Recent content by Geofleur

Yes, thank you for that very detailed explanation. It's almost painfully obvious now, but I just couldn't see it before for some reason.

I'll try to keep that in mind in future!

I think maybe I finally understand the meaning of the result I got. I started with the four-vectors ## V^\mu = \gamma(v,c) ##, and ## A^\mu = \left(\gamma \frac{d\gamma}{dt}v + \gamma^2 \frac{dv}{dt}, c\gamma \frac{d\gamma}{dt} \right), ## where I assume that both the three-acceleration and...

@vanhees71 I'm not sure I understand - it seems to me that I did calculate the four-acceleration using the proper time, just like so: ## A^\mu = \frac{dV^{\mu}}{d\tau}=\gamma\frac{dV^{\mu}}{dt}##, so that ## A^\mu = \gamma \left( \frac{d\gamma}{dt}\textbf{v} +...

Thanks for your responses! I'm going to need to chew on this for a day or so; once it seems like I understand what's going on, I'll post another attempt at deriving ## A^\mu = (d\textbf{v}/dt,0) ## as well as an attempt at interpreting my previous result correctly.

The four-velocity and four-acceleration of a particle may be written as ## (V^\mu) = \gamma(\textbf{v},c) ## and ## (A^\mu) = \gamma \left(\frac{d\gamma}{dt}\textbf{v}+\gamma\frac{d\textbf{v}}{dt},c\frac{d\gamma}{dt} \right)##. where ## \gamma ## is the Lorentz factor, ## \textbf{v} ## is...

What I understand from these answers is the following: The neutron, left to itself, decays; even inside a nucleus it tends to decay if the nucleus is too neutron-rich. Protons, on the other hand, do not decay on their own (well, some speculative theories say they do, after a really, really long...

I've been trying to understand why adding neutrons to a nucleus will eventually destabilize it; I would like to know if the following explanation is correct: The neutron has a slightly higher mass than the proton. But higher mass translates into higher energy because ## E = mc^2 ##. However...

Does that mean, instead of saying that the photon has spin 1, we should instead say that it has helicity 1? I didn't realize that the representations of the Poincare group for massless particles cannot be labeled by spin!

There's a really good video over at 3Blue1Brown that explains the quantum mechanics of photons passing through polarization filters: They don't mention it, but when light is circularly polarized, each photon has a spin angular momentum of ##+\hbar## or ##-\hbar##, depending on whether the...

Perhaps it's also worth pointing out that quantization (for both particles and fields) works by replacing a classical Poisson bracket with ##1 / i\hbar ## times the commutator, making quantum mechanics and classical mechanics very similar theories from an algebraic standpoint. Also, Lagrangian...

Alright, that's how I'll approach it - thanks!

I really like the idea of presenting the modern formalism the first time around and getting to history later, and starting with light polarization makes a lot of sense to me as well (maybe I'll try that this time around!). In fact, I've been reading all these different presentations of QM...

Now we just say that a particle corresponds to a localized wave, right? I have to admit, I'm still a bit fuzzy about one aspect of this idea. Back in college, I was taught that we only ever measure particles, as when a detector screen lights up only at individual points in the double slit...

What didn't you like about that section?