# Lagrangian Problem (Find Relation between Amplitude and Momentum)

• Wannabe Physicist
In summary, the Lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.I think you're almost there. You have an expression for ##H## in terms of ##p## and ##q##. The definition of "ampl
Wannabe Physicist
Homework Statement
Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is ##A## . If ##p## denotes the momentum of
the particle, then ##4p^2## is

a) ##(A^2-q^2)(4+A^2-q^2)##
b) ##(A^2+q^2)(4+A^2-q^2)##
c) ##(A^2-q^2)(4+A^2+q^2)##
d) ##(A^2+q^2)(4+A^2+q^2)##
Relevant Equations
1) ##p = \frac{\partial L}{\partial \dot{q}}##
2) [not sure if this is relevant] ##H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L##
The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for ##p=p(\dot{q})## in the form ##\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}## and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.

Last edited:
Delta2
Wannabe Physicist said:
Homework Statement:: Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is ##A## . If ##p## denotes the momentum of
the particle, then ##4p^2## is

a) ##(A^2-q^2)(4+A^2-q^2)##
b) ##(A^2+q^2)(4+A^2-q^2)##
c) ##(A^2-q^2)(4+A^2+q^2)##
d) ##(A^2+q^2)(4+A^2+q^2)##
Relevant Equations:: 1) ##p = \frac{\partial L}{\partial \dot{q}}##
2) [not sure if this is relevant] ##H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L##

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for ##p=p(\dot{q})## in the form ##\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}## and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.
I think you're almost there. You have an expression for ##H## in terms of ##p## and ##q##. The definition of "amplitude" here is that ##A = q_{max}##, the maximum value of ##q##, which occurs when ##p = 0##. So you can relate ##A## to ##H##, which allows you to relate ##A##, ##p## and ##q##.

Wannabe Physicist and Delta2
Yes! I had thought of that! But then am I not presuming that ##q## has dimensions of position?

Wannabe Physicist said:
Yes! I had thought of that! But then am I not presuming that ##q## has dimensions of position?
I see it now. I was assuming the amplitude itself has dimensions of position. Thanks for helping @stevendaryl !

Wannabe Physicist said:
Yes! I had thought of that! But then am I not presuming that ##q## has dimensions of position?

I don't think it matters what the units of ##q## are. It only matters that ##A## means the maximum value of ##q##. I actually had never heard of the word "amplitude" applied to situations other than sines and cosines, but I suppose it makes sense for more complicated functions, as well.

Wannabe Physicist said:
Homework Statement: Consider the Lagrangian $$L = 1-\sqrt{1-\dot{q}^2}-\frac{q^2}{2}$$ of a particle executing oscillations whose amplitude is ##A## . If ##p## denotes the momentum of
the particle, then ##4p^2## is

a) ##(A^2-q^2)(4+A^2-q^2)##
b) ##(A^2+q^2)(4+A^2-q^2)##
c) ##(A^2-q^2)(4+A^2+q^2)##
d) ##(A^2+q^2)(4+A^2+q^2)##
Relevant Equations: 1) ##p = \frac{\partial L}{\partial \dot{q}}##
2) [not sure if this is relevant] ##H = \left(\Sigma_{i} p_i \dot{q}_i\right) - L##

The given lagrangian doesn't seem to correspond to any of the basic systems (like simple/ coupled harmonic oscillators, etc). So I calculated the momentum ##p## which is the partial derivative of ##L## with respect to generalized velocity ##\dot{q}##. Doing so I obtain
$$p = \frac{\dot{q}}{\sqrt{1-\dot{q}^2}}$$
Now, I am facing this problem: I do not know how to relate this to the amplitude of the oscillations.

In the hopes of getting a direction or line of thought to follow, I tried writing the Hamiltonian of this system, which is obtained by rewriting the above equation for ##p=p(\dot{q})## in the form ##\dot{q} = \dot{q}(p) = \displaystyle{\frac{p^2}{1+p^2}}## and substituting it in the equation (2) in the "relevant equations" listed above. I found
$$H = \sqrt{1+p^2} +\frac{q^2}{2}$$

But I am still not sure how to proceed.

After trying all of the above I looked up the answer key and the correct option is (a). But I am not sure how to get the answer.
It is not said that the momentum is canonical therefore you can not use the canonical momentum formula

sandeep_bhnaja said:
It is not said that the momentum is canonical therefore you can not use the canonical momentum formula
It may be worth noting that the thread is well over 2 years old!

jim mcnamara

## What is the Lagrangian problem?

The Lagrangian problem is a mathematical problem in classical mechanics that involves finding the relationship between the amplitude and momentum of a system. It is named after Joseph-Louis Lagrange, who developed the method for solving this problem.

## Why is the Lagrangian problem important?

The Lagrangian problem is important because it provides a powerful and elegant method for solving complex problems in classical mechanics. It allows for the calculation of the equations of motion for a system using a single function, the Lagrangian, instead of multiple equations of motion.

## What is the Lagrangian function?

The Lagrangian function, denoted as L, is a mathematical function that represents the difference between the kinetic and potential energies of a system. It is used in the Lagrangian problem to find the equations of motion for a system.

## How is the Lagrangian problem solved?

The Lagrangian problem is solved using the principle of least action, which states that the path taken by a system between two points in time is the one that minimizes the action integral, given by S = ∫L dt. This leads to the Euler-Lagrange equations, which can be solved to find the equations of motion for the system.

## What are some applications of the Lagrangian problem?

The Lagrangian problem has many applications in physics and engineering, including in the study of celestial mechanics, fluid dynamics, and quantum mechanics. It is also used in the design and analysis of mechanical systems, such as robots and spacecraft.

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