# Algebra letting me down?

• Apr 25th 2013, 08:28 PM
Kiwi_Dave
Algebra letting me down?
I am studying this paper:
Noether's theorem and the work-energy theorem for a charged particle in an electromagnetic field | DeepDyve

and he makes a step that seems wrong to me, but the final answer can't be wrong so I am confused.

$\displaystyle \dot F=\frac{\partial L}{\partial \underline x} \cdot \underline x + \frac{\partial L}{\partial \dot {\underline x}} \cdot \underline {\dot x}+2t \frac{\partial L}{\partial t} - 2h$

define $\displaystyle \underline p= \frac{\partial L}{\partial \dot{\underline x}}$

he then effectively says

$\displaystyle \dot F=\dot {\underline p} \cdot \underline x + \underline p \cdot \dot{\underline x}+2t \frac{\partial L}{\partial t} - 2h$

Whereas I get

$\displaystyle \dot F=\frac{\partial L}{\partial \underline x} \cdot \underline x + \underline p \cdot \underline {\dot x}+2t \frac{\partial L}{\partial t} - 2h$

I have attached a piece of the paper incase you want to see where this came from. Here, Zeta is the position vector x and tau is the time t.

Am I missing something here?
• Apr 26th 2013, 07:11 AM
topsquark
Re: Algebra letting me down?
For a cyclic coordinate x (one that does not appear in the kinetic energy term) we have for the equation of motion:
$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0$

So
$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}$

-Dan
• Apr 26th 2013, 10:10 PM
Kiwi_Dave
Have I Correctly argued that this is a function of time?
Now I have the equation:

$\displaystyle \dot F=\frac{\partial L}{\partial \underline x} \cdot \underline x + \frac{\partial L}{\partial \dot {\underline x}} \cdot \underline {\dot x}+2t \frac{\partial L}{\partial t} - 2h$

and I am told that F is a total derivative with respect to time, but it does not seem immediately obvious to me.

$\displaystyle L=L(x(t),\dot x(t),t)$

Am I correct that if I can show the equation to be un-changed under a change of coordinates, then it must be total derivative with respect to time?

I can show that x/dx is unchanged under a scaling of the x axis (and the same for y and z). I can show that t/dt is un-changed under a scaling of time. I can show that $\displaystyle \frac{\dot x}{\dot dx}$ is un-changed under a scaling of x and/or t. Is this sufficient to say that F is a total derivative with respect to time?

Since my equation is the Rund-Trautman identity, perhaps it is sufficient to demonstrate F dot is invariant for a change of coordinates of the type that lead to the Rund Trautman equation?
• Apr 27th 2013, 08:52 PM
topsquark
Re: Algebra letting me down?
According to your notation F(dot) is already a time derivative. Perhaps you should tell me what your Lagrangian is and define F and h.

-Dan
• Apr 27th 2013, 09:56 PM
Kiwi_Dave
Re: Algebra letting me down?
Thanks Dan.

I have attached an image of the lagrangian which comes from electromagnetism.

I am studying a text book on Noether's theorem, a text which I like in so many ways but the author makes too many mistakes (according to many reviews) and I don't trust it to be accurate. He uses this Lagrangian as an example and comes to a different conserved quantity (same generators) to the paper that I referenced at the start of this thread. I am convinced that the paper is correct and the text book is wrong (he seems to conclude that this Lagrangian is absolutely invariant).

But MY REAL PROBLEM is that every example I look at seems to be invariant. Perhaps this is because my examples all come from classical mechanics (perhaps they are trivial examples of Noether's theorem?). Even if I take the generators derived in the Kobe paper (tau = t and zeta = 2x) I can change them arbitrarily (I tried tau = t and zeta = 3x) and it is still invariant. So I doubt myself and think that I may not recognise something that is invariant.

When I search for examples on-line people seem to use the definition of invariance rather than using the Rund-Trautman identity (as my text and the Kobe paper do). And no one seems to work an example that demonstrates that something is NOT invariant under a specified infintesimal transformation.

I think everything would probably come clear for me if I saw a worked example that showed an infantesimal transformation of some Lagrangian from classical mechanics IS NOT invariant.

My second, smaller problem is that I don't know a rigorous way of defining a total derivative other than in a formal way from the partial derivatives and the chain rule.

I am trying to understand the process rather than to solve any one particular problem.
• Apr 27th 2013, 10:12 PM
Kiwi_Dave
Re: Algebra letting me down?

F is defined by the right hand side. The object of the exercise as I understand it is to demonstrate that the right hand side is equal to the total derivative of some function (which I have called F, if it exists). This would demonstrate that the Lagrangian is divergence invariant under the specified infintesimal transformation.

I don't have an intuitive understanding of what "divergence invariant" means. Perhaps the divergence of F must be zero?
• Apr 28th 2013, 01:25 AM
topsquark
Re: Have I Correctly argued that this is a function of time?
An intermediate comment:
$\displaystyle \frac{\partial L}{\partial \underline x} \cdot \underline x + \frac{\partial L}{\partial \dot {\underline x}} \cdot \underline {\dot x}+2t \frac{\partial L}{\partial t} - 2h$

$\displaystyle = \frac{d}{dt} ( \bold{p} \cdot \bold{x} ) + 2t \frac{\partial L}{\partial t} - 2H$

Let's suppose that this is supposed to be a total derivative. Then we may ignore the momentum term and just consider the last part:
$\displaystyle 2t \frac{\partial L}{\partial t} - 2H$

So for this to work the expression above must be a total time derivative. I'm not getting anything so far. Still working....

-Dan
• Apr 28th 2013, 01:33 AM
topsquark
Re: Algebra letting me down?
Quote:

Originally Posted by Kiwi_Dave

F is defined by the right hand side. The object of the exercise as I understand it is to demonstrate that the right hand side is equal to the total derivative of some function (which I have called F, if it exists). This would demonstrate that the Lagrangian is divergence invariant under the specified infintesimal transformation.

I don't have an intuitive understanding of what "divergence invariant" means. Perhaps the divergence of F must be zero?

Divergence invariant?? Any F that would exist here would be a scalar! (You can't take the divergence of a scalar.) Where does this come into play?

In that article (at the bottom of your original post), I'd appreciate it if you would give me more that comes before equation 24. Perhaps that will help explain things. I feel like I'm whistling in the dark.

-Dan
• Apr 28th 2013, 01:53 AM
Kiwi_Dave
Re: Algebra letting me down?
Thank's for persevering!

I have attached the entire Kobe paper.

Armed with the entire paper a slightly different question comes to mind.

How do we know that the Fs in equations 19a, 19b, 19c and 19d are total derivatives?