The h Bar

5:00 PM

Another analogy: centrifugal forces in Newtonian physics are ‘fictitious’ insofar as they arise from working in a noninertial reference frame

You don’t have to postulate them; they’re there in the theory already

The same is true for gravity in GR. But it is not true for gravity in SR

More Anonymous

@ACuriousMind you will have clarify something further:
1.You say " laws of SR hold around every observer in SR" ...
2.So every obsever includes non-inertial frames too?
3. "freely-falling" means something special in GR,
4. I think Im asking why is this non-intertial frame special in GR

Semiclassical

4. presumes that non-inertial frames are a thing in GR like they are in Newtonian mech. I’m not sure that’s true?

More Anonymous

@Semiclassical kinematics or dynamics in Newtonian mech? Also I am aware GR doesnt have a seperation of kinematics or dynamics

skullpetrol

@MoreAnonymous Einstein presents SR and GR concisely together in his little book, may i suggest you have a look at it?

ACuriousMind

@MoreAnonymous (fwiw, the numbering there is more confusing than helpful :P) It's special in GR because it is the GR equivalent of "inertial frame" - as I said, gravity is a not a "real" force in GR, it's more like changing the notion of what a "straight line" (aka geodesic) is. Freely-falling observers follow geodesics, i.e they are the generalization of inertial frames travelling in straight lines.

More Anonymous

5:12 PM

@skullpetrol sure but I feel concisely might work against me :P

skullpetrol

yeah, it is too concise

but he warns the reader in the preface

Semiclassical

Here’s a question I don’t know: In Newtonian mechanics, you can always choose to work in an inertial frame. If you now compare that with a non-inertial frame, some forces in the latter will not appear in the former

So some forces in the non-inertial frame have a kinematic origin, whereas others really do have an origin in dynamics

Can one always find such a frame in GR? My guess would be no

skullpetrol

does "inertia" even belong in GR :P

Slereah

You have Fermi frames in GR

More Anonymous

@skullpetrol stress energy tensor?

Slereah

5:18 PM

Fermi–Walker transport

Fermi–Walker transport is a process in general relativity used to define a coordinate system or reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame. == Fermi–Walker differentiation == In the theory of Lorentzian manifolds, Fermi–Walker differentiation is a generalization of covariant differentiation. In general relativity, Fermi–Walker derivatives of the spacelike vector fields in a frame field, taken with respect to the timelike unit vector field in the frame field, are used to define non-inertial...

Semiclassical

Hmm

Slereah

The metric is ~ flat around the observer

ie there's no felt acceleration

Semiclassical

Too technical for my limited knowledge of GR, alas

Sir Cumference

@JohnRennie Yeah I suppose. I was kinda hoping mathematical physics would have some use in e.g. stellar physics though

Slereah

That's because GR is bad

Don't do it

ACuriousMind

5:20 PM

@Semiclassical I think you're describing the "momentarily co-moving reference frame"

Semiclassical

My intuition for “no” was based on analogy with electromagnetism in SR

More Anonymous

It's special in GR because it is the GR equivalent of "inertial frame"

@ACuriousMind I feel I'm getting confused because both GR and SR can use:

$$ ds^2 = ...$$

How does when distinguish when both use this ^

Slereah

By looking at what's on the right side

skullpetrol

...

ACuriousMind

^that. The metric is static in SR (and usually flat), but dynamic (through the Einstein equations) and usually curved in GR.

Semiclassical

5:22 PM

Namely: if you start in a reference frame with both electric and magnetic fields at some point, then you can either find a frame where E=0 or one where B=0.

But you cannot do both!

That is, if you can find a frame where E=0, then in general you can’t find one where B=0 and vice versa

More Anonymous

@ACuriousMind So if someone gave me the Rindler metric then?

Semiclassical

which is why it’s not correct to say that magnetic fields are ‘merely’ the result of a Lorentz transformation

In some cases, they can be constructed as such. In others, though, they can’t

So there’s no neat division of electric/magnetic fields into kinematic/dynamic, like there is for forces in Newtonian mech

ACuriousMind

The Rindler "metric" is just flat space as seen from an accelerating observer, it's not a different metric from flat space.

More Anonymous

So Rindler metric = SR

ACuriousMind

@Semiclassical I feel this is just a manifestation of the distinction between "electric" and "magnetic" being unnatural from the relativistic viewpoint, where these are just components of the field strength tensor.

More Anonymous

5:29 PM

@ACuriousMind I guess u can't accelerate yourself and see a Schwarzschild metric

?

ACuriousMind

No, certainly not

More Anonymous

Hmm ... I remember it being difficult but possible to see if 2 metrics were coordinate transformable in 4D ... Is there something similar for what u can see in an accelerated frame?

@ACuriousMind ^

Semiclassical

@ACuriousMind agreed

More Anonymous

@ACuriousMind For example if I give you one set of coordinates and then give you another and ask you if I can accelerate one and see the other?

ACuriousMind

@MoreAnonymous It's difficult but not impossible: The Cartan-Karthede algorithm allows you to determine for two arbitrary metrics to determine whether they are the same in different coordinates or not.

It's often much easier to tell when they are not - only one scalar invariant (such as the Ricci scalar) needs to be different.

More Anonymous

5:37 PM

Yes ... but I'm looking for an algorithm which "if I give you one set of coordinates and then give you another and ask you if I can accelerate one and see the other?"
For example if I give you the Lorentz metric and then give you the Rindler Metric is there an algorithm that will say from Lorentz u can accelerate to Rindler?

Semiclassical

@acm one thing I want to pick your brains on at some point is this business of unitary vs non-unitary evolution in QM

ACuriousMind

@MoreAnonymous I just gave you that algorithm.

But if it's just acceleration you don't need to go through it, of course - you can just do the coordinate transformation into the accelerated frame directly on one metric and see if you get the other metric.

Semiclassical

My sense was that, if you take the collapse postulate as writ, then you are thereby committed to that step being non-unitary

More Anonymous

@ACuriousMind u also followed it with: allows you to determine for two arbitrary metrics to determine whether they are the same in different coordinates or not.

I was under the impression Rindler and Lorentz were not same metrics (in the sense one cannot do a coordinate transformation and get the other)???

skullpetrol

@Semiclassical the first three letters have to match :P

ACuriousMind

5:42 PM

@MoreAnonymous Going to an accelerated frame is just a coordinate transformation.

Semiclassical

Since a unitary evolution can’t turn a mixed state into a pure state

More Anonymous

@ACuriousMind OMG! My brain is dead

How could I forget something so basic

Talk tomorrow?

Semiclassical

@skullpetrol yes, but I’m lazy and ACM is right here :P

skullpetrol

true dat pal

More Anonymous

@Semiclassical ... I feel ur going in the danger zone of physics.stackexchange.com/questions/502221/…

skullpetrol

5:44 PM

acm vs acu

ACuriousMind

@Semiclassical Yes. This problem (without involving collapse!) also shows up when you try to explain how unitary evolution can turn a pure state of some system into a mixed thermal state. (Just picking a non-unitary evolution operator doesn't cut it, either - the standard notion of applying an operator to a pure state simply cannot produce a mixed state)

Semiclassical

Sounds legit

More Anonymous

I wonder what @bolbteppa has to say about this^

ACuriousMind

For thermal systems, you can say that the unitary evolution acts on the system + heat bath, which is much akin to describing measurements as unitary evolutions of system + apparatus.

More Anonymous

Ah ...

ACuriousMind

5:48 PM

tracing out the heat bath/apparatus, you get mixed states for the original systems if the pure state of the total system is entangled.

Slereah

Ahah

I know why Minguzzi chose a static metric for his simultaneity bundle

Because otherwise the Riemannian space $ \Sigma = M / G$ doesn't have a metric induced!

it's a pure topological manifold!

More Anonymous

@bolbteppa I wanted to know ur thoughts on "Explicit symmetry breaking" ... u know the context ;)

Semiclassical

The more subtle point is whether that non-unitary evolution should bother you. If you’ve got an epistemic view of the density matrix, then it should presumably not bother you a wit

ACuriousMind

@Semiclassical Yes, this is a "problem" only if you are a "$\psi$-ont" ;P

More Anonymous

nooooo .... Semiclassical by the time ur done with this kind of subtlety ... u will need to change ur name to classical :P

Semiclassical

5:51 PM

A psi-ontologist, if you will

ACuriousMind

That's much better than what I said!

Semiclassical

Lol. It’s a turn of phrase I’ve heard from others

A tricky point in this is how Bohr/Heisenberg would’ve replied to the question

Slereah

I'm a psychic

ACuriousMind

@Slereah I'm entirely willing to believe that you spend your time communing with the ghosts of long dead general relativists

4

Slereah

@ACuriousMind I won't leave them alone until they give me those "unpublished" papers dammit

5

Semiclassical

5:57 PM

The notion of the Heisenberg cut is probably relevant here as well

More Anonymous

@Slereah I see the spirits trying to give u their unpublished papers its just spirit paper they have :P

U must stop digging their graves for these papers

boooooo

u have been warned

Slereah

Hm

I wonder if, given a topologically trivial spacetime, you can foliate it with light clocks

Well, for synchronized points, anyway

I'm worried about caustics I suppose

The word may apparently be more "cut locus"

More Anonymous

@ACuriousMind what do u mean by it? (sorry re-reading our messages )

https://chat.stackexchange.com/transcript/message/51706216#51706216

Semiclassical

@ACuriousMind I guess another way of putting the epistemic reply is that, during a measurement, the question “what is the quantum state” has no meaning

More Anonymous

what do u mean by "it"?

ACuriousMind

6:10 PM

@MoreAnonymous "It" = "the freely-falling frame"

You can reply to chat messages directly if you click on the little arrow on the right when you mouse over them, by the way

It's much more convenient than reposting the link to the message.

@Semiclassical Well, the "truly epistemic" viewpoint is that that question never has an unambiguous meaning, no? Different observers have different knowledge of the same system and therefore may ascribe different states to it.

More Anonymous

@ACuriousMind So I can equate an "inertial frame" to a "non-intertial frame" in GR ... But can't in do so in SR?

^I'm getting this from "free-falling frame" = "non-inertial" ??

Semiclassical

@ACuriousMind point. But even just looking at the experience of one observer, you’d already run into the issue

ACuriousMind

@MoreAnonymous I'm saying that "a frame of an observer only under the influence of gravity" is a non-inertial frame in SR (the metric is flat and gravity is a force, hence the path of the observer is not a geodesic), but the equivalent of an "inertial frame" in GR (the metric is curved and the path of the observer follows a geodesic).

Semiclassical

I think there’s a word or two missing in that sentence

skullpetrol

1 hour ago, by skullpetrol

yeah, it is too concise

:P

ACuriousMind

6:21 PM

@Semiclassical I...don't think so?

Maybe that construction (using the same verbal phrase for two clauses connected by 'but') isn't idiomatic in English, though :P

Semiclassical

“I’m saying that X is Y in SR (because...), but the equivalent of Z in GR (because...)”

More Anonymous

@ACuriousMind I'm under the impression if I have 0 stress energy tensor in GR that makes me go to flat spacetime which is SR ... I'm think over there ... GR and SR interpretation of geodesic is the same? In which case if there is an accelerating obsever in this situation he can also predict the bending of light? (alone by SR) I'm thinking of this in the context of Einstien's famous elevator thought experiment

Semiclassical

Something doesn’t parse in that sentence

ACuriousMind

@Semiclassical Your preprocessor is supposed to expand it into "I'm saying that X is Y is SR (...), but that X is the equivalent of Z in GR (...)

More Anonymous

@Semiclassical my language becomes worse the more mentally fatigued I become

Semiclassical

6:26 PM

Gotcha

ACuriousMind

@MoreAnonymous No, vanishing stress-energy does not equal flat space - the Schwarzschild metric is a vacuum solution!

Semiclassical

“but the former is the equivalent of an inertial frame in GR” is how I’d render it fwiw

ACuriousMind

Yeah, that sentence was not a masterpiece of comprehensibility on my part, I guess

More Anonymous

@ACuriousMind interestingly in Schwarzschild I was of the opinion u still need the point (the singularity) which is definitely not 0 ... but thats a digression ... can we focus on GR talking about a situation in GR where Stress energy solution subspace refers to flat space time

ACuriousMind

On the contrary, even with the singularity, the stress-energy is zero everywhere where it is defined at all - at the singularity the metric diverges so you don't have a value of stress-energy there at all.

More Anonymous

6:35 PM

@ACuriousMind we are digressing ... So I start with a star with finite stress energy tensor and get a blackhole with none?? I was of the opinion there the derivative of the stress energy tensor is 0 ... So where did it go??? Anyhow one concept of GR at a time unless u prefer 2 simultaneously?

ACuriousMind

@MoreAnonymous A true Schwarzschild black hole cannot actually form in finite time, so the stress-energy doesn't go anywhere (but also energy is not, in general, conserved in GR due to a lack of time translation invariance).

You're trying to reason about things like QM and GR with intuition, which is a Bad Idea(TM) unless you have trained it not to lead you astray.

These theories are very much not intuitive - they are precise theoretical frameworks you need to carefully study before applying them to anything.

More Anonymous

@ACuriousMind I think my real folly is not spending enough time with them and being cocky while I have excessive mental fatigue

ACuriousMind

Go to sleep then, or do something else relaxing - tomorrow is another day :)

More Anonymous

Thanks ... Please be patient with me tomorrow as well :)

I'll be re-reading alot

Goodnight chat

skullpetrol

cya pal

Slereah

7:32 PM

Yoy can describe sxhwarzchild as having a streds energy tensor :p

But then it is very bad

ACuriousMind

It's probably some derivative of $\delta$? :P

Slereah

It is a delta yeah

$\delta(r)$

With some factos of mass and 4-velocities

Novalium Company

8:12 PM

@MoreAnonymous omg u actually sent me real spoilers lol

More Anonymous

8:33 PM

@NovaliumCompany Please tell me u found out they were real as u watched the movie :P

I'm sure as a dank meme worshipper my actions are acceptable :P

Novalium Company

9:11 PM

@MoreAnonymous they were real

@MoreAnonymous Let's create a new religion - Memedoism.

Worshiped only by real dank memers

Physics Meta

10:55 PM

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I am not going to link the specific answer that prompted me to write this post because I want to make a general point. I am aware of the policy of not forcing users to comment when they downvote, as best explained this answer. And I agree with it, mainly for the privacy concern, in order to avoi...

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