# Post History

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**#2: Post edited**

~~Let's first remember how you check that you are at an extremum (minimum, maximum, saddle point) of a normal differentiable function. You do so by checking that the first derivative is zero (plus further tests that are irrelevant).~~- Now what does the derivative being zero mean? Well, it means that if you move a very small distance away from the point, the function value in first order doesn't change, that is, if $x_0$ is a minimum of $f$, then
- $$f(x+h) = f(x) + O(h^2)$$
- Now when we seek an extremum of $S(q) = \int_{t_0}^{t_1} L(q, \dot q, t)\mathrm dt$, over *paths* q(t), we do essentially the same. One important thing to know is that the paths we consider all begin and end at the same point.
- So let's evaluate the function at a path $\tilde q(t) = q(t) + \delta q(t)$ where $\delta q(t)$ is a small deviation, and of course in order to start and end at the same point, we must have $\delta q(t_0) = \delta q(t_1) = 0$.
- We will also need the time derivative of $\tilde q(t)$, but that's easy: The derivative of a sum is just the sum of derivatives, that is
- $$\dot{\tilde q} = \dot q + \delta\dot q$$
- (here and in the following I just write $q$ instead of $q(t)$ and so on; just remember that every $q$ has an implicit $(t)$).
- So let's insert that into the action integral:
- $$S(\tilde q) = \int_{t_0}^{t_1} L(q+\delta q, \dot q+\delta\dot q, t)\mathrm dt$$
- Now we can do a tailor expansion of $L$ in both arguments to first order (that is, basically ignore anything having a product of $\delta $ quantities):
- $$S(\tilde q)= \int_{t_0}^{t_1} \big( L(q,\dot q,t) +
- \frac{\partial L}{\partial q}(q\dot q,t)\delta q +
- \frac{\partial L}{\partial\dot q}(q\dot q,t)\delta\dot q\big)
- \mathrm dt$$
- Now the integral of the sum can be split into a sum of the integral of the terms. Then we find that the first integral is just $S(q)$, and we get:
- $$S(\tilde q) = S(q) + \int_{t_0}^{t_1} \frac{\partial L}{\partial q}\delta q\mathrm dt + \int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt$$
- Now we'd like to get rid of the $\delta\dot q$. But that's easy by using the partial integration rule
- <p>$$\int_a^b u\dot v\mathrm dt = [uv]_a^b - \int_a^b \dot uv\mathrm dt$$</p>
- In this case, $a=t_0$, $b=t_1$, $u=\frac{\partial L}{\partial\dot q}$ and $v=\delta q$.
- Nor remember that $\delta q(t_0)=\delta q(t_1) = 0$, therefore the first term on the right hand side vanishes, and we get
- $$\int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt = -\int_{t_0}^{t_1} \left(\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot q}\right)\delta\dot q\mathrm dt\delta q\\,\mathrm dt$$
- Inserting this into the formula for $S(\tilde q)$ and combining the two remaining integrals, we therefore get to first order in $\delta q$:
- $$S(\tilde q) = S(q) + \int_{t_0}^{t_1}\left(\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}\right)\delta q\\,\mathrm dt$$
- However out demand is that, to first order, $S(\tilde q)=S(q)$, which means that the integral above must be zero. And since $\delta q$ is supposed to be arbitrary (except at the end points), this means that $$\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}=0$$
- But that is exactly the Euler-Lagrange equation.

- Let's first remember how you check that you are at an extremum (minimum, maximum, saddle point) of a normal differentiable function. You do so by checking that the first derivative is zero (plus further tests that are irrelevant
**for the current question**). - Now what does the derivative being zero mean? Well, it means that if you move a very small distance away from the point, the function value in first order doesn't change, that is, if $x_0$ is a minimum of $f$, then
- $$f(x+h) = f(x) + O(h^2)$$
- Now when we seek an extremum of $S(q) = \int_{t_0}^{t_1} L(q, \dot q, t)\mathrm dt$, over *paths* q(t), we do essentially the same. One important thing to know is that the paths we consider all begin and end at the same point.
- So let's evaluate the function at a path $\tilde q(t) = q(t) + \delta q(t)$ where $\delta q(t)$ is a small deviation, and of course in order to start and end at the same point, we must have $\delta q(t_0) = \delta q(t_1) = 0$.
- We will also need the time derivative of $\tilde q(t)$, but that's easy: The derivative of a sum is just the sum of derivatives, that is
- $$\dot{\tilde q} = \dot q + \delta\dot q$$
- (here and in the following I just write $q$ instead of $q(t)$ and so on; just remember that every $q$ has an implicit $(t)$).
- So let's insert that into the action integral:
- $$S(\tilde q) = \int_{t_0}^{t_1} L(q+\delta q, \dot q+\delta\dot q, t)\mathrm dt$$
- Now we can do a tailor expansion of $L$ in both arguments to first order (that is, basically ignore anything having a product of $\delta $ quantities):
- $$S(\tilde q)= \int_{t_0}^{t_1} \big( L(q,\dot q,t) +
- \frac{\partial L}{\partial q}(q\dot q,t)\delta q +
- \frac{\partial L}{\partial\dot q}(q\dot q,t)\delta\dot q\big)
- \mathrm dt$$
- Now the integral of the sum can be split into a sum of the integral of the terms. Then we find that the first integral is just $S(q)$, and we get:
- $$S(\tilde q) = S(q) + \int_{t_0}^{t_1} \frac{\partial L}{\partial q}\delta q\mathrm dt + \int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt$$
- Now we'd like to get rid of the $\delta\dot q$. But that's easy by using the partial integration rule
- <p>$$\int_a^b u\dot v\mathrm dt = [uv]_a^b - \int_a^b \dot uv\mathrm dt$$</p>
- In this case, $a=t_0$, $b=t_1$, $u=\frac{\partial L}{\partial\dot q}$ and $v=\delta q$.
- Nor remember that $\delta q(t_0)=\delta q(t_1) = 0$, therefore the first term on the right hand side vanishes, and we get
- $$\int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q}\delta\dot q\mathrm dt = -\int_{t_0}^{t_1} \left(\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot q}\right)\delta\dot q\mathrm dt\delta q\\,\mathrm dt$$
- Inserting this into the formula for $S(\tilde q)$ and combining the two remaining integrals, we therefore get to first order in $\delta q$:
- $$S(\tilde q) = S(q) + \int_{t_0}^{t_1}\left(\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}\right)\delta q\\,\mathrm dt$$
- However out demand is that, to first order, $S(\tilde q)=S(q)$, which means that the integral above must be zero. And since $\delta q$ is supposed to be arbitrary (except at the end points), this means that $$\frac{\partial L}{\partial q}-\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q}=0$$
- But that is exactly the Euler-Lagrange equation.