Finding rate of relativistic work, special relativity.

[T-chef]

Hello all,

Homework Statement

Given $$\frac{dE}{dt}=\frac{d(m(u))}{dt}\cdot u$$
show that $$\frac{dE}{dt}=\frac{m_0}{(1-u^2)^{\frac{3}{2}}}u\frac{du}{dt}$$
where u is velocity, m(u) is relativistic mass, and $m_0$ is rest mass.

Homework Equations

$$m(u)=\frac{m_0}{\sqrt{1-u^2}}$$

The Attempt at a Solution

Substituting in m(u) gives $$\frac{dE}{dt} = \frac{d}{dt}(\frac{m_0}{\sqrt{1-u^2}})\cdot u$$
Using the chain rule
$$= \frac{d}{du}(\frac{m_0}{\sqrt{1-u^2}})\frac{du}{dt}\cdot u$$
$$= \frac{m_0 u}{(1-u^2)^{3/2}}\frac{du}{dt} \cdot u$$
From here I get into trouble. My first thought was since u should be parallel to $\frac{du}{dt}$ the dot product just becomes multiplication, but then there's an extra u factor. Any help from here would be greatly appreciated.

 

[Chestermiller]

The problem statement is ambiguous. The quantity in parenthesis should be written as mu, and not m(u). More precisely, it should be m_oγu.

 

[clamtrox]

I don't think the relativistic mass should be a vector, so that dot must be just an ordinary product between two scalars. I do agree that the problem seems a little strange.

 

[T-chef]

That makes much more sense, thanks for the help guys