Consider the first law of thermodynamics, $\Delta U = Q-W$. How come one can write this in "infinitesimal" form, that is $dU = dQ-dW$? Since $Q$ and $W$ aren't "delta quantities", what is the mathematical justification behind this?

@schn it's ad hoc

mathematicians don't use infinitesimals

well not in the same situations that physicists do

I would say that the 'infinitesimal' version is just a shorthand for the Delta U = Q-W version

that said, what -is- legit is $dU=T dS-pdV$

You sometimes see dQ = TdS with a little bar over the d in dQ

to emphasize that it's something you integrate along a given path to get the heat transfer, but for which there's no state function Q

Why so silence here?

Does anyone has some cool topic to talk here?

What's your Sunday plans?

anything special on sundays?

@AbhasKumarSinha It's quiet on Sundays because the people procrastinating at work aren't here ;)

@ACuriousMind That's good, if you are not wasting your Sunday, then you are wasting your Sunday...

what are some good "bibles" textbooks in computer science? As physics student outsider, of course

@thermomagneticcondensedboson google, wikipedia i think

@thermomagneticcondensedboson What sort of computer science? It's a broad field

This Sunday is basically a Saturday for me. A lot of Canada has a holiday tomorrow for Remembrance day.

@JMac holidays in India - timeanddate.com/holidays/india

@AbhasKumarSinha Oh yes! I had something special this Sunday.

@ACuriousMind Just to get some knowledge outside physics

Anybody got any ideas with this?

i think I have to pull the Hamiltonian in but I’m not sure how...

@JakeRose Differentiate the given equation w.r.t. time and use Heisenberg's e.o.m.

Once you simplify the commutators, you'll get an equation that involves p but not x.

hello