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