The Two Equations

Classical Mechanics (non-relativistic):

\[E = \frac{p^2}{2m} + V\]

Special Relativity:

\[E^2 = p^2 c^2 + m^2 c^4\]

or equivalently

\[E = \sqrt{p^2 c^2 + m^2 c^4}\]

At first glance these look incompatible. Only one can be "true," right? Actually, both are true — the classical equation is an excellent approximation of the relativistic one when \( v \ll c \).

The Key Insight: Different Definitions of "E"

The symbol E does not mean the same thing in both equations.

  • In relativity: E is the total energy (includes rest energy \( mc^2 \)).

  • In classical mechanics: E actually corresponds to what relativity calls kinetic energy K (or kinetic + potential). Rest energy is ignored because it was unknown and is constant anyway.

Correct modern notation:

\[K = E_\text{total} - m c^2 \approx \frac{p^2}{2m} \quad (\text{low speed})\]

So the classical equation should really be read as:

\[K + V \approx \frac{p^2}{2m} + V\]

Proper Derivation of the Classical Limit

Start with the full relativistic expression:

\[E_\text{total} = m c^2 \sqrt{1 + \frac{p^2}{m^2 c^2}}\]

In the non-relativistic limit: \( p \ll m c \) (i.e., \( v \ll c \)), so let \( x = \frac{p2}{m2 c^2} \ll 1 \).

Using the binomial expansion \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \):

\[E_\text{total} \approx m c^2 \left(1 + \frac{p^2}{2 m^2 c^2}\right) = m c^2 + \frac{p^2}{2 m}\]

Therefore the kinetic energy is:

\[K = E_\text{total} - m c^2 \approx \frac{p^2}{2 m}\]

This matches classical mechanics perfectly.

Why You Cannot Just Drop \( m^2 c^4 \) at the Beginning

If you naively set \( m^2 c^4 = 0 \) in the relativistic equation, you get:

\[E \approx p c\]

This is the ultra-relativistic (or massless particle) limit, valid only when \( p c \gg m c^2 \) (i.e., \( v \approx c \)). It is the opposite regime from classical mechanics. You cannot use this approximation and then expect it to match the low-speed formula \( p^2/2m \).

Numerical Example (Electron)

Rest energy of electron: \( m c^2 \approx 511 \) keV.

Low momentum case (\( p c = 1 \) keV, very non-relativistic):

  • Exact relativistic total energy:

\[E_\text{total} = \sqrt{(1)^2 + (511)^2} \approx 511.001 \text{ keV}\]

  • Relativistic kinetic energy: \( K \approx 0.00098 \) keV

  • Classical: \( \frac{p^2}{2m} \approx 0.00098 \) keV → Excellent agreement

If you had wrongly used \( E = p c = 1 \) keV as "kinetic energy", you would be off by a factor of ~1000!

Only when \( p c \) is many times larger than 511 keV does the \( E \approx p c \) approximation become valid.

Summary Table

Regime Relation What "E" means

Classical (\(v \ll c\))

\( E \approx p^2/2m + V \)

Kinetic + potential

Full Relativity

\( E_\text{total} = \sqrt{p^2 c^2 + m^2 c^4} \)

Total energy (incl. rest)

Relativity, low speed

\( K = E_\text{total} - mc^2 \approx p^2/2m + V \)

Kinetic + potential

Conclusion

There is no paradox. Special relativity reduces smoothly to classical mechanics in the low-speed limit through a proper Taylor (binomial) expansion. The apparent contradiction came from inconsistent use of the symbol E and mixing different speed regimes.