Get all possible constants of motion given an explicit Hamiltonian

In summary: If you have different ##\alpha_i## for each coordinate, then you will get some complicated constants of the motion. There must be two constants of motion from Euler-Lagrange because ##V## ultimately depends on only one independent...
  • #1
JD_PM
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Homework Statement
I will not use summation sign: repeated pair of (upper and lower) indices are summed over: [itex]\sum_{a} A_{a} B^{a} \equiv A_{a}B^{a} = A_{c}B^{c}[/itex] (summed over indices are dummy indices so you can rename them as you like).

Consider a Hamiltonian of ##3## degrees of freedom ##q_i, i = 1,...3##. Momentum is given by ##p_i##. Suppose that the Hamiltonian has the following form:



$$H = \alpha^i (p_i)^2 + V(q)$$



The only possible combination of ##q_i## given by the potential function ##V(q)## is ##q_1^2 +q_2^2 +q_3^2##. We can apply the following change of variables* ##X = q_1^2 +q_2^2 +q_3^2## and obtain the potential in function of the new variable: ##V(X)##



If all ##\alpha^i## are the same, then there is extra symmetry and corresponding constants of motion.



a) Find all constants of motion in case all ##\alpha^i## are the same. HINT: Find the Lagrangian first.



EXTRA (which means I added it; we can deal with this after solving a)):



b) Why all ##\alpha^i## need to be the same?



*I learned this technique in PF! :) More: https://www.physicsforums.com/threads/hamiltons-equation-and-euler-lagranges-equation-comparison.982951/
Relevant Equations
$$H = \alpha^i (p_i)^2 + V(q)$$
I do not understand the following sentence (particularly, the concept of extra symmetry): 'If all ##\alpha^i## are the same, then there is extra symmetry and corresponding constants of motion'.

OK so let's find the Lagrangian; we know it has to have the form:

$$L(q, \dot q) = T(q, \dot q) - V(q)$$

The idea is to apply a change of variables to the given ##H(q, p)## to get ##L(q, \dot q)##

This should be straightforward; we pick the change of variables ##p = \dot q (p, q)## and then apply it to the Hamiltonian to get the Lagrangian. But to do so we need an equation for ##\dot q## in function of ##p## and ##q##

The issue is that I do not see how to get such an equation...

Any hint is appreciated.

Thanks.
 
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  • #2
I suppose you know what Hamilton Equations are right?
 
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  • #3
You could always take a guess at the Lagrangian.
 
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  • #4
Gaussian97 said:
I suppose you know what Hamilton Equations are right?

PeroK said:
You could always take a guess at the Lagrangian.

Ahh so using the Hamilton's equation:

$$\dot q_i = \frac{\partial H}{\partial p_i}$$

I indeed get ##p = \dot q (p, q)##

$$\dot q_i = 2 \alpha^i p_i$$

$$p_i = \frac{\dot q_i}{2 \alpha^i}$$

Applying such a change of variables I get the Lagrangian:

$$L(q, \dot q) = \alpha^i \Big( \frac{\dot q_i}{2 \alpha^i} \Big)^2 - V(q) = \frac{\dot q_i^2}{4 \alpha^i} - V(q)$$

OK once we have the Lagrangian we are set to find the constants of motion.

By using the Euler-Lagrange equation:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}$$

$$\frac{\ddot q_i}{2 \alpha^i} = 2(q_1 + q_2 + q_3)$$

But can we get constants of motion out of the Euler-Lagrange equation? I would go for Noether's theorem at this point.
 
  • #5
JD_PM said:
Ahh so using the Hamilton's equation:$$\dot q_i = \frac{\partial H}{\partial p_i}$$I indeed get ##p = \dot q (p, q)##
$$\dot q_i = 2 \alpha^i p_i$$
Are you trying to solve the case where all ##\alpha^i## the same? If so, then drop the index on ##\alpha^i## and just call it ##\alpha##. If not, you need to be more careful about matching up free indices on both sides of an equation. Don't mix up free and dummy indices.
(Btw, I think you can safely keep all your indices downstairs in this problem.)

[...] By using the Euler-Lagrange equation:$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) ~=~ \frac{\partial L}{\partial q^k}$$
$$\frac{\ddot q_i}{2 \alpha^i} = 2(q_1 + q_2 + q_3)$$
I think your RHS is wrong. Denote ##r^2 := q_1^2 + q_2^2 + q_3^2## and ##V = V(r^2)##.
The chain rule can be used to compute ##\partial V/\partial q_k##, but maybe you should switch to spherical polar variables first (if all ##\alpha^i## are the same).

Also, did this exercise come from a textbook or lecture note? If so, please give a precise reference and/or link.
 
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  • #6
strangerep said:
Are you trying to solve the case where all ##\alpha^i## the same? If so, then drop the index on ##\alpha^i## and just call it ##\alpha##.

Yes, I am. That's a fair point, let's drop the ##i## upper index for ##\alpha##

strangerep said:
I think your RHS is wrong. Denote ##r^2 := q_1^2 + q_2^2 + q_3^2## and ##V = V(r^2)##.
The chain rule can be used to compute ##\partial V/\partial q_k##

I should have included more steps. Let me shed some light on what I did to compute the RHS.

I used the change of variables given by the original exercise: ##X := q_1^2 +q_2^2 +q_3^2##. We note that only the potential term depends explicitly on ##q_i##. Thus, in this problem:

$$\frac{\partial L}{\partial q_i} = \frac{\partial V}{\partial q_i}$$

Then, by the chain rule, we know that:

$$\frac{\partial V}{\partial q_i} = \frac{\partial V}{\partial X}\frac{\partial X}{\partial q_i} = \frac{\partial V}{\partial X}\Big( 2(q_1 + q_2 + q_3) \Big)$$

But ##\frac{\partial V}{\partial X} = 1##, as ##V(X) = X##

So I end up with the RHS:

$$2(q_1 + q_2 + q_3)$$

Please feel free to point out any mistake if you see any.

strangerep said:
Also, did this exercise come from a textbook or lecture note? If so, please give a precise reference and/or link.

This is an old exam question. I am working out exercises to practice for a test. The original copy (in Dutch though...):

Screenshot (1005).png
 
  • #7
I am still thinking how to find all constants of motion out of the Lagrangian.

I'll post what I get.
 
  • #8
JD_PM said:
I am still thinking how to find all constants of motion out of the Lagrangian.

I'll post what I get.

What is the difference between this problem and one where the ##q_i## are specifically Cartesian coordinates?

Hint: the same equations have the same solutions!

Note: taking all the ##\alpha_i## to be the same reduces the problem to the usual central force Lagrangian.

If you have different ##\alpha_i## for each coordinate, then you will get some complicated constants of the motion. There must be two constants of motion from Euler-Lagrange because ##V## ultimately depends on only one independent coordinate.
 
  • #9
PeroK said:
What is the difference between this problem and one where the ##q_i## are specifically Cartesian coordinates?

Hint: the same equations have the same solutions!

Ahh so the constants of motion (taking all the ##\alpha_i## to be the same) simply are: ##q_1 + q_2 + q_3## (the scaling factor doesn't matter).

PeroK said:
There must be two constants of motion from Euler-Lagrange because ##V## ultimately depends on only one independent coordinate.

But I do not get this; I get ##3## constants instead of ##2##.
 
  • #10
JD_PM said:
But I do not get this; I get ##3## constants instead of ##2##.
Yes, generally you get two from the E-L equations and one from the Hamiltonian (energy). Or, perhaps by integrating the third E-L equation in ##r##?
 
  • #11
PeroK said:
Yes, generally you get two from the E-L equations and one from the Hamiltonian (energy). Or, perhaps by integrating the third E-L equation in ##r##?

Mmm I am not sure, I will read more about it.

However I think the exercise itself is solved; we found the constants of motion ##q_1 + q_2 + q_3##
 
  • #12
JD_PM said:
Mmm I am not sure, I will read more about it.

However I think the exercise itself is solved; we found the constants of motion ##q_1 + q_2 + q_3##

That can't be right! That's no motion at all!

My hint was that you can define new coordinates, ##(r, \theta, \phi)##:

##r = (q_1^2 + q_2^2 + q_3^2)^{1/2}##
##q_1 = r \sin \theta \cos \phi## etc.

The constants of motion are then expressed in these coordinates (which you could transform back to the ##q_i## if you wanted).
 
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  • #13
PeroK said:
That can't be right! That's no motion at all!

My hint was that you can define new coordinates, ##(r, \theta, \phi)##:

##r = (q_1^2 + q_2^2 + q_3^2)^{1/2}##
##x = r \sin \theta \cos \phi## etc.

The constants of motion are then expressed in these coordinates (which you could transform back to the ##q_i## if you wanted).

So you suggest we use spherical coordinates. We have:

$$r = (q_1^2 + q_2^2 + q_3^2)^{1/2}$$

$$x = r \sin \theta \cos \phi \ \ \ \ (1)$$

$$y = r \sin \theta \sin \phi \ \ \ \ (2)$$

$$z = r cos \theta \ \ \ \ (3)$$

Squaring both sides and adding up equations ##(1), (2), (3)## we get:

$$r = (x^2 + y^2 + z^2)^{1/2}$$

Mmm but I do not see why we are switching to these coordinates (by the way, strangerep also suggested it).
 
  • #14
JD_PM said:
Mmm but I do not see why we are switching to these coordinates (by the way, strangerep also suggested it).

What else is there? Note that it should be ##q_1 = r\sin \theta \cos \phi##. I've edited that above.

The point is that if you find a system with the same symmetries - this system is analagous to the usual central force potential - then you can use the same mathematical techniques to analyse it.

The new coordinates here are analogies of the spherical ##r, \theta, \phi##. With ##\theta, \phi## some sort of generalised angles. Depending on the system, they may be physical angles or things that play the same role as physical angles.

In this case, you can get two conserved quantities that are analagous to angular momentum and a conserved quantity that is analagous to total mechanical energy.

If the equations look the same, then you can use the same mathematical techiques to solve them. This is the at heart of generalised mechanics, like the Euler-Lagrange equations.
 
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  • #15
OK so we have the (given) Hamiltonian:

$$H(q, p_i) = \alpha (p_i)^2 + V(q)$$

And the Lagrangian

$$L(q, \dot q_i) = \frac{\dot q_i^2}{4 \alpha} - V(q)$$

So the idea now is to define a new variable ##r^2 := q_1^2 + q_2^2 + q_3^2##. Then, the Hamiltonian and Lagrangian become:

$$H(r^2, p_i) = \alpha (p_i)^2 + V(r^2)$$

$$L(r^2, \dot q_i) = \frac{\dot q_i^2}{4 \alpha} - V(r^2)$$

Let's get the two constants of motion out of the Lagrangian:

The Euler-Lagrange equation is:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_i }\Big) = \frac{\partial L}{\partial q_i}$$

The LHS of the Euler-Lagrange equation is:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_i }\Big) = \frac{\ddot q_i}{2 \alpha}$$

The RHS of the Euler-Lagrange equation is:

$$\frac{\partial V}{\partial q} = \frac{\partial V}{\partial r}\frac{\partial r}{\partial q} = 2r \Big( \frac{q_1 + q_2 + q_3}{\sqrt{q_1^2 + q_2^2 + q_3^2}}\Big)$$

Do you agree at this point?
 
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  • #16
Not really. The idea is to change coordinates to (generalised) spherical coordinates. Whatever you do it should make sense in the special case where ##q_1 = x, q_2 = y, q_3 = z##. With the appropriate ##\alpha## that is an example of your Lagrangian.

Alternatively, if you know about Noether's theorem, you could look at invariance under infinitesimal rotations of ##q_1, q_2, q_3##.

Or, by simply recognising the symmetry here you could write down some constants of motion, such as ##q_1\dot{q_2} - q_2\dot{q_1}##! Perhaps that's what you were supposed to do?
 
  • #17
PeroK said:
Not really. The idea is to change coordinates to (generalised) spherical coordinates. Whatever you do it should make sense in the special case where ##q_1 = x, q_2 = y, q_3 = z##. With the appropriate ##\alpha## that is an example of your Lagrangian.

Thanks, I will think about it and post an attempt.

PeroK said:
Alternatively, if you know about Noether's theorem, you could look at invariance under infinitesimal rotations of ##q_1, q_2, q_3##.

Or, by simply recognising the symmetry here you could write down some constants of motion, such as ##q_1\dot{q_2} - q_2\dot{q_1}##! Perhaps that's what you were supposed to do?

I think that I am supposed to change coordinates to (generalised) spherical coordinates, so I will use that method.
 
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  • #18
The Lagrangian is:

$$L(q, \dot q) = \frac{1}{4 \alpha} \Big( \dot q_1^2 + \dot q_2^2 + \dot q_3^2\Big) - V(q)$$

Let's change to spherical coordinates. We know:

$$r = (q_1^2 + q_2^2 + q_3^2)^{1/2}$$

$$q_1 = r \sin \theta \cos \phi \ \ \ \ (1)$$

$$q_2 = r \sin \theta \sin \phi \ \ \ \ (2)$$

$$q_3 = r cos \theta \ \ \ \ (3)$$

The derivatives wrt time of ##q_1, q_2## and ##q_3## are:

$$\dot q_1 = r (\cos \theta \cos \phi - \sin \theta \sin \phi)$$

$$\dot q_2 = r (\cos \theta \cos \phi + \sin \theta \cos \phi)$$

$$\dot q_3 = -r \sin \theta$$

Then, we just have to square and plug the time-derivatives into the Lagrangian to get (I skip typing the calculation for the moment; the result is weird, so I'll type the calculation if there is a point we see it is wrong):

$$\dot q_1^2 + \dot q_2^2 + \dot q_3^2 = r^2(2 sin^2 \theta + cos^2 \theta)$$

OK so the Lagrangian in spherical coordinates is:

$$L(r, \theta, \phi) = \frac{1}{4 \alpha} \Big( 2 sin^2 \theta + cos^2 \theta\Big) - V(r^2)$$

This is what you meant, right?
 
  • #19
What I would expect is:

$$L(r, \theta, \phi) = \frac{1}{4 \alpha} \Big( (r \dot \theta \sin \phi)^2 + \dot r^2 + (r \dot \phi)^2 \Big) - V(r^2)$$

But we could assume the above Lagrangian and move on (it is just calculation)

But what to do when we have the Lagrangian in spherical coordinates?
 
  • #20
JD_PM said:
What I would expect is:

$$L(r, \theta, \phi) = \frac{1}{4 \alpha} \Big( (r \dot \theta \sin \phi)^2 + \dot r^2 + (r \dot \phi)^2 \Big) - V(r^2)$$

But we could assume the above Lagrangian and move on (it is just calculation)

But what to do when we have the Lagrangian in spherical coordinates?
Your Lagrangian above should have terms in ##\dot r^2##.

You use Euler-Lagrange. As ##L## is independent of ##\theta, \phi##, you get two constants of motion, plus an equationn of motion from the ##r## equation.
 
  • #21
JD_PM said:
I used the change of variables given by the original exercise: ##X := q_1^2 +q_2^2 +q_3^2##. We note that only the potential term depends explicitly on ##q_i##. Thus, in this problem:
$$\frac{\partial L}{\partial q_i} = \frac{\partial V}{\partial q_i}$$
OK so far (though I would have chosen a variable named ##r^2## instead of ##X##).

Then, by the chain rule, we know that:
$$\frac{\partial V}{\partial q_i} = \frac{\partial V}{\partial X}\frac{\partial X}{\partial q_i} = \frac{\partial V}{\partial X}\Big( 2(q_1 + q_2 + q_3) \Big)$$
No, that's totally wrong.

Let's rename variables temporarily, and use ##x,y,z## instead of ##q_1, q_2, q_3##. So ##r^2 = x^2 + y^2 + z^2##.
Now, suppose ww have some (arbitrary) function ##f## which is a function only of the radius ##r##, i.e., ##f = f(r)##.
What is $$ \frac{\partial f}{\partial x} ~=~ ? $$(Hint: first work out ##\partial r^2/\partial x##.)

But ##\frac{\partial V}{\partial X} = 1##, as ##V(X) = X##
Totally wrong again. Where does it say that? ##V## is an arbitrary function of ##X##.

[...] I think the exercise itself is solved; we found the constants of motion q1+q2+q3
No -- totally wrong. You are nowhere near solving this exercise and would be marked 0 if you submitted the above in an exam.

(I've got to go now, but I'll try to post some extra stuff later today.)
 
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  • #22
JD_PM said:
The derivatives wrt time of ##q_1, q_2## and ##q_3## are:
$$\dot q_1 = r (\cos \theta \cos \phi - \sin \theta \sin \phi)$$
$$\dot q_2 = r (\cos \theta \cos \phi + \sin \theta \cos \phi)$$
$$\dot q_3 = -r \sin \theta$$
All these time derivatives are wrong. Where are the ##\dot r##, ##\dot\theta##, and ##\dot\phi## ?
 
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  • #23
@JD_PM I was assuming that you would have previously encountered one or both of these:

a) A Newtonian analysis of the general central force problem, with arbitrary potential ##V(r)##.

b) The Euler-Lagrange analysis of the general central force problem.

Although in principle this more general problem with generalised coordinates is not any different - and not technically any harder - I thought the idea of the question was to get you to apply what you already knew in a more specific context to a more general scenario.

It seems odd that you may not have encountered the core ideas before.
 
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  • #24
PeroK said:
It seems odd that [JD_PM] may not have encountered the core ideas before.
It seems odd to me too. It's unrealistic to try and teach him a large slab of central force theory here on PF.
 
  • #25
strangerep said:
No, that's totally wrong.

Let's rename variables temporarily, and use ##x,y,z## instead of ##q_1, q_2, q_3##. So ##r^2 = x^2 + y^2 + z^2##.
Now, suppose ww have some (arbitrary) function ##f## which is a function only of the radius ##r##, i.e., ##f = f(r)##.
What is $$ \frac{\partial f}{\partial x} ~=~ ? $$(Hint: first work out ##\partial r^2/\partial x##.)

OK by the chain rule we know that

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x}$$

Let's first work out ##\partial r^2/\partial x## as you suggested:

$$\partial r^2/\partial x = 2x$$

Now let's work out ##\partial r/\partial x##

$$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{ x^2 + y^2 + z^2 }}$$

So we get:

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \Big( \frac{x}{\sqrt{ x^2 + y^2 + z^2 }} \Big)$$

But what's ##\partial f/\partial r##? As you said, ##f## is an arbitrary function of ##r##; we have no explicit formula to compute such a derivative.

Besides, I did not use the computation of your hint, so I guess I am missing something.

strangerep said:
Totally wrong again. Where does it say that? ##V## is an arbitrary function of ##X##.

Right, I did not recall V is an arbitrary function of ##X##. My bad, thanks.

strangerep said:
You are nowhere near solving this exercise and would be marked 0 if you submitted the above in an exam.

Thanks for your honesty.

That's only one of the reasons why I posted the question in PF. But the main reason why I am posting such a question (and others) is that I feel I really really learn discussing problems here. I appreciate the patience of all users who have helped me out so far.
 
  • #26
strangerep said:
All these time derivatives are wrong. Where are the ##\dot r##, ##\dot\theta##, and ##\dot\phi## ?

You are right, my bad. Let's start over. We have:

$$r = (q_1^2 + q_2^2 + q_3^2)^{1/2}$$

$$q_1 = r \cos \theta \sin \phi$$

$$q_2 = r \sin \theta \sin \phi$$

$$q_3 = r \cos \phi$$

The derivatives wrt time of ##q_1, q_2## and ##q_3## are:

$$\dot q_1 = (\cos \theta \sin \phi)\dot r + (r \cos \theta \cos \phi) \dot \phi - (r \sin \theta \sin \phi)\dot \theta$$

$$\dot q_2 = (\sin \theta \sin \phi) \dot r + (r \sin \theta \cos \phi) \dot \phi + (r \cos \theta \sin \theta)$$

$$\dot q_3 = \cos \phi \dot r - r \sin \phi \dot \phi$$

These indeed lead to the Lagrangian:

$$L(r, \theta, \phi) = \frac{1}{4 \alpha} \Big( (r \dot \theta \sin \phi)^2 + \dot r^2 + (r \dot \phi)^2 \Big) - V(r^2)$$
 
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  • #27
PeroK said:
@JD_PM I was assuming that you would have previously encountered one or both of these:

a) A Newtonian analysis of the general central force problem, with arbitrary potential ##V(r)##.

b) The Euler-Lagrange analysis of the general central force problem.

Although in principle this more general problem with generalised coordinates is not any different - and not technically any harder - I thought the idea of the question was to get you to apply what you already knew in a more specific context to a more general scenario.

It seems odd that you may not have encountered the core ideas before.

I have encountered a). Not b) though. However I am really interested in this problem. My approach to it won't be the neatest but I really feel I am going to solve it eventually (I am kind of relentless).

Of course, if there is a point it is evident I need to read more, I will acknowledge it and stop.

Thank you for your patience.
 
  • #28
JD_PM said:
$$L(r, \theta, \phi) = \frac{1}{4 \alpha} \Big( (r \dot \theta \sin \phi)^2 + \dot r^2 + (r \dot \phi)^2 \Big) - V(r^2)$$

You can get a constant of motion using the Lagrangian's independence from ##\theta##.

After that you might need a bit of inspiration!
 
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  • #29
PeroK said:
You can get a constant of motion using the Lagrangian's independence from ##\theta##.

After that you might need a bit of inspiration!

Alright, I think I got this one!

Since ##\frac{\partial L}{\partial \theta} = 0##, the coordinate ##\theta## is cyclic. There's a corollary in my book that states the following:

If ##\theta## (in this case) is a cyclic coordinate (in the sense that it does not appear in the Lagrangian), then ##p_{\theta}##, the generalized momentum conjugate to ##p_{\theta}##, is constant in any motion.

Thus, the momentum ##p_{\theta}## is conserved, where:

$$p_{\theta} = \frac{\partial L}{\partial \dot \theta} = \frac{1}{2 \alpha} (r \sin \phi)^2 \dot \theta$$

OK we've got a constant of motion! But there are more...

@PeroK back in #20 you said:

PeroK said:
You use Euler-Lagrange. As ##L## is independent of ##\theta, \phi##, you get two constants of motion, plus an equationn of motion from the ##r## equation.

But I only see L to be independent of ##\theta##. Thus we would expect to get only one constant of motion (instead of two as you suggested in #20)

Do you agree? :smile:
 
  • #30
JD_PM said:
Alright, I think I got this one!

Since ##\frac{\partial L}{\partial \theta} = 0##, the coordinate ##\theta## is cyclic. There's a corollary in my book that states the following:

If ##\theta## (in this case) is a cyclic coordinate (in the sense that it does not appear in the Lagrangian), then ##p_{\theta}##, the generalized momentum conjugate to ##p_{\theta}##, is constant in any motion.

Thus, the momentum ##p_{\theta}## is conserved, where:

$$p_{\theta} = \frac{\partial L}{\partial \dot \theta} = \frac{1}{2 \alpha} (r \sin \phi)^2 \dot \theta$$

OK we've got a constant of motion! But there are more...

@PeroK back in #20 you said:
But I only see L to be independent of ##\theta##. Thus we would expect to get only one constant of motion (instead of two as you suggested in #20)

Do you agree? :smile:

Yes, it's harder to get the other one out. You get equations of motion for ##r## and ##\phi##.

One trick, which relies on some physical reasoning, is that you must get motion in a plane. Therefore, you can change your initial coordinates to new (primed) coordinates so that ##q'_3 = 0##, which is equivalent to ##\phi' = \frac{\pi}{2}##.

If you don't do that, then spherical coordinates don't simplify things as much as you might hope.

The alternative is to take the constant you have found and transform it back to your original ##q## coordinates. Hint: it helps if you know the answer in advance!
 
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  • #31
To recap: I still have to find a constant of motion related to conservation of angular momentum and a constant of motion related to conservation of energy. I guess the former can be obtained out of the Lagrangian and the later can be obtained out of the Hamiltonian.

Let's focus on the Lagrangian first.

OK so I see you propose two methods to get the other constant of motion related to the Lagrangian:

1) Method 1 (which I do not see right now):

PeroK said:
One trick, which relies on some physical reasoning, is that you must get motion in a plane. Therefore, you can change your initial coordinates to new (primed) coordinates so that ##q'_3 = 0##, which is equivalent to ##\phi' = \frac{\pi}{2}##.

2) Method 2:

PeroK said:
take the constant you have found and transform it back to your original ##q## coordinates. Hint: it helps if you know the answer in advance!

Alright, but I have an issue:

$$p = \frac{1}{2 \alpha} (q_1^2 + q_2^2 + q_3^2) \sin^2 \phi \dot \theta$$

How can I transform ##\sin^2 \phi \dot \theta## to ##q## coordinates?

I was trying to use:

$$\dot q_1 = (\cos \theta \sin \phi)\dot r + (r \cos \theta \cos \phi) \dot \phi - (r \sin \theta \sin \phi)\dot \theta$$

But things get messy...

It is OK for me if you prefer to use method 1 instead.
 
  • #32
JD_PM said:
To recap: I still have to find a constant of motion related to conservation of angular momentum and a constant of motion related to conservation of energy. I guess the former can be obtained out of the Lagrangian and the later can be obtained out of the Hamiltonian.

Let's focus on the Lagrangian first.

OK so I see you propose two methods to get the other constant of motion related to the Lagrangian:

1) Method 1 (which I do not see right now):
2) Method 2:
Alright, but I have an issue:

$$p = \frac{1}{2 \alpha} (q_1^2 + q_2^2 + q_3^2) \sin^2 \phi \dot \theta$$

How can I transform ##\sin^2 \phi \dot \theta## to ##q## coordinates?

I was trying to use:

$$\dot q_1 = (\cos \theta \sin \phi)\dot r + (r \cos \theta \cos \phi) \dot \phi - (r \sin \theta \sin \phi)\dot \theta$$

But things get messy...

It is OK for me if you prefer to use method 1 instead.

The hint I gave was that it's better if you know what you are aiming for. In this case, we are dealing with generalised angular momentum. Euler-Lagrange has produced a constant of motion, which we can call:
$$l_3 = l_z = r^2 \dot \theta \sin^2 \phi$$
Which is angular momentum per unit mass. The ##\alpha## in this problem is playing the role of mass.

The first approach is to take a guess that this must equal $$q_1 \dot q_2 - q_2 \dot q_1$$
Which is the appropriate component of angular momentum in the original ("Cartesian") coordinates.

One you have that, you could deduce the others by symmetry. Or, deduce this by considering different spherical coordinates using ##q_1## or ##q_2## instead of ##q_3## as the "z-axis".

The second approach is to guess what the other components of angular momentum are and use Euler-Lagrange equations to show they are constants of motion. This probably gets messy.

However, you should be able to show that$$l^2 = r^4(\dot \phi^2 + \dot \theta^2 \sin^2 \phi)$$ is constant. And this quantity should show up in your spherical Hamiltonian.

Note that this is all mathematically identical to the case where the ##q## coordinates are ##x, y, z##. There's nothing significantly different here.
 
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  • #33
PeroK said:
The second approach is to guess what the other components of angular momentum are and use Euler-Lagrange equations to show they are constants of motion. This probably gets messy.

However, you should be able to show that$$l^2 = r^4(\dot \phi^2 + \dot \theta^2 \sin^2 \phi)$$ is constant. And this quantity should show up in your spherical Hamiltonian.

Note that this is all mathematically identical to the case where the ##q## coordinates are ##x, y, z##. There's nothing significantly different here.

Alright, let's try that approach.

The other two components of angular momentum (which are not cyclic):

$$p_{\phi} = l_y = \frac{\partial L}{\partial \dot \phi} = \frac{1}{2 \alpha} (r^2 \dot \phi)$$

$$p_{r} = l_x = \frac{\partial L}{\partial \dot r} = \frac{1}{2 \alpha} (\dot r)$$

The total generalized momentum (per unit mass) is:

$$p = p_{r} + p_{\phi} + p_{\theta} = \frac{1}{2 \alpha}\Big( (r \sin \phi)^2 \dot \theta + r^2 \dot \phi + \dot r\Big)$$

But why is ##\dot r## missing from your equation

$$l^2 = r^4(\dot \phi^2 + \dot \theta^2 \sin^2 \phi)$$

?
 
  • #34
JD_PM said:
Alright, let's try that approach.

The other two components of angular momentum (which are not cyclic):

$$p_{\phi} = l_y = \frac{\partial L}{\partial \dot \phi} = \frac{1}{2 \alpha} (r^2 \dot \phi)$$

$$p_{r} = l_x = \frac{\partial L}{\partial \dot r} = \frac{1}{2 \alpha} (\dot r)$$

The total generalized momentum (per unit mass) is:

$$p = p_{r} + p_{\phi} + p_{\theta} = \frac{1}{2 \alpha}\Big( (r \sin \phi)^2 \dot \theta + r^2 \dot \phi + \dot r\Big)$$

But why is ##\dot r## missing from your equation

$$l^2 = r^4(\dot \phi^2 + \dot \theta^2 \sin^2 \phi)$$

?
The component of motion in the radial direction makes no contribution to angular momentum. I took the components of momentum in the angular directions. Note also that angular momentum is a vector, so you can't sum its components. You must sum the square of its components - to get the total angular momentum squared.
 
  • #35
PeroK said:
Note also that angular momentum is a vector, so you can't sum its components.

My bad, what I meant was:

$$\vec p = \vec p_{r} + \vec p_{\phi} + \vec p_{\theta} = \frac{1}{2 \alpha}\Big( \dot r \hat r + r^2 \dot \phi \hat \phi + (r \sin \phi)^2 \dot \theta \hat \theta \Big)$$

I now have an idea on how to get the constant of motion related to angular momentum, thanks.

To get the constant of motion related to conservation of energy I just have to use the Hamiltinian. But could you give me a hint on how should I use it?
 
<h2>1. What is a constant of motion in Hamiltonian mechanics?</h2><p>A constant of motion, also known as a conserved quantity, is a physical quantity that remains constant over time in a system described by a Hamiltonian. In other words, it is a quantity that does not change as the system evolves.</p><h2>2. How do you find the constants of motion for a given Hamiltonian?</h2><p>To find the constants of motion for a given Hamiltonian, you can use the Hamiltonian equations of motion. These equations describe the time evolution of the system and can be used to determine which quantities remain constant.</p><h2>3. Why are constants of motion important in Hamiltonian mechanics?</h2><p>Constants of motion are important in Hamiltonian mechanics because they provide insights into the behavior and properties of a system. They can help us understand the symmetries and conservation laws of a system and can simplify the equations of motion.</p><h2>4. Can all Hamiltonians have constants of motion?</h2><p>No, not all Hamiltonians have constants of motion. In order for a Hamiltonian to have constants of motion, it must have certain symmetries or invariances. If a Hamiltonian has no symmetries, then it will not have any constants of motion.</p><h2>5. How do constants of motion relate to the principle of least action?</h2><p>Constants of motion are related to the principle of least action through Noether's theorem. This theorem states that for every continuous symmetry of a physical system, there exists a corresponding conserved quantity. Therefore, constants of motion can be derived from the symmetries of a system, which are related to the principle of least action.</p>

1. What is a constant of motion in Hamiltonian mechanics?

A constant of motion, also known as a conserved quantity, is a physical quantity that remains constant over time in a system described by a Hamiltonian. In other words, it is a quantity that does not change as the system evolves.

2. How do you find the constants of motion for a given Hamiltonian?

To find the constants of motion for a given Hamiltonian, you can use the Hamiltonian equations of motion. These equations describe the time evolution of the system and can be used to determine which quantities remain constant.

3. Why are constants of motion important in Hamiltonian mechanics?

Constants of motion are important in Hamiltonian mechanics because they provide insights into the behavior and properties of a system. They can help us understand the symmetries and conservation laws of a system and can simplify the equations of motion.

4. Can all Hamiltonians have constants of motion?

No, not all Hamiltonians have constants of motion. In order for a Hamiltonian to have constants of motion, it must have certain symmetries or invariances. If a Hamiltonian has no symmetries, then it will not have any constants of motion.

5. How do constants of motion relate to the principle of least action?

Constants of motion are related to the principle of least action through Noether's theorem. This theorem states that for every continuous symmetry of a physical system, there exists a corresponding conserved quantity. Therefore, constants of motion can be derived from the symmetries of a system, which are related to the principle of least action.

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