Discrepancy in Lagrangian to Hamiltonian transformation?

In summary, the conversation discusses the relationship between the Lagrangian (L) and Hamiltonian (H) equations and their corresponding forms (equations [1] and [2]). However, the equations [3a] and [3b] derived from these forms do not hold true in all cases, as shown by the example in equations [4]-[6]. This leads to a discussion about the need for a different equation (equation [6]) and the mismatch between results obtained from equations [3] and [6]. The conversation also mentions the use of a Legendre Transform to convert the Lagrangian to the Hamiltonian, resulting in different arguments for the functions.
  • #1
JALAJ CHATURVEDI
2
0
I know,
$$ L=T-V \;\;\; \; \;\;\; [1]\;\;\; \; \;\;\; ( Lagrangian) $$
$$ H=T+V \;\;\; \; \;\;\;[2] \;\;\; \; \;\;\; (Hamiltonian)$$
and logically, this leads to the equation,
$$ H - L= 2V \;\;\; \; \;\;\; [3a]$$
$$ H + L= 2T \;\;\; \; \;\;\; [3b]$$
but,when it comes to :
$$\widehat{T}=\frac{-{\left(2\pi h\right)}^{2 } }{2 m} \frac{\partial^2 }{\partial x^2 }\ \;\;\; \; \;\;\; [4]$$

and
$$\widehat{V}=V\left(x\right)\ \;\;\; \; \;\;\; [5]$$
equation [3] doesn't seem to hold. And we need an altogether different equation which is :
$$ H=\dot{q}\frac{\partial L}{\partial \dot{q} }-L\ \;\;\; \; \;\;\; [6]$$

1. Why is this happening?2. Even if equation [6] does hold, the results obtained by equation [3] & [6] must coincide, which is certainly not the case?For example, given a Hamiltonian

$$\widehat{H}=\frac{-{\left(2\pi h\right)}^{2 } }{2 m} \frac{\partial^2 }{\partial x^2 }\ + V\left(x\right)\ \;\;\; \; \;\;\; [7]$$
why not can I use equation [3a] and get the corresponding Lagrangian instead of going through equation [6] ?

An example in favour of my question :

$$\ H=\frac{p_{\theta}^{2}}{2 ml^{2 }}+mgl\left(1-\cos\theta\right)\ $$and was asked to find $$ \frac{\text{d}L}{\text{d}t}\ $$ and the answers didn't match!
 
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  • #2
The Langrangian is an equation of the generalised coordinates ##q## and velocities ##\dot q##. To get from the Lagrangian to the Hamiltonian you perform a Legendre Transform and the Hamiltonian becomes a function of ##q## and ##p = \frac{\partial L}{\partial \dot q}##. So you need to consider that ##H## and ##L## are functions of different arguments.
 

Related to Discrepancy in Lagrangian to Hamiltonian transformation?

1. What is the discrepancy between the Lagrangian and Hamiltonian formulations?

The discrepancy between the Lagrangian and Hamiltonian formulations lies in their approach to describing the behavior of a physical system. The Lagrangian formulation uses generalized coordinates and velocities to track the motion of a system, while the Hamiltonian formulation uses generalized coordinates and momenta. This can result in differences in the equations of motion and the resulting trajectories of the system.

2. Why is it important to be aware of the discrepancy in the Lagrangian to Hamiltonian transformation?

Understanding the discrepancy in the Lagrangian to Hamiltonian transformation is important because it allows for a deeper understanding of the underlying principles of classical mechanics. It also allows for the use of different mathematical tools and techniques to solve problems and analyze systems, providing a more comprehensive understanding of physical phenomena.

3. How is the discrepancy between the Lagrangian and Hamiltonian formulations addressed?

The discrepancy between the Lagrangian and Hamiltonian formulations can be addressed through a process known as the Legendre transformation. This mathematical tool allows for the transformation of the Lagrangian to the Hamiltonian and vice versa, providing a bridge between the two formulations and allowing for a more complete understanding of a system.

4. Can the discrepancy in the Lagrangian to Hamiltonian transformation be observed in real-world systems?

Yes, the discrepancy in the Lagrangian to Hamiltonian transformation can be observed in real-world systems. For example, in classical mechanics, the Lagrangian and Hamiltonian formulations may result in slightly different predictions for the motion of a pendulum. In quantum mechanics, the two formulations can result in different energy levels for a system.

5. How does the discrepancy in the Lagrangian to Hamiltonian transformation relate to other areas of physics?

The discrepancy in the Lagrangian to Hamiltonian transformation is a fundamental aspect of classical mechanics, which is the foundation of many other areas of physics. It is also relevant in quantum mechanics, where the Hamiltonian operator plays a critical role in determining the evolution of a system. Additionally, the concepts of the Lagrangian and Hamiltonian are used in other areas of physics, such as field theory and thermodynamics.

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