Recent content by Mr-R

Cheers vela. Will edit it.

I think you just have to expand the wave function to first order in ##\Delta x## and then $$P(\Delta x)=|\Psi(\Delta x)|^2 \Delta x$$ Imagine the area under an infinitesimal interval, in this limit you can substitute multiplication for the integral. I am not entirely sure though. See the post...

I second this.

You are not supposed to see those. Are you sure that your browser can render Latex? I gave an example for you, but if your browser can't render it then there is no point of typing again. Please check that first. Meanwhile, read about Einstein summation convention. (it seems that this is your...

Fix ##i##, ##j## and then run over ##k##. For example for ##i=x## and ##j=y## $$\epsilon_{xy,kk}=\epsilon_{xy,xx}+\epsilon_{xy,yy}+\epsilon_{xy,zz}$$ Where a comma denotes a partial derivative. (##A_{,x}=\frac{\partial A}{\partial x}##) Hope this helps.

I know what you mean. To be honest usually one can solve a problem 'quickly' if he/she has encountered it before, or at least something similar. Otherwise, it would take considerably more time to be solved depending on the complexity of the problem. I don't have much to say except keep...

Got it :biggrin: Cheers

As already stated by the members above, you can raise and lower it the indices on the stress energy tensor as you like. Why the usual stress tensor ##T^{\alpha \beta}## has two upper (or lower indices)? Maybe because the way they are sometimes defined. For a perfect fluid its defined as...

The expression for momentum for photons in QM is ##p=\hbar k##. Where ##k## is the wave number (##k=\frac{2\pi}{\lambda}##) and ##\hbar## is the Planck constant divided by ##2\pi## . Or equivalently, ##p=\frac{h}{\lambda}## . Your equation for ##p## is for massive particles with non zero rest...

I think that this is only possible for a conservative force. Got nothing against the rest of your post :P

This! I was going to ask why the Riemann tensor is a Tensor although it is made up of Christoffel symbols but then remembered that it is obvious from it's derivation. Or from this definition as well ##[\nabla_i,\nabla_j] \zeta^k=R^k_{~cij}~\zeta^c##. Thanks for your help and also for the references.

Thanks to you, I think that I understand now,. The Ricci scalar ##R=g_{ik}R^{ik}## is a scalar because even if we chose an inertial frame it might not vanish due to the derivatives of Christoffel symbols contained in ##R##, ##\partial \Gamma##. Which means ##R\propto \partial^2g##. So even a...

haushoer, that's what I am saying. Maybe the author means it's not a Lorentz invariant scalar? I doubt it though.

##G## was calculated to and found to be ##G=g^{ik}( \Gamma^m_{il}\Gamma^l_{km}-\Gamma^l_{ik}\Gamma^m_{lm})##. Isn't this a scalar?? Your explanation makes sense. But I can't relate it to my question. Please refer to the form of ##G## I provided above.

Book: Landau Lifshitz, The Classical Theorey of Fields, chapter 11, section 95. I have gone through the derivation of Einstein field equations but not without holes to fill and fix in my understanding. Let's start with the action for the grtavitational field ##S_g## which after some explanation...