Help Me Understand This Author's Point: Noether's Theorem

In summary, the conversation is about understanding the Noether theorem and specifically, how the author gets to equation 32 in the source document. The conversation includes a step-by-step explanation of how to derive equation 32, including Taylor expanding the Lagrangian and using the Lagrange equation. The conversation ends with the question of how the author came up with the equation. The person asking for help mentions that they are a A'level student and sometimes struggle with remembering concepts.
  • #1
TimeRip496
254
5
I don't understand how the author get to these point. Please help me as i have been spending so much time trying to figure this out but to no avail. Thanks for your help
Source: http://phys.columbia.edu/~nicolis/NewFiles/Noether_theorem.pdf
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  • #2
TimeRip496 said:
I don't understand how the author get to these point. Please help me as i have been spending so much time trying to figure this out but to no avail. Thanks for your help
Source: http://phys.columbia.edu/~nicolis/NewFiles/Noether_theorem.pdf
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Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
[tex]L(x + \epsilon , y+ \eta) - L(x,y) = L(x,y) + \frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial y} \eta - L(x,y)[/tex]
The left hand side is the variation in L, i.e., [itex]\delta L[/itex]. Now take [itex]x = q[/itex], [itex]y = \dot{q}[/itex], [itex]\epsilon = \gamma[/itex] and [itex]\eta = \dot{\gamma}[/itex], you get
[tex]\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .[/tex]
If the transformation [itex]q^{'} = q + \gamma[/itex] is a symmetry, then [itex]\delta L = 0[/itex]. So, you have
[tex]\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.[/tex]
Now, use the Lagrange equation
[tex]\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,[/tex]
in the first term and combine the two terms
[tex]\frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) \gamma + (\frac{\partial L}{\partial \dot{q}} ) \frac{d}{dt}\gamma = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \gamma \right) = 0 .[/tex]
 
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  • #3
samalkhaiat said:
Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
[tex]L(x + \epsilon , y+ \eta) - L(x,y) = L(x,y) + \frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial y} \eta - L(x,y)[/tex]
The left hand side is the variation in L, i.e., [itex]\delta L[/itex]. Now take [itex]x = q[/itex], [itex]y = \dot{q}[/itex], [itex]\epsilon = \gamma[/itex] and [itex]\eta = \dot{\gamma}[/itex], you get
[tex]\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .[/tex]
If the transformation [itex]q^{'} = q + \gamma[/itex] is a symmetry, then [itex]\delta L = 0[/itex]. So, you have
[tex]\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.[/tex]
Now, use the Lagrange equation
samalkhaiat said:
Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
[tex]L(x + \epsilon , y+ \eta) - L(x,y) = L(x,y) + \frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial y} \eta - L(x,y)[/tex]
The left hand side is the variation in L, i.e., [itex]\delta L[/itex]. Now take [itex]x = q[/itex], [itex]y = \dot{q}[/itex], [itex]\epsilon = \gamma[/itex] and [itex]\eta = \dot{\gamma}[/itex], you get
[tex]\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .[/tex]
If the transformation [itex]q^{'} = q + \gamma[/itex] is a symmetry, then [itex]\delta L = 0[/itex]. So, you have
[tex]\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.[/tex]
Now, use the Lagrange equation
[tex]\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,[/tex]
in the first term and combine the two terms
[tex]\frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) \gamma + (\frac{\partial L}{\partial \dot{q}} ) \frac{d}{dt}\gamma = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \gamma \right) = 0 .[/tex]

in the first term and combine the two terms
[tex]\frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) \gamma + (\frac{\partial L}{\partial \dot{q}} ) \frac{d}{dt}\gamma = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \gamma \right) = 0 .[/tex]

Thanks! But can you show me taylor expansion of the Lagrangian? I am kind of new to this.
 
  • #4
TimeRip496 said:
Thanks! But can you show me taylor expansion of the Lagrangian? I am kind of new to this.
This is calculus problem! You should be able to Taylor expand function of two variables:
[tex]L(q + \epsilon \gamma , \dot{q} + \epsilon \dot{\gamma}) = L(q,\dot{q}) + \epsilon \gamma \frac{\partial L}{\partial q} + \epsilon \dot{\gamma} \frac{\partial L}{\partial \dot{q}} + \frac{1}{2} \epsilon^{2} \gamma^{2} \frac{\partial^{2}L}{\partial q^{2}} + \cdots[/tex]
Infinitesimal transformations means that [itex]\epsilon \ll 1[/itex], so you can take [itex]\epsilon^{2} = \epsilon^{3} = \cdots \approx 0[/itex] and keep only the linear terms in [itex]\epsilon[/itex]:
[tex]L(q + \epsilon \gamma , \dot{q} + \epsilon \dot{\gamma}) = L(q,\dot{q}) + \epsilon \gamma \frac{\partial L}{\partial q} + \epsilon \dot{\gamma} \frac{\partial L}{\partial \dot{q}}[/tex]
 
  • #5
samalkhaiat said:
Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
[tex]L(x + \epsilon , y+ \eta) - L(x,y) = L(x,y) + \frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial y} \eta - L(x,y)[/tex]
The left hand side is the variation in L, i.e., [itex]\delta L[/itex]. Now take [itex]x = q[/itex], [itex]y = \dot{q}[/itex], [itex]\epsilon = \gamma[/itex] and [itex]\eta = \dot{\gamma}[/itex], you get
[tex]\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .[/tex]
If the transformation [itex]q^{'} = q + \gamma[/itex] is a symmetry, then [itex]\delta L = 0[/itex]. So, you have
[tex]\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.[/tex]
Now, use the Lagrange equation
[tex]\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,[/tex]
in the first term and combine the two terms
[tex]\frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) \gamma + (\frac{\partial L}{\partial \dot{q}} ) \frac{d}{dt}\gamma = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \gamma \right) = 0 .[/tex]
Thanks! But how do I get to equation 32?
 
  • #6
TimeRip496 said:
Thanks! But how do I get to equation 32?
Don't you know how to take the total time derivative of [itex]L(x(t),y(t);t)[/itex]? What is your formal Education level?
[tex]\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt} [/tex]
Now take [itex]x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}[/itex], so [itex]\frac{dy}{dt} = \ddot{q}[/itex].
 
  • #7
samalkhaiat said:
Don't you know how to take the total time derivative of [itex]L(x(t),y(t);t)[/itex]? What is your formal Education level?
[tex]\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt} [/tex]
Now take [itex]x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}[/itex], so [itex]\frac{dy}{dt} = \ddot{q}[/itex].
Sorry sometimes when I do too much my brain gets fuzzy and I tend to forget all my stuff. I just have one last question, as to how the author just come up with this equation
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Thanks again for your help! Besides i am a A'level student.
 
  • #8
samalkhaiat said:
Don't you know how to take the total time derivative of [itex]L(x(t),y(t);t)[/itex]? What is your formal Education level?
[tex]\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt} [/tex]
Now take [itex]x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}[/itex], so [itex]\frac{dy}{dt} = \ddot{q}[/itex].
Sorry. Forget my previous post which was a stupid question. My last question is that in this case, is the degree of freedom referring to the type of axis(e.g. x, y or z) or the position along one axis for the euclidean translational symmetry on this paper?
 
  • #9
TimeRip496 said:
My last question is that in this case, is the degree of freedom referring to the type of axis(e.g. x, y or z)
It may or may not be. As you are still an A-level student, I would suggest you wait till you are mathematically more able. Or, if you like, you can start by reading about the meaning of generalized coordinates in analytical Mechanics.
Good Luck
 

FAQ: Help Me Understand This Author's Point: Noether's Theorem

What is Noether's Theorem?

Noether's Theorem is a fundamental principle in physics discovered by mathematician Emmy Noether. It states that for every continuous symmetry in a physical system, there is a corresponding conservation law.

Why is Noether's Theorem important?

Noether's Theorem is important because it helps to explain the fundamental laws of physics and provides a deeper understanding of the relationship between symmetries and conservation laws in nature.

What is the significance of Noether's Theorem in modern physics?

Noether's Theorem has had a major impact on modern physics, particularly in the fields of quantum mechanics, general relativity, and particle physics. It has been used to derive the laws of conservation of energy, momentum, and angular momentum from symmetries in physical systems.

How does Noether's Theorem relate to Einstein's theory of relativity?

Noether's Theorem played a crucial role in the development of Einstein's theory of relativity. It helped to explain the principle of covariance, which states that the laws of physics should be the same for all observers regardless of their frame of reference.

Can you provide an example of Noether's Theorem in action?

One example of Noether's Theorem in action is the conservation of angular momentum in a rotating system. The symmetry of rotation in the system leads to the conservation of angular momentum, which can be mathematically proven using Noether's Theorem.

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