(Requires calculus)
Prove the relativistic work-energy theorem in one dimension. The force exerted on a particle is given not by F=ma, but F=dp/dt. Using the expression for the relativistic momentum of a particle p=γmv, integrate the force exerted on the particle over the distance it would take to get from speed 0 to a speed v, and show that this is equal to the relativistic formula for the kinetic energy of a particle, K=(γ-1)mc2. Why do we use E=γmc2 rather than K in the principle of conservation of energy? (Compare to the classical case)
To prove the relativistic work-energy theorem, we need to integrate the force exerted on a particle over the distance it takes for the particle to accelerate from speed 0 to speed v. The force is given by F = dp/dt, where p represents the relativistic momentum of the particle.
First, we need to determine the expression for the relativistic momentum. In special relativity, the momentum of a particle is given by p = γmv, where γ represents the Lorentz factor, m is the particle's rest mass, and v is its velocity.
Let's consider a particle starting from rest (v = 0) and accelerating to a speed v. The work done on the particle is the integral of the force with respect to distance, so we need to express dp/dt in terms of velocity.
Using the chain rule, we can write dp/dt = dp/dv * dv/dt. In this case, dp/dv represents the derivative of momentum with respect to velocity, which is the relativistic mass times the derivative of velocity with respect to time, dp/dv = (d(γmv)/dv) = γm(dv/dt).
Now, we can substitute dp/dt in terms of velocity and then integrate the force over the distance required to reach speed v:
∫F dx = ∫(dp/dt) dx
= ∫γm(dv/dt) dx
= ∫γm dv
= γm ∫dv
= γm(v - 0)
= γmv
The integral of the force over distance yields γmv, which represents the relativistic work done on the particle.
To relate this to the relativistic kinetic energy of the particle, we use the equation for kinetic energy, K = (γ - 1)mc². Comparing this to the work done, we have:
K = (γ - 1)mc²
We know that γ = (1 - v²/c²)^(-1/2), where c represents the speed of light. For v << c, the term (γ - 1) becomes negligibly small compared to γ.
In the classical case, the kinetic energy is expressed as K = (1/2)mv². However, in the relativistic case, the relativistic mass m takes into account increases in energy as the particle accelerates towards the speed of light. Using E = γmc² in the principle of conservation of energy ensures that the full relativistic energy of the system is accounted for, rather than only considering the kinetic energy term.
Therefore, in the relativistic work-energy theorem, we use E = γmc² rather than K to incorporate the effects of relativistic mass increase and accurately account for the total energy of the system.