Conservation laws from Lagrange's equation

In summary: So just by the definition of the Taylor series, we can choose ##a = x##.In summary, the conversation discusses the derivation of the conservation of momentum and energy from Lagrange's equation in Taylor's book Classical Mechanics. The equation in question is derived by considering the displacement of all particles in a system and observing that the Lagrangian remains unchanged. The first equation is derived using the chain rule and the Euler-Lagrange equations. The concept of displacing all particles in a system is mathematically useful in ensuring that the change in Lagrangian is zero. The constant "a" in the Taylor series can be chosen as the initial position "x" or as a variable "y".
  • #1
Phylosopher
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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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  • #2
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.
 
  • #3
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.
 
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  • #4
stevendaryl said:
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.

stevendaryl said:
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##?

b43af001b691a52034c46ff67dd15b4133285961


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

stevendaryl said:
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

You are correct. I should have added that to the thread, when I was asking.
 
  • #5
Phylosopher said:
Fantastic, I get it. But for the Taylor series we are using, what would be the constant ##a##?

b43af001b691a52034c46ff67dd15b4133285961


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##
 
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FAQ: Conservation laws from Lagrange's equation

What are conservation laws in physics?

Conservation laws in physics are fundamental principles that describe the behavior of physical systems. They state that certain quantities, such as energy, momentum, and angular momentum, remain constant over time and cannot be created or destroyed.

What is Lagrange's equation?

Lagrange's equation is a mathematical formula used in classical mechanics to describe the motion of a system. It is derived from the principle of least action and relates the system's kinetic and potential energies to its generalized coordinates and their time derivatives.

How are conservation laws related to Lagrange's equation?

Lagrange's equation is based on the principle of least action, which states that a system will follow the path that minimizes the action, a quantity related to the system's energy. This leads to the conservation of energy, and other conservation laws can be derived from this equation using Noether's theorem.

What are some examples of conservation laws from Lagrange's equation?

Some examples of conservation laws derived from Lagrange's equation include the conservation of energy, momentum, and angular momentum. These laws apply to a wide range of physical systems, from particles in motion to complex systems like planets in orbit.

Why are conservation laws important in physics?

Conservation laws are important in physics because they provide fundamental principles that govern the behavior of physical systems. They allow us to make predictions and understand the underlying mechanisms of natural phenomena. They also help us to conserve resources and protect the environment by providing a framework for understanding energy and matter conservation.

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