What property of lagrangian is used here?

aaaa202
Messages
1,144
Reaction score
2
Hi, the attached picture shows a derivation of what I can only assume to be the property that the lagrange equations are invariant under a transformation of the coordinates.
But I have some trouble understanding how you go from the term pointed out the rear of the arrow to the point pointed out by the front of the arrow. Can anyone explain?
 

Attachments

  • lagrangian.png
    lagrangian.png
    21.3 KB · Views: 507
Physics news on Phys.org
It's really easy

\frac{\partial q^{i}}{\partial \dot{Q}^{j}} = 0

then he used the equality written on the first line just before the text written (apparently) in some Nordic language.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top