J. Rogers, SE Ohio
Course: Physics II or Modern Physics
Duration: 50 minutes
Prerequisites: Basic algebra, classical mechanics (F=ma, KE=½mv²)
Learning Objectives
By the end of this class, students will:
- Understand that all of special relativity emerges from one fundamental equation
- Derive the Lorentz factor, relativistic energy, momentum, and mass-energy equivalence
- See how classical mechanics emerges as a low-speed limit
- Recognize the unity underlying seemingly separate relativistic "effects"
Materials Needed
- Whiteboard/projector for live derivations
- Calculator for numerical examples
- Optional: Graphing tool for visualizing γ vs v
Lesson Plan
Opening Hook (3 minutes)
"I'm going to show you something that will completely change how you think about physics. Everything you've heard about Einstein's relativity - time dilation, length contraction, E=mc² - isn't a collection of separate weird effects. It's all the same thing. One equation. We're going to derive all of it in the next 47 minutes."
Write on board:
Special Relativity = ?
The Foundation (7 minutes)
Present the fundamental axiom:
E² = (pc)² + (m₀c²)²
Explain: "This is the energy-momentum invariant. It's not derived from anything else - it's a fundamental law of nature, like conservation of energy. Every particle in the universe obeys this relationship."
Key points:
- E = total energy of the particle
- p = momentum of the particle
- m₀ = rest mass (mass when particle is at rest)
- c = speed of light (constant of nature)
Physical meaning: "This equation says that energy and momentum are fundamentally linked. You can't change one without affecting the other in a very specific way."
Live Derivation Session (30 minutes)
Set up the challenge: "Now we're going to solve this one equation for different quantities and see what happens. Each solution will give us a famous result from relativity."
Derivation 1: Total Energy (5 minutes)
"What if we want to express energy in terms of velocity?"
Starting from: E² = (pc)² + (m₀c²)²
Use the work-energy principle: dE/dp = v
- Differentiate the invariant: 2E(dE/dp) = 2pc²
- Therefore: dE/dp = pc²/E
- Equating: v = pc²/E
- Solve for momentum: p = Ev/c²
Substitute back:
E² = (Ev/c² · c)² + (m₀c²)²
E² = E²v²/c² + (m₀c²)²
E² - E²v²/c² = E²v²/c² + (m₀c²)² - E²v²/c²
E² - E²v²/c² = (m₀c²)²
E²(1 - v²/c²) = (m₀c²)²Result:
E = m₀c²/√(1 - v²/c²)
"We just derived relativistic energy from first principles!"
Derivation 2: The Lorentz Factor (3 minutes)
"That square root term keeps appearing. Let's give it a name."
Define:
γ = 1/√(1 - v²/c²)
Therefore:
E = γm₀c²
"This γ (gamma) is the famous Lorentz factor. It's not some mysterious mathematical construction - it's just the factor that makes energy and momentum consistent!"
Derivation 3: Relativistic Momentum (3 minutes)
"Now let's find momentum in terms of velocity."
From p = Ev/c² and E = γm₀c²:
p = γm₀c²v/c²
p = γm₀v
"Classical momentum p = mv becomes p = γmv. The γ factor accounts for relativistic effects."
Derivation 4: Mass-Energy Equivalence (2 minutes)
"What happens when the particle is at rest?"
When v = 0:
- γ = 1/√(1-0) = 1
- p = 0
From E² = (pc)² + (m₀c²)²:
E² = 0 + (m₀c²)²
E² = (m₀c²)²
E = m₀c²
"This is E = mc²! It falls right out when we set velocity to zero."
Derivation 5: Kinetic Energy (5 minutes)
"How much energy comes from motion?"
Total energy = Rest energy + Kinetic energy
E = KE + m₀c² γm₀c² = KE + m₀c²γm₀c² - m₀c² = KE + m₀c² - m₀c² γm₀c² - m₀c² = KE
KE = γm₀c² - m₀c²
Therefore:
KE = m₀c²(γ - 1)
Check the low-speed limit: When v << c, use binomial approximation:
γ ≈ 1 + ½(v²/c²)Why is this the low-speed approximation?Start with: γ = 1/√(1 - v²/c²)Step 1: When v << c, then v²/c² << 1 (a very small number)Step 2: We can rewrite γ as:
γ = (1 - v²/c²)^(-1/2)Step 3: When the term inside parentheses is close to 1,
we can use the binomial approximation:
(1 + x)^n ≈ 1 + nx when x << 1Step 4: In our case: x = -v²/c² and n = -1/2, so:
γ = (1 - v²/c²)^(-1/2) ≈ 1 + (-1/2)(-v²/c²) = 1 + ½(v²/c²)Step 5: Now substitute into KE = m₀c²(γ - 1):
KE = m₀c²[(1 + ½(v²/c²)) - 1] = m₀c² × ½(v²/c²) = ½m₀v²
The key insight: the binomial approximation is only valid
when v²/c² << 1, which means v << c
(low speeds compared to light speed).That's why it's the "low-speed limit" -
the approximation breaks down at high speeds!
KE ≈ m₀c² × ½(v²/c²) = ½m₀v²
"We recover classical kinetic energy! Relativity reduces to classical mechanics at low speeds."
Derivation 6: General Momentum Form (5 minutes)
"Let's solve the original equation directly for kinetic energy in terms of momentum."
From E² = (pc)² + (m₀c²)² and E = KE + m₀c²:
(KE + m₀c²)² = (pc)² + (m₀c²)²
KE² + 2KE(m₀c²) = (pc)²
Using quadratic formula:
KE = √[(pc)² + (m₀c²)²] - m₀c²
"This is the completely general form - kinetic energy in terms of momentum."
Quick Numerical Example (2 minutes)
"Let's see how this works with real numbers."
For an electron (m₀ = 9.1×10⁻³¹ kg) at v = 0.9c:
- γ = 1/√(1-0.81) = 2.29
- Classical KE would be ½mv² = 0.405 m₀c²
- Relativistic KE = m₀c²(γ-1) = 1.29 m₀c²
"At 90% light speed, relativistic effects more than triple the kinetic energy!"
The Big Reveal (8 minutes)
Return to the opening question:
Special Relativity = E² = (pc)² + (m₀c²)²
Summary on board:
ONE EQUATION → ALL OF RELATIVITY
E² = (pc)² + (m₀c²)²
Solve for E → E = γm₀c² (relativistic energy)
Define γ → γ = 1/√(1-v²/c²) (Lorentz factor)
Solve for p → p = γm₀v (relativistic momentum)
Set v=0 → E = m₀c² (mass-energy equivalence)
Find KE → KE = m₀c²(γ-1) (relativistic kinetic energy)
Take v→0 → KE ≈ ½m₀v² (classical limit)
Key insight: "What we call 'special relativity' isn't a theory with multiple parts. It's one mathematical relationship viewed from different angles. Time dilation, length contraction, relativistic mass - these aren't separate phenomena that mysteriously work together. They're different projections of the same underlying reality."
"Traditional physics courses spend weeks on this, presenting each result as a separate mystery. But you just derived everything in 30 minutes because you understood the fundamental unity from the start."
Closing Challenge (2 minutes)
"Next time someone tells you relativity is complicated, show them this. One equation. Everything else is just algebra."
"The deeper lesson: Look for the simple principles underlying complex-seeming phenomena. Nature is usually more elegant than it first appears."
Assessment Questions
Quick Check (during class):
- What happens to γ when v = 0?
- What happens to γ as v approaches c?
- Why does KE = ½mv² work for everyday speeds?
Homework Problems:
- Calculate γ for v = 0.5c, 0.9c, 0.99c
- Find the kinetic energy of a proton moving at 0.8c
- At what speed does relativistic KE equal classical KE?
- Show that p = γm₀v reduces to p = m₀v when v << c
Teaching Notes
Common Student Misconceptions:
- "Relativity only applies at very high speeds" → Show that the equations are always correct; classical mechanics is just an approximation
- "γ is just a mathematical trick" → Emphasize that γ emerges from requiring physical consistency
- "These are separate effects" → Repeatedly emphasize the unity of the single equation
Extension Activities:
- Graph γ vs v to visualize the relativistic regime
- Derive length contraction and time dilation from the same invariant
- Connect to particle accelerator physics or cosmic ray observations
- Show that the photon momentum is when m₀ is zero.
Differentiation:
- Advanced students: Challenge them to derive time dilation from the invariant
- Struggling students: Focus on the pattern recognition - same equation, different solutions
- Mathematical support: Provide step-by-step algebra worksheets
Materials for Follow-up Classes
This foundation enables rapid coverage of:
- Spacetime intervals and four-vectors
- Relativistic collisions and conservation laws
- Connection to electromagnetic field theory
- Introduction to general relativity concepts
The key advantage: Students now see relativity as fundamentally simple rather than mysteriously complex, making advanced topics much more accessible.
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