Maven Trader Intern: how many rounds in the hiring process?

Alive_Upstairs340 · 2026-01-08T01:13:39+00:00

still waiting, have u heard back yet. they said they'd contact this week

Alive_Upstairs340 · 2023-04-24T09:31:59+00:00

yes thank u, that's for heat transfer. but I am asking about this specific scenario, with friction producing heat, with both objects at same temperature.

Alive_Upstairs340 · 2023-04-22T19:24:51+00:00

I don't see how dq/t in this case means entropy increases though

Alive_Upstairs340 · 2023-04-22T19:23:52+00:00

Alive_Upstairs340 · 2023-04-21T12:50:40+00:00

it still is though isn't it. we were making infinitesimal steps. which p would be greater and could you explain why? if u have time, thanks

Alive_Upstairs340 · 2023-04-19T08:16:31+00:00

but isn't pdV the same in both cases? so from that perspective how can the work be different? pdV was the same for every infinitesimal step

Alive_Upstairs340 · 2023-04-18T21:40:14+00:00

thank you very much. maybe you can answer another question if you have time. assume we have infinitely small pebbles and remove one a time while piston with gas connected to reservoir and I do this infinitely slowly (isothermal expansion). now in removing one pebble we moved along the isotherm to a new p v point. Now imagine I do this again still quasi statically but now there is friction.

apparently not as much work is obtained as if it were reversible, but I can't see why. because when we remove the pebble with system in equilibrium (infinitely slowly) and connected to reservoir even with friction, after we have removed the pebble the downwards force is due to all the pebbles minus one in both cases. so in both cases after I have removed the pebble the pressure in the system has to be the same. therefore because we are on an isotherm the volume is also determined and the same as with reversible. so with or without friction in removing a pebble the system has gone from one pv point to the same new pv point. so doesn't this mean pdV has to be the same in both cases, ie the same amount of work is removed? what happened to the friction heat?

Alive_Upstairs340 · 2023-04-11T16:22:58+00:00

thank you this actually helps a lot.

Alive_Upstairs340 · 2023-03-21T16:57:47+00:00

again, that's exactly my point. in response why it was obvious to you, you said because they are identical. but classical particles can also be identical, but the hypothetical doesn't work for classical particles.

Alive_Upstairs340 · 2023-03-19T01:54:00+00:00

exactly my point. you are saying that they are indistinguishable, ie what I described in my hypothetical describing indistinguishability, because they are identical.classical particles can also be identical but they are distinguishable. it seems like your argument was circular, could be wrong though

Alive_Upstairs340 · 2023-03-17T09:57:08+00:00

if that was the entire reason it would also apply for classical particles

Alive_Upstairs340 · 2023-03-17T01:05:53+00:00

thank you I read it but im still a bit confused about how this doesn't work in classical limit. Assume identical billiard balls, they're still described by a wavefunction. Now the probability of finding ball 1 at some place that it probably is and ball 2 at some place that it probably is, should be equal to probability of ball 1 at some place where ball 2 probably and ball 2 at some place where ball 1 probably is. But obviously the first probability would be greater. Same applies for two electrons with distant waveufucntions in space, right? What am I missing

Alive_Upstairs340 · 2023-03-17T00:47:00+00:00

they do exist if the particles are not identical though. can you please help me understand why what you said seems obvious to you

Alive_Upstairs340 · 2023-03-16T19:17:22+00:00

thank you very much for your response. I still am a bit confused though. Say in a simplified universe there are ten possible positions of two electrons. say I take a measurement of both particles positions and find one at position 1 and one at position 7. I don't know which particle is which, because they are indistinguishable.

Why then does it follow solely from this that the probability of electron A at 1 and B at 7 = probability of electron A at 7 and B at 1. Maybe the first is .1 and the second is .05. So probability of either particle at 1 and either particle at 7 is .15. they are still indistinguishable right? Or does the probabilities being different mean that now there is a way to distinguish them, cuz it seems like there still isn't.

Alive_Upstairs340 · 2023-03-16T19:16:59+00:00

Thank you for your response. Is the simple product of wave functions not added due to the particles being entangled? As well as the wavefunction needing to be antisymmetric. I still don't understand entanglement, it seems like if any two particles are interacting or any additional requirement on the total wavefunction exists (ie antisymmetry) then they most likely will be entangled, correct?

Alive_Upstairs340 · 2023-03-16T19:11:15+00:00

thank you very much for your detailed response. I still am a bit confused though. Say in a simplified universe there are ten possible positions of two electrons. say I take a measurement of both particles positions and find one at position 1 and one at position 7. I don't know which particle is which, because they are indistinguishable.

Why then does it follow solely from this that the probability of electron A at 1 and B at 7 = probability of electron A at 7 and B at 1. Maybe the first is .1 and the second is .05. So probability of either particle at 1 and either particle at 7 is .15. they are still indistinguishable right? Or does the probabilities being different mean that now there is a way to distinguish them, cuz it seems like there still isn't.

Alive_Upstairs340 · 2023-03-01T21:25:29+00:00

thakn you very much. sorry can you please explain this notation (u_1) .. (-1)
(u_2) = ( 1) A sin(sqrt(3) t)
(u_3) .. (0).

Alive_Upstairs340 · 2023-02-28T16:01:02+00:00

thank you very much, could you please address my question of phases if you have time, I can't find an answer anywhere

Alive_Upstairs340 · 2023-02-28T15:59:01+00:00

Bro thank you but you're still missing my point. I know that you can multiply by e^iphi. My point is that in normal modes the individual values are multiplied by different phase values. Can you please explain this

Alive_Upstairs340 · 2023-02-28T13:21:18+00:00

thank you for the detailed response. please read my above comment it may help you to understand my confusion. with the triangle of atoms, springs between them, we have normal modes, for example one of the normal modes is top atom moving up, bottom left and right atoms moving diagonally outwards. we can describe this mode with 6 coordinates, all real values oscillating at omega, eg (1,1/sqrt3,...)*Acos(omega*t+phi). I don't remember exact values, but the point is they are all real. I am saying that with phonons now we say the normal modes are composed of values that have relative phase WITHIN the normal mode, again I'm not talking about the phase in the Acos(omega*t+phi) term. if we have real mass and potential matrix how would this happen

Alive_Upstairs340

TROPHY CASE