Relativistic Kinetic Energy

[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.

 

[vela]

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.