Using Noether's theorem to get a constant of motion

[JD_PM]

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).

 

[Orodruin]

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.

 

[JD_PM]

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).

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

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

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

 

[Orodruin]

In (3'), why are you writing the same term twice?

You are also being inconsistent with your covariant and contravariant indices.

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/

 

[JD_PM]

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.

https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

That's a nice insight! I will have a look at it and edit my equations.

 

[Orodruin]

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.

 

[JD_PM]

I tried to be funny and failed XD

 

[Orodruin]

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.

 

[JD_PM]

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?

 

[Orodruin]

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?

 

[Orodruin]

I tried to be funny and failed XD

This is from my GR lecture notes:

 

[Orodruin]

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.

 

[JD_PM]

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.

 

[JD_PM]

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):

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

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

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

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

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

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

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

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

 

[JD_PM]

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.

 

[Orodruin]

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.

 

[Orodruin]

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 $Q$).

 

[JD_PM]

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.

 

[Orodruin]

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.

 

[JD_PM]

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?

 

[JD_PM]

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 ):

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

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:

$$v^a \dot q^b G_{ab}\Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big)$$

 

[Orodruin]

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.

 

[JD_PM]

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?

 

[Orodruin]

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).

 

[JD_PM]

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:

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).

 

[Orodruin]

This is essentially saying that you can choose coordinates such that your Killing field has components $\delta_1^a$. This part of the argument was missing before. Either way, it is not necessary to pick those coordinates in order to arrive at the result, even if it is a possible way forward.

Can you summarize your current argument?

 

[JD_PM]

Can you summarize your current argument?

Let's recap.

Essentially, this exercise is about showing that the following equation holds:

$$Q_v = \sum_{k=1}^n p_k \Big( \frac{\partial (q^j)^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} = \sum_{a b}v^a \dot q^b G_{a b} \ \ \ \ (6)$$

OK, I know how to get $p_k$ (equation $(**')$):

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

The issue here is how to compute $\frac{\partial (q^j)^{\lambda}}{\partial \lambda}_{\lambda = 0}$

I definitely think that the previous method isn't valid for this exercise because it doesn't involve the Killing vector field.

I am sure that equation $(2)$ is involved in the computation of $\frac{\partial (q^j)^{\lambda}}{\partial \lambda}_{\lambda = 0}$

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

But I do not see how to proceed... any hint would be appreciated.

 

[Orodruin]

The derivative with respect to $\lambda$ is dependent on the actual symmetry transformation you are considering. To compute it, you need to choose a symmetry transformation.

 

[JD_PM]

you need to choose a symmetry transformation.

I picked translation in #21

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 meant a hint in how to compute it for translational symmetry.

 

[Orodruin]

I picked translation in #21

But you have not shown that translation is a symmetry. This is absolutely necessary in order to use Noether’s theorem.

Caveat: A coordinate system is always going to exist where the required transformation is a symmetry. We are not necessarily working in those coordinates.

What do you need $dq^a/d\lambda$ to be to reproduce the correct constant of motion? You should pick that as your transformation and show that it corresponds to a symmetry of the action. You cannot just assume that a given transformation is a symmetry.

 

[JD_PM]

Please let me use Einstein's notation from now on.

But you have not shown that translation is a symmetry. This is absolutely necessary in order to use Noether’s theorem.

OK so you're asking to show that

$$L(q^{\lambda}, \dot q^{\lambda}, t) = L(q, \dot q, t)$$

The given Lagrangian is:

$$L = G_{ab} \dot q^a\dot q^b$$

The translation is:

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

If we take the derivative with respect to time on both sides we get

$$\dot q^a \to \dot q^a$$

The above justifies that $\dot q^a$ terms do not change under translation.

But we are not done yet because $G_{ab}$ depends on $q$. Thus we still have to show that:

$$G_{ab}(q^c) = G_{ab}(q^c + \lambda)$$

But how? ... (we would have assumed the above to be true in class, but let's show it).

What do you need $dq^a/d\lambda$ to be to reproduce the correct constant of motion?

What I know is that $dq^i/d\lambda$ equals to the Kronecker-Delta function, because when we take $dq^a/d\lambda$ on $q^a \rightarrow q^a + \lambda$ we end up with $1$ while taking $dq^b/d\lambda$ on $q^b \rightarrow q^b$ yields $0$.

My point is that, if I the above reasoning is OK, how can the following derivation be wrong?

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

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

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

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

Note I have used $i$ as the dummy index.

You cannot just assume that a given transformation is a symmetry.

That's the right way to proceed, indeed.

Please note: this thread may be becoming too long so I would perfectly understand if the best is to stop.

 

[Orodruin]

But we are not done yet because GabGabG_{ab} depends on qqq. Thus we still have to show that:

Gab(qc)=Gab(qc+λ)Gab(qc)=Gab(qc+λ)​

G_{ab}(q^c) = G_{ab}(q^c + \lambda)
But how?

In general you cannot because it may not be true. That is the entire point. You need to choose a transformation for which the Lagrangian is invariant. You cannot just choose any transformation. The translation you have chosen is not going to be a symmetry for most Lagrangians.

 

[JD_PM]

In general you cannot because it may not be true. That is the entire point. You need to choose a transformation for which the Lagrangian is invariant. You cannot just choose any transformation. The translation you have chosen is not going to be a symmetry for most Lagrangians.

Could we please postpone the prove that $G_{ab}(q^c) = G_{ab}(q^c + \lambda)$ holds and just assume by now that $q^a \to q^a + \lambda$ is a symmetry of the given Lagrangian?

I would like to focus on the computation of $dq^a/d\lambda$, which is my big struggle.

What I propose is the test and error method; we assume that $q^a \to q^a + \lambda$ and see if we get the given constant of motion out of Noether's Theorem

 

[Orodruin]

What I propose is the test and error method; we assume that qa→qa+λqa→qa+λq^a \to q^a + \lambda and see if we get the given constant of motion out of Noether's Theorem

You won’t because your symmetry needs to be based on the relevant Killing field. Without that, how do you suppose the Killing field would end up in the conserved quantity?

 

[Orodruin]

I would like to focus on the computation of dqa/dλdqa/dλdq^a/d\lambda, which is my big struggle.

By definition of a continuous transformation these are the components of the vector field generating the transformation. There is nothing else to it.

 

[JD_PM]

You won’t because your symmetry needs to be based on the relevant Killing field. Without that, how do you suppose the Killing field would end up in the conserved quantity?

I think I need a better understanding of what's a Killing vector field. I will study more about it.

Thank you for your help Orodruin! :)

 

[Orodruin]

I think I need a better understanding of what's a Killing vector field. I will study more about it.

That the field is a Killing field really does not have much to do with things at this moment. It will only be relevant in showing that the given transformation indeed is a symmetry of the action. What is relevant at this point is that you understand how a vector field (any vector field) generates a transformation through its flow, i.e., the integral curves of that vector field maps a point in the manifold curves in the manifold in such a way that the vector field is tangent to the curves everywhere. If the curve parameter is $\lambda$ and the field has components $X^a$, then the tangent at curve parameter $\lambda = 0$ is given by
$$
\left.\frac{dq^a}{d\lambda}\right|_{\lambda=0} = X^a.
$$

 

[JD_PM]

That the field is a Killing field really does not have much to do with things at this moment. It will only be relevant in showing that the given transformation indeed is a symmetry of the action. What is relevant at this point is that you understand how a vector field (any vector field) generates a transformation through its flow, i.e., the integral curves of that vector field maps a point in the manifold curves in the manifold in such a way that the vector field is tangent to the curves everywhere. If the curve parameter is $\lambda$ and the field has components $X^a$, then the tangent at curve parameter $\lambda = 0$ is given by
$$
\left.\frac{dq^a}{d\lambda}\right|_{\lambda=0} = X^a.
$$

Thank you for the explanation. I think that I should work out an example to understand it better. For instance, let's go back to the simplest Lagrangian I brought to the table back in #21

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

Could you please explain your argument above based on this specific example I know how to work out? If I see it applied I think I may understand it better.

 

[Orodruin]

Instead of that Lagrangian, let us consider the two-dimensional case, and instead of translations, let us consider rotations (to get your mind away from fixating on translations, which I believe is stopping you from reaching insight).

The 2D Lagrangian of relevance is
$$
L = \dot x^2 + \dot y^2.
$$
We can add a rotationally invariant potential term to this, but let us keep from doing that at the moment.

A rotation is defined by $x \to x \cos(\lambda) + y \sin(\lambda)$, $y \to -x\sin(\lambda) + y \cos(\lambda)$. The components of the tangent vector field to this rotation is given by taking the derivative with respect to $\lambda$:
$$
v^1 = \left.\frac{dx}{d\lambda}\right|_{\lambda = 0} = y, \qquad
v^2 = \left.\frac{dy}{d\lambda}\right|_{\lambda = 0} = -x.
$$
The vector field generating rotations is therefore given by $\vec v = d\vec x/d\lambda = y\vec e_1 - x \vec e_2$. Note that, $\lambda$, $\vec x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2)$.

Is this a symmetry of the action? The time derivative of $\vec x$ transforms as
$$
\dot{\vec x} \to \dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2).
$$
It follows that
$$
L = \dot x^2 + \dot y^2 = \dot{\vec x}^2 \to [\dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)]^2
= \dot{\vec x}^2 + 2\lambda \dot{\vec x} \cdot \dot{\vec v} + \mathcal O(\lambda^2).
$$
We find
$$
\dot{\vec x} \cdot \dot{\vec v} = (\dot x \vec e_1 + \dot y \vec e_2) \cdot (\dot y \vec e_1 - \dot x \vec e_2)
= \dot x \dot y - \dot y \dot x = 0
$$
and thus
$$
L \to L + \mathcal O(\lambda^2).
$$
Thus, the derivative of the Lagrangian is zero at $\lambda = 0$ (and therefore everywhere) and it follows that the rotations indeed are a symmetry of the action.

The corresponding conserved quantity is given by
$$
Q = \vec v \cdot \vec p = (y\vec e_1 - x \vec e_2) \cdot 2(\dot x \vec e_1 + \dot y \vec e_2) = 2(y\dot x - x \dot y),
$$
which in essence is angular momentum. Since our transformation was a symmetry of the action, angular momentum is conserved.Edit: Note that you could start from $d\vec x/d\lambda = \vec v$ with $\vec v$ given by the expression above and integrate this to find the (finite) transformation. However, to show that rotations are a symmetry, we only needed the infinitesimal transformations.

 

[JD_PM]

Instead of that Lagrangian, let us consider the two-dimensional case, and instead of translations, let us consider rotations (to get your mind away from fixating on translations, which I believe is stopping you from reaching insight).

Thank you for posting this example; it is what I needed to move forward

OK so I am going to use your structure. I will work out the translations.

We consider the Lagrangian

$$L = G_{ab} \dot q^a\dot q^b $$

We could add a translationally invariant potential term to the Lagrangian but let's stick to the given problem, which lacks of it.

Translation is given by:

$$q^a \rightarrow q^a + \lambda$$

The components of the tangent vector field to this translation are given by taking the derivative with respect to $\lambda$

$$k^a = \left.\frac{dq^a}{d\lambda}\right|_{\lambda = 0} = 1, \qquad k^b = \left.\frac{dq^b}{d\lambda}\right|_{\lambda = 0} = 0$$

Thus, the vector field generating translations is given by:

$$\vec k = d\vec q^a/d\lambda = \vec e_1$$

We note that:

$$\vec q^a \to \vec q^a + \lambda \vec k + \mathcal O(\lambda^2)$$

We now check translational symmetry. Taking the derivative with respect to time of the above equation:

$$\dot{\vec q^a} \to \dot{\vec q^a} + \lambda \dot{\vec k} + \mathcal O(\lambda^2)$$

It follows that (note that the derivative with respect to time of $\vec k$ is zero; $\lambda \dot{\vec k} = \lambda (0) =0$):

$$L = G_{ab} \dot{\vec q^a}\dot{\vec q^b} = G_{ab} \Big( \dot{\vec q^a} + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) \Big( \dot{\vec q^b} + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) = G_{ab} \dot{\vec q^a}\dot{\vec q^b} + \mathcal O(\lambda^2) $$

Thus:

$$L \to L + \mathcal O(\lambda^2)$$

Thus we have shown that the translations indeed are a symmetry of the action.

Now it is time to get the constant of motion $p_ak^a$

We already computed $p_a$

$$p_a = 2 G_{ba} \dot q^b$$

Thus:

$$Q = p_ak^a = 2 G_{ba} \dot q^b$$

Mmm we are almost there! I am still missing the Killing vector $v^a$ term. Well, as this quantity (which turns out to be the momentum) is conserved, could we simply multiply $v^a$ on both sides as follows?

$$2 G_{ba} \dot q^b=0$$

$$v^a G_{ba} \dot q^b=(v^a)0$$

Thus we get (note I've dropped the scaling factor):

$$Q_v = v^a \dot q^b G_{ab}$$

 

[JD_PM]

Note I have not used

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

Mmm I should have used it while checking that translation is a symmetry of the Lagrangian...

I am thinking...

EDIT: I see no mistake in my reasoning on why translations are a symmetry of the action.

Please let me know if I am wrong .

PS: This is so fun!

 

[Orodruin]

Translation is given by:

qa→qa+λ​

First of all, this is a very particular translation, adding $\lambda$ to each coordinate. It is not the transformation we want. Instead, you should be looking at the infinitesimal transformation as I did above:

$x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2).$

but with a suitably chosen vector field $v^a$. This will generally not be a coordinate translation just as the rotation above was not.

The components of the tangent vector field to this translation are given by taking the derivative with respect to λλ\lambda

ka=dqadλ∣∣∣λ=0=1,kb=dqbdλ∣∣∣λ=0=0​

a and b are just arbitrary indices that can take any value. You cannot have $k^a\neq 0$ and $k^b = 0$ at the same time.

You have also assumed that $G_{ab}$ does not depend on the coordinates when you wanted to check the invariance of the action, which generally is wrong (if it does not you will have translational symmetry in all directions but this is not what is relevant to this problem).

 

[JD_PM]

First of all, this is a very particular translation, adding $\lambda$ to each coordinate. It is not the transformation we want. Instead, you should be looking at the infinitesimal transformation as I did above:

I get your point. You mean:

$$q^a \rightarrow q^a + \epsilon k^a$$

a and b are just arbitrary indices that can take any value. You cannot have $k^a\neq 0$ and $k^b = 0$ at the same time.

I see. So:

$$k^1 = \left.\frac{dq^1}{d\lambda}\right|_{\lambda = 0} = 1, \qquad k^2 = \left.\frac{dq^2}{d\lambda}\right|_{\lambda = 0} = 0$$

You have also assumed that $G_{ab}$ does not depend on the coordinates when you wanted to check the invariance of the action, which generally is wrong (if it does not you will have translational symmetry in all directions but this is not what is relevant to this problem).

OK so:

$$L = G_{ab}(q) \dot{\vec q^a}\dot{\vec q^b} = G_{ab}(q+\epsilon k^a) \Big( \dot{\vec q}^a + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) \Big( \dot{\vec q^b} + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) = G_{ab}(q) \dot{\vec q^a}\dot{\vec q^b} + \mathcal O(\lambda^2)$$

But we still have to prove that (where I got stuck back in #31):

$$G_{ab}(q^a) = G_{ab}(q^a+\epsilon k^a)$$

I know it follows from equation $(2)$; but how?

Unfortunately I need another hint for this one...

 

[vanhees71]

I've not followed the entire discussion, which I find very hard to understand. Here are nevertheless some thoughts. I use analytical mechanics as described by the Lagrangian version of Hamilton's action principle with the action given as
$$S[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t),$$
where $q=(q^k)$ are arbitrary generalized coordinates.

In Noether's theorem one considers continuous transformations of the form
$$t'=t'(t,q,\dot{q}), \quad q'=q'(t,q,\dot{q}).$$
Such a transformation is called a symmetry, if the variation of the action is invariant, from which after some calculations involving infinitesimal transformations
$$t'=t+\epsilon T(t,q), \quad q'=q + \epsilon Q(t,q)$$
and evaluating the symmetry condition to first order in $\epsilon$ follows that there must exist a function $\Omega(q,t)$ such that
$$Q^k \frac{\partial L}{\partial q^k}+ \left (\frac{\mathrm{d} Q^k}{\mathrm{d}t} -\frac{\mathrm{d} T}{\mathrm{d}t} \dot{q}^k \right)\frac{\partial L}{\partial \dot{q}^k} + L \frac{\mathrm{d} T}{\mathrm{d} t} + T \frac{\partial L}{\partial t} + \frac{\mathrm{d} \Omega}{\mathrm{d} t}=0, \qquad \text{(SC)}$$
where $\mathrm{d}/\mathrm{d} t$ denotes the total time derivative with $q^k$ taken as functions of $t$ and $\partial_t=\partial/\partial_t$ as the time derivative concerning the explicit time dependence (with the $q$ and $\dot{q}$ held fixed).

Looking now at the solutions of the equations of motion,
$$\dot{p}_k=\frac{\partial L}{\partial q^k}, \quad p_k=\frac{\partial L}{\partial \dot{q}^k},$$
this symmetry condition implies that
$$\frac{\mathrm{d}}{\mathrm{d} t} (Q^k p_k - T H+\Omega)=0, \qquad \text{(CE)}$$
where
$$H=p_k \dot{q}^k-L$$
is the Hamilton function of the system.

So to apply Noether's theorem just consider your case. Here you obviously check the special case of "point transformations" only, i.e., a change in the generalized coordinates only, i.e.,
$$Q=v(q)$$.
Now, using your Lagrangian, just derive what follows from the symmetry condition (SC) for the $v^k$ and then use the conservation equation (CE) to get what's the corresponding conserved "Noether quantity".

 

[JD_PM]

Thank you for your post vanhees71, I will read it carefully.

I've not followed the entire discussion, which I find very hard to understand.

I perfectly understand you.

Notice I have been trying to approach the problem and had several misconceptions that Orodruin kindly has been correcting.

I think you can get the picture of where I am at this point only having a look at #40 and #43. I am only missing showing that the symmetric matrix $G_{ab}$ doesn't change under translation

$$G_{ab}(q^a) = G_{ab}(q^a+\epsilon k^a)$$

Please let me know if you wish if there's anything in particular you do not understand from my posts.

 

[vanhees71]

That condition would mean that all matrix elements don't change along a certain direction, given by the vector $k^a$ (supposed these $k^a$ are constant). Then it's simply translational invariance, and the canonical momentum in this direction is conserved, because from your Lagrangian it follows that
$$k^a \frac{\partial L}{\partial q^a}=0 \; \Rightarrow \; k^a \dot{p}_a=0 \; \Rightarrow \; k^a p_a=\text{const}.$$
In your case
$$p_a = \frac{\partial L}{\partial \dot{q}^a}=2 G_{ab} \dot{x}^b$$
and thus indeed for this special case of a symmetry
$$G_{ab} k^a \dot{x}^b=\text{const}.$$
That's of course not the most general case asked for in your question in #1, which asks for possible symmetries under point transformations, leading to the Killing vectors of the metric $G_{ab}$ under consideration.

 

[MathematicalPhysicist]

Note that capitalization is important here, it should be Killing vector after German mathematician Wilhelm Killing. It has nothing to do with murderous vectors.

But I am sure Wilhelm was quite the killer...

 

[JD_PM]

That condition would mean that all matrix elements don't change along a certain direction

Ahh so $G_{ab}(q^a)$ changes to $\Big(\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a\Big)$ due to the transformation!

Thus we get:

$$L = G_{ab}(q) \dot{\vec q^a}\dot{\vec q^b} =\Big(\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a\Big) \Big( \dot{\vec q}^a + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) \Big( \dot{\vec q^b} + \lambda \dot{\vec k} + \mathcal O(\lambda^2)\Big) = v^a G_{cb} \dot{\vec q^c}\dot{\vec q^b} + \mathcal O(\lambda^2)$$

Then:

$$L \to v^a L + \mathcal O(\lambda^2)$$

So key is that the vector $v^a$ shows up due to the transformation! Then the conserved quantity stated by Noether's theorem $p_a k^a$ has to be multiplied by $v^a$.

Then we end up with:

$$Q_v = v^a \dot q^b G_{ab}$$

-----------------------------------------------------------------------------------------------------

What I do not understand now is how to get $v^a$ out of $\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a$

I obviously got the right answer because I already knew it beforehand.

 

[Orodruin]

Ahh so Gab(qa)Gab(qa)G_{ab}(q^a) changes to (∂aGbcva+Gba∂cva+Gca∂bva)(∂aGbcva+Gba∂cva+Gca∂bva)\Big(\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a\Big) due to the transformation!

No, it does not. Be careful in comparing the Lagrangians before and after transformation. In particular when v^a are not constant.

 

[JD_PM]

Ahh so $G_{ab}(q^a)$ changes to $\Big(\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a\Big)$ due to the transformation!

No, it does not. Be careful in comparing the Lagrangians before and after transformation. In particular when v^a are not constant.

So may you please explain me what does $G_{ab}(q^a)$ become after the transformation?

I know that after the transformation we end up with the Lagrangian $L = v^a G_{cb} \dot{\vec q^c}\dot{\vec q^b}$, but I do not know how to deal with the transformation of the matrix $G_{cb}$.

Please consider I've never dealt with this kind of problem.