$

Or, using the common abbreviation $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$,

$E_{\text{total}}=\gamma m{c}^2$

Then we have:

$E_{\text{kinetic}}=E_{\text{total}}-E_{\text{intrinsic}}$

Controversy about "relativistic mass"

There has been controversy about the term "mass" and its symbol $m$. Many older textbooks use the term intrinsic mass or rest mass or invariant mass for what we call mass, and use the symbol $m_0$ to denote it. They use $m$ to denote what they call relativistic mass or effective mass, with the equation:

$m=\frac{m_0}{\sqrt{1-v^2/c^2}}$

Doing this preserves the formula $p=mv$, 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 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 $\gamma$.

we have:

$E_{\text{total}}=\gamma m{c}^2$

and

$\overrightarrow{p}=\gamma m\overrightarrow{v}$

The energy/momentum formula

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

$\frac{m^2{c}^4}{1-v^2/c^2}=\frac{m^2{v}^2{c}^2}{1-v^2/c^2}+m^2{c}^4$

This means that

$E_{\text{total}}^2=(pc{)}^2+E_{\text{intrinsic}}^2$

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

• An alternative derivation, from physicsforums.com

• Special relativity/space, time, and the Lorentz transform
• Special relativity/momentum
• Special relativity/E = mc²
• Special relativity/spacetime diagrams and vectors