Recent content by PhysicsGirl90

Ok, I will try it and see what i get

Hello all, I am having some trouble calculating a convolution. For the question, and my attempt at a solution, please take a look at the pic. Can someone please point me in the right direction?

Thanks again tiny-tim and thank you DrClaude for your help

Hey tiny-tim, Thanks for your suggestion. I tried it but i get stuck trying to get the same equation as the text. I have included what i got so far in the picture. Can you give it a look and tell me what I am doing wrong?

Reading through my QM text, I came across this short piece on ladder operators that is giving me trouble (see picture). What I am struggling with is how to get to equations 2 and 3 from equation 1. Can someone point me in the right direction? Where does the i infront of the x go?

While practicing for an upcoming exam, i ran into this question that has truly got me stumped. A phonon dispersion graph is given for NiAl (attached), and then they ask you to schematicaly plot the density of modes graph. I have the result of what is should look like, but i do not understand how...

I was reading through a book on statistical physics when i came across this sentence: "An Einstein solid has two degrees of freedom for every oscillator." How is this possible? I picture an oscillator (ex. mass on spring) to move only in one dimension, thus one degree of freedom. Where does...

(1/44)^(10^5) = (10^log(1/44))^(10^5)...Thanks for helping me think it through.

Thank you for your comment Bruce...i figured it out...convert (1/44) to a power of 10 and the rest follows easily.

They have a different base. So if we wanted to compare them we they would both have to have the same base.

Working on statistical physics i came across this expression: p = (1/44)^(10^5) = 10^(-164345) However TI-83 calculator is unable to verify it (gives answer 0). Can someone tell me how to get from (1/44)^(10^5) to 10^(-164345) analytically?

I don't think this excersise can be done...however i flip or turn it i always end up with M= Hi/Ho = 5...but with x/y = 5 we have two variables and one equation. I think the teacher made a mistake when he gave this excersise out.

That is the second formula M = s'/s = Hi/Ho...but i only have s' and s, and i have to determine Hi

Homework Statement With a lens with f = 25 cm, we want to project an image that is 5 times bigger than object, so M = 5 - Find the image distance s' - Find the image height Hi Homework Equations 1/f = 1/s + 1/s' M = s'/s = Hi/Ho With: f = focus distance of lens M =...

Thanks Bruce, with your hints about the vectors I finally understood where everything comes from.