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

[TimeRip496]

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

 

[samalkhaiat]

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:
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)
The left hand side is the variation in L, i.e., \delta L. Now take x = q, y = \dot{q}, \epsilon = \gamma and \eta = \dot{\gamma}, you get
\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .
If the transformation q^{'} = q + \gamma is a symmetry, then \delta L = 0. So, you have
\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.
Now, use the Lagrange equation
\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,
in the first term and combine the two terms
\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 .

 

[TimeRip496]

Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
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)
The left hand side is the variation in L, i.e., \delta L. Now take x = q, y = \dot{q}, \epsilon = \gamma and \eta = \dot{\gamma}, you get
\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .
If the transformation q^{'} = q + \gamma is a symmetry, then \delta L = 0. So, you have
\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.
Now, use the Lagrange equation

Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
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)
The left hand side is the variation in L, i.e., \delta L. Now take x = q, y = \dot{q}, \epsilon = \gamma and \eta = \dot{\gamma}, you get
\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .
If the transformation q^{'} = q + \gamma is a symmetry, then \delta L = 0. So, you have
\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.
Now, use the Lagrange equation
\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,
in the first term and combine the two terms
\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 .

in the first term and combine the two terms
\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 .

Thanks! But can you show me taylor expansion of the Lagrangian? I am kind of new to this.

 

[samalkhaiat]

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:
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
Infinitesimal transformations means that \epsilon \ll 1, so you can take \epsilon^{2} = \epsilon^{3} = \cdots \approx 0 and keep only the linear terms in \epsilon:
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}}

 

[TimeRip496]

Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
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)
The left hand side is the variation in L, i.e., \delta L. Now take x = q, y = \dot{q}, \epsilon = \gamma and \eta = \dot{\gamma}, you get
\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .
If the transformation q^{'} = q + \gamma is a symmetry, then \delta L = 0. So, you have
\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.
Now, use the Lagrange equation
\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,
in the first term and combine the two terms
\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 .

Thanks! But how do I get to equation 32?

 

[samalkhaiat]

Thanks! But how do I get to equation 32?

Don't you know how to take the total time derivative of L(x(t),y(t);t)? What is your formal Education level?
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt}
Now take x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}, so \frac{dy}{dt} = \ddot{q}.

 

[TimeRip496]

Don't you know how to take the total time derivative of L(x(t),y(t);t)? What is your formal Education level?
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt}
Now take x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}, so \frac{dy}{dt} = \ddot{q}.

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

Thanks again for your help! Besides i am a A'level student.

 

[TimeRip496]

Don't you know how to take the total time derivative of L(x(t),y(t);t)? What is your formal Education level?
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt}
Now take x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}, so \frac{dy}{dt} = \ddot{q}.

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?

 

[samalkhaiat]

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