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

 

[asteriatic]

It seems like you are on the right track with using the chain rule to solve this problem. However, you are correct in noting that there is an extra u factor that needs to be accounted for. This can be done by using the product rule, which states that the derivative of a product is equal to the first factor multiplied by the derivative of the second factor plus the second factor multiplied by the derivative of the first factor. In this case, the first factor is \frac{m_0 u}{(1-u^2)^{3/2}} and the second factor is u. Applying the product rule, we get:

\frac{dE}{dt} = \frac{d}{du}(\frac{m_0}{\sqrt{1-u^2}})\frac{du}{dt}\cdot u + \frac{m_0 u}{(1-u^2)^{3/2}}\cdot \frac{d}{dt}(u)

The first term on the right side can be simplified using the chain rule, just as you did before. The second term can be simplified using the fact that \frac{d}{dt}(u) = \frac{du}{dt}. This gives us:

\frac{dE}{dt} = \frac{m_0 u}{(1-u^2)^{3/2}}\frac{du}{dt}\cdot u + \frac{m_0 u}{(1-u^2)^{3/2}}\cdot \frac{du}{dt}

Combining like terms, we get:

\frac{dE}{dt} = \frac{m_0 u}{(1-u^2)^{3/2}}\left(\frac{du}{dt}\cdot u + \frac{du}{dt}\right)

Factoring out \frac{du}{dt}, we get:

\frac{dE}{dt} = \frac{m_0 u}{(1-u^2)^{3/2}}\frac{du}{dt}\cdot (u+1)

Finally, using the fact that u = \frac{dx}{dt} and \frac{dx}{dt} = \frac{du}{dt}, we can rewrite this as:

\frac{dE}{dt} = \frac{m_0}{(1-u^2)^{3/2}}\frac{du}{dt