Transforming Cone Line Elements: Proving Flat Geometry?

  • Thread starter Thread starter joshua2112
  • Start date Start date
  • Tags Tags
    Elements Line
joshua2112
Messages
1
Reaction score
0

Homework Statement


Consider the cone defined by z = √ x 2 + y 2 . By defining new coordinates ρ, φ where x = ρ cos φ, y = ρ sin φ, show that the line element on the cone is ds^2 = 2dρ^2 + ρ^2dφ^2 . Why does this result prove the cone has flat geometry?

Homework Equations


x=rcosφ
y=rsinφ

The Attempt at a Solution


ive tried several times but end up with cos and sin in my answer
 
Last edited:
Physics news on Phys.org
No one can help you if you don't show your attempted solution. (And yes, now would be a good time to start learning how to do latex on this forum.)
 
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