1:05 AM
-3

Before you downvote me please read. Lets say someone uses Einstein equations to calculate c as non constant using this https://en.m.wikipedia.org/wiki/Einstein_field_equations#Mathematical_form Would that explain cosmic expansion? And would that fix the "dark energy" problem?

7 hours later…
8:24 AM
-1

Some time ago I went down a long (10 kilometres) smooth concrete pipe about 3 metres in diameter. What I noticed from a handclap some distance away, the high-frequency components arrived first and the lower ones later. This descending 'whistle' made speech unintelligible over this distance. Since...

2 hours later…
9:59 AM
@JohnRennie good afternoon sir.

@JohanLiebert hi :-)

Sir I have a question. Can you have a look at this: physics.stackexchange.com/q/565035/249968

The collisions of gas molecules with the walls are elastic on average.
That is, sometimes the gas molecule transfers some of its KE to vibrational energy of atms in the walls, and sometimes the atoms in the walls transfer some of their vibrational energy to the gas atom.
The vibrations of the atoms in the walls are what we call the heat stored in the walls. At any temperature above absolute zero the atoms in a solid are vibrating with an energy of order $kT$.
@JohanLiebert OK so far?

@JohnRennie yes, sir.
@JohnRennie Sir, then why was it necessary to make this a postulate of the KTG?

Suppose the walls are very cold so the vibrational energy is much less than the KE of the gas atoms. In that case it is most probable that gas atoms colliding with the wall will transfer KE to the wall. This means the wall heats up due to increased vibrational energy, and it means the gas cools down due to reduced KE.
So the end result is that the hot gas heats up the cold walls.
@JohanLiebert you mean why do we assume the collisions are elastic on average?

10:14 AM
@JohnRennie sorry sir, incorrect wording. I meant what would be the implications hadn't we assumed that the collisions are elastic (in KTG)?
I mean what is the necessity of that postulate then?

Are we still talking about collisions with the wall, or do you mean collisions between gas atoms?

@JohnRennie any type of collisions.

Consider gas molecules first because that's simpler.

@JohnRennie ok

Energy has to be conserved in collisions between gas atoms because there is nowhere else for it to go. Energy can't just disappear so the total energy before the collision must be the same as the total energy after the collision.

10:18 AM
@JohnRennie yes

So for atoms collisions have to be elastic. It isn't an assumption so much as a statement of the obvious.
The difference with solids (and liquids) is they have vibrational modes that can store energy. Energy is still conserved, but it can seem to disappear because it gets transferred into vibrational energy.
That's why when we consider collisions of gas atoms with a solid surface the KE of the gas atom is not conserved.

@JohnRennie ok

On average the collisions of gas atoms with the wall are elastic if, and only if, the gas and the wall have the same temperature.

Ok, understood.
Sir I have a bit of general question.

Yes ... ?

10:24 AM
When making a theoretical model what criteria must a postulate follow to be a part of our model?

I think that's to general for me to give a useful answer
When you are building a model you can make any postulates you want.
If your model contradicts experiment, or if it turns out to have internal inconsistencies then your postulates must be wrong, but otherwise anything goes.

Sir are you available on Telegram?

No

Sir to contact via hangout what do I require to know about you. Do I need to know your contact number?

@JohanLiebert You usually don't go "I want to make a model" and then pick postulates from some random list from which you could select based on criteria. You have a problem of some sort you want to solve and you look for postulates from which you can derive a solution, or you have experimental data (e.g. "particles seem to behave like waves in the double slit") that you then encode that as a bit more general postulate to see what else might follow from it.

10:31 AM
@JohanLiebert just open Hangouts, click the start new conversation link and type John Rennie.

@JohnRennie ok.
@ACuriousMind so for example, I am making a model and have laid out some postulates but there are somethings which are obvious to other physicist but they aren't implicit from my earlier postulates. Then should I consider to add them as postulate?
Or would it be verbose to do so?
@JohnRennie sir, I tried doing that but it says that no such name found in "my contacts".

10:48 AM
@JohanLiebert Try my email. Make a note of this because I'll delete it in a few seconds. Ready?

Yes
Done.

I'm in Hangouts now but I don't see a message from you ...

@JohnRennie it says you have not accepted my invitation yet.

11:11 AM
@JohanLiebert That's too general to be an example. Why are you making a model? What do you want to explain or explore? What are your goals?
Newtonian mechanics is a model. It does not contain things which are obvious to physicists today, such as that the world is quantum. That does not mean we should always use quantum mechanics, because we conceive of Newtonian mechanics as a model intended to explain and predict the movement of macroscopic matter in situations where the effects of the things we leave out are neglegible.
Models are contextual, they are not about what the world "really" is like - whatever that means - but about being useful in a specific set of situations
Often this is expressed as "All models are wrong, some are useful", which is a nice aphorism but implicitly contains an idea of "right" and "wrong" as different from "useful" and "not useful" which remains vague

11:31 AM
I'm tempted to go on another rant about epistemology but I've done that here often enough :P

11:44 AM
I have a question that I asked in here a long time ago and got an answer that I didn't expect but assumed would make sense eventually. If the two metric conventions differ by a sign, $\text{diag}(+,-,-,-)=-\text{diag}(-,+,+,+)$, why doesn't the sign in the field equations depend on the metric convention being used? $$R_{\mu\nu}\pm \frac{1}{2}R g_{\mu\nu}.$$
I was told it does not change the sign, and there is always a negative sign

Why do you think it should change?

It might not be a particularly compelling argument, but I just thought $$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\rightarrow R_{\mu\nu}-\frac{1}{2}R(-g_{\mu\nu})=R_{\mu\nu}+\frac{1}{2}Rg_{\mu\nu}$$

@Charlie $R_{\mu\nu}$ depends on the metric too, doesn't it?

If I had to take a guess at the answer, I'd say that changing the sign on the metric also affects $R_{\mu\nu}$
aha
If that's the case, my only follow up issue is that $T_{\mu\nu}$ isn't built from the metric
so do we still not pick up an overall $-$?
Unless the most general form of $T_{\mu\nu}$ does come from the metric, I guess the only form I've been show so far is $$T_{\mu\nu}=\vec p \otimes n\vec U.$$

@Charlie There is more to sign conventions in GR than the sign of the metric, see en.wikipedia.org/wiki/Einstein_field_equations#Sign_convention

12:02 PM
Ah I see
ok I'm happy with that, I was under the impression that there was no flexibility but it seems that was wrong
ty

My professor makes strange statements like you can't proof that all EM waves travel at the speed of light I'm asking if this is common in other universities, I mean to have these types of the so called "professors"

@AbdullahO.Alfaqir It is impossible to measure anything in the real world to absolute precision, there is always some margin of error, maybe that's what he's referring to

I was just surprised because maxwell has already shown that all EM waves travel at C

If we can't exactly measure anything, aren't we like, floating in a universe of vagueness and blurriness... lol I dont' know what I'm saying

Predictions made by models in science should not be interpreted as "this is exactly how the world works" the prediction that the speed of light is a constant is consistent with all experimental evidence so far (to my knowledge), but it is not inconceivable that this is simply because our experiments can't measure precisely enough the deviation from $c$.

12:10 PM
Is the concept of something being accurate even exist?

that's probably metaphysics territory @JingleBells

Hmm, but if we can't measure something precisely, does that mean it doesn't have a precise value... pff my brain hurts
Alright, i'm out

12:58 PM
In the classical world, the concept of accuracy was easy - you could believe in realism, that all physical quantities have definite real values and that any deviations from the theoretical prediction would be due to the measurement apparatus not being ideal.
In the quantum world, you have Bell's theorem and the problems for realism that come with it - if you continue to believe in definite real values you have to give up your belief in locality. And then we're firmly in quantum interpretation territory.

1:38 PM
Are fields material or are they only mathematical tools? For example, if an electromagnetic field is only a mathematical tool then the force it generates on a particle is acting at a distance, but if it's a property of the space it lies on then there's no action at a distance

0

i humbly request this question: Thoughts on neural networks "discovering" physical concepts to be reopened. i realize that there was a "what is your opinion..." line in it earlier. i have removed it. please have a look at it in its current form and i hope you will agree that it admits concrete an...

8 hours later…
9:26 PM
@Sophie This question doesn't have a satisfying answer, the bridge between maths and physics is that the relationships that exist between measurable, physical quantities can be modelled by the relationship between certain mathematical objects. It doesn't make sense to ask whether fields are pretty much everything in physics is a "mathematical tool".