How to find integrals of motion for a particle on a surface?

In summary: Anyway, you're right. You don't need to write the Lagrange equations. You just need to find the coordinates that the Lagrangian does not depend on. Then the corresponding momenta will be the constants of motion.In summary, the conversation discusses a problem related to finding the integrals of motion for a system with a cylindrical constraint. The suggested approach is to first write the Lagrangian in Cartesian coordinates and then transform it to cylindrical coordinates. It is also mentioned that the Lagrange equations do not need to be written, and instead the coordinates that the Lagrangian does not depend on should be identified to find the constants of
  • #1
penguin46
19
0
Homework Statement
Find the integrals of motion for a particle constrained to the surface (in cylindrical coordinates) r = a + b cos(z/c) where a b and c are constant.
Relevant Equations
The Lagrangian maybe? I don't know tbh
I have no idea where to even start with this, please help. I barely even know what integral of motion means.
 
Physics news on Phys.org
  • #2
ShayanJ said:
If it's actually a homework, it means at some point someone was trying to teach you this. So why don't you take a deep breath and take a look at your course material. Any notes you may have made, or maybe books that were introduced. Find a place where things make sense to you and start from there.

Then if you don't understand something,. try to ask a specific question here. After that you can make an effort to solve the problem yourself and if you get stuck, just post what you've done here and people will help you to make progress.
I've looked over my course material multiple times, and I've been trying to solve this problem for a couple days now
 
  • #3
penguin46 said:
I've looked over my course material multiple times, and I've been trying to solve this problem for a couple days now
Sorry, I just thought that reply wasn't helpful and deleted it, but you saw it. Doesn't matter.
Anyway, forget about this problem for a moment. Do you know how to write the Lagrangian of a system and then use it to write the Lagrange equations?
 
  • #4
ShayanJ said:
Sorry, I just thought that reply wasn't helpful and deleted it, but you saw it. Doesn't matter.
Anyway, forget about this problem for a moment. Do you know how to write the Lagrangian of a system and then use it to write the Lagrange equations?
Yes, but honestly it's been about a year since I've done it. That's where I started but I wasn't sure that was the right tack, even just writing the energy was starting to look quite complicated
 
  • #5
penguin46 said:
Yes, but honestly it's been about a year since I've done it. That's where I started but I wasn't sure that was the right tack, even just writing the energy was starting to look quite complicated
To make it easier, first write the Lagrangian in Cartesian coordinates and then transform to cylindrical coordinates.
But remember you also have the constraint r - a + b cos(z/c) = 0. So you should make take that into account when writing the Lagrangian too.
 
  • #6
ShayanJ said:
To make it easier, first write the Lagrangian in Cartesian coordinates and then transform to cylindrical coordinates.
But remember you also have the constraint r - a + b cos(z/c) = 0. Do you know how to write the Lagrange equation for a system with constraints?
What I did is write the position of the particle in terms of the radial angle and the z-coordinate and then write x= r* cos(theta) y= r*sin(theta) z=z where r is the constraint given, then attempted to differentiate each term to find xdot, ydot, and zdot and then square them to find kinetic energy. I got xdot ydot and zdot but xdot and ydot have three terms each so after squaring would have like 6 terms each, and I honestly wasn't sure that finding the equations of motion was the right idea. As I said I basically have no idea how to approach this problem, I have a vague idea of what integrals of motion are (we just covered noethers theorem) but no idea how to actually find them.
 
  • #7
penguin46 said:
What I did is write the position of the particle in terms of the radial angle and the z-coordinate and then write x= r* cos(theta) y= r*sin(theta) z=z where r is the constraint given, then attempted to differentiate each term to find xdot, ydot, and zdot and then square them to find kinetic energy. I got xdot ydot and zdot but xdot and ydot have three terms each so after squaring would have like 6 terms each, and I honestly wasn't sure that finding the equations of motion was the right idea. As I said I basically have no idea how to approach this problem, I have a vague idea of what integrals of motion are (we just covered noethers theorem) but no idea how to actually find them.
It's not a good idea to actually use Cartesian coordinates to solve the problem. Of course things are going to be complicated when you ignore the symmetry of the problem which is a cylindrical symmetry. It's better to write e.g. ## T=\frac 1 2 m(\dot x ^2 +\dot y ^2 + \dot z ^2) ## and then transform to cylindrical coordinates.

Anyway, you're right. You don't need to write the Lagrange equations. You just need to find the coordinates that the Lagrangian does not depend on. Then the corresponding momenta will be the constants of motion.
 
  • #8
ShayanJ said:
It's not a good idea to actually use Cartesian coordinates to solve the problem. Of course things are going to be complicated when you ignore the symmetry of the problem which is a cylindrical symmetry. It's better to write e.g. ## T=\frac 1 2 m(\dot x ^2 +\dot y ^2 + \dot z ^2) ## and then transform to cylindrical coordinates.
I don't know how to do this
ShayanJ said:
Anyway, you're right. You don't need to write the Lagrange equations. You just need to find the coordinates that the Lagrangian does not depend on. Then the corresponding momenta will be the constants of motion.
So that would be the radial angle, right? But shouldn't there be 3 integrals of motion for a system with 2 degrees of freedom?
 
  • #9
penguin46 said:
I don't know how to do this
Remember ## x=r \cos \phi, y=r\sin \phi, z=z ## and also the chain rule.

penguin46 said:
So that would be the radial angle, right? But shouldn't there be 3 integrals of motion for a system with 2 degrees of freedom?
Well, write it down and find out!
 
  • #10
ShayanJ said:
Remember ## x=r \cos \phi, y=r\sin \phi, z=z ## and also the chain rule.Well, write it down and find out!
My professor was explaining this today and tbh I'm more confused than I started out. He did something with Noether's Theorem? I don't understand any of it. I have an exam on this next week.
 
  • #11
penguin46 said:
My professor was explaining this today and tbh I'm more confused than I started out. He did something with Noether's Theorem? I don't understand any of it. I have an exam on this next week.
Noether's theorem just establishes the connection between symmetries and conservation laws. But just finding the conserved quantities is not that hard. You just need to write down the Lagrangian in the appropriate coordinates.
Let me give an example of how to use the chain rule here:
## x=r \cos \phi \Rightarrow \dot x=\dot r \cos \phi-r\sin \phi \dot \phi ##
Just do the same for other coordinates and substitute them in ## T=\frac 1 2 (\dot x^2+\dot y ^2+\dot z ^2) ##.
 
  • #12
ShayanJ said:
Noether's theorem just establishes the connection between symmetries and conservation laws. But just finding the conserved quantities is not that hard. You just need to write down the Lagrangian in the appropriate coordinates.
Let me give an example of how to use the chain rule here:
## x=r \cos \phi \Rightarrow \dot x=\dot r \cos \phi-r\sin \phi \dot \phi ##
Just do the same for other coordinates and substitute them in ## T=\frac 1 2 (\dot x^2+\dot y ^2+\dot z ^2) ##.
Well then zdot would ust be zdot and ydot would be rdot*sin(phi) +r*cos(phi)*phidot.

But isn't that what you told me not to do? Because doing this will just lead to a very big equation once you put sub in the expression for r and square every term
 
  • #13
penguin46 said:
Well then zdot would ust be zdot and ydot would be rdot*sin(phi) +r*cos(phi)*phidot.

But isn't that what you told me not to do? Because doing this will just lead to a very big equation once you put sub in the expression for r and square every term
Maybe I misunderstood what you were doing because this actually produces a very simple equation. You just have to simplify it using the fact that after the substitution you will get similar terms with opposite signs and also using ## \sin^2 \vartheta+\cos^2 \vartheta = 1 ##. You don't need to write down the Lagrange equations, only the Lagrangian.
 
  • #14
okay So i got L = rdot^2 + r^2*phidot^2. Is that correct? I really should brush up on taylors book because a lot of this is vague to me. No idea what to do next.

The book we are using for my course now is landau and lifshitz btw, but he does deviate from it (for noethers theorem e.g.)
 
  • #15
penguin46 said:
okay So i got L = rdot^2 + r^2*phidot^2. Is that correct? I really should brush up on taylors book because a lot of this is vague to me. No idea what to do next.

The book we are using for my course now is landau and lifshitz btw, but he does deviate from it (for noethers theorem e.g.)
Yes, that's corerct. Just remember the ## \dot z^2 ##.
Now you can substitute ## r = a + b \cos (\frac z c )##. Remember you need to use the chain rule again to calculate ##\dot r##.
 
  • #16
ShayanJ said:
Yes, that's corerct. Just remember the ## \dot z^2 ##.
Now you can substitute ## r = a + b \cos (\frac z c )##. Remember you need to use the chain rule again to calculate ##\dot r##.
(b/c)^2*zdot^2*sin^2(z/c) + (a+b*cos(z/c))^2*phidot^2 + zdot^2
 
  • #17
penguin46 said:
(b/c)^2*zdot^2*sin^2(z/c) + (a+b*cos(z/c))^2*phidot^2 + zdot^2
It seems correct. Now you can see that this is independent of r, ## \phi ## and t. So the corresponding momenta(including energy), are the constants of motion.
 
  • #18
ShayanJ said:
It seems correct. Now you can see that this is independent of r, ## \phi ## and t. So the corresponding momenta(including energy), are the constants of motion.
Okay. First of all, thank you for the help so far. Second, so when doing questions such as these, you simply have to find which of the coordinates the Lagrangian is independent of? I suppose that does make sense. Third, in the beginning of Landau and Lifshitz they kind of treat time invariance as a trivial conserved quantity and state that there should be 2s-1 additional conserved quantities, so should there not be 3 conserved quantities on top of energy conservation in this case?

Also, isn't it trivial that momentum is conserved in r since there can't be any motion in the r direction ever?
 
  • #19
CHAPTER II
CONSERVATION LAWS
§6. Energy
During the motion of a mechanical system, the 2s quantities qi and q,
(i = 1, 2, ... , s) which specify the state of the system vary with time. There
exist, however, functions of these quantities whose values remain constant
during the motion, and depend only on the initial conditions. Such functions
are called integrals of the motion.
The number of independent integrals of the motion for a closed mechanical
system with s degrees of freedom is 2s-1. This is evident from the following
simple arguments. The general solution of the equations of motion contains
2s arbitrary constants (see the discussion following equation (2.6}). Since the
equations of motion for a closed system do not involve the time explicitly,
the choice of the origin of time is entirely arbitrary, and one of the arbitrary
constants in the solution of the equations can always be taken as an additive
constant to in the time. Eliminating t +to from the 2s functions qi = qi(t +to,
Ct, C2, ... , C2s-I}, qi = qi(t +to, C1, C2, ... , Czs-1}, we can e:ll..-press the 2s-1
arbitrary constants Ct, c2, ... , C2s-I as functions of q and q, and these functions
will be integrals of the motion.
Not all integrals of the motion, however, are of equal importance in mech-
anics. There are some whose constancy is of profound significance, deriving
from the fundamental homogeneity and isotropy of space and time. The
quantities represented by such integrals of the motion are said to be conserved,
and have an important common property of being additive: their values for a
system composed of several parts whose interaction is negligi,ble are equal
to the sums of their values for the individual parts.
-
It is to this additivity that the quantities concerned owe their especial
importance in mechanics. Let us suppose, for example, that two bodies
interact during a certain interval of time. Since each of the additive integrals
of the whole system is, both before and after the interaction, equal to the
sum of its values for the two bodies separately, the conservation laws for these
quantities immediately make possible various conclusions regarding the state
of the bodies after the interaction, if their states before the interaction are
known.
Let us consider first the conservation law resulting from the homogeneity
of time. By virtue of this homogeneity, the Lagrangian of a closed system
does not depend explicitly on time.
...
As you can see, Landau and Lifshitz explain that there are 2s-1 constants of motion and then goes on to say that the first one is energy.
And there is motion in r direction because r is not constant, it's just a function of z.
And when you want all the constants of motion, you will get the trivial ones too, like the energy in this case, which is a constant and equal to the Lagrangian itself because there is no interaction.
 

1. What is an integral of motion for a particle on a surface?

An integral of motion for a particle on a surface is a mathematical quantity that remains constant throughout the motion of the particle. It is a conserved quantity that describes the overall behavior of the particle on the surface.

2. How do I find the integrals of motion for a particle on a surface?

The integrals of motion can be found by solving the equations of motion for the particle on the surface. This involves using mathematical techniques such as Lagrange's equations or Hamilton's equations to determine the conserved quantities that describe the motion.

3. What is the significance of integrals of motion for a particle on a surface?

Integrals of motion provide valuable information about the behavior of a particle on a surface. They can help us understand the overall motion of the particle, predict its future trajectory, and determine any constraints or symmetries that exist in the system.

4. Can the integrals of motion change during the motion of the particle on a surface?

No, the integrals of motion are conserved quantities and therefore remain constant throughout the motion of the particle on the surface. This is due to the fact that the equations of motion are time-invariant, meaning they do not change with time.

5. Are there different types of integrals of motion for a particle on a surface?

Yes, there are different types of integrals of motion depending on the specific system and equations of motion being studied. Some common types include energy, momentum, and angular momentum integrals, but there can also be more complex integrals that involve multiple conserved quantities.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
911
  • Advanced Physics Homework Help
Replies
7
Views
991
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
19
Views
1K
  • Advanced Physics Homework Help
Replies
26
Views
4K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
879
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
862
Replies
4
Views
574
Back
Top