Hello~ Thanks for your opinions.
One of my ideas is that for Ricci and Riemann both have 6 independent components and Ricci can be expressed as the linear combinations of Riemann, then I just have to check if the equation right when multiplying ##g^{\alpha\gamma}## and take sum for ##\alpha##...
Hello~
For usual Riemann curvature tensors defined: ##R^i_{qkl},## I read in the book of differential geometry that in 3-dimensional space, Ricci curvature tensors, ##R_{ql}=R^i_{qil}## can determine Riemann curvature tensors by the following relation...
Thanks for your help first!
I noticed that too! But then I felt hard to solve it. In the Riemannian case the complex number help me to separate the equation, and now I don't know how to have further result...
I'll keep trying~ thank you!
Hello everyone:
I studied in differential geometry recently and have seen a statement with its proof:
Suppose there is a Riemannian metric: ##dl^2=Edx^2+Fdxdy+Gdy^2,## with ##E, F, G## are real-valued analytic functions of the real variables ##x,y.## Then there exist new local coordinates...
I think in your question the radius of the sphere is not important(? Maybe it just regards the sphere as a point, and just use the centripetal force to find its speed?
At first I'm not sure what you want also, and guess maybe you hope to get the ratio can be bounded with some cool number, maybe ##Ce^R##... likewise haha.
For me the fraction cannot be simplified XD
Ummm... Then I think his viewpoints are that the value ##m\in \mathbb{R}\setminus\{0\}## cannot make all the ##x## fit the equation, in which case we cannot say ##y=x^m## is a solution to the ODE.
Since ##r_i## is the largest and ##r_R## is the smallest, the ratio in the log must smaller than ##\frac{(r_iR)!}{(r_R!)^R},## which may depend on ##r_i##, ##r_R## and ##R.## Does the problem require a fixed constant?
The problem required to find the value of ##m## such that ##y=x^m## is the solution, so I think your ##m## has to make all the ##x## and ##y## fit the equation.
Besides, ##x## varies and what is given is a function. I think ##x=0## just meets a special case. (my opinion)
That's for the centro of the motion is at the bottom of the roller. Your formula should be applied as ##v=2r\cdot\omega,## where ##2r## is the distance from the bottom to the top.
First, this must make no sense. The correct one should be ##E=hf.##
Since you can get the energy of a photon, then given the amount of light you can calculate the number of the photons. Like what you said, the energy transferred to light per second is ##300mJ\cdot 66\%.##