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Get all possible constants of motion given an explicit Hamiltonian

  • Thread starter 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.
 

Answers and Replies

  • #2
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I suppose you know what Hamilton Equations are right?
 
  • #3
PeroK
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You could always take a guess at the Lagrangian.
 
  • #4
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I suppose you know what Hamilton Equations are right?
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
strangerep
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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.
 
  • #6
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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##

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.

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
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I am still thinking how to find all constants of motion out of the Lagrangian.

I'll post what I get.
 
  • #8
PeroK
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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
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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).

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
PeroK
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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
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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
PeroK
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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
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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
PeroK
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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.
 
  • #15
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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
PeroK
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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
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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.

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.
 
  • #18
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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
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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
PeroK
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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
strangerep
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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 gotta go now, but I'll try to post some extra stuff later today.)
 
  • #22
strangerep
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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## ?
 
  • #23
PeroK
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@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.
 
  • #24
strangerep
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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
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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.

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.

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.
 

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