How to determine correct Lagrangian?

AI Thread Summary
The Lagrangian is defined as the quantity that, when used in the Euler-Lagrange equations, yields the correct equations of motion for a system. To determine if a given expression for T-U is the correct Lagrangian, one must validate the model against experimental results. The relationship between the Lagrangian and equations of motion is not circular; rather, it illustrates the equivalence between Newtonian and Lagrangian mechanics. The Lagrangian allows for easier model creation and testing compared to Newton's laws, which require explicit force modeling. Ultimately, the choice between Lagrangian and Newtonian methods depends on the specific requirements of the problem at hand.
better361
Messages
24
Reaction score
1
First, let me take as the definition of a Lagrangian the quantity that when put into the Euler Lagrange equations, it gives the correct equation of motion.

It sounds like we need to know the equations of motion first. For example. the Lagrangian for a particle subject to a constant magnetic field. It is not your standard L=T-U.

1. With this in mind, when I write down T-U for a system, how do I know if it is also the Lagrangian of a system?
2. Also, this seems somewhat circular as to get the equations of motion we use the Lagrangian, but the Lagrangian is defined by the correct equations of motion. Can someone clarify this for me?
 
Physics news on Phys.org
better361 said:
1. With this in mind, when I write down T-U for a system, how do I know if it is also the Lagrangian of a system?

This is a matter of how you write down models. Writing down a Lagrangian is the Lagrange mechanics equivalent of writing down the inertia and force relations in Newtonian mechanics. You can do this however you like (you could introduce a gravitational force proportional to the distance instead of the inverse square law), but ultimately you must test the model against experiments.

better361 said:
2. Also, this seems somewhat circular as to get the equations of motion we use the Lagrangian, but the Lagrangian is defined by the correct equations of motion. Can someone clarify this for me?
What you are talking about here is just the proof of equivalence between Newtonian and Lagrange mechanics. You are showing that you can get the equations of motion from the variation of the action and that you can get the Lagrangian from the equations of motion. In itself, Lagrange mechanics does not require your Lagrangian to be of a particular form. The Lagrangian defines your model.
 
So what the Lagrangian does is that it gives us an ability to create and test models for systems in a way that is easier than using Newton's law?
 
If you are using Newton's law you need to model the forces. The Lagrangian approach has some advantages and the Newtonian (and also the Hamiltonian approach) has some. What is better suited really depends on what you want to do. Things such as symmetries and constants of motion are more apparent in the Lagrangian approach.
 
  • Like
Likes vanhees71
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...

Similar threads

Back
Top