Using Noether's theorem to get a constant of motion

[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.