Recent content by davidge

@kimbyd, there's a little problem with your explanation in post #11. That's it the given photo was taken between 2003 and 2004. The value 100 billion I referred to was mentioned in the Cosmos Series in the early 80's. So there must be a source for that total number other than the Hubble's Ultra...

It's probably too advanced for a beginner, no? Perhaps it's better to look for introductory books on differential geometry.

I totally agree with you. The problem arises when you try to explain something that only makes sense with mathematics, to people who have not learned the mathematics. In such case, I wonder if it's better not to talk about the given phenomenum at all.

Usually, people trying to explain quantum entanglement, uses the scenario where two particles were created and one of them ends up very far away from the other, and then a measurement is made, etc. The problem I see is that they seem to assume the two particles are classical particles, like two...

That's right (you just need a minus sign in the exponential sum) . Re-write it as ##\int d|p| |p| \exp(i|p||x|) \bigg(\exp \bigg(i|p| \sqrt{|p|^2+m^2} \bigg) - \exp \bigg(-i|p| \sqrt{|p|^2+m^2} \bigg) \bigg)## and make the substitution ##|p| = iz##.

The first integral in your post, originally with the limits ##-\infty## to ##\infty## can be re-expressed as a sum of two integrals, one from ##-\infty## to ##0## plus other one from ##0## to ##\infty##. The first one, ##-\infty## to ##0##, can be re-expressed as being from ##0## to ##-\infty##...

The extra exponential term is because $$\int_{-\infty}^{0} d|p| \ ... + \int_{0}^{\infty} d|p| \ ... \ = \int_{0}^{-\infty} d(-|p|) \ ... + \int_{0}^{\infty} d|p| \ ... \ = \ ... \ \\ = \int_{-\infty}^{\infty} d|p| |p| \exp(i |p| |x|) \bigg(\exp(it \sqrt{|p|^2 + m^2}) - \exp(-it \sqrt{|p|^2 +...

It seems you are unconsciously using concepts of Newtonian theory into Einstein's theory. Note that in Einstein's theory, it doesn't make sense to talk about attraction in the way you seem to be thinking of. Gravity is, in Einstein's theory, the curvature in the underlying space-time which is...

This answer my question. Thanks

That ##E## is the electric field, actually. I wanted to give it in a form given in Maxwell's theory for the OP to compare the two.

Ah, ok

In QFT, consider for simplicity the Number Operator acting on a general-single-particle state $$N_a (k) \{ket \} \bigg(\sum_{k'}A_{k'} \exp(-ik'x) \frac{1}{V} \bigg) = |A_k|^2 \{ket \} \bigg(\sum_{k'}A_{k'} \exp(-ik'x) \frac{1}{V} \bigg)$$ Take the free Maxwell equations ##\partial_\alpha...

Thanks, sandy. I didn't want to post a message in that thread, because they are masters and I'm a noob. From what I read, it seems that those calculations are computer-made. The calculation goes for, say, 430 loops. Is it correct?

So should we use expansion in all orders in any and every case we are eventually considering?

I have been reading about QFT amplitudes. It seems that difficulty increases as we consider more and more terms in the Dyson's expansion for the Scattering operator, and we need to normalize each of them if we want to get a sensible result. My question is, nature usually uses what order? I...