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I am having trouble understanding the $ \dfrac{\partial L}{\partial y'} $ part in Euler-Lagrange Equation.

For example, if $ L = y^2(z) $, what is the symbolic expression for $ \displaystyle\frac{\partial (y^2(z))}{\partial (\frac{\partial y}{\partial z})} $?

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    $\begingroup$ $y$ and $y'$ are treated as independent variables in the Lagrangian so $\frac{dy'}{dy} = 0$ and $\frac{dy}{dy'} = 0$. $\endgroup$
    – Winther
    Jul 24, 2015 at 9:07

3 Answers 3

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To expand Rasmus' answer, a lagrangian can be seen as a function $\mathcal{L}=\mathcal{L}(x,y,p)$ of three independent variables. Then you will mainly work with $p=y'$, but the notation $$ \frac{\partial\mathcal{L}}{\partial y'} $$ should be understood as a shorthand for $$ \frac{\partial \mathcal{L}}{\partial p}. $$ More concretely, if (say) $\mathcal{L}(x,y,y')=\frac{1}{2}|y'|^2 + \frac{1}{2}y^2$, then $$ \frac{\partial\mathcal{L}}{\partial y'} = y', $$ since $$ \frac{\partial}{\partial p} \left( \frac{1}{2}|p|^2 + \frac{1}{2}y^2 \right) = p. $$

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That would simply be $0$ because $y'$ is regarded as a variable independent of $y$.

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    $\begingroup$ So if my $L$ expression does not contain first-order differential term, the Euler Lagrange expression would just reduce to:$$\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}} = \frac{\partial \mathcal{L}}{\partial y} = 0$$ ? $\endgroup$
    – Hong
    Jul 24, 2015 at 16:44
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When extremizing

$$I\left[y,\dot{y}\right]=\int\mathcal{L}(y,\dot{y},t)dt$$

One finds the difference between two near values and declares that the functional has been minimized when there is no change to first order. (assume fixed end points)

$$\Delta I\left[y,\dot{y}\right]=I\left[y+\delta y,\dot{y}+\delta \dot{y}\right]-I\left[y,\dot{y}\right]$$

$$\Delta I\left[y,\dot{y}\right]=\int(\mathcal{L}(y+\delta y,\dot{y}+\delta\dot{y},t)-\mathcal{L}(y,\dot{y},t))dt$$

Taking a taylor expansion of the integrand to first order, (note we will need to differentiate with respect to $\dot{y}$.

$$\delta I\left[y,\dot{y}\right]=\int(\delta y\frac{\partial \mathcal{L}}{\partial y}+\delta\dot{y}\frac{\partial \mathcal{L}}{\partial \dot{y}})dt$$

Integrating by parts (with vanishing boundary term)

$$\delta I\left[y,\dot{y}\right]=\int\delta y(\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}})dt$$

Now if $\delta I=0$ for all admissible variations $\delta y$ then we have that the integrand must vanish identically $$\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{y}}=0$$

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