User talk:Siveshwar

Welcome to Wikipedia
Hello shiveswar....Welcome to wikipedia....Hope that you learn a lot from wikipedia as well contribute and share your knowledge with others, thus helping for the betterment of the society.....Ganeshsashank (talk) 17:01, 8 February 2010 (UTC)

Solve these questions
Verify green's theorem for the following by evaluating both the integrals:

Q.1) $$\oint_C x^3dy - y^3dx\,\!$$

where C is the circle $$x= 2 cos\theta \,\!$$ and $$y= 2sin\theta \,\!$$ where $$0\le \theta \le 2\pi \,\!$$

Q.2) $$\oint_C(x+y)dx + x^2dy\,\!$$ where C is the triangle with the vertices $$ (0,0); (2,0); (2,4)\,\!$$ taken in that order.

Q.3) Evaluate the integral for the lemniscate along X-axis.

$$\int \int \frac{r}{\sqrt{(a^2+r^2)}} dr d\theta \,\!$$

where $$\frac{-\pi}{4} \le \theta \le \frac{\pi}{4}\,\!$$ and  $$ 0 \le r \le acos2\theta \,\!$$ Ganeshsashank (talk) 08:39, 21 February 2010 (UTC)

Doubt On schrödinger wave equation
Hai shiva, when we derive the Schrödinger wave equation, we derive it with repect to time. But, during the explanation for $$\alpha\,\!$$-Decay, we take the wave equtaion with respect to distance. So, my doubt is
 * Can we take the wave funtion $$\psi \,\!$$'s derivative with respect to any fundamental parameter??

Schrödinger's time dependent wave equation is:

$$\nabla^2 \psi + \frac{2m(E-V)}{\hbar^2} \psi =0 \,\!$$

And the schrödinger's equation taken during evaluation of single step is:

$$\frac{\partial^2 \psi}{\partial^2 x^2} + \frac{2m(E-V_0)}{\hbar^2} \psi =0 \,\!$$

Answer me: Can we do like this?? Ganeshsashank (talk) 14:43, 23 February 2010 (UTC)

Ok dude, I got it...

When we use $$\nabla^2\,\!$$, it can be with respect to any thing whether $$\frac{\partial^2}{\partial t^2}\,\!$$ or $$\frac{\partial^2}{\partial x^2}\,\!$$. Here $$\nabla ^2 \,\!$$ is just an operator. —Preceding unsigned comment added by Ganeshsashank (talk • contribs) 09:42, 25 February 2010 (UTC)