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I added it in the 8th post .

##(e^{3i\pi}+e^{\frac{i\pi}{3}})=-1+e^{\frac{i\pi}{3}}= -1 +1/2 + \frac{(i \sqrt(3))}{2}=-1/2 + \frac{(i \sqrt(3))}{2} =e^{\frac{2i\pi}{3}}##

Yes the mistake is in the exponent##(e^{3i\pi}+e^{\frac{\pi}{3}})##. It's ##(e^{3i\pi}+e^{\frac{i\pi}{3}})##. I just plugged ##\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}## into ##\sqrt[3\,]...

I think I've clocked it. ## \sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}= \sqrt[3\,]{e^{\frac {2i\pi}{3}}}## ##\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}## ##=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)## ##=e^{\frac...

If I write ##e^{3i\pi}=(e^{i\pi})^3=(-1)^3=-1## Then ##z=-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6} ##

##(-\frac 32 - i \frac{\sqrt 3}{2})^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}## and ##(-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}## When trying to solve ##(-3/2 - \frac{i}{2} \sqrt{3})=(a+bi)^3## I get ##a(a^2-b-2b^2)=\frac{3}{2}## and ##b(2a^2+a-b^2)=\frac{\sqrt{3}}{2}##. How...

How do I find the real part of ## (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}## ? The cubic root makes it hard for me to see what to do here.

My attempt : multiplying numerator and denominator by ## (-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)} wehave \frac{(-1)^{4/9} +2(-1)^{2/9}(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}+ (-3/2 - \frac{i}{2} \sqrt{3})^{(2/3)}}{(-1)^{4/9}+ (-3/2 - \frac{i}{2} \sqrt{3})^{(2/3)}} ##

I don't know how to start with the factorization. $$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$$ Any hints would be nice. Thank you.

So giving a different metric gives us the right topology to make H a Lie group. Thank you.

How can we write the topology change for H as a topological group ?

"The group can, however, be given a different topology, in which the distance between two points is defined as the length of the shortest path in the group joining to " What does this give us in formal math ?

How do we change the topology to make H a Lie group ?

I can't find a proper homeomorphism in ##U_2##.

1-a finite-dimensional real smooth manifold 2- the group operations of multiplication and inversion are smooth maps.

I'm a bit lost, is ##[0,2\pi]-(\mathbb{Q}\cap [0,2\pi])## equal to H or ##\mathbb{P}## ?I've never worked with topological groups,how does a disconnected topological group or set become connected?

How do we change the topology of the subgroup from the original group? Could you show a simple proof of why it becomes connected ?

"The group given by ## H = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0 } \\ { 0 } & { e ^ { 2 \pi i a \theta } } \end{array} \right) | \theta \in \mathbb { R } \right\} \subset \mathbb { T } ^ { 2 } = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0...

I am failing to understand the way to get this answer . Is there a simple computation or property that easily gives this answer ? I tried the formula for ##v_x(Q)## without success.

We have a sample of X, a Normalized Gaussian random variable.We divide the data into positive and negative. Each will have a conditional variance of ## 1−\frac{2}{π}## . Can someone show how to get this result ? I found this problem here (page 3) ...

Yes, so we do have a gain in entropy.Thanks again.

##dS_2+dS_1=-\frac{dmL_f} {T_0}+\frac{dmL_f} {T_0+∆T}=-\frac{dmL_f} {T_0}+\frac{dmL_f} {T_0}(1-\frac{∆T} {T_0})= -dmL_f\frac{∆T}{{T_0}^2}=-\frac{dm(v_l-v_s)∆P} {T_0}=\frac{dm(v_s-v_l)2Mg} {ST_0}=\frac{(v_s-v_l)2Mgh} {v_s T_0} =∆S\; \; \;## since ##v_s=\frac{Sh}{dm}## I made another mistake while...

It's a mistake, it's supposed to be ##v_s##, the ##L##'s are ##L_f##'s also.

##dmL_f= Q \; \; ##,##∆T=\frac{T(v_l-v_s)∆P}{L} \; \;##,##\frac{dmL_f}{T_0}= dS_2 \; \;##,##\frac{dmL_f}{T_1}= dS_1 ##

I guess so.Thank you for these clarifications.

Question 2 : Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma## first then as a function of ##\delta T_v## and ##\gamma##

Since δ##T_V## is a function of T and P there should be a way to change variables P and T in the formula for ##\delta T_S## and express it as a function of only δ##T_V## and ##\gamma##.

Thank you but now I don't see a way to remove T and have ##\delta T_S## as a function of only ##\delta T_V## and ##\gamma ##.

##\delta P_V =(\frac{\partial P}{\partial T} )_V \delta T_V =\frac{nR}{V}\delta T_V##. I don't know how to use ##TP^{\frac{1-\gamma }{\gamma}}=cst## to make ##\delta T_S## a function of ##\delta P_V##,T,P and ##\gamma##.

1- ##\delta P_V =(\frac{\partial P}{\partial T} )_V \delta T_V## 2-##\delta V =(\frac{\partial V}{\partial T})_P \delta T+(\frac{\partial V}{\partial P})_T \delta P## so ##C_v \delta T=-P\delta V=-P((\frac{\partial V}{\partial T})_P \delta T+(\frac{\partial V}{\partial P})_T \delta P)## I can...

If we change the pressure condition below a certain pressure some of the liquid will turn into gas no ?

Perhaps it's an unstable equilibrium.I guess T is close to the evaporation temperature.

We consider a system composed of liquid helium in equilibrium with its vapor at very low temperature T, each phase being considered extensive. We neglect the mass of the gas compared to that of the liquid, as well as the heat capacities of the gas and the walls compared to that of the liquid...

Indeed,Is ##\Delta S_{th}=\mp\frac{Q}{T_0}=\mp\int_{T_i}^{T_0}\frac{C(T)}{T_0}dT ## correct ? If so then question b) is ## \Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T}-\frac{1}{T_0})dT ## when the water heats up and ## \Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T_0}-\frac{1}{T})dT ##...

for a)##\Delta S=\mp \int_{T_i}^{T_0}\frac{C(T)}{T}dT## and ##\Delta S_{th}=\int_{T_i}^{T_0}\frac{dQ}{T_0}dT## so ##S_{univ}=\Delta S_{th}+\Delta S##. What is ##dQ## equal to ? I don't know how to answer question b). Thank you for your help.

Thank you for this explanation,it's clear now.

Hello, I am having trouble understanding the two friction force terms from the ball rolling on page 4 of this physical model: http://www.dewtronics.com/tutorials/roulette/documents/Roulette_Physik.pdf What is the reason for the cos\theta term ? I think the frame of reference is made up of the...

Hello, The state | W \rangle = \frac { 1 } { \sqrt { 3 } } ( | 001 \rangle + | 010 \rangle + | 100 \rangle ) is entangled. The Schmidt decomposition is : What would the Schmidt decomposition be for | W \rangle ? I am also intersted in writing the reduced density matrix but I need the basis...

Hello 1. Homework Statement The dipole moment of an ammonia molecule is ##d_0=5*10^{-30} C.m##.If we apply a static electric field of ##\mathcal { E }=1*10^{6 }V*m^{-1}## to an ammonia molecule initially in the state ## |ψG⟩## where the nitrogen molecule is considered to be on the left,we make...

I think I have an explanation : The generator gives the system {resitance,capacitor} the electric energy : The capacitor charges itself by en absorbing half ( the work ##W=\int vdq##). The difference represents the lost energy through the Joule effect : This energy is received...

I want to know how to express the entropy change , I think the entropy change of the universe is 0 but I'm not sure what the equation is.The RC circuit is getting heat from the reservoir and work from the generator,I'm guessing it's to make it a reversible process.

Hello,I would like some help for a problem Homework Statement Initially:At t=0 [/B]the cylindrical capacitor of capacitance c=\frac{\epsilon s}{d} (d the distance between the 2 electrodes and s their surface; \epsilon = \epsilon(T) is the dielectric permittivity) is discharged and we close the...

Mentor note: Thread moved from the Technical Math section, so there is no template. @yamata1, , in the future, please post homework problems or exercises in the Homework & Coursework sections, not in the Technical Math sections. I have moved your post. Hello, I would like some help with an...

Thank you for your explanations.The same test asked what the wavelength of the photon emmited for the transition between first and second excited states of the hydrogen atom was.The choices were : ##a##- < 652 nm ##b##- >652 nm ##c##- equal to 652 nm ##d##- >241 nm ##e##- equal to 241nm The...

So ##e## is also correct because ##\mp2\hbar## implies j can only be ##2\hbar## but "if,and only if" would exclude ##b##.

If j can take values 0,1or 2 and ##-j<m<j## ,how can knowing m be enough to determine a one-dimensional subspace of the state space ?

since ##J_z|j,m>=m|j,m>## and ##J^2|j,m>=j(j+1)|j,m>## regardless of ##m## we still don't have ##j## ,it cannot be enough to determine a one-dimensional subspace of the state space ?

Measuring ##J_z## does not give us the eigenvalues for the other angular momentum operators,so I think not.

Could ##a##,##c## or ##e## ever be correct ?