Working through the numbers, I have got an answer of $$c_{1,0}=0$$ $$c_{1,1}=-\frac{1}{\sqrt{2}}$$ and $$c_{1,-1}=\frac{1}{\sqrt{2}}$$
Which means the answer to i. is 1/2, ii. is 1/2+1/2 = 1
But how do I calculate the probability of measuring Lz=0?
The probability of m=1 AND l =1 was 1/2 (from...
Sorry! Edited my post, it was meant to be 2ħ2.
I am having trouble understanding the integral though... What are Yl,m... and dΩ :(
I have a table with values of Y0,0, Y1,0, Y1,1 etc...
So for example $$Y_{0,0} = \sqrt{\frac{1}{4 \pi}}$$
$$c_{0,0} = \int \sqrt{\frac{1}{4 \pi}} \sqrt{\frac{3}{4...
Homework Statement
Consider an electron in a state described by angular wavefunction $$\psi(\theta,\phi)=\sqrt{\frac{3}{4 \pi}}\sin \theta \cos \phi$$ Here θ and φ are the polar and azimuthal angles, respectively, in the spherical coordinate system.
i. Calculate the probability that a...
Yeah, that seems to my solution too, though I divided it by ##c^2## so my answer was $$\frac{28GeV}{(3*10^8)^2}*i*7.06*10^{-3}=i*2.20*10^{-18} GeV/c^2$$
Fair enough, I know the calculations I did after it were wrong though, so ignore them, I'm sure you have the right answer ! I will look through what I have done in the morning to try and work it out. Thank you for all your help, both of you !
If the neutrino arrived ##60.7*10^{-9}s## before light would, then ##t_2-t_1=\frac{d}{v}-\frac{d}{c}=-60.7*10^{-9}s##
In which case ##t_2-t_1+\frac{d}{c}=\frac{d}{v}##
$$v=\frac{d}{t_2-t_1+\frac{d}{c}}=\frac{731*10^3}{-60.7*10^{-9}+\frac{731*10^3}{3*10^8}} = 300 007 473.5 m.s^{-1}$$
Atleast...
Do you mean for part (b) ?
I have done $$v=c\sqrt{1-\frac{m^2c^4}{E^2}}$$
$$v^2=c^2-\frac{m^2c^6}{E^2}$$
$$m^2=\frac{E^2(c^2-v^2)}{c^6}$$
$$m=\frac{1}{c^2}\sqrt{E^2-\frac{E^2v^2}{c^2}}$$
Plugged in numbers and got ...
Ok, even then I'm not sure how to differentiate this expression with respect to p
I can re-arrange to get ##E=\sqrt{m^2c^2+p^2c^2}## but I don't know how to differentiate that wrt p
I do understand that, which is partly why I don't understand what the question is asking me...
Any ideas on how to solve this? I don't even understand how to do part (a) at this point
Homework Statement
In 2011, researchers at the OPERA experiment thought they had seen neutrinos with mass m and energy E = 28 GeV moving faster than light. The baseline between the source and the detector was 731 km, and the neutrinos seemed to arrive 60.7 ns early, compared to the maximum...
Is it simply ##x'=\frac{x}{\gamma}=\frac{9.3*10^{20}}{1*10^{12}}=9.3*10^{8}m = 3*10^{-8} pc## then?
In which case (c) would also simply be ##t=\frac{t'}{\gamma} = \frac{(3.1*10^{12}+1.55*10^{-12})}{1*10^{12}} ≈ 3.1 s## ?
Doesn't seem right to me..
EDIT ...
Homework Statement
The highest energy protons have gamma factors around ##1.0*10^{12}##.
(a) Our galaxy has a disk diameter of 30 kpc, which is ##9.3*10^{20}m##. If a photon and one of these high energy protons start traversing the galaxy at the same time, by how long will the arrival of the...
I follow the reasoning for this.
The radioactive decay equation gives us ##N=N_0e^{\frac{-t'}{t_0}} \rightarrow t'=2.21*10^{-8}s##
This is the time in the rest frame of the muon.
Then using the Lorentz transformations : ##t = \gamma t'##
##v=\frac{d}{t} \rightarrow v=\frac{d}{\gamma t'}##
How...
##\gamma mc^2-mc^2 = \frac{1}{2}mv^2 \rightarrow \gamma mc^2 -mc^2=\frac{1}{2}m\beta^2c^2##
where ##\gamma = \frac{1}{2}\beta^2+1##
Therefore ##\frac{1}{2}\beta^2=\gamma-1##
Which means that $$\frac{1}{2}mv^2=\frac{1}{2}m\beta^2c^2=mc^2(\gamma-1)$$
$$\gamma mc^2-mc^2 = mc^2(\gamma-1)$$
Is that...