Confused about applying the Euler–Lagrange equation

[Malamala]

Hello! I have a Lagrangian of the form:

$$L = \frac{mv^2}{2}+f(v)v$$
where $f(v)$ is a function of the velocity. I would like to derive the equation of motion in general, without writing down an expression for $f(v)$ yet. I have that $\frac{\partial L}{\partial x} = 0$. However, what is $\frac{\partial L}{\partial v}$? Is it $mv+f(v)$ or $mv+f(v)+\frac{\partial f}{v}v$? Thank you!

 

[Ibix]

I guess you meant $\frac{\partial f}{\partial v}$.

For the purpose of experiment, assume $f(v)=\alpha v$ where $\alpha$ is a constant with appropriate units (or you may assume something else simple if you prefer). You can now calculate your Lagrangian explicitly. Which of your approaches gives the right answer in that case?

 

[TSny]

However, what is $\frac{\partial L}{\partial v}$? Is it $mv+f(v)$ or $mv+f(v)+\frac{\partial f}{v}v$?

It is essentially the second expression. But the notation is a bit off in the last term where you wrote $\frac{\partial f}{v}v$. A partial derivative should have the symbol $\partial$ in both the numerator and denominator: $\frac{\partial f}{\partial v}$. However, note that $f(v)$ is a function of the single variable $v$. So, a partial derivative is not really appropriate. Instead, the notation should express an ordinary derivative $\frac{df}{dv}$ or $f'(v)$. Thus, the last term would be $f'(v)v$.

I assume that you are dealing with a one-dimensional problem with spatial coordinate $x$ and where $v = \frac{dx}{dt}$.

 

[PeroK]

Hello! I have a Lagrangian of the form:

$$L = \frac{mv^2}{2}+f(v)v$$
where $f(v)$ is a function of the velocity. I would like to derive the equation of motion in general, without writing down an expression for $f(v)$ yet. I have that $\frac{\partial L}{\partial x} = 0$. However, what is $\frac{\partial L}{\partial v}$? Is it $mv+f(v)$ or $mv+f(v)+\frac{\partial f}{v}v$? Thank you!

Let $g(v) = f(v)v$. Rewrite your Lagrangian using $g(v)$. What do you do now that the lone $v$ has gone?