Yes, that clears it up completely, thank you.
I can now actually see the part in the book where the asterisk denoting that the initial setup involved a complex conjugate has been worn away due to shoddy printing, so I feel a lot better about this whole thing in general now.
Thanks so much!
\langle Z| should just equal
\langle V| - \frac{\langle W|V\rangle}{|W|^2} \langle W|
right? So if I look at \langle Z|Z \rangle, I have to distribute, but I can't see how he's getting the terms he's getting, namely because I don't think I'm too comfortable with Dirac notation yet.
1...
Homework Statement
This isn't really a problem so much as me not being able to see how a proof has proceeded. I've only just today learned about Dirac notation so I'm not too good at actually working with it. The proof in the book is:
|Z> = |V> - <W|V>/|W|^2|W>
<Z|Z> = <V - ( <W|V>/|W|^2 ) W|...
This change gave me 3,000,000 \frac{kg}{yr} adjusting for sig. figures, which is a much more believable answer.
Doing the math for part ii tells me that it'll take ~10^{18} years for the Earth to increase its mass by 50%, a much less worrisome time frame.
Thank all of you for your time.
Then I'd adjust the mass to 3×10^{-6} and do what I was doing before,
particle number density × mass per particle
(\frac{1×10^{17}}{km^3}) × (3×10^{-6}g)
= \frac{0.3kg}{m^3} the amount of mass the Earth gains per m^{3}
Then I multiply this by the volume of the Earth-cylinder...
then is this right?
\frac{4}{3}\pi r^{3} = \frac{4}{3}\pi 50\mu m^{3}
= 5×10^{-7}cc
\frac{5g}{1cc} = \frac{xg}{5.0×10^{-7}}
x = 5×(5.0×10^{-7})
x = 2.6 × 10^{-6}g = the mass per particle
Because this seems a lot more like what the mass of a particle should be.
So would it be right to say that since the particle number density is 100cm^{-3}:
\frac{m}{v} = \frac{5g}{1cc}
Then there must be 5 grams of mass per 100 particles? That gives .05g per particle which is different but I don't think that completely fixes the answer, and I still can't see...
Homework Statement
i) Estimate (in kg/yr) the rate of Earth accreting inter-planetary dust.
Assume a particle number density of n=1e-8/m3, a particle mass density of 5 gm/cc, a particle radius 50 microns and a relative velocity equal to Earth's orbital velocity (so the dust particles are, on...
Oh okay I see now. I was so sure e was the base of the natural log. I actually tried looking around for other things it could stand for, but putting it in as the electron charge on wolfram made it work.
Thanks!
I think this is maybe where I'm missing something?
A tesla divided by an electron mass is giving me 1 per second per ampere, and wolfram seems to agree with this.
\nu_{0} + s^{-1}A^{-1} = Hz + s^{-1}A^{-1} is adding incompatible units, so I'm pretty sure I can't do it.
If I calculate the frequency based off the given wavelength, I get 4.56*10^14 hertz, but that doesn't get me any further I don't think.
The problem is the difference between the initial frequency and the satellite lines. I can't add or subtract the difference because the difference isn't a...
Homework Statement
"The normal Zeeman effect splits a spectral line at frequency \nu_{0} and two satellite lines at \nu_{0} ± eB/(4\pi m_{e}). By what amount (in angstroms) are the satellite lines of the hydrogen Balmer \alpha line (\lambda_{0} = 6562.81 Å) split from the central component...