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RewCie

5:48 AM

Good Morning

I hope Elon Musk doesn't gets to know how much energy does a traditional transaction in USD gets consumed

woke

123

6:16 AM

Hi All...

What is the meaning of First Law of Thermodynamics in this form. U = Q + W if there is not change in internal energy.

If U is internal energy of system at thermodynamic equilibrium than why we represent this current state of system in term of work and heat because both are process function and there is no work and heat at equilibrium.

2 hours later…

Slereah

8:09 AM

the changes in internal energy happen outside of equilibrium

Thermodynamics also deal with out of equilibrium transformations

it's trickier to deal with but still within its domain

123

@Slereah How we understand U = Q + W, i can understand dU = dQ + dW

Slereah

The actual formula is $$dU = T dS - P dV$$

So that $Q = T dS$ and $W = P dV$

$Q$ cannot be expressed generally as a total derivative, though

123

8:25 AM

@Slereah It means it is incorrect to write U = Q + W

Slereah

I mean you can integrate both sides if you want, but yeah beware of the usual warning

Integrating $T dS$ and $P dV$ is path dependent

123

I wanted to understand U is the existing internal energy of thermodynamic system and dU is infinitesimal change in internal energy. dU is caused by dQ and dW separately or simultaneously. My question is that if there is no change in internal energy we can write U. How U = Q + W

I assume Q and dQ, W and dW are different things

How existing internal energy U in thermodynamic equilibrium can be equal to work and heat. because in equilibrium there should be no heat and work.

Slereah

Well, it's an integration

Therefore, there is a constant of integration

If you integrate it out it's gonna be something like $U_0 + \Delta U$

123

@Slereah Pls explain me first what is Q and dQ, W and dW, U and dU. then i am able to understand what is going on in the system. Also i understand $U_o$ should be integration constant.

Let say Q is constant heat supplied/exhaust within the system e.g: 400J over time and dQ is change in heat e.g: 400J to 400.01J or 400J to 399.99J in infinitesimal time? Am i correct or wrong?

Slereah

8:40 AM

Well thermodynamics isn't usually done with time as a variable, the integration is usually done over the thermodynamic variables

ie temperature, volume, etc

123

@Slereah Okayy... what about my Q and dQ perspective.

Slereah

Well depending on the equation of state, $\delta Q$ will be some function of the thermodynamics variables

123

because Q = TdS it means there change in entropy happening. It means constant heat is entering or exiting from the system.

Slereah

For instance, if you take an ideal gas with an isobaric change of temperature, the heat is $$\delta Q = \frac{5}{2} Nk dT$$

123

Does $Q$ and $\delta{Q}$ means the same thing? $Q = \delta{Q}$

Slereah

8:45 AM

Well

That is where things are a bit vague usually

Physicists tend to write a bit confusingly with respect to the "inexact differentials"

123

Does $\delta{Q}$ means virtual heat like virtual work?

Slereah

Some don't bother and just write $dQ = T dS$, some differentiate it with $\delta Q = T dS$, and some just write it as if it were a differential form as $Q = T dS$

123

Now i am more confused now there are three representation of heat $Q$ , $dQ$ , $\delta{Q}$. Does all are equal or different.

Slereah

as I said, it's kind of a wash

Different physicists use different notations

You just have to watch out for it

123

It means equation of state tell us what is the meaning? because we used it interchangeably. Book can use any notation there is no clear distinction between notation?

If this is the case it makes thermodynamic subject more difficult to understand.

Slereah

8:51 AM

May 12 at 12:28, by bolbteppa

"Every mathematician knows it is impossible to understand an elementary course in thermodynamics." - V. I. Arnold

See this

123

:P i think this sentence should not only applied to mathematician it is also applied to physicist.

Slereah

I think people are overcomplicating things a bit overall

Don't think about $Q$ at all, in my opinion, just write $dU = T dS - P dV$

123

What about $dU = TdS - dPV$ . Is it also correct represent the work ($W = dPV$)

Pls correct me. Physicist use $Q$ , $dQ$ , $\delta{Q}$ interchangeably any time but equation of state tells us what is that means. Am i correct or wrong?

2 hours later…

Slereah

10:55 AM

Why do some physicists have such bold claims

"We definately live in a $(4,4)$ supermanifold"

ACuriousMind

@Slereah I think the overconfidence of many theorists in the accuracy of their speculative theories is a by-product of the competitive academic environment - if you don't believe in what you're doing, you won't be as motivated to go to the lengths that are necessary to compete.

and by the time you have "succeeded" and have tenure or whatever, you have internalized that mode of overconfidence so much you can't easily turn it off again

Slereah

I can assert with confidence that the universe is most definately a triangle

Pythagoras can probably back me up on this

ACuriousMind

depends, do you like beans?

Slereah

I remember that some cosmologist used a peanut as a basic shape

I like how in the 19th century there was the idea that if you found some limiting process where the idea made sense, it would work

But then again I guess that's still a thing and now we call it renormalization

4 hours later…

bolbteppa

3:11 PM

@Slereah "1500 years ago, everybody knew that the earth was the center of the universe. 500 years ago, everybody knew that the earth was flat. And 15 minutes ago, you knew that humans were alone on this planet. Imagine what you'll know tomorrow"

Slereah

Hawking was kind of like that

One of his paper was like "Any good theory of quantum gravity needs euclidian wormholes!"

bolbteppa

Q: Do spinors form a vector space?

In contradiction to a number of other authors (sample ref below), Gerrit Coddens, at France’s prestigious Ecole Poytechnique, asserts that: 2.2 Preliminary caveat: Spinors do not build a vector space As we will see, spinors in $SU(2)$ do not build a vector space but a curved manifold. This is alm...

This is a weird one

ACuriousMind

gotta love people redefining words and then claiming everyone else is using them wrong :P

whatever Codden is doing, calling his things "spinors" is just silly and designed to produce maximal confusion among people using the standard definition

bolbteppa

3:31 PM

I can see some (very) iffy statements but there is something weird going on here for sure

Pure spinor

In the domain of mathematics known as representation theory, pure spinors (or simple spinors) are spinors that are annihilated under the Clifford action by a maximal isotropic subspace of the space V {\displaystyle V} of vectors with respect to the scalar product determining the Clifford algebra. They were introduced by Élie Cartan in the 1930s to classify complex structures. Pure spinors were a key ingredient in the study of spin geometry and twistor theory, introduced by Roger Penrose in the 1960s. == Definition == Consider a complex vector space...

2 hours later…

John Rennie

5:06 PM

From this week's New Scientist.

It's an alarming accurate description of life as an experimental physicist, marred only by its failure to mention that the graph is invariably a log-log plot.

Physics Meta

Q: Could we link the "Image equation to TeX" question from TeX SE somewhere?

It has been discussed before that one should not post images of text or formulas. However, many users still do – probably because they consider typing all this text too much work, which to some degree is understandable. I (probably) had the same problem and then found this question on TeX SE: Ima...