The h Bar

barrycarter

12:05 AM

@Obliv I'm off. If you need more help, feel free to google chat me at carter.barry@gmail.com

skill patrol

12:22 AM

@DanielSank how about a star?

:-)

0celo7

1:39 AM

@ACuriousMind Really skinny people are so smug

Like where do they put their organs

Anthony

Hey @0celo7, temperature is usually called avg energy per degree of freedom- is this true in quantum as well?

Obliv

Is this a correct definition of the image of a set A onto a set B? $f(A) = \{b \in B | b = f(a), $ for some $ a \in A\}$

If so, is it essentially describing this: the elements of $A$ that are transformed by the function $f$ ,which are a subset of B, are mapped to the set $f(A)$?

DanielSank

2:01 AM

@skillpatrol Heh.

@Anthony That's not a particularly great definition.

You are likely referring to the equipartition theorem.

Note that the usual form only works for degrees of freedom for which the energy is quadratic.

user116211

2:15 AM

user116211

Check this pulse....what are these blue and red arrows?

user116211

I've found it in PSE post.... it was written that they are restoring forces.....

user116211

But aren't they transverse velocity distribution of the pulse?

Anthony

@DanielSank could you give me a better definition?

0celo7

@Obliv I think $f(A)=\{f(a)\mid a\in A\}$ is better.

DanielSank

2:46 AM

$1/T = dS/dE$ is a great definition.

0celo7

@DanielSank what is $S$

what is $E$

can you guarantee $S$ is $C^1$?

3075

Is it frowned upon to use sans serif fonts for lecture notes?

user116211

@0celo7 Entropy and internal energy resp.?

0celo7

@MAFIA36790 what's that

user116211

Are you trolling me now?

0celo7

3:00 AM

Am I ever not trolling?

What is the Socratic method if not trolling?

user116211

Ya! Ya!

Anthony

3:26 AM

@DanielSank But then can't I ask what entropy is?

DanielSank

3:37 AM

@Anthony Yes but that's simple: it's the logarithm of the number of possible states a physical system can have given whatever conditions you want to specify, i.e. volume, energy, etc.

@Anthony You may find this interesting:

14

Q: Is the Boltzmann constant really that important?

I read a book in which one chapter gave a speech about the fundamental constants of the Universe, and I remember it stated this: If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even just slightly different (smaller or bigger) than what they ac...

Anthony

Gracias senor.

@DanielSank But then the avg energy per dof is just a classical coincidence, as it were?

DanielSank

@Anthony Well, I'd say it's largely due to the fact that near equilibrium almost any energy potential is quadratic.

The equipartition theorem and the notion that temperature is proportional to the average energy in each dof is very widely applicable.

You're first "definition" is pretty reasonable, it's just that it fails at really low temperature or when you have a system which is not much like its classical equivalent.

Spins, for example, are pretty non-classical at any temperature.

Er, well, perhaps I shouldn't say that.

user116211

The famous lines by CuriousOne:

user116211

Second law of thermodynamics: heat flows only from hot to cold (unless something else happens... like the refrigerator is being supplied with electricity from the power plant). Two temperatures are equal when there is no heat flow. No need for entropy and all that. Just measure heat flow. In practice we use small thermodynamic machines called thermocouples. When they show zero voltage difference between two thermal baths, then the temperatures are equal. — CuriousOne Feb 9 at 8:32

user116211

@Anthony: ^^^

Anthony

3:52 AM

@DanielSank But why is it that the change in energy with respect to entropy is the same as this classical definition? I can't make sense of how the two relate.

@MAFIA36790 I'm not sure how that helps. :P

user116211

hmmm.....

DanielSank

@MAFIA36790 Yeah well, I strongly disagree with CuriosOne's opinions on how to think about and teach statistical/thermal physics.

CuriousOne goes around this site making strong statements about all kinds of stuff and I am compelled so say that I very often disagree. Not always, but often.

user116211

@DanielSank Yes, that's why I mentioned famous.... when he sees temperature, he always comes up with these same words.

DanielSank

@Anthony That's what the equipartition theorem tells you.

Have you ever tried to derive that theorem? It's surprisingly easy.

Anthony

I saw it the other day, I can look at it again. I guess it's just weird to me. I should probably think about some other things for a little while.

DanielSank

3:55 AM

@Anthony Here's how I personally think about temperature:

As I said, 1/T = dS/dE, but what does that mean?

Entropy is the number of possible arrangements of a physical system.

Often, but not always, if I give a system more energy, the number of arrangements it can have goes up.

user116211

@Anthony: What's exactly bothering you? I'm not seeing any wrong in Daniel's argument so far?

DanielSank

For example, a particle free in space has more possible momenta if it has more energy (because the number of momenta is the surface of a sphere of radius p = sqrt(2Em).

So, normally, if I give a system more energy, it's S goes up.

Now suppose I have two systems.

Now consider taking some energy from system A and giving it to B.

Normally, B's entropy goes up and A's goes down.

If B's would go up by more than A's would go down, then this exchange of energy will happen statistically speaking, because it leads to more possible arrangements of the coupled systems.

Anthony

@MAFIA36790 Nothing, just kind of mental haze from thinking empty thoughts.

DanielSank

So basically we just said that if dS_B / dE_B > dS_A / dE_A, then energy does from A to B.

Anthony

That's not statistically, is it? That follows from assuming entropy is maximized, right?

DanielSank

3:59 AM

But that translates to 1/T_B > 1/T_A which is equivalent to T_B < T_A. In other words, the hotter thing gives energy to the cooler thing.

@Anthony Yeah but entropy maximizes only in a statistical sense. It doesn't have to happen, it's just extremely overwhelmingly likely to happen.

Basically, the point of this is that "hot" means "likely to give off energy to something else".

...which comes down to how to distribute energy to maximize entropy.

For me, this way of thinking is pretty useful.

Anthony

Alright.

Thank you!

One more question.

DanielSank

Sure.

go for it

Anthony

Can the maxwell boltzman distribution come from using boltzman coefficients?

DanielSank

@Anthony What are Boltzmann coefficients?

Anthony

er, the exponentials in the partition function

probably not the right name

DanielSank

4:01 AM

Oh right.

user116211

I've to google this :(

user116211

oh! partition function!

Anthony

my confusion is in that lower velocities and higher velocities are both unlikely to occur

DanielSank

@Anthony Well, you can derive pretty much anything from the Boltzmann factors.

Anthony

I feel like it should be monotonic

user116211

4:02 AM

@Anthony right?

Anthony

@MAFIA36790 Hmm?

DanielSank

@Anthony Hahaha, nope! Un-intuitive isn't it!

Anthony

@DanielSank But why isn't it?

user116211

@Anthony That's what the distribution looks like, isn't it?

DanielSank

@Anthony Let me give you a very simple example to illustrate why.

Suppose I have a random variable in a 2D plane such that the x and y coordinates are both Gaussian distributed (and independent).

Anthony

4:03 AM

@MAFIA36790 I'm not sure what you're saying

DanielSank

Can you imagine what I'm describing?

Anthony

Sure

it's just a 2d gaussian right?

DanielSank

@Anthony Sure, call it that if you want.

exp(-(x^2 + y^2) / 2 sigma^2)

Anthony

Yeah

DanielSank

Ok, now suppose we draw a bunch of points from this distribution and then plot a histogram of the square of the radius of each point.

What do you think that distribution looks like?

user116211

4:04 AM

You said the lower velocity and higher velocity are unlikely..... well, this is what the distribution function says; that's what I'm saying.

Anthony

@MAFIA36790 Oh yeah

@DanielSank I'm not very quick, I expect it would look gaussian

DanielSank

@Anthony Actually, consider the radius, not the radius squared.

user116211

BTW, @Daniel , I want to ask you one thing on the derivation of Maxwell's Distribution function....

DanielSank

It turns out it's not Gaussian.

Anthony

Well it's still bell shaped, isn't it?

DanielSank

4:06 AM

@Anthony Not really.

Hang on, I'll make you a plot.

Anthony

Oh I'm being silly

DanielSank

The thing is, as you go to larger radius, you have more angle available.

Anthony

Oh, really?

That's uncomfortable

DanielSank

At very small radius there's vanishingly small angle, so the amount of probability for small radius goes to zero.

Anthony

no

that's horrible

DanielSank

4:07 AM

What?

Anthony

Nothing, I just don't like it lol

DanielSank

@Anthony Oh, well, that's exactly what's going on in the Maxwell Boltzmann distribution of speeds of a particle!

Interestingly, Nature doesn't really care what we like :P

Anthony

So it's just because we're taking the speeds?

DanielSank

@Anthony Yes.

It's because you're integrating over the angles.

I shouldn't have said "vanishingly small angle", I should have said that at small radius the area of a ring is small.

Annoyingly, my file with this function is on my other computer and I never pushed it so I don't have access to it right now.

Anthony

But if we were to plot the energies, then we would have monotonic decay?

That doesn't seem right. Energy is proportional to the speed squared.

DanielSank

4:11 AM

@Anthony I believe that is correct.

Actually that may not be the case.

Maxwell–Boltzmann distribution

In statistics the Maxwell–Boltzmann distribution is a particular probability distribution named after James Clerk Maxwell and Ludwig Boltzmann. It was first defined and used in physics (in particular in statistical mechanics) for describing particle speeds in idealized gases where the particles move freely inside a stationary container without interacting with one another, except for very brief collisions in which they exchange energy and momentum with each other or with their thermal environment. Particle in this context refers to gaseous particles (atoms or molecules), and the system of particles...

This page has everything you want to know.

Apparently the energy distribution is not monotonic.

c'est la vie

Anthony

Yeah, I was just looking at that.

I don't understand why. It should just be related to $e^{-\beta E}/Z$

DanielSank

Well, if you want the average energy you have to do $\sum_i \exp(-\beta E_i) E / Z$, right?

Oh wait, that's the average.

Anthony

Yeah.

DanielSank

We're talking about the full distribution now though.

How are you proposing to compute the distribution?

There are rules for how to manipulate probability distributions and if you haven't taken time to understand them things can be rather unintuitive.

Anthony

Maybe I'm not actually that sure.

I was thinking more in the quantum case where you just sum.

DanielSank

4:17 AM

For example, suppose I have two variables $X$ and $Y$ distributed according to $P_x(x)$ and $P_y(y)$. Do you know how to compute the distribution of $Z \equiv X + Y$?

Anthony

But I guess this is classical.

DanielSank

@Anthony Sum what though?

If you want to compute a probability distribution you don't just sum the distributions of the variables you're summing! :)

That is completely incorrect, it turns out.

Anthony

Yeah, that makes sense. I actually can't seem to remember how to add distributions like that.

DanielSank

@Anthony It's not trivial.

You convolve the two functions!

Anthony

Oh, that's good.

Oh dear.

DanielSank

4:19 AM

@Anthony Well it's not that bad either :)

It's funny you're asking about this because I just recently had to compute something very similar to what you're asking and I had to learn how to compute the distribution of the product of two random variables.

The result was a Bessel function o_O

First time I ever saw a Bessel function outside of a homework problem.

Anthony

In any case, this isn't what I was saying. For energy levels in QM the probabilities are monotonically decreasing because there we can just calculate $e^{-\beta E_i}/Z$.

and lol

But in the classical case, I guess that's not how we calculate probabilities, and that's why I'm confused?

DanielSank

@Anthony Dude, it depends what you're computing. That boltzmann factor is the probability of a state, not the probability distribution of the energy.

Anthony

I mean in the QM case, the distribution of states is synonymous with the distribution of energies, right?

user116211

In Schroeder's book he writes, the distribution function as

DanielSank

@Anthony Definitely not.

Suppose I have two states, A and B.

Suppose state A has energy 0 and probability 1/2, state B has energy 100 and probability 1/2.

user116211

4:23 AM

$$D(v)\propto \left(\textrm{probability of a molecule having velocity} \, \vec v\right) \times \left(\textrm{no. of molecules having speed}\, v\right)$$

Anthony

Oh, this isn't what I meant. Oh dear. I suppose I'm thinking entirely in terms of canonical ensemble. Is this the issue?

DanielSank

How is the probability of the states related to the probability distribution of the energy, which in our case is 50% 0 and 50% 100?

user116211

Now, the former is proportional to the Boltzmann factor $e^{-mv^2/2kT}$

DanielSank

@Anthony Not sure.

user116211

In deducing the term for the second one, he says:

Anthony

4:25 AM

I'm confused by your example also.

DanielSank

@MAFIA36790 Yes but what's the proportionality factor :)

Anthony

It's the same? Isn't it?

DanielSank

@Anthony Ok forget the example. It's not a great one.

Anthony

Alright.

user116211

@DanielSank I'm writing... I'm not Flash :(

DanielSank

4:25 AM

@MAFIA36790 Ok but that factor is the critical part of all of this.

Also, I don't understand what $D(v)$ means and I don't understand why you're multiplying by the number of molecules having a speed.

The number of molecules having a speed is the probability of that speed multiplied by the total number of particles.

user116211

@DanielSank That could be found out later by normalising, isn't it?

user116211

@DanielSank Not me, but Schroeder :/

user116211

@DanielSank Boltzmann Distribution function.

DanielSank

@MAFIA36790 Uh ok... that doesn't make sense to me.

The "Boltzmann distribution" usually means precisely "the probability of having a particular speed".

The probability of a particular state with energy $E$ is $\exp(- \beta E)/Z$.

but you have to realize that there are many states with the same velocity!

That's the critical element.

There's a so-called "degeneracy factor" which is proportional to the surface area of the sphere in momentum space which has energy $E$.

See what I mean?

user116211

> To evaluate the second factor, imagine a three dimensional vector-space in which every point represents a velocity vector. The set of velocity vectors corresponding to any given speed $v$ lives on the surface of sphere with radius $v$

DanielSank

4:31 AM

@MAFIA36790 Ok good. That's exactly the right way to think. I'm dismayed, however, that the book calls this "no. of molecules having speed $v$".

That's terrible language. It should say "number of states having speed $v$".

This kind of misuse of language is incredibly confusing, at least to me.

0celo7

@DanielSank Would you happen to know about the Born stripping criterion?

user116211

> The larger the $v$ is, bigger the sphere and more the possible velocity vectors there are.

user116211

@DanielSank My mistake.

DanielSank

@MAFIA36790 I thought you were transcribing from the book.

user116211

$$\left(\textrm{number of vectors} \, \vec v \, \textrm{corresponding to }\, v \right) = 4\pi v^2$$

user116211

4:34 AM

@DanielSank Absence of mind; I'm copying from the book though.

DanielSank

@MAFIA36790 Yep, that makes sense.

You're now computing the degeneracy factor.

skill patrol

in The Clubhouse, 35 mins ago, by Michael Myers

Villanova vs. North Carolina: Final minutes of national title game

►

user116211

What bothers me is the bold statement above.... why is it that the the larger the sphere becomes, the larger is the number of radii? I thought the number of radii were constant ;/

skill patrol

What a wild cat finish!!!

DanielSank

@MAFIA36790 Ah, well for example suppose we chop up the space of available states into set of little 3D cells.

Each one has a size $dp \times dp \times dp$, see what I mean?

user116211

4:41 AM

@DanielSank got this statement atleast.

DanielSank

@MAFIA36790 Ok, well now suppose you draw a radius out from the origin.

user116211

@skillpatrol You probably don't watch cricket.

DanielSank

Ask yourself, how many little cells are there at a given distance from the origin?

If you go further, surely there are a lot more.

It's just the surface area of a sphere.

So if you ask how many states there are for a given $v$, you find that the number is proportional to $v^2$.

user116211

@DanielSank I thought it's the number of radii, damn.

skill patrol

@MAFIA36790 only the championship games :-)

DanielSank

4:43 AM

@MAFIA36790 This is why I keep insisting on saying the number of states.

user116211

@DanielSank Ha!

DanielSank

The way your book is saying this is confusing, but technically correct.

It says "number of vectors $\vec{v}$...".

user116211

@DanielSank That's what the author wrote indeed... I mean no matter how large the sphere is, the number of radii is always constant.

DanielSank

@MAFIA36790 I'm not sure what "number of radii" means.

The thing you care about is how many physical states there are for a given velocity.

Always think of the set of states of your system first.

It's just a big block of states, one dimension per degree of freedom (or two depending on if you count position and its associated momentum as the same degree of freedom or separate ones).

user116211

@DanielSank That's correct indeed and a million-dollar worthy advice.

DanielSank

4:46 AM

Then you start asking questions about properties of those states, e.g. their velocity.

If you pick a fixed velocity you're selecting all the states which sit at a fixed radius from the origin in the momentum coordinate.

Anthony

@DanielSank I still don't under stand why the distribution isn't given as a monotonically decreasing function

DanielSank

@Anthony Because geometry, dude.

Anthony

the canonical ensemble gives the correct long term behaviour

Maxwell–Boltzmann statistics

In statistical mechanics, Maxwell–Boltzmann statistics describes the average distribution of non-interacting material particles over various energy states in thermal equilibrium, and is applicable when the temperature is high enough or the particle density is low enough to render quantum effects negligible. The expected number of particles with energy for Maxwell–Boltzmann statistics is where: where: is the i-th energy level is the number of particles in the set of states with energy is the degeneracy of energy level i, that is, the number of states with energy which may nevertheless be...

like I was saying

@DanielSank No, I understand what you're saying.

DanielSank

@Anthony What does "long term" mean?

Anthony

Like high energy, sorry.

In any event, why doesn't the lowvelocity=low energy=morelikely logic, work?

DanielSank

4:52 AM

@Anthony Because there aren't as many low energy states.

There are less ways for a particle to have low energy than there are for it to have high energy!

You have two factors: 1) The probability to have a certain energy, 2) The number of states that actually have that energy.

Anthony

oh

counting multiplicity

ay captain?

DanielSank

One of those functions is monotonically increasing, but the other is monotonically decreasing, so when you multiply them you get a bump.

@Anthony Aye aye!

Si, senor.

Anthony

Thanks @DanielSank.

What do you do, anyway.

user116211

Ah! me too:

DanielSank

@Anthony Quantum computing.

Anthony

4:55 AM

OH SHIT

Google Quantum AI Lab in Santa Barbara.

DanielSank

@Anthony Yep.

Anthony

Okay @DanielSank, can we meet some day?

user116211

@DanielSank: Thank you, sir ;P

Anthony

Can you hook me up with $$$?

lol

DanielSank

@MAFIA36790 No problem.

@Anthony Uh, I don't know. Do you live in SB?

Anthony

4:56 AM

I'm going to be working at Rigetti in Berkeley this next year if you've heard of them, and then hopefully grad school for QC.

DanielSank

@Anthony Oh yes, I know about them.

Anthony

@DanielSank I was joking a bit, but I'll be around there someday.

QuAIL sounds sick.

DanielSank

@Anthony I'm enjoying myself thoroughly.

Anthony

So are you working with DWAVE then?

whoops

DanielSank

@Anthony I do not personally work with them.

Anthony

4:58 AM

D-Wave

I meant are you using google's machine, or are you developing algorithms, or systems, or what?

user116211

@Anthony: Are you grad?

DanielSank

@Anthony Heh, we're trying to build a Google machine.

Google bought a D-Wave machine a while back.

Part of my group uses that.

However, we're also trying to build our own stuff. I'm an experimentalist so that's where I come in.

Anthony

Damn, that's awesome~

user116211

His life is awesome.

Anthony

Anyway thanks for the help, @DanielSank.

Nice talking @MAFIA36790.

See ya'll later.

user116211

5:01 AM

o/

DanielSank

@Anthony Sure.

\o

user116211

@DanielSank: bye too; see you later; today is session, BTW; do come.

DanielSank

@MAFIA36790 wut?

user116211

Physics Chat Session.

DanielSank

@MAFIA36790 Oh... when's that and why should I come?

Big important Issues?

user116211

5:03 AM

@DanielSank Well, they would decide at that moment, I guess ;P

user116211

@JohnRennie: o/

John Rennie

thanks :-)
The mood seems to have swung against posting canonical Q/As at the moment, but that one was worth the effort.

@MAFIA36790 just popped in to respond to a ping - can't stay ...

user116211

@JohnRennie Come in to the session.

John Rennie

@MAFIA36790 I'll be back for the chat session later

user116211

@JohnRennie Hmmm... when did we become anti-canonical ;/

DanielSank

5:13 AM

When is this session?

user116211

@DanielSank 10 hrs later

DanielSank

@MAFIA36790 Hm, 8 o'clock in the morning for me.

I might be able to attend.

user116211

@DanielSank Here, at 23:30hrs , pretty good time :P

DanielSank

@MAFIA36790 Depends on your schedule.

@MAFIA36790 your profile quote is rather... heavy.

user116211

@DanielSank Ha! well, yes.

user116211

5:17 AM

The best is dmckee's.

Danu

5:56 AM

@MAFIA36790 Why is that your profile quote, by the way?

Do you truly believe it?

@DanielSank How do they score on the fun-scale?

user116211

6:14 AM

@Danu Which one?

DanielSank

6:50 AM

@Danu I've enjoyed myself so far.

I met @ChrisWhite, @BernardMeurer, and some duder whose user name is a bunch of numbers. In each case we met up and consumed food.

Nothing to complain about there.

BTW @Danu I made more chili.

skill patrol

7:10 AM

You met the Chris White?

:O

ACuriousMind

7:54 AM

@0celo7 I have no idea what that is.

John Rennie

8:14 AM

@barrycarter: I've fleshed out the discussion about the Earth/twin scenario in your question and I've posted it as an answer:

0

A: Rapid acceleration does what to inertial frame clocks?

This is more of an extended comment than an answer since it's really more a discussion of related issues. However I wanted to capture some of thje discussion we've had in the chat room. To briefly reprise the question, the accelerating twin starts at rest $10$ lyrs from Earth, accelerates to $0....

skill patrol

The Backwards Brain Bicycle - Smarter Every Day 133

►

Do not try this at home.

:P

Slereah

Don't tell me how to live my life

ramsay

11:45 AM

hello,
my sir asked me: if light falls on mirror it reflects but when it falls on any glass(like a window) it just passes why does this happen?

Slereah

It doesn't.

There is always some reflection, some absorption and some transmission

In all materials

ramsay

hmm, but why do we see our face in case of mirrors but not in case of glasses

Slereah

you do

The reflection isn't as good, though

Look closely at a glass

ramsay

ok, now my hand, i don't see my face on my hand :D

Slereah

well

To see a reflection, the surface should also be smooth, ideally

Otherwise, the rays are going to be reflected in random directions

and it will smudge out anything

user116211

11:52 AM

@Yuggib is not coming ;/

ramsay

but why reflection is so bright on mirrors than glasses

Slereah

Mirrors have a layer of metal on the back

ramsay

honestly saying, i didn't get you Slereah! but thank you for your help(*i think you remember my internet ban*)
if anyone knows the answer please do PING me

David Z

It's just what Slereah said

For further reading, I suggest looking up "diffuse reflection" (the kind you get from e.g. hands) and "specular reflection" (the kind you get from mirrors and glass)

user116211

@DavidZ: Is it compulsory to make canonical posts CW?

David Z

12:09 PM

@MAFIA36790 Nope, not as far as I know.

I believe the only rule we have about CW is that it should be used for book recommendation posts.

(Personally I don't like that rule; I think CW mode should pretty much never be used, but I'm in the minority.)

user116211

@DavidZ You are not in the minority... at least there are many people who think so.... there is a same discussion at the chem chat.... the mod there also feels the same.

David Z

Well, it's not such a big deal. Anyway, we had a meta discussion about this, and the community was in favor of making book recommendation questions CW.

Slereah

What is even the community wiki

Is there a page or a tag for it

or is it just an abstract concept

ACuriousMind

@Slereah It means you don't get rep from a post.

Which, in turn, means everyone is invited to even heavily edit it, since that means the post doesn't actually belong to any particular user

David Z

There's also a lower reputation threshold for editing without getting your edits reviewed. I think it's like 500 or something, instead of the normal 2000.

Slereah

12:15 PM

Why would I do something I don't get rep from tho :V

You crazy man

How am I supposed to win internet points

ACuriousMind

There's a checkbox at the bottom right of every "post your answer" form where you can decide to make your post community wiki.

@Slereah I'm not crazy, I don't think I have ever created a CW answer :P

skill patrol

12:34 PM

crazy like a fox :P

0celo7

@ACuriousMind Apparently it's a way of ionizing stuff by accelerating the nucleus and the electrons just "fall off"

I don't know how inertia works for "standing waves"

user116211

@Slereah There are some good people who work for the sake of betterment of SE without thinking of profit....

Slereah

Those tools

0celo7

dammit Springer

I must feed my book addiction

skill patrol

addiction is a terrible thing

0celo7

12:42 PM

well I gave up alcohol and I have to substitute it with something

skill patrol

Orly, when?

0celo7

After Spring Break

skill patrol

Too many girls gone wild on spring break ;-)

0celo7

@ACuriousMind What does the author mean by "internal direct sum"

ACuriousMind

@0celo7 I'd say they don't mean much more than is written there - that it is a direct sum of subspaces. Which is kind of a silly thing to say, because every summand is a subspace of the sum, always.

0celo7

12:49 PM

@ACuriousMind Ok

@ACuriousMind So you wouldn't happen to know how inertia works in QM

ACuriousMind

I would not even regard that as a meaningful question

0celo7

@ACuriousMind Well you can strip off electrons by accelerating the nucleus

How does that work?

ACuriousMind

No idea, it sounds ridiculous to me. How are you even going to "accelerate the nucleus" without accelerating the electrons?

0celo7