# Relativistic Kinetic Energy

• kudoushinichi88
In summary, the conversation discusses the derivation of the relativistic kinetic energy and the combination of two integrals. The lecturer presents an expression that can be used to combine the integrals and suggests using differentiation to derive the proper limits. The final result is given as K = \int c^2\,dm.

#### kudoushinichi88

In the derivation of the relativistic kinetic energy,

$$K=\int_{x_1}^{x_2}F\,dx = \int_{0}^{v}\frac{d}{dt}(mv)\,dx = \int_{0}^{v}(mv\,dv+v^2\,dm)$$

here, my lecturer told us without showing that

$$mv\,dv+v^2\,dm = c^2\,dm$$

Can someone please give me hints on how to combine these two integrals? I have no idea how to start.

kudoushinichi88 said:
In the derivation of the relativistic kinetic energy,

$$K=\int_{x_1}^{x_2}F\,dx = \int_{0}^{v}\frac{d}{dt}(mv)\,dx = \int_{0}^{v}(mv\,dv+v^2\,dm)$$
You need to be a bit more careful with the limits. The integral with respect to dm doesn't have as limits 0 and v.
here, my lecturer told us without showing that

$$mv\,dv+v^2\,dm = c^2\,dm$$

Can someone please give me hints on how to combine these two integrals? I have no idea how to start.
You can show by differentiating the expression for the relativistic mass

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

with respect to v. The LHS of the result the lecturer gave you is the integrand, so just substitute it into get

$$K = \int c^2\,dm$$

I'll leave it to you to figure out the proper limits.