Using Noether's theorem to get a constant of motion

[Orodruin]

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

It becomes $G_{ab}(q+\epsilon v)$ for small $\epsilon$. Series expand this in $\epsilon$.

 

[JD_PM]

It becomes $G_{ab}(q+\epsilon v)$ for small $\epsilon$. Series expand this in $\epsilon$.

If we expand as a power series in $\epsilon$ we get

$$G_{ab}(q+\epsilon v) = G_{ab}(q+\epsilon v + \epsilon^2 v + ...)$$

But why to do so?

NOTE: I will be online the following hours.

 

[Orodruin]

If we expand as a power series in ϵϵ\epsilon we get

Gab(q+ϵv)=Gab(q+ϵv+ϵ2v+...)​

No, we do not. What is the series expansion of $f(x+\epsilon)$?

NOTE: I will be online the following hours.

I am going to bed, it is 23:40 here.

 

[JD_PM]

Orodruin I am enjoying so much this discussion, but I'd like to postpone it if you do not mind. I have more questions I'd like to ask in the forum the following days, that is why.

At this point I know the following:

According to Noether's theorem, the corresponding conserved Noether's charge $Q$ is equal to the momentum times the generator of the symmetry, which in this case is $v^a$

We still have to prove it.

Thank you for this insight so far.

it is 23:40 here.

Also here but I am afraid I am becoming PFaholic...

 

[JD_PM]

Back at it! :)

To recap: I was trying to understand what does $G_{ab} (q)$ become after the transformation. Next, I had to series expand it.

It becomes $G_{ab}(q+\epsilon v)$ for small $\epsilon$. Series expand this in $\epsilon$.

OK

$$G_{ab}(q+\epsilon v) = G_{ab}(q) + G'_{ab}(q) \epsilon v + G''_{ab}(q) \frac{ (\epsilon v)^2}{2!} + ... $$

So $G_{ab}(q+\epsilon v) = G_{ab}(q)$ because we drop the following terms.

 

[JD_PM]

OK so let me correct post #48. I will rewrite it carefully.

Ahh so $G_{ab}(q)$ changes to $G_{ab}(q + \epsilon v)$.

Then we get (let me avoid typing vector arrows this time)

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

Do you agree at this point?

 

[Orodruin]

Back at it! :)

To recap: I was trying to understand what does $G_{ab} (q)$ become after the transformation. Next, I had to series expand it.
OK

$$G_{ab}(q+\epsilon v) = G_{ab}(q) + G'_{ab}(q) \epsilon v + G''_{ab}(q) \frac{ (\epsilon v)^2}{2!} + ... $$

So $G_{ab}(q+\epsilon v) = G_{ab}(q)$ because we drop the following terms.

No, you cannot drop the order $\epsilon$ terms. They are crucial and completely necessary for you to compute the derivative. (However, you may ignore terms of order $\epsilon^2$ or higher.) Also, $q$ is generally a set of coordinates and comes with an index and $G$ is not a function of one variable only so you really cannot write $G'$.

 

[JD_PM]

No, you cannot drop the order $\epsilon$ terms. They are crucial and completely necessary for you to compute the derivative. (However, you may ignore terms of order $\epsilon^2$ or higher.) Also, $q$ is generally a set of coordinates and comes with an index and $G$ is not a function of one variable only so you really cannot write $G'$.

Ahh so I better write

$$G_{ab} (q^c + \epsilon v^a) = G_{ab} (q^c) + \partial_c G_{ab} (q^c) \epsilon v^a + \partial^2_c G_{ab} (q^c) \frac{ (\epsilon v^a)^2 }{2!} + ...$$

 

[Orodruin]

I would leave out the indices from the argument, they are just confusing you (better not write the argument at all when the argument is just q actually). Also note that the indices in your argument on the LHS do not match.

 

[JD_PM]

I would leave out the indices from the argument, they are just confusing you (better not write the argument at all when the argument is just q actually). Also note that the indices in your argument on the LHS do not match.

OK so we have

$$G_{ab} (q + \epsilon v) = G_{ab} + \partial_c G_{ab} \epsilon v^a + \partial^2_c G_{ab} \frac{ (\epsilon v^a)^2 }{2!} + ...$$

 

[JD_PM]

Let me start from scratch.

I will work out translation.

Translation is given by:

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

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

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

Thus, the vector field generating translations is given by:

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

Thus we have:

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

Here comes (at least for me) the tricky part: showing that translation is a symmetry of the transformation.

I guess we will have to use the provided Killing equation $\partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a = 0$ at some point to do so. But let's go step by step.

Taking the time derivative of translation equation we get:

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

Applying the transformation to the given Lagrangian:

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

And here is where I am stuck. My guess here is that the Killing vector should come up after applying the transformation (i.e. $L \to v^a L + \mathcal O(\lambda^2)$). Then everything would make sense.

Do you agree with my guess? am I on the right track?

 

[Orodruin]

OK so we have

$$G_{ab} (q + \epsilon v) = G_{ab} + \partial_c G_{ab} \epsilon v^a + \partial^2_c G_{ab} \frac{ (\epsilon v^a)^2 }{2!} + ...$$

No, you need to check your indices.

 

[JD_PM]

No, you need to check your indices.

OK so after checking your insight again (I think) my equation doesn't violate neither commandment 3 nor commandment 5.

Commandment 3: You shall not have different free indices on opposite sides of an equality or in different terms of the same expression

I have $b$ and $c$ as free indices on the LHS and in each term on the RHS. So everything should be OK with this one.

Commandment 5: You shall not have more than two of anyone index in an index expression

Only index $a$ appears twice (which is allowed) and $b$ and $c$ only appear once in each term. So everything should be OK with this one.

I suspect I am breaking down commandment 7: When there are two occurrences of an index in your expression, you shall not rename them to something that already exists in your expression. However, I do not see it...

 

[strangerep]

Only index $a$ appears twice (which is allowed)...

...which makes it a dummy summation index. But your lhs has 'a' as a free index.

and $b$ and $c$ only appear once in each term

... so 'c' is a free index on your rhs, but not your lhs.

I'm guessing your $v^a$ should be $v^c$ ?

 

[JD_PM]

...which makes it a dummy summation index. But your lhs has 'a' as a free index.

... so 'c' is a free index on your rhs, but not your lhs.

I'm guessing your $v^a$ should be $v^c$ ?

Thank you strangerep! I see what happened. I was keeping track of the indices on the LHS argument even though I was not showing them.

So is it allowed if we simply ignore indeces in the LHS argument? Thus we end up with

$G_{ab} (q + \epsilon v) = G_{ab} + \partial_c G_{ab} \epsilon v^c + \partial_c^2 G_{ab} \frac{(\epsilon v^c)^2}{2!}$

 

[Orodruin]

Thank you strangerep! I see what happened. I was keeping track of the indices on the LHS argument even though I was not showing them.

So is it allowed if we simply ignore indeces in the LHS argument? Thus we end up with

$G_{ab} (q + \epsilon v) = G_{ab} + \partial_c G_{ab} \epsilon v^c + \partial_c^2 G_{ab} \frac{(\epsilon v^c)^2}{2!}$

You have 4 cs in the second order term ... however, that term is irrelevant for the rest of the problem.

 

[JD_PM]

You have 4 cs in the second order term

Mm true. Should we simply not write the index $c$ in the second order term then?

OK I am afraid I do not see where you are driving me with this series expansion. May you please explain it to me?

Are my guesses at #61 correct?

Thank you for your help so far.

 

[Orodruin]

You have chosen both $\epsilon$ and $\lambda$ as your transformation variable. Otherwise it is progressing in the correct direction. Also note that $d\vec v/ds$ can be rewritten through the chain rule $d\vec v/ds = \dot q^a \partial_a \vec v$.

Now keep only terms up to linear order in $\lambda$ (or $\epsilon$, whichever you choose to keep).

 

[strangerep]

So is it allowed if we simply ignore indices in the LHS argument?

Yes, it's a common notation shortcut to write $f(q)$ to mean that the function $f$ depends on the entire vector $q$, not just one of its components. Sometimes, when dealing with 3-vectors, one makes this more explicit by using bold font for the vector, i.e., $f({\mathbf q})$.

Thus we end up with
$G_{ab} (q + \epsilon v) = G_{ab} + \partial_c G_{ab} \epsilon v^c + \partial_c^2 G_{ab} \frac{(\epsilon v^c)^2}{2!}$

Yes, except that (as Orodruin already mentioned) your 2nd-order derivative term is wrong because it contains 4 $c$'s (though this doesn't matter if only a 1st-order expansion is needed).

But, for future reference, the 2nd-order term should be $\frac{\epsilon^2}{2!} \, v^c v^d \, \partial_c \partial_d G_{ab}$. Review the topic of "multivariate Taylor expansions" to understand why.

 

[JD_PM]

Thank you both. I think I got it.

You have chosen both $\epsilon$ and $\lambda$ as your transformation variable. Otherwise it is progressing in the correct direction.

True, I mixed them up. I will use $\lambda$ from now on.

Let's show that translation is a symmetry of the action

Taking the time derivative of translation equation we get:

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

Applying the transformation to the given Lagrangian (and plugging the Taylor series expansion into it) we get:

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

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

So I finally get:

$$L \to G_{ab} \dot{\vec q^a} \dot{\vec q^b} + \partial_c G_{ab} \lambda v^c \dot{\vec q^a} \dot{\vec q^b} + \mathcal O(\lambda^2)$$

Thus, the derivative of the Lagrangian is zero at $\lambda = 0$ (and therefore everywhere) and it follows that translation is a symmetry of the action.

OK I have shown it! (I am quite happy now!).

The corresponding conserved quantity is given by:

$$Q = \vec v \cdot \vec p = v^a G_{ab} \dot{q^b}$$

Are you satisfied with my answer?

 

[Orodruin]

You have chosen both $\epsilon$ and $\lambda$ as your transformation variable. Otherwise it is progressing in the correct direction.

Thus, the derivative of the Lagrangian is zero at λ=0λ=0\lambda = 0 (and therefore everywhere)

Wait, what? How did you get that from your expression? The derivative is the coefficient of the linear term, which certainly is not zero in your expression. You made a mistake earlier when you did not keep the terms proportional to lambda in the other two terms.

Edit: Also, I am not even sure what $\vec q^a$ is supposed to mean ...

 

[JD_PM]

You have chosen both $\epsilon$ and $\lambda$ as your transformation variable. Otherwise it is progressing in the correct direction.

But I have chosen $\lambda$ in my post #70 (and dropped $\epsilon$).

Wait, what? How did you get that from your expression? The derivative is the coefficient of the linear term, which certainly is not zero in your expression. You made a mistake earlier when you did not keep the terms proportional to lambda in the other two terms.

Edit: Also, I am not even sure what $\vec q^a$ is supposed to mean ...

You are absolutely correct. I forgot to take the derivative.

Mm alright. I missed some terms.

I am sure that the following is correct:

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

Where (we note that $\lambda^2 \dot{\vec v} \cdot \dot{\vec v} = 0$):

$$\Big( \dot{\vec q^a} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)\Big) \Big( \dot{\vec q^b} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)\Big) = \dot{\vec q^a} \dot{\vec q^b} + \dot{\vec q^a} \lambda \dot{\vec v} + \dot{\vec q^b} \lambda \dot{\vec v} + \mathcal O(\lambda^2)$$

So I finally get:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \partial_c G_{ab} \lambda v^c \dot{q^a} \dot{q^b} + G_{ab} \dot{q^a} \lambda \dot{v^b} + G_{ab} \dot{q^b} \lambda \dot{v^a} + \partial_c G_{ab} \lambda v^c \dot{q^a} \lambda \dot{v^b} + \partial_c G_{ab} \lambda v^c \dot{q^b} \lambda \dot{v^a} + \mathcal O(\lambda^2)$$

Which can be simplified to:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \partial_c G_{ab} \lambda v^c \dot{q^a} \dot{q^b} + 2G_{ab} \dot{q^a} \lambda \dot{v^b} + 2\partial_c G_{ab} \lambda^2 v^c \dot{q^a} \dot{v^b} + \mathcal O(\lambda^2)$$

Mm but we have the same problem: when taking the partial derivative wrt $\lambda$ we end up with the non zero terms $\partial_c G_{ab} v^c \dot{q^a} \dot{q^b}$ and $2G_{ab} \dot{q^a} \dot{v^b}$ so I guess I am still missing something...

Please let me know if there is something I did not clarify well enough above.

Thank you for your help.

 

[JD_PM]

Mm but we have the same problem: when taking the partial derivative wrt $\lambda$ we end up with the non zero terms $\partial_c G_{ab} v^c \dot{q^a} \dot{q^b}$ and $2G_{ab} \dot{q^a} \dot{v^b}$ so I guess I am still missing something...

Mm I suspect $2G_{ab} \dot{q^a} \dot{v^b}$ term may be zero because the derivative of the Killing vector field may be zero. But honestly I do not know why that would be the case.

 

[Orodruin]

I am sure that the following is correct:

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

Where (we note that $\lambda^2 \dot{\vec v} \cdot \dot{\vec v} = 0$):

There should be no vector arrows here and your $v$s should have an index on them. The $\lambda^2$ term is not zero, it is just $\mathcal O(\lambda^2)$ ...

$$\Big( \dot{\vec q^a} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)\Big) \Big( \dot{\vec q^b} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)\Big) = \dot{\vec q^a} \dot{\vec q^b} + \dot{\vec q^a} \lambda \dot{\vec v} + \dot{\vec q^b} \lambda \dot{\vec v} + \mathcal O(\lambda^2)$$

You have kept a term that is second order in lambda. You have also baked two terms together that are not equal (the linear lambda terms where you inserted an arbitrary 2 instead of keeping the terms separate, they are not generally the same because q and v are different).

Once you have corrected that you need to use what I have mentioned already in #68,

Also note that $d\vec v/ds$ can be rewritten through the chain rule $d\vec v/ds = \dot q^a \partial_a \vec v$.

 

[JD_PM]

Eureka!

You have kept a term that is second order in lambda. You have also baked two terms together that are not equal (the linear lambda terms where you inserted an arbitrary 2 instead of keeping the terms separate, they are not generally the same because q and v are different).

Once you have corrected that you need to use what I have mentioned already in #68,

Thanks.

So without keeping second order terms in lambda we get:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \partial_c G_{ab} \lambda v^c \dot{q^a} \dot{q^b} + G_{ab} \dot{q^a} \lambda \dot{v^b} + G_{ab} \dot{q^b} \lambda \dot{v^a} + \mathcal O(\lambda^2)$$

Knowing that $\dot v^b = \partial_c v^b \dot q^c$ and $\dot v^a = \partial_d v^a \dot q^d$ (where $c$ and $d$ are dummy indeces, so we can call them whatever we want), we can write the following:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \partial_c G_{ab} \lambda v^c \dot{q^a} \dot{q^b} + G_{ab} \dot{q^a} \lambda \partial_c v^b \dot q^c + G_{ab} \dot{q^b} \lambda \partial_d v^a \dot q^d + \mathcal O(\lambda^2)$$

Let's go step by step here. Let me swap $c$ by $b$ in the term $G_{ab} \dot{q^a} \lambda \partial_c v^b \dot q^c$ and swap $a$ by $c$ and swap $d$ by $a$ in the term $G_{ab} \dot{q^b} \lambda \partial_d v^a \dot q^d$ (note I am allowed to do that because these are dummy indices). Thus we end up with:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \lambda \Big( \partial_c G_{ab} v^c + G_{ac} \partial_b v^c + G_{cb} \partial_a v^c \Big) \dot{q^a} \dot{q^b} + \mathcal O(\lambda^2)$$

Alright! By swapping $c$ by $b$ on $\Big( \partial_c G_{ab} v^c + G_{ac} \partial_b v^c + G_{cb} \partial_a v^c \Big)$ we get $\Big( \partial_a G_{bc} v^a + G_{ba} \partial_c v^a + G_{ca} \partial_b v^a \Big)$. This is the Killing equation we were given in #1, which equals zero (I do not know why it does though).

Thus we end up with:

$$L \to G_{ab} \dot{q^a} \dot{q^b} + \mathcal O(\lambda^2)$$

Thus, the derivative of the Lagrangian is zero at $\lambda = 0$ (and therefore everywhere) and it follows that translation is a symmetry of the action.

How do you see it now Orodruin? :)

 

[Orodruin]

This is the Killing equation we were given in #1, which equals zero (I do not know why it does though).

It does so by the assumption that $v$ is a Killing field. If a field is a Killing field, it generates symmetry transformations, that is the point.

and it follows that translation is a symmetry of the action.

No it does not, it follows that the transformation generated by the Killing field $v$ is a symmetry of the action. This does not need to be a translation (even if you can find coordinates where it is).

 

[JD_PM]

It does so by the assumption that $v$ is a Killing field. If a field is a Killing field, it generates symmetry transformations, that is the point.

Sorry Sr but I've never studied Killing fields. Do you know of any good source I could read to learn? I am also wondering if your lecture notes are available online.

No it does not, it follows that the transformation generated by the Killing field $v$ is a symmetry of the action. This does not need to be a translation (even if you can find coordinates where it is).

Mmm I will think about it. I do not completely get it right now.

However, do you consider that the original question has been solved?

 

[samalkhaiat]

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.

Consider the infinitesimal transformations \delta q^{c}(t) \equiv \bar{q}^{c}(t) - q^{c}(t) = \epsilon v^{c}(q) , \ \ \ |\epsilon | \ll 1 . \ \ \ (1) This leads to \delta \dot{q}^{c} = \epsilon \ \partial_{a}v^{c} \ \dot{q}^{a} , \ \ \ \ \ \partial_{a} \equiv \frac{\partial}{\partial q^{a}} . \ \ \ \ \ \ \ \ \ (2) The induced infinitesimal change in L (q , \dot{q}) is then calculated from \delta L = \frac{\partial L}{\partial q^{c}} \ \delta q^{c} + \frac{\partial L}{\partial \dot{q}^{c}} \ \delta \dot{q}^{c} . Substituting (1), and (2) for L = G_{ab} (q) \dot{q}^{a}\dot{q}^{b}, we find \delta L = \epsilon \left( v^{c}\partial_{c}G_{ab} + G_{cb} \ \partial_{a}v^{c} + G_{ac} \ \partial_{b}v^{c}\right) \dot{q}^{a} \dot{q}^{b} , or \delta L = \epsilon \left( \mathcal{L}_{v}G_{ab}\right) \dot{q}^{a} \dot{q}^{b} . So, if v^{a} is such that the functional form of G_{ab} remains invariant, i.e., \mathcal{L}_{v}G_{ab} = 0, then \delta L = 0 (i.e., (1) is a symmetry transformation). But, for an arbitrary infinitesimal transformation \delta q^{a}, the Lagrangian changes according to \delta L = \left( \frac{\partial L}{\partial q^{a}} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}}\right)\right) \delta q^{a} + \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}} \delta q^{a}\right) . So, if \delta q^{a} is given by our symmetry transformation (1) (i.e., \delta L = 0), then on actual trajectories (solutions of the E-L equation) we have the following conservation law \frac{d}{dt}Q \equiv \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}} \delta q^{a}\right) = 0.

 

[Orodruin]

Sorry Sr but I've never studied Killing fields.

You are studying Killing fields right now. This is the role they have. You were told by the task what you need to know about them.

 

[JD_PM]

Consider the infinitesimal transformations \delta q^{c}(t) \equiv \bar{q}^{c}(t) - q^{c}(t) = \epsilon v^{c}(q) , \ \ \ |\epsilon | \ll 1 . \ \ \ (1) This leads to \delta \dot{q}^{c} = \epsilon \ \partial_{a}v^{c} \ \dot{q}^{a} , \ \ \ \ \ \partial_{a} \equiv \frac{\partial}{\partial q^{a}} . \ \ \ \ \ \ \ \ \ (2) The induced infinitesimal change in L (q , \dot{q}) is then calculated from \delta L = \frac{\partial L}{\partial q^{c}} \ \delta q^{c} + \frac{\partial L}{\partial \dot{q}^{c}} \ \delta \dot{q}^{c} . Substituting (1), and (2) for L = G_{ab} (q) \dot{q}^{a}\dot{q}^{b}, we find \delta L = \epsilon \left( v^{c}\partial_{c}G_{ab} + G_{cb} \ \partial_{a}v^{c} + G_{ac} \ \partial_{b}v^{c}\right) \dot{q}^{a} \dot{q}^{b} , or \delta L = \epsilon \left( \mathcal{L}_{v}G_{ab}\right) \dot{q}^{a} \dot{q}^{b} . So, if v^{a} is such that the functional form of G_{ab} remains invariant, i.e., \mathcal{L}_{v}G_{ab} = 0, then \delta L = 0 (i.e., (1) is a symmetry transformation). But, for an arbitrary infinitesimal transformation \delta q^{a}, the Lagrangian changes according to \delta L = \left( \frac{\partial L}{\partial q^{a}} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}}\right)\right) \delta q^{a} + \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}} \delta q^{a}\right) . So, if \delta q^{a} is given by our symmetry transformation (1) (i.e., \delta L = 0), then on actual trajectories (solutions of the E-L equation) we have the following conservation law \frac{d}{dt}Q \equiv \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{a}} \delta q^{a}\right) = 0.

Samalkhaiat thank you for your insight. I still remember your reply on the thread 'Deriving the Equation of Motion out of the Action' from which I learned a lot.

It looks to me like you are using a different approach: calculus of variations. I am afraid I still do not have enough knowledge to appreciate your reply here, as I get lost considering the infinitesimal transformations:

$$\delta q^{c}(t) \equiv \bar{q}^{c}(t) - q^{c}(t) = \epsilon v^{c}(q) , \ \ \ |\epsilon | \ll 1$$

I will read Goldstein and if I do not understand your reply after doing so, I will ask.

 

[JD_PM]

You are studying Killing fields right now. This is the role they have. You were told by the task what you need to know about them.

Yes. But what I mean is that here we are just applying the equation

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

If I am not mistaken Killing fields are also present in GR. That is why I thought that studying your lecture notes (if they happened to be available online of course) would be a great idea.

To conclude: do you consider that the original question has been solved?

 

[Orodruin]

Yes. But what I mean is that here we are just applying the equation

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

If I am not mistaken Killing fields are also present in GR. That is why I thought that studying your lecture notes (if they happened to be available online of course) would be a great idea.

The role of Killing fields in GR is just an application of what you have seen here. That a field is a Killing field means that it satisfies the given equation. It is useful for describing the symmetries of a manifold (including spacetime). The lecture notes are not publicly available at the moment.

To conclude: do you consider that the original question has been solved?

Yes.