Confused by Notation: Need Help Understanding c!

  • Thread starter Thread starter tigger88
  • Start date Start date
  • Tags Tags
    Confused Notation
tigger88
Messages
17
Reaction score
0

Homework Statement



I'm just confused by notation in this question, because the textbook I use has different notation than my course notes and now I'm just not sure what anything means.

Homework Equations



Group velocity x Phase velocity = c^2

The Attempt at a Solution



What is c in this case? Is it the speed of light, or some other velocity?
I know it's a really stupid question...

Ta for any help though.
 
Physics news on Phys.org
Well, I assume that it is the speed of light. But really, the equation is not true in general.
Yes, photons move at c in free space, so phase velocity is c in free space (but inside midiums, the speed would get lower.).
However, the group velocity, could actually be lower than c. even 0. Just for reference, this is called EIT.
So if it is someother speed, well, normally it would be denoted as v, instead of c. c is normally reserved for the speed of light, or constant.
I guess what the most important part of this equation is that group velocity and phase velocity are the same in most of the cases: so if it is light, then it is c^2; if it is something else, then it is v^2. It is kinda open interpretation.
 
According to Wikipedia it is indeed the speed of light. I've never seen that particular equation before myself, but the derivation on Wikipedia looks reasonable...
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top