Solving Lagrangian Mechanics: Hamilton's Eqns & EL Eqns

In summary, the Lagrangian of a particle with charge q and mass m is given by L = \frac{1}{2} m \dot{r}^2 - q \phi(r,t) + \frac{q}{c} r \bullet A(r,t). The corresponding Hamiltonian function H(r,p) is given by H = \frac{p^2}{2m} + q \phi(r,t), and in order for it to be a constant of motion, the electric and magnetic field intensities must satisfy the conditions E = -\frac{\partial \phi}{\partial r} and B = \frac{\partial}{\partial r} \times A. Hamilton's equations of motion are first-order
  • #1
stunner5000pt
1,461
2
The Lagrangian of a aprticle of charge q and mass m is given by
[tex] L = \frac{1}{2} m \dot{r}^2 - q \phi(r,t) + \frac{q}{c} r \bullet A(r,t) [/tex]

a) determine the Hamiltonain function H(r,p) (in terms of r and its conjugate momentum p) and explain waht conditions the electric and mangetic field intensities
[tex] E = - \frac{\partial \phi}{\partial r} - \frac{1}{c} \frac{\partial A}{\partial t} [/tex]
[tex] B = \frac{\partial}{\partial r} \times A [/tex]
must satisfy if the Hamiltonain is to be a constant of motion


I am not sure if the Hamiltonian is
[tex] H = \dot{r} \frac{\partial L}{\partial \dot{r}} -L [/tex]
OR [tex] H = r \frac{\partial L}{\partial r} -L [/tex]
[tex] \dot{r} \frac{\partial L}{\partial \dot{r}} = m \dot{r}^2 + \frac{q}{c} \dot{r} \bullet A(r,t) [/tex]

[tex] H = \dot{r} \frac{\partial L}{\partial \dot{r}} -L = m \dot{r}^2 + \frac{q}{c} \dot{r} \bullet A(r,t) - \frac{1}{2} m \dot{r}^2 - q \phi(r,t) + \frac{q}{c} r \bullet A(r,t) [/tex]
[tex] H = \frac{1}{2} m\dot{r}^2 + q \phi(r,t) [/tex]
[tex] H(r,p) = \frac{1}{2} p \dot{r} + q \phi(r,t) [/tex]

for H to be a constant of motion, then dh/dt = 0 ,yes?

[tex] \frac{dH}{dt} = \frac{1}{2} \dot{p} \dot{r} + \frac{1}{2} p \ddot{r} + q \frac{\partial \phi}{\partial t} [/tex] (not sure about the last term)
[tex] \frac{dH}{dt} = F \frac{\dot{r}}{2} + p \frac{\ddot{r}}{2} + q \frac{\partial \phi}{\partial t} [/tex]

im not sure how the E and B field come into play because none of their terms appear in the equation for dH/dt

b) Work out Hamilton's equations of motion for this system (i.e. the first order equations of r(t),p(t)), and show taht htey are equivalent to the second order Euler Lagrange equation for this system.

I am currently on this...
Your help is greatly appreciated
 
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  • #2
It is a general result in Hamiltonian mechanics that the total derivative with respect to time of the Hamiltonian is just the negative of the partial derivative with respect to time of the Lagrangian. This can be shown by taking the derivative of H, and then plugging in the definitions of p and the Lagrangian equations of motion.
 
  • #3
Hi stunner,

1) The Hamiltonian is always given by [tex] H = \dot{q} \frac{\partial L}{\partial \dot{q}} - L = p \dot{q} - L [/tex] for a single coordinate [tex] q [/tex]. If you have more than one coordinate then you generalize to [tex] H = \sum_i \dot{q}_i p_i - L [/tex] where as always the definition of the canonical momentum is [tex] p_i = \frac{\partial L}{\partial \dot{q}_i} [/tex].

2) The Hamiltonian is constant provided the Lagrangian contains no explicit time dependence. This fact is probably proved in your book somewhere, but it is easy to get if you just use the equations of motion. What does this requirement translate into when you include electric and magnetic fields?

3) Hamilton's equations of motion are first order differential equations while Lagrange's equations are second order. Schematically you have [tex] \dot{q} = \frac{\partial H}{\partial p} [/tex] and [tex] \dot{p} = - \frac{\partial H}{\partial q} [/tex]. Hint: differentiate the first equation with respect to time.
 
  • #4
Physics Monkey said:
Hi stunner,

1) The Hamiltonian is always given by [tex] H = \dot{q} \frac{\partial L}{\partial \dot{q}} - L = p \dot{q} - L [/tex] for a single coordinate [tex] q [/tex]. If you have more than one coordinate then you generalize to [tex] H = \sum_i \dot{q}_i p_i - L [/tex] where as always the definition of the canonical momentum is [tex] p_i = \frac{\partial L}{\partial \dot{q}_i} [/tex].

2) The Hamiltonian is constant provided the Lagrangian contains no explicit time dependence. This fact is probably proved in your book somewhere, but it is easy to get if you just use the equations of motion. What does this requirement translate into when you include electric and magnetic fields?

3) Hamilton's equations of motion are first order differential equations while Lagrange's equations are second order. Schematically you have [tex] \dot{q} = \frac{\partial H}{\partial p} [/tex] and [tex] \dot{p} = - \frac{\partial H}{\partial q} [/tex]. Hint: differentiate the first equation with respect to time.

i don't think I am interpreting this right

text says that the hamilton (for this problem) is usually not time dependant since dL/dt is not zero. (unless phi, A are not explicitly time dependant)

so then if we look at hte Elefctric field... since A is not explicitly time dependant, then E should be
[tex] E = - \frac{\partial \phi}{\partial r}[/tex]

also i m taking from from StatusX said taht [tex]- \frac{dH}{dt} = - m \dot{r} \ddot{r} - q \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial t} - q \frac{\partial \phi}{\partial t} [/tex]

and this shouldbe equal to
[tex] \frac{dL}{dt} = m \dot{r} \ddot{r} - q \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial t} - q \frac{\partial \phi}{\partial t}+ \frac{q}{c} \frac{\partial r}{\partial t} + \frac{q}{c} \frac{\partial A}{\partial r} \frac{\partial r}{\partial t} + \frac{q}{c} \frac{\partial A}{\partial t} [/tex]

so then these terms should e zero
[tex] m \dot{r} \ddot{r} - q \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial t} - q \frac{\partial \phi}{\partial t} [/tex]

so r shouldn't be time dependant?? And A shouldn't be time dependant?
 
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  • #5
I think you may be confused so let me repeat the correct fomulation. If the Lagrangian contains no explicit time dependence [tex] \partial L/ \partial t = 0 [/tex] (not [tex] d L/ dt = 0 [/tex] !) then the Hamiltonian is time independent [tex] dH/dt = 0 [/tex].
 
  • #6
Actually the particle's generalized coordinates are x,y,z and of course everything needs to be adapted.

Daniel.
 
  • #7
ok ok
so the Hamilton is
[tex] H(r,p) = \frac{p^2}{2m} + q \phi(r,t) [/tex]
[tex] \frac{\partial H}{\partial t} = \frac{p}{m} \dot{p} + q \frac{\partial \phi}{\partial t} [/tex]

for the Lagrangian
[tex] \frac{\partial L}{\partial t} = \frac{p}{m} \dot{p} - q \frac{\partial \phi}{\partial t} + \frac{q}{c} \dot{r} A + \frac{q}{c} r \frac{\partial A}{\partial t} [/tex]

ok so then in order for dl/dt = - dh/dt then r and A need to be constant wrt time.
so that [tex] E = -\frac{\parital \phi}{\partial r} [/tex]
and [tex] B = \frac{\partial}{\partial r} \times A [/tex]

for the B part
[tex] \dot{r} = \frac{\partial H}{\partial p} = \frac{p}{m} [/tex]
so then [tex] p = m \dot{r} [/tex]
[tex] r(t) = \frac{p}{m} t + r_{0} [/tex]

[tex] \dot{p} = -\frac{\partial H}{\partial r} = - q \frac{\partial \phi}{\partial r} [/tex]
not sure how to do this part ...
 
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  • #8
If [itex] H\left(q^{i},p_{i},t\right) [/itex] is given by

[tex] H=\frac{1}{2m}\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\right) +q\varphi \left(x,y,z,t\right) -\frac{q}{c}\left(x A_{x}(x,y,z,t)+y A_{y}(x,y,z,t)+z A_{z}(x,y,z,t)\right) [/tex]

what is

[tex] \frac{dH}{dt} [/tex]

equal to?

HINT: Use Hamilton's equations of motion.

Daniel.
 
  • #9
illl do it for the x and t only
but isn't what u wroteh texpression for hte Lagrangian and not hte Hamiltonian?
isnt hte hamilton [tex] H = \frac{p^2}{2m}+ q \phi(r,t) [/tex]

[tex] \frac{dH_{x}}{dt} = \frac{p_{x}\dot{p_{x}}}{m} + q \frac{\partial \phi}{\partial x} \frac{\partial x}{\parital t} + q \frac{\partial \phi}{\partial t} - \frac{q}{c} \dot{x}A - \frac{q}{c} x \frac{\partial A}{\partial x} \frac{\partial x}{\partial t} - \frac{\partial A}{\partial t} [/tex]

but i don't hink i need the totla derivative... in my text it says if the partial of H wrt time is equal to the negative of the partial of L wrt time are equal then H is constnat

its quite clear that A sohuld be indepednant on time and that the variables of motion x,y,z should be independnt of time as well
 
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  • #10
is this fine?

how do i solve for the hamiltonian equations of motion, however?
 
  • #11
Hamilton's equations are

[tex] \dot{q}^{i}\left(t\right)=\frac{\partial H}{\partial p_{i}} [/tex]

[tex] \dot{p}_{i}\left(t\right)=-\frac{\partial H}{\partial q^{i}} [/tex]

U can compute them quite easily using the Hamiltonian's expression written above.

As for

[tex] \frac{dH}{dt}=\frac{\partial H}{\partial t} + \frac{\partial H}{\partial q^{i}} \dot {q}^{i} +\frac{\partial H}{\partial p_{i}} \dot{p}_{i} =0 [/tex]

i'm sure you know how to compute each term of the equation above...

Daniel.
 

1. What is the difference between Hamilton's equations and the Euler-Lagrange equations?

Hamilton's equations and the Euler-Lagrange equations are both used to solve Lagrangian mechanics problems, but they approach the problem from different perspectives. Hamilton's equations are derived from the principle of least action and describe the dynamics of a system in terms of its coordinates and momenta. The Euler-Lagrange equations, on the other hand, are derived from the Lagrangian function and describe the equations of motion for a system in terms of its generalized coordinates.

2. How do Hamilton's equations and the Euler-Lagrange equations relate to each other?

Hamilton's equations are essentially a more compact and elegant form of the Euler-Lagrange equations. They can be derived from the Euler-Lagrange equations by introducing the concept of conjugate momenta. Therefore, both sets of equations are equivalent and can be used interchangeably to solve Lagrangian mechanics problems.

3. What is the significance of the Lagrangian function in Lagrangian mechanics?

The Lagrangian function is a mathematical representation of the kinetic and potential energies of a system. It plays a crucial role in Lagrangian mechanics as it allows us to derive the equations of motion for a system using the principle of least action. It also provides a more concise and general description of a system's dynamics compared to traditional Newtonian mechanics.

4. How are Hamilton's equations and the Euler-Lagrange equations used in practical applications?

Hamilton's equations and the Euler-Lagrange equations are used in a wide range of applications, including classical mechanics, quantum mechanics, and control theory. In classical mechanics, they are used to solve problems involving systems with multiple degrees of freedom. In quantum mechanics, they are used to describe the dynamics of wave functions. In control theory, they are used to design optimal control strategies for systems.

5. Are there any limitations to using Hamilton's equations and the Euler-Lagrange equations?

While Hamilton's equations and the Euler-Lagrange equations are powerful tools for solving Lagrangian mechanics problems, they do have some limitations. They are not applicable to non-conservative systems, and they cannot handle systems with constraints. In these cases, other methods such as the Lagrange multiplier method may be more suitable.

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