papucisse

They are observables. You can measure anyone of them. you're right, it's S+ and S- that are not observables .

Yet Another Question: since Sx, Sy, Sz are not observables, are you sure their hermitian ?

[goodjob] Congratulations on your score, and on your offer from LSU. You should be partying by now [dance]

Kooki3, Most of those schools don't care much about you score after you reach the 70%; Indeed another "top" school (not caltech though) even told me that most of their admitted students in average score in the 40%, especially if you are an american student (or did your undergrad in the US), If you were from china and if you want to do Theory in grad-school it may be a different story. with that said, HOW THE HELL DID U SCORE 800 WITHOUT HAVING E&M? :crazy::crazy: This is unheard of! I know many graduating seniors who have taken Grad-level E&M and Quantum and have scored a lot less (they' re all going to some of the schools you may be considering, just to let you know). Since you seem to be good at this, you should definitely take it again after your first semester E&M, maybe you'll be our next yevgeny (for fun, and of course if you have $115 to spare for the GRE Fee). But I think all you need now is some research experience (possibly this summer through an REU) and a publication will be the icing on the cake. I also understand that nothing or noone is going to make you feel at ease with your "excellent" score, because just a few months ago I was in a similar situation hoping to get in the best school in my field and thinking that my GRE score was going to be an obstacle. But believe me, you have crossed the biggest obstacle by scoring over 70%; and I'm sure you'll be just fine. As for me I can't really tell you my score ( fear embarrasment :drunk:), but I am now attending the #1 graduate program in my field. Hope this note helps a little.

Hello, I m an international student too, studying in the US, with somewhat of a similar situation (different field though. The school I got in turned out to be the #1 in my field, and they're offering full financial coverage, so I guess there is no thinking twice.:p. I m a sole believer that things like that happen for a reason, so enjoy the lemonade you would have made out of life's lemons :cool: But coming back to your question, when you will do your interview (like I did, I dont know if the requirement is the same for your interview) the ambassador will only have information about the one I-20 from the University you will be attending because that's the Info that need to go on your visa... so I don't believe it would make a difference to the embassy if u recieved one I-20 or 20 I-20's because they won't know, and probably won't care:shy:.

This is a slight variation of problem#5 in chapter 3 of Kittel's Intro to solid State Physics (page 93); I modified the problem to adress the specific question I am trying to understand: Consider a line of 2N ions of alternating charge +/- q with a repulsive potential energy A/(R^n) between nearest neighbors (where R represents the distance between particles). determine the energy per particle and the total energy of the crystal. hint: energy per particle would be u®= (attractive term) +(repulsive term) I hope someone can help.

ok thanks ecm; another question: was it from the problem or is it a property of Spin (S)? (i m using griffiths "intro to QM", and Spin is on page 154 if you want to make a reference) Thanks

Hey ECM, welcome back! yeah it makes sense now, thanks; it sort of like the question what is the constant of a spring after the length has been cut in half. I sent you and yevgeny a PM about I will be going for grad school, have you recieved it?

a question: since S=Sx, shouldnt ==0 (no dependence on Sy, and Sz)?

the rest of your reasoning makes sense, but I have a hard time understanding why k' is greater than k, shouldnt it be k'=k*dq/L (since dq/L is the ratio by length representing the small piece) ?

I am not too familiar with this but here is what comes to my mind, perhaps it would be helpful perhaps not. 1- I would try to use the distance traveled (D2-D1) to determine the speed of the source relative to earth (or time if needed ). 2- I would use only D1 for the distance travelled by the radiation. Now, what am I assuming wrong? where can I read more on this?

I finally called ETS last night to get my resultsform the november 8th test. I did somewhat worse than I expected:drunk:, But I still did "OK"; I think I can officially say now that I m done with the physics GRE[dance]. Thanks Ecm and yevgeny for the multiple posts and for solving most of the difficult problems on this forum. I will still be arround cause solving other questions like the GRE's will be a good preparation for the prelims and qualifying exams in grad school. banana rules [banana]!

Read your pm yevgeny

I guess they already took "her" away ( It didnt occur to me to look up your profile and hers :shy:, so to be fair I also updated mine). But where are you ECM, we miss you already :( yevgeny, no I don't have a telescope?; but there is an observatory nearby. what is it in reference to?( or was it a joke that I missed :o)

Felicitaciones ECM[bounce]! I very very happy for you[dance]; For this offer to top going to caltech, it must be really really big! I want to know all about it! (or am I being too noisy? :o ), PM me! I also have something to tell you guys about, it's also a confusing good news :D

Congrats yevgeny, I new you would well[goodjob]! For me it was a bit difficult because there were way too many topics on the test that I had no clue what they were talking about. But I was surprised how concentrated and ready I felt while taking the test. I was able to read over all the test, and solved all the problems I thought were easy (but nothing near the 90 some problems you solved:drunk:). I think I did ok. But now I'm even busier than last week, because I have less than 3 weeks to send my applications to those schools (most of them have a deadline of December 1st for the fall 2004 admission:crazy:). When are you planning to start? I might be taking the general GRE this weekend, that's no biggie (except for the verbal part [xx(]). Meanwhile I still have tons of homework to complete for class, so read you two later. Ps: what are your research experiences?

Thanks ecm, So what did you think yevgeny? I m really glad it's over:D; though I thought it was particularly difficult (compared to the practice exam):crazy:. Ecm, now clock starts rolling for you, but I m convinced you will do well:); so good luck. I will still come to post, to keep you company :cool:

thanks ecm, Here's question 79: It's based on the fact that at high tempereature you have all three groups: Translational, Rotational and vibrational energies that contribute to the internal energy; and at low temperature there is only the translational energy. so for high temperature: Translational: 3 Degrees of Freedom, each contributes (1/2)RT to energy Rotational: 2 DOF(beacuse diatomic=linear shape), each contributes (1/2)RT to the energy Vibrational:(3N-5)DOF (linear shape); since N=2 we get(3*(2)-5)=1 DOF each vibrational DOF contributes (1)RT to the energy. so at high T the internal energy is: U=U(trans) +U(rot) +U(vib)= (3/2)RT+(2/2)RT+(1)RT=(7/2)RT since heat capacity at constant V is Cv=dU/dT (derivative wrt to T) so for high temps Cv1=(7/2)R. And for low temp the internal energy is: U=U(trans)= (3/2)RT; and since Cv=dU/dT; Cv2=(3/2)R so the ratio (Cv1/Cv2)=[(7/2)R/(3/2)R]=7/3 which I believe is the answer in the practice test[dance]. So Remember: at Low temperature only Translational energy, if you boost the temperatur a bit, you get both Translational and Rotational; then only at high temperature do you have all three( Translational, Rotational, and Vibrational). I assume if they change the question to put it on the GRE again, they will tell you at exaclty what temperatures you start getting different energies, since it varies with the gaz. read you soon

Thanks to all of you for all those helpful posts. And physicsgre, your website has been my "homepage" for the more than a month now; Thank you for taking the time to put it all together, and Looks like it is the main website for people who actively study for the Physics GRE, and it's going to get even bigger [goodjob]!

Thanks to all of you for all those helpful posts. And physicsgre, your website has been my "homepage" for the more than a month now; Thank you for taking the time to put it all together, and Looks like it is the main website for people who actively study for the Physics GRE, and it's going to get even bigger [goodjob]!

Thank you, I m a physics major (senior) and, like you, I m trying to work with professors in some very good graduate schools( sorry caltech is not (yet:shy:) one of them but stanford is one in california); So perhaps we will all see each other at the orientations of those schools we're applying to:). Oh yeah, I m from an african country but I'm doing my undergrad in the US. I am sorry, what was I thinking[|)]?; you're right! indeed even for molecules that have more than two atoms and that are linear in shape ( I believe CO2 is an example) there are 2 rotational DOF. so let me rephrase it in case someone else ( like me :p) will be reading the post: Rotational DOF: 0 for Single atoms; 2 for linear molecules, and 3 for non linear (and non-single) molecules. Thank you, Somehow the first time I took the practice test ( and did so badly:() I couldn't understand what most questions were asking. Problem 79 was one of them. But after I read your reply, I went back to redo it and got the right answer (7/3)[dance]; Perhaps I should Take the practice exam again tomorrow, now that I have memorized my basic fomulas, hopefully It will boost my confidence for the exam a bit before the test this saturday You guys have changed my perspectives on the exam, (you, ecm, romad, and physicsgre), thanks a lot! and i'll try to be here more often before this saturday!:D Ps: I enjoyed your question on physicsgre.com [goodjob] . It threw me away at first though cause you said "assume velocity and wavelength are not related". Ok I m going back to memorize some more formulas, read you soon.

I'm not sure I understand the question well :shy:, but it looks like you get the best insulation in case 2 because heat is not transmitted in vacuum. so my guess is: case 1: the heat flows from Highest temperature to lowest case 2: practically no heat flow. Let me know if it helps any or if you already found the answer already;).

Looks like you've been studying hard for this test; I m taking it too this saturday(Nov 8th), and after reading all the messages sent between you and ecm; looks like you two will do in the 90 percentile; For me I m trying to do above average since there are several topics I have not yet covered in class ( I m a first semester senior, in a US institution). But back to your question: Like the energies, you have three groups that contribute to the DOF(Degrees of Freedom): - Translational DOF : 3 (always); - Rotational DOF : 0 (if it is a single atom),1 (for diatomic compounds) and 3 (for compounds made of more than 2 atoms) - Vibrational DOF: (3N-5) for linear molecules, and (3N-6) for non linear molecules; N is the number of atoms in the molecules. remark that if N the total DOF is the addition of all three contributions. May I ask which question on the GRE sample this one is pertinent too? PS:You may want to memorize the corresponding contributions to the internal energy also for the GRE.