Using Noether's theorem to get a constant of motion

In summary: I don't know how to call it... "notation"?In summary, the conversation discusses Noether's theorem and its application to finding a constant of motion, represented by the equation ##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 approach to solving this problem is described as calculating four terms and simplifying to get the final result. The use of summation
  • #1
JD_PM
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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).
 
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  • #2
JD_PM said:
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.
 
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  • #3
Orodruin said:
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).

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

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

$$Q_v = \sum_a 2 G_{aa} \dot q^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + \sum_b 2 G_{ab} \dot q^a \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4')$$
 
  • #6
scarily called killing vector

Note that capitalization is important here, it should be Killing vector after German mathematician Wilhelm Killing. It has nothing to do with murderous vectors. :oldeyes:
 
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  • #7
I tried to be funny and failed XD
 
  • #8
JD_PM said:
Because the conserved quantity is a linear combination of the momenta of the system with coefficients depending on ##q##. As we have ##a## and ##b## I thought we should have ##2##.

But I will check it.
Note that both terms are the same. It does not matter what you call the summation index.
 
  • #9
Orodruin said:
Note that both terms are the same. It does not matter what you call the summation index.

But we expect two terms, don't we?
 
  • #10
JD_PM said:
But we expect two terms, don't we?
Why? And why would you call it two terms if they are exactly the same and not just put a 2 in front?
 
  • #11
JD_PM said:
I tried to be funny and failed XD
This is from my GR lecture notes:
83006CEF-20BC-4260-9732-5C061518A895.png
 
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  • #12
As for Noether’s theorem, it states that if ##q^a \to q^a + \epsilon k^a## is a symmetry of the Lagrangian, then ##p_a k^a## is a constant of motion. There is only one term here but note that, due to the summation convention, it is a linear combination of the canonical momenta. Also note the index placements.
 
  • #13
Orodruin said:
Edit: I strongly suggest rewriting everything using the summation convention and keeping these rules in mind:
https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/
OK After having a look at the rules, I am going to type all equations in both notations:

Using summation notation:

The given action:

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

The given Killing vector equation:

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

The given conservative quantity we have to prove:

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

The constant of motion given by Noether's theorem:

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

The computation of momentum (generally):

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

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

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

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

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

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

Thus:

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

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

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

This means I am doing something wrong.
 
  • #14
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):

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

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

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

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

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

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

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

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

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

##(5')## follows from ##(3')## of course.
 
  • #16
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.
 
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  • #17
Orodruin said:
Also, drop the factor of 2, it is just confusing.
In fact, inserting a factor 1/2 instead could be beneficial here. Of course, if ##Q## is conserved then ##Q/2## is conserved (or indeed any constant multiplied by ##Q##).
 
  • #18
Orodruin said:
In (3) you are assuming what you need to prove. Also, drop the factor of 2, it is just confusing.

OK

Orodruin said:
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.

Orodruin said:
Also note that your equation (5) has more than two a indices and refer back to the Insight.

Mmm you are right.

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

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

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

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

I will think about it and post what I get.
 
  • #19
JD_PM said:
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.
 
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  • #20
Ready? Set? Go! (yeah, I trying to be funny again)...

Orodruin said:
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.

Orodruin said:
##q^a \to q^a + \epsilon k^a##

Just to make sure I do not mess up with notation; what you wrote above is equivalent to the following notation:

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

Where:

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

Isn't it?
 
  • #21
Orodruin said:
As for Noether’s theorem, it states that if ##q^a \to q^a + \epsilon k^a## is a symmetry of the Lagrangian, then ##p_a k^a## is a constant of motion.

Once at this point what we have to do next is to specify the type of symmetry we are dealing with.

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

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

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

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

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

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

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

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

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

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

So we end up with:

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

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

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

So the conserved quantity is:

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

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

OK so the conserved quantity is:

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

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

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

Playing a bit with the dummy indices we get:

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

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

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

You again should check all your equations against the basic rules in the Insight. As soon as you find an equation that does not comply, something has gone wrong and continuing from there is useless until you put it right.
 
  • #23
Orodruin said:
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?

Orodruin said:
You cannot pick k=3

I think there is a misunderstanding. I meant ##n=3##.

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

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

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

And I worked out:

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

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

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

Why cannot use the same idea here then?
 
  • #24
JD_PM said:
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.

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

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

OK so it is going to be a Killing vector field.

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

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

But this is what I have done above...

This is coming from here:

Captura de pantalla (868).png

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

Orodruin said:
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

Orodruin said:
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).
 
  • #26
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?
 
  • #27
Orodruin said:
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.
 
  • #28
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.
 
  • #29
Orodruin said:
you need to choose a symmetry transformation.

I picked translation in #21

JD_PM said:
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.
 
  • #30
JD_PM said:
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.
 
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  • #31
Please let me use Einstein's notation from now on.

Orodruin said:
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).

Orodruin said:
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.

Orodruin said:
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.
 
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  • #32
JD_PM said:
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.
 
  • #33
Orodruin said:
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
 
  • #34
JD_PM said:
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?
 
  • #35
JD_PM said:
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.
 
<h2>1. What is Noether's theorem?</h2><p>Noether's theorem is a fundamental principle in physics that relates the symmetry of a physical system to a conserved quantity, also known as a constant of motion.</p><h2>2. How does Noether's theorem work?</h2><p>Noether's theorem states that for every continuous symmetry of a physical system, there exists a corresponding conserved quantity. This means that if a physical system remains unchanged under certain transformations, then there is a quantity that remains constant throughout the system's evolution.</p><h2>3. Can Noether's theorem be applied to all physical systems?</h2><p>Yes, Noether's theorem can be applied to all physical systems, as long as the system has a continuous symmetry. This includes classical mechanics, quantum mechanics, and field theories.</p><h2>4. What is the significance of using Noether's theorem to get a constant of motion?</h2><p>Using Noether's theorem to obtain a constant of motion allows us to simplify the equations of motion for a physical system. It also provides a deeper understanding of the symmetries that govern the behavior of the system.</p><h2>5. Are there any limitations to using Noether's theorem to get a constant of motion?</h2><p>While Noether's theorem is a powerful tool, it does have some limitations. It only applies to systems with continuous symmetries, and it does not provide a direct method for finding the conserved quantity. Additionally, Noether's theorem does not apply to systems that are not time-reversible.</p>

1. What is Noether's theorem?

Noether's theorem is a fundamental principle in physics that relates the symmetry of a physical system to a conserved quantity, also known as a constant of motion.

2. How does Noether's theorem work?

Noether's theorem states that for every continuous symmetry of a physical system, there exists a corresponding conserved quantity. This means that if a physical system remains unchanged under certain transformations, then there is a quantity that remains constant throughout the system's evolution.

3. Can Noether's theorem be applied to all physical systems?

Yes, Noether's theorem can be applied to all physical systems, as long as the system has a continuous symmetry. This includes classical mechanics, quantum mechanics, and field theories.

4. What is the significance of using Noether's theorem to get a constant of motion?

Using Noether's theorem to obtain a constant of motion allows us to simplify the equations of motion for a physical system. It also provides a deeper understanding of the symmetries that govern the behavior of the system.

5. Are there any limitations to using Noether's theorem to get a constant of motion?

While Noether's theorem is a powerful tool, it does have some limitations. It only applies to systems with continuous symmetries, and it does not provide a direct method for finding the conserved quantity. Additionally, Noether's theorem does not apply to systems that are not time-reversible.

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