# I Conservation laws from Lagrange's equation

#### Phylosopher

My question is related to the book: Classical Mechanics by Taylor. Section 7.8

So, In the book Taylor is trying to derive the conservation of momentum and energy from Lagrange's equation. I understood everything, but I am struggling with the concept and the following equation:

$$\delta\mathcal{L}= \epsilon \frac{\partial \mathcal{L}}{\partial x_{1}}+\epsilon \frac{\partial \mathcal{L}}{\partial x_{2}}+...=0$$

He derived this by saying that in theory, displacing the position of all N particles from $\vec{r_{i}} \rightarrow \vec{r_{i}}+\vec{\epsilon}$ will make $\delta \mathcal{L}=0$ "Unchanged". Which I understand.

Also this means that:

$$\mathcal{L}(\vec{r_{1}},....) = \mathcal{L}(\vec{r_{1}}+\vec{\epsilon},....)$$

Then from this he derived the equation that I am confused by, by letting $\vec{\epsilon}$ be in the direction of $x$ only.

I am not sure how to derive the first equation. It seems to be just the chain rule! But I didn't succeed with it.

In short I have two questions:

1- How to derive the first equation.
2- Why should all particles be moved by $\vec{\epsilon}$. I know that this is mathematically useful to have $\delta \mathcal{L}=0$. If each is displaced differently then there is an energy from outside the system. But what if all were displaced by an energy from outside the system exactly by the same amount $\vec{\epsilon}$ that he assumes, how can I be sure that $\delta \mathcal{L}=0$.

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#### Phylosopher

For the chain rule I did the following:

$$d \mathcal{L}= \frac{\partial \mathcal{L}}{\partial(x_{1}+\epsilon)} d(x_{1}+\epsilon)+...$$

I couldn't modify more. But I was considering something like maybe this:

$$d(x_{1}+\epsilon)=\frac{d(x_{1}+\epsilon)}{dx_{1}}dx_{1}= dx_{1}$$

Which makes sese. $d$ is just $\Delta$, and $\epsilon$ is just a constant.

#### stevendaryl

Staff Emeritus
In talking about a transformation such as $\vec{x} \rightarrow \vec{x} + \vec{\epsilon}$, we're just studying the mathematical properties of the lagrangian. Nobody's really talking about moving everything over by an amount $\vec{\epsilon}$.

So you have an initial function: $\mathcal{L}(x_1, x_2, ..., x_n, \dot{x}_1, \dot{x}_2 ...)$ (where $\dot{}$ means time derivative). We're just treating it as a mathematical function. Functions can be expanded in a Taylor series. So

$\mathcal{L}(x_1 + \delta x_1, x_2 ..., x_n, \dot{x}_1, \dot{x}_2 ...) \approx \mathcal{L}(x_1, x_2 ..., x_n, \dot{x}_1, \dot{x}_2 ...) + \frac{\partial \mathcal{L}}{\partial x_1} \delta x_1 +$ higher-order terms. So the change in $\mathcal{L}$ is:

$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial x_1} \delta x_1 +$ higher order terms

But the Euler-Lagrange equations tell us that:

$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}_1} = \frac{\partial \mathcal{L}}{\partial x_1}$

So we can write:

$\delta \mathcal{L} = (\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}_1}) \delta x_1 + ...$

At this point, we're saying that if $\delta \mathcal{L} = 0$, then it follows that:

$(\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x_1}}) \delta x_1 = 0$

So $\frac{\partial \mathcal{L}}{\partial \dot{x_1}} =$ a constant.

It's not true for all Lagrangians that $\delta \mathcal{L} = 0$ when you shift the coordinates. But if the Lagrangian doesn't depend on the coordinates, then $\delta \mathcal{L} = 0$, and so the momentum $p_1 \equiv \frac{\partial \mathcal{L}}{\partial \dot{x_1}}$ will be constant.

#### Phylosopher

In talking about a transformation such as →x→→x+→ϵ\vec{x} \rightarrow \vec{x} + \vec{\epsilon}, we're just studying the mathematical properties of the lagrangian. Nobody's really talking about moving everything over by an amount →ϵ\vec{\epsilon}.
I was thinking like (gedanken): If you have an infinite space where no background is available and you only see particles floating in space (So you cannot really relate the position of the particles but to there own). By displacing all the particles the same way by $\epsilon$, the system in the observer perspective did not change its energy(Nothing actually changed).

Which I think is a cool idea. Not just mathematically. I hope you get my point.

L(x1+δx1,x2...,xn,˙x1,˙x2...)≈L(x1,x2...,xn,˙x1,˙x2...)+∂L∂x1δx1+\mathcal{L}(x_1 + \delta x_1, x_2 ..., x_n, \dot{x}_1, \dot{x}_2 ...) \approx \mathcal{L}(x_1, x_2 ..., x_n, \dot{x}_1, \dot{x}_2 ...) + \frac{\partial \mathcal{L}}{\partial x_1} \delta x_1 + higher-order terms. So the change in L\mathcal{L} is:

δL=∂L∂x1δx1+\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial x_1} \delta x_1 + higher order terms
Fantastic, I get it. But for the Taylor series we are using, what would be the constant $a$?

We have $x+\delta x$, $(x+\delta x-a)^{n}$. I would guess that $a$ is $x$, but it must be a constant.

It's not true for all Lagrangians that δL=0\delta \mathcal{L} = 0 when you shift the coordinates. But if the Lagrangian doesn't depend on the coordinates

#### stevendaryl

Staff Emeritus
Fantastic, I get it. But for the Taylor series we are using, what would be the constant $a$?

We have $x+\delta x$, $(x+\delta x-a)^{n}$. I would guess that $a$ is $x$, but it must be a constant.
Well, it's also true when $a$ is a variable:

$f(y) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} (y-x)^n$

or letting $y - x = \delta x$,

$f(x+\delta x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} (\delta x)^n$