Recent content by Tertius

That makes sense. And then, from what i've found in literature, it is common to have the spreading parameter ##\sigma = \frac{h}{2 \lambda}##. That should, in combination with the momentum being high enough, give physically real and consistent results, I would hope.

The goal is to have accurate 1D numerical results for tunneling probabilities through an arbitrary barrier without relying on analytic approximations such as WKB. If there is a more ideal approach to this, I am happy to change tactics. Time independent, for example, but I am not sure how to...

Again, the solutions are not plane waves. AS set up, the problem assumes hard walls at the extremities of the grid, so the eigenfunctions should go to zero at both ends. You can solve the full eigenvalue problem ##H\psi=E\ps## and get the corresponding discrete set of eigenstates. Thanks...

I thought I solved the problem in answering my own post a few days ago, but the tunneling probability vs. energy trend is clearly wrong. I've remade the post because I have totally changed my approach and need a better understanding of the boundary setup. Overall description: a plane wave...

I figured out where my approach was wrong. The boundary conditions for this are a little tricky because it is not a bound system. So I discretized it manually and just solved $$H\psi = E\psi$$, where E is the kinetic energy of the incoming particle. E in this case is just a single eigenvalue...

I am doing this to have my own solution for customization and understanding. I also want to manually check the WKB approximation accuracy at various energies against this static solution. I've split the problem into 3 regions and am solving it in 1D, but am having problems with how to define...

I realized the problem was quite simple. The covariant derivative is "correcting" for changes to the metric. The changes in the FRW metric look volumetric because ##\sqrt{g}=a^3##.

I'm studying Carroll's section on covariant derivatives, which shows that the covariant divergence of a vector ##V^\mu## is given by $$\nabla_\mu V^\mu = \partial_\mu V^\mu + \Gamma^\mu_{\mu\lambda}V^\lambda$$. Because ##\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{g}}\partial_\lambda \sqrt{g}## we...

@PeterDonis, @vanhees71 I see. I neglected the negative sign, if you can get different answers with different metric conventions then it can't be physical. The quantity I am really after is the matter density that is timelike. From my understanding timelike matter should always have a...

@PeterDonis Ok, I've had some time to evaluate my thinking here. Using Noether's theorem results in a stress energy momentum tensor that is defined using translational symmetries. These are not only usually non-existent in curved spacetime, but there are other symmetries, such as gauge...

That's fair. Just didn't want the thread to die. I'll do the maths and hopefully have a more detailed thread later on.

I hadn't thought of it in terms of needing isolation before. But I see what you are saying. I should have specified my flat space 4-momentum comment was just an example (a mostly not precise relationship) of what I am trying to understand about curved spacetime. To be precise, I want to find...

Right, so the trace with a perfect fluid is the expected $$-\rho + 3p$$ For dust, we use ##p=0##, and for photons (traceless) we use ##\rho=3p## So a timelike KVF, if it exists, can give us the total energy. A spacelike KVF, which does exist in FRW spacetimes, can give us momentum. So in FRW...

I see. I originally thought about the trace as the invariant mass density of the system because of this paper: https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-38/issue-4/The-connection-between-the-energy-momentum-tensor-and-the-tensor/cmp/1103860087.pdf But I...

I should have specified 'the spatial curvature changes' I was referring to a Schwarzschild metric, which has a timelike KVF, but has spatial curvature. From my understanding, the trace is a coordinate independent quantity, and should be observer independent because if any timelike observers...