Using Noether's theorem to get a constant of motion

[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.

 

[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?