• Support PF! Buy your school textbooks, materials and every day products Here!

Using Noether's theorem to get a constant of motion

  • Thread starter JD_PM
  • Start date
  • #1
432
35

Homework Statement:

I will not use summation notation; repeated pair of (upper and lower) indices are summed over (Einstein's notation).

Given the following action (note there's no potential term):



$$S = \int dt \Big( G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1)$$



Let us define a vector ##v^a## (scarily called killing vector) which makes the following equation true:



$$\Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big) = 0 \ \ \ \ (2)$$



Prove that ##Q_v = v^a \dot q^b G_{a b}## is a constant of motion.

HINT: You can solve the problem using Noether's Theorem

Relevant Equations:

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1)$$

$$\sum_a \Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big) = 0 \ \ \ \ (2)$$
Noether's theorem tells us that an invariance of the Lagrangian yields a constant of motion. In this problem, that constant is:

$$Q_v = p^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + p^b \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0}= v^a \dot q^b G_{a b} \ \ \ \ (3)$$

So the approach is simple: calculate the 4 terms you see in the middle part of equation ##(3)## and you should get ##v^a \dot q^b G_{a b}##

OK so I will first calculate ##p^a## and ##p^b##

For ##p^k## we have

$$p^k = \frac{\partial L}{\partial \dot q_k} = G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a = G_{kb} \dot q^b + G_{ak} \dot q^a = 2 G_{ak} \dot q^a$$

Thus:

$$p^a = \frac{\partial L}{\partial \dot q_a} = 2 G_{aa} \dot q^a$$

$$p^b = \frac{\partial L}{\partial \dot q_b} = 2 G_{ab} \dot q^a$$

Then:

$$Q_v = 2 G_{aa} \dot q^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + 2 G_{ab} \dot q^a \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4)$$

But how to proceed with ##\frac{\partial q_a^{\lambda}}{\partial \lambda}## and ##\frac{\partial q_b^{\lambda}}{\partial \lambda}## terms?

I think that the way I used Noether's theorem is OK though (AKA I think that equation (4) is OK).
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
Homework Statement:: I will not use summation notation; repeated pair of (upper and lower) indices are summed over (Einstein's notation).
You really really should. Even writing out some sums, you have ended up with some confusions. Not using the conventions available to simplify notation also makes your posts harder to read and therefore makes it more difficult to help you.


Noether's theorem tells us that an invariance of the Lagrangian yields a constant of motion. In this problem, that constant is:

$$Q_v = p^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + p^b \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0}= v^a \dot q^b G_{a b} \ \ \ \ (3)$$
The RHS here does use the summation convention! You cannot just not use it.

$$p^k = \frac{\partial L}{\partial \dot q_k} = G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a = G_{kb} \dot q^b + G_{ak} \dot q^a = 2 G_{ak} \dot q^a$$
You seem to get things mixed up here because of your aversion to the summation convention. The indices do not just takes two values ##a## and ##b##. The indices ##a## and ##b## generally take any values between 1 and N where N is the dimensionality of the problem.

I strongly suggest learning to use the summation convention properly.
 
  • #3
432
35
You really really should. Even writing out some sums, you have ended up with some confusions. Not using the conventions available to simplify notation also makes your posts harder to read and therefore makes it more difficult to help you.
I think that it depends on the person. I started using summation convention but now I feel more comfortable using Einstein's notation (by the way, I started using it inspired by samalkhaiat).

I will now type the equations with summations so that everyone feels comfortable (by the way, I noticed I used it in ##(2)## without noticing).

$$Q_v = \sum_a p^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + \sum_b p^b \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0}= \sum_{ab}v^a \dot q^b G_{a b} \ \ \ \ (3')$$

$$p_k = \frac{\partial L}{\partial \dot q_k} = \sum_{ab} \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = \sum_{b} G_{kb} \dot q^b + \sum_{a} G_{ak} \dot q^a = 2\sum_a G_{ak} \dot q^a$$

$$Q_v = \sum_a 2 G_{aa} \dot q^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + \sum_b 2 G_{ab} \dot q^a \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4')$$
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
  • #5
432
35
In (3'), why are you writing the same term twice?
Because the conserved quantity is a linear combination of the momenta of the system with coefficients depending on ##q##. As we have ##a## and ##b## I thought we should have ##2##.

But I will check it.

That's a nice insight! I will have a look at it and edit my equations.
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
scarily called killing vector
Note that capitalization is important here, it should be Killing vector after German mathematician Wilhelm Killing. It has nothing to do with murderous vectors. :oldeyes:
 
  • #7
432
35
I tried to be funny and failed XD
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
Because the conserved quantity is a linear combination of the momenta of the system with coefficients depending on ##q##. As we have ##a## and ##b## I thought we should have ##2##.

But I will check it.
Note that both terms are the same. It does not matter what you call the summation index.
 
  • #9
432
35
Note that both terms are the same. It does not matter what you call the summation index.
But we expect two terms, don't we?
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
But we expect two terms, don't we?
Why? And why would you call it two terms if they are exactly the same and not just put a 2 in front?
 
  • #11
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
I tried to be funny and failed XD
This is from my GR lecture notes:
83006CEF-20BC-4260-9732-5C061518A895.png
 
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
As for Noether’s theorem, it states that if ##q^a \to q^a + \epsilon k^a## is a symmetry of the Lagrangian, then ##p_a k^a## is a constant of motion. There is only one term here but note that, due to the summation convention, it is a linear combination of the canonical momenta. Also note the index placements.
 
  • #13
432
35
Edit: I strongly suggest rewriting everything using the summation convention and keeping these rules in mind:
https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

OK After having a look at the rules, I am going to type all equations in both notations:

Using summation notation:

The given action:

$$S = \int dt \sum_{ab} \Big( G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1')$$

The given Killing vector equation:

$$\sum_a \Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big) = 0 \ \ \ \ (2')$$

The given conservative quantity we have to prove:

$$Q_v = \sum_{ab} v^a \dot q^b G_{a b} \ \ \ \ (*')$$

The constant of motion given by Noether's theorem:

$$Q_v = 2\sum_a p_a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} = \sum_{ab}v^a \dot q^b G_{a b} \ \ \ \ (3')$$

The computation of momentum (generally):

$$p_k = \frac{\partial L}{\partial \dot q^k} = \sum_{ab} \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = \sum_{b} G_{kb} \dot q^b + \sum_{a} G_{ak} \dot q^a = 2\sum_a G_{ak} \dot q^a \ \ \ \ (**')$$

Applying ##(**')## to our case we get:

$$p_a = \frac{\partial L}{\partial \dot q^a} = 2 \sum_{a} G_{ab} \dot q^a \ \ \ \ (***')$$

Plugging ##(***')## into ##(3')## we get:

$$Q_v = 4 \sum_{a} G_{ab} \dot q^a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4')$$

But as Orodruin points out in his insight # 3: You shall not have different free indices on opposite sides of an equality or in different terms of the same expression.

Thus:

$$Q_v = 4 \sum_{a} G_{ab} \dot q^a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \neq \sum_{ab}v^a \dot q^b G_{a b} \ \ \ \ (5')$$

Note that in ##(5')##, on the LHS ##b## is not a dummy index (it is not summed over) while on the RHS it is. Thus they are different. Let me label the dummy index ##b## on the RHS as ##w## for the sake of clarity:

$$Q_v = 4 G_{ab} \dot q^a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \neq \sum_{aw}v^a \dot q^w G_{a w} \ \ \ \ (5')$$

This means I am doing something wrong.
 
  • #14
432
35
Note that this post adds nothing new; you can ignore it if you wish.

Using Einstein's notation (non-primed and primed equation labels are equivalent):

$$S = \int dt \Big( G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1)$$

$$ \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a = 0 \ \ \ \ (2)$$

$$Q_v = v^a \dot q^b G_{a b} \ \ \ \ (*)$$

$$Q_v = 2 p_a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} = v^a \dot q^b G_{a b} \ \ \ \ (3)$$

$$p_k = \frac{\partial L}{\partial \dot q^k} = G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a = G_{kb} \dot q^b + G_{ak} \dot q^a = 2 G_{ak} \dot q^a \ \ \ \ (**)$$

$$p_a = \frac{\partial L}{\partial \dot q^a} = 2 G_{ab} \dot q^a \ \ \ \ (***)$$

$$Q_v = 4 G_{ab} \dot q^a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4)$$

$$Q_v = 4 G_{ab} \dot q^a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \neq v^a \dot q^b G_{a b} \ \ \ \ (5)$$
 
Last edited:
  • #15
432
35
I think the following equation is wrong:

$$Q_v = 2\sum_a p_a \Big( \frac{\partial (q^a)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} = \sum_{ab}v^a \dot q^b G_{a b} \ \ \ \ (3')$$
##(5')## follows from ##(3')## of course.
 
  • #16
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
In (3) you are assuming what you need to prove. Also, drop the factor of 2, it is just confusing.

You have also not specified what the continuous transformation leaving the action invariant is. This is extremely important in order to specify what the derivative ##dq/d\lambda## is.

Also note that your equation (5) has more than two a indices and refer back to the Insight.
 
  • #17
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
Also, drop the factor of 2, it is just confusing.
In fact, inserting a factor 1/2 instead could be beneficial here. Of course, if ##Q## is conserved then ##Q/2## is conserved (or indeed any constant multiplied by ##Q##).
 
  • #18
432
35
In (3) you are assuming what you need to prove. Also, drop the factor of 2, it is just confusing.
OK

You have also not specified what the continuous transformation leaving the action invariant is. This is extremely important in order to specify what the derivative ##dq/d\lambda## is.
The invariance of the Lagrangian is under translation. I know this because ##q^a## is cyclic (##\frac{\partial L}{\partial q^a} = 0##). It follows from that fact that the momentum of the system is conserved and the continuous transformation linked to conservation of momentum is translation.

Also note that your equation (5) has more than two a indices and refer back to the Insight.
Mmm you are right.

OK so equation (5) is wrong because of two reasons:

1) Having two different free indices on opposite sides of the equality (Orodruin's # 3).

2) Having more than two equal indices on one side of the equation (Orodruin's recurrent # 5).

The issue now is how to compute ##dq/d\lambda##

I will think about it and post what I get.
 
  • #19
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
The invariance of the Lagrangian is under translation. I know this because ##q^a## is cyclic (##\frac{\partial L}{\partial q^a} = 0##). It follows from that fact that the momentum of the system is conserved and the continuous transformation linked to conservation of momentum is translation.
This is not completely true. The case of a cyclic coordinate is just a special case of Noether's theorem where ##k^a## mentioned in #12 is constant and equal to one for a particular coordinate and zero for the others. Noether's theorem is more general than that. In particular, it is not (necessarily) the case for the symmetry being considered in this scenario.

Edit: Also note that there is nothing in the question telling you that there exist cyclic coordinates. In general ##G_{ab}## depends on the coordinates.
 
  • #20
432
35
Ready? Set? Go! (yeah, I trying to be funny again)...

Also note that there is nothing in the question telling you that there exist cyclic coordinates. In general ##G_{ab}## depends on the coordinates.
Absolutely! I've just noticed I assumed ##G_{ab}## to be independent of ##q##. That's why I got ##\frac{\partial L}{\partial q^a} = 0##. If we assume ##G_{ab}## does depend on ##q## and of course we are not guaranteed to get ##\frac{\partial L}{\partial q^a} = 0## and thus ##q## does not have to be cyclic.

##q^a \to q^a + \epsilon k^a##
Just to make sure I do not mess up with notation; what you wrote above is equivalent to the following notation:

$$q^i \rightarrow (q^i)^{\lambda}$$

Where:

$$\lambda = \epsilon k^a$$

Isn't it?
 
  • #21
432
35
As for Noether’s theorem, it states that if ##q^a \to q^a + \epsilon k^a## is a symmetry of the Lagrangian, then ##p_a k^a## is a constant of motion.
Once at this point what we have to do next is to specify the type of symmetry we are dealing with.

Let's check translation.


I've already calculated ##p_k## (equation ##(**)## at #14. By the way, I see now why you want to put the 1/2 factor; you want to cancel out the number 2 that the calculation of momentum yields :wink:):

$$p_k = \frac{\partial L}{\partial \dot q^k} = 2 G_{ak} \dot q^a$$

OK so if the Lagrangian is invariant under translation, then the component of linear momentum in the ##n##-direction will be conserved.

To simplify our job, let me pick ##k=3##. Then we have to deal with ##q^a##, ##q^b## and ##q^c##.

I am going to work out only the translation involving ##q^a##.

$$q^a \to q^a + \lambda$$

$$q^b \to q^b$$

$$q^c \to q^c$$

I think here's the trick which solves the problem; we notice that we have:

$$\sum_{j=1}^{k} \Big[ \frac{\partial (q_j^k)^{\lambda}}{\partial \lambda}\Big] = \delta_{k j}$$

So we end up with:

$$\frac{\partial (q^a)^{\lambda}}{\partial \lambda} |_{\lambda=0} = 1$$

$$\frac{\partial (q^b)^{\lambda}}{\partial \lambda} |_{\lambda=0} = 0$$

$$\frac{\partial (q^c)^{\lambda}}{\partial \lambda} |_{\lambda=0} = 0$$

So the conserved quantity is:

$$Q_{q^a} = 2G_{aa} \dot q^a$$

In fact, inserting a factor 1/2 instead could be beneficial here. Of course, if ##Q## is conserved then ##Q/2## is conserved (or indeed any constant multiplied by ##Q##).
OK so the conserved quantity is:

$$Q_{q^a} = G_{aa} \dot q^a$$

If we do the same with ##q^b## and ##q^c## we get ##Q##:

$$Q = \dot q^a\Big( G_{aa} + G_{ab} + G_{ac} \Big)$$

Playing a bit with the dummy indices we get:

$$Q = \dot q^b\Big( G_{ab} + G_{bb} + G_{bc} \Big)$$

OK The above looks like I am on the right track but still missing something. What I expect to get is an equation like the following:

$$v^a \dot q^b G_{ab}\Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big)$$
 
  • #22
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
To simplify our job, let me pick k=3k=3k=3. Then we have to deal with qaqaq^a, qbqbq^b and qcqcq^c.
You cannot pick k=3, k is a vector field. You need a particular vector field and you need to show that the action is invariant under the given transformation in order to be able to use Noether’s theorem. The form of the wanted conserved quantity should give you a hint about which vector field to choose.

You again should check all your equations against the basic rules in the Insight. As soon as you find an equation that does not comply, something has gone wrong and continuing from there is useless until you put it right.
 
  • #23
432
35
which vector field to choose.
I am sorry but I do not see what you mean when you suggest picking a vector field in this context. May you explain it, please?

You cannot pick k=3
I think there is a misunderstanding. I meant ##n=3##.

What I did was thinking about a simpler example, work it out and then apply the same idea to this problem:

I asked myself 'what is the simplest Lagrangian I can think of?' OK that is:

$$L = \sum_{i = 1}^{n} (\dot q^i)^2$$

And I worked out:

$$q^i \rightarrow q^i + \lambda$$

And I used Noether's theorem applying the idea above and I got the right conserved quantity(the scaling factor doesn't matter):

$$Q = 2( \dot x + \dot y + \dot z)$$

Why cannot use the same idea here then?
 
  • #24
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,572
6,335
I am sorry but I do not see what you mean when you suggest picking a vector field in this context. May you explain it, please?
If you want to have a symmetry transformation, such as the one required to apply Noether's theorem, then you need a vector field that generates that symmetry transformation. In essence, that vector field is what the derivatives ##dq^a/d\lambda## are the components of.

Please tell write down in your own words the entire statement of Noether's theorem as you are familiar with it.

I think there is a misunderstanding. I meant n=3.
What is n?

Why cannot use the same idea here then?
The idea is the same but what was a symmetry for your simplified Lagrangian is not a symmetry of the general Lagrangian. You need the transformation of the coordinates to be a symmetry of the Lagrangian in order for Noether's theorem to apply. You cannot use it for any transformation (at least not if you want to derive a conservation law).
 
  • #25
432
35
If you want to have a symmetry transformation, such as the one required to apply Noether's theorem, then you need a vector field that generates that symmetry transformation. In essence, that vector field is what the derivatives ##dq^a/d\lambda## are the components of.
OK so it is going to be a Killing vector field.

Suppose we have a collection of curves, so that the collected tangent vectors form a Killing vector field. I've read that we can choose coordinates so that ##\lambda## is one of the coordinates, ##q^{a_1}##, for ##a_1## a single fixed direction. Let us define ##\xi = \frac{d}{d \lambda}##. Then:

$$\xi^j = \frac{d (q^a)^j}{d \lambda} = \delta_{a_1}^{j}$$

But this is what I have done above...

This is coming from here:

Captura de pantalla (868).png

More details: http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf

Please tell write down in your own words the entire statement of Noether's theorem as you are familiar with it.
Let a set of functions ##{f}## map

$$q \rightarrow q^{\lambda}$$

If such mappings leave the Lagrangian invariant (where ##q^{\lambda}##), then the quantity ##\sum_{j=1}^n p_j [dq_j^{\lambda}/d\lambda]|_{\lambda = 0}## is conserved

What is n?
It is the number of coordinates we choose. For instance, in my much simpler example ##L = \sum_{i = 1}^{n} (\dot q^i)^2## I picked ##n=3## (##n=1## corresponding to the ##x## coordinate, ##n=2## corresponding to the ##y## coordinate and ##n=3## corresponding to the ##z## coordinate).
 

Related Threads for: Using Noether's theorem to get a constant of motion

Replies
2
Views
561
Replies
7
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
576
Replies
3
Views
810
Replies
2
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
868
Top