Derivation: From Energy-Momentum Relation to Relativistic Kinetic Energy

 J. Rogers, SE Ohio

Starting Point: The Energy-Momentum Relation

For any particle with rest mass m₀, the fundamental relation is:

E² = (pc)² + (m₀c²)²

Step 1: Express Total Energy in Terms of Kinetic Energy

The total energy E consists of rest energy plus kinetic energy:

E = K_E + m₀c²

Substituting into the energy-momentum relation:

(K_E + m₀c²)² = (pc)² + (m₀c²)²

Step 2: Expand and Simplify

Expanding the left side:

K_E² + 2K_E(m₀c²) + (m₀c²)² = (pc)² + (m₀c²)²

The (m₀c²)² terms cancel:

K_E² + 2K_E(m₀c²) = (pc)²

Step 3: Solve for Kinetic Energy

This is a quadratic equation in K_E. Using the quadratic formula (taking the positive root since K_E ≥ 0):

K_E = -m₀c² + √[(m₀c²)² + (pc)²]

Rearranging:

K_E = √[(pc)² + (m₀c²)²] - m₀c²

Step 4: Factor Out Rest Energy

K_E = m₀c²√[(pc/m₀c²)² + 1] - m₀c²

K_E = m₀c²[√(1 + (pc/m₀c²)²) - 1]

This is the general relativistic kinetic energy formula in terms of momentum.

Step 5: Express in Terms of Velocity

For a particle with velocity v, the relativistic momentum is:

p = γm₀v where γ = 1/√(1 - v²/c²)

Therefore: pc/m₀c² = γv/c = v/c / √(1 - v²/c²)

So: (pc/m₀c²)² = (v²/c²)/(1 - v²/c²)

Substituting back:

K_E = m₀c²[√(1 + v²/c²/(1 - v²/c²)) - 1]

K_E = m₀c²[√((1 - v²/c² + v²/c²)/(1 - v²/c²)) - 1]

K_E = m₀c²[√(1/(1 - v²/c²)) - 1]

K_E = m₀c²[1/√(1 - v²/c²) - 1]

K_E = m₀c²(γ - 1)

The Low-Speed Limit (v << c)

When v << c, we can expand γ using the binomial approximation:

γ = (1 - v²/c²)^(-1/2) ≈ 1 + ½(v²/c²) + ...

Therefore: K_E = m₀c²(γ - 1) ≈ m₀c² × ½(v²/c²) = ½m₀v²

This recovers the classical kinetic energy formula as expected.

Deriving Lorentz directly from the energy-momentum relation

Instead of starting with γ as some mysterious factor introduced through Lorentz transformations, you can derive γ directly from the energy-momentum relation.

The standard approach:

1. Introduce γ = 1/√(1-v²/c²) from coordinate transformations
2. Then define p = γmv and E = γmc²
3. Show these satisfy E² = (pc)² + (mc²)²

Our approach:

1. Start with E² = (pc)² + (mc²)² as fundamental
2. Use the classical relations E = K_E + mc² and p = mv as low-energy approximations
3. Demand these be consistent → solve for what γ must be

From E² = (pc)² + (mc²)², if we write E = γmc² and p = γmv, then:

(γmc²)² = (γmvc)² + (mc²)²

γ²m²c⁴ = γ²m²v²c² + m²c⁴

γ²c² = γ²v² + c²

γ²(c² - v²) = c²

Now this gets complicated, so lets slow down and go step by step.

Step 1: Factor out c² from inside the square root √(c² - v²) = √[c²(1 - v²/c²)] = c√(1 - v²/c²)

Step 2: Substitute this back γ = c/√(c² - v²) = c/[c√(1 - v²/c²)]

Step 3: Cancel the c's γ = c/[c√(1 - v²/c²)] = 1/√(1 - v²/c²)

γ = c/√(c² - v²) = 1/√(1 - v²/c²)

This is huge pedagogically - γ isn't some arbitrary mathematical construction from coordinate geometry. It's the necessary factor that makes energy and momentum consistent with the fundamental spacetime invariant.

We've found a way to derive the Lorentz factor from physical principles rather than geometric postulates. That's a much deeper foundation.

Summary

From the fundamental energy-momentum relation E² = (pc)² + (m₀c²)², we derived:

• General form: K_E = √[(pc)² + (m₀c²)²] - m₀c²
• Velocity form: K_E = m₀c²(γ - 1)
• Low-speed limit: K_E ≈ ½m₀v² when v << c
• Lorentz factor γ =  1/√(1 - v²/c²)

The energy-momentum relation thus provides a unified framework connecting classical and relativistic mechanics.

Derivation from the Energy-Momentum Invariant

1. Axiom: The fundamental relationship for a particle is E² = (pc)² + (m₀c²)².

2. Principle of Work-Energy: The rate of change of energy with respect to momentum is the particle's velocity. This comes from dE = F·dx and dp = F·dt, which gives dE/dp = v.

3. Find  Differentiate the axiom with respect to p:
   d/dp [E²] = d/dp [(pc)² + (m₀c²)²]
   2E (dE/dp) = 2p c²
   dE/dp = pc²/E

4. Equate and Solve: Now we equate the two expressions for dE/dp:
   v = pc²/E
   This gives us a crucial link: p = Ev/c².

5. Derive the Expression for Energy  Substitute this expression for p back into the axiom:
   E² = ( (Ev/c²)c )² + (m₀c²)²
   E² = E²v²/c² + (m₀c²)²
   E² - E²v²/c² = (m₀c²)²
   E²(1 - v²/c²) = (m₀c²)²
   E² = (m₀c²)² / (1 - v²/c²)
   Taking the positive root, we get:
   E = m₀c² / √(1 - v²/c²)

6. Define the Lorentz Factor  We see a recurring term, so we define γ ≡ 1/√(1 - v²/c²). This gives us the famous formula:
   E = γm₀c²

7. Derive the Expression for Momentum  Now substitute our derived formula for E back into the p = Ev/c² relation:
   p = (γm₀c²)v / c²
   p = γm₀v