The h Bar

Silly Goose

06:55

I shall embark on the journey of L&L ENM!

bolbteppa

07:13

@DIRAC1930 Your statement about verifying energy conservation is very iffy because in the time-energy section they directly say energy conservation can only be verified up to some error. When the particle is free ($t \to \pm \infty$) we can measure things, that's what it's all saying, i.e. only then can we talk about measuring energy to check energy conservation, I think you are not happy that one can only do it when the particle is free at $\pm \infty$.

About this distribution thing, he means it is a trick they use to analyze Green functions when studying the Landau pole and strong interactions, check the papers he refers to there, it's not challenging/impacting this Section 1 thing if that makes sense, he separately refers to that result in your paper and assumes the result is true to say something else (which turned out to be a big mistake) about fields.

Slereah

08:35

Looking for some natural bundle information and i find some math stack exchange post by Michor

Mr. Feynman

11:59

Is it possible that Carroll does not define observers?

Slereah

12:14

It wouldn't be that strange

It's not a common thing for GR books to not go that deep into observers

Ryder Rude

@Slereah I think of an observer in GR as a worldline, and each point on the worldline is attached to a set of basis vectors. The temporal basis vector is aligned to the tangent to the worldlien. Is this correct?

Slereah

it is a common one yes

Ryder Rude

So when we switch to this basis, we r suposd to get the observer's local "view" of spacetime

Slereah

You can also attach a clock to it

Ryder Rude

Yea

Is this basis completely determined by the wordline? @Slereah

The time vector of this basis is determined by the tangent, yes

I'm thinking that, to deduce wut the observer wud observe, we wud switch to this basis

I mean one basis at each point of the wordline

@Slereah I'm thinking that, once we hav determined the temporal basis vector using the worldline tangent, we can choose any three basis vector in the space orthogonal to this tangent as our spatial basiss

Slereah

12:27

@RyderRude It is not

Even for an orthonormal basis you can still rotate it

Ryder Rude

i now think only the time direction is determined by the tangent to the worldline. We hav choice of the spatial vectors

But all choices r equally good

@Slereah And i shud also note that performing this procedure on a worldline will not necessarily give a locally minkowski metric along the worldline. We can hav other metrics too. Is this correct?

I mean... The observer will not necessarily experience minkowski spacetime locally. Is this really tru? @Slereah

Slereah

I mean you're just defining a frame along a curve

by itself that does not define a coordinate system

@RyderRude Would be weird if it wasn't since that's one of the basic axiom of GR

That's what the equivalence principle is

There exists such a coordinate system

Fermi coordinates

In the mathematical theory of Riemannian geometry, there are two uses of the term Fermi coordinates. In one use they are local coordinates that are adapted to a geodesic. In a second, more general one, they are local coordinates that are adapted to any world line, even not geodesical.Take a future-directed timelike curve γ = γ ( τ ) {\displaystyle \gamma =\gamma (\tau )} , τ {\displaystyle \tau } being the proper time along γ {...

Ryder Rude

Yeah, thats tru. But wut about the metric that the observwr wud actually experinece locally

We can find local minkowski metric at all points of spacetime. But wud the observer really experience this metric locally?

Slereah

Try to mind your spelling

it's a bit exhausting to read

Ryder Rude

I'm sorry

I mean that acceleration is absolute. So, a non-inertial observer would locally observe a non-minkowski metric? @Slereah

Slereah

12:39

No you can always diagonalize the metric locally

Ryder Rude

@Slereah I mean suppose we talk about the set of locally minkowski bases at a point of spacetime. Is it possible for the temporal basis vector in any of these to be tangent to a non-geodesic at the point?

Slereah

I mean it's just one vector

If you want to do any basis, given one vector, you can always form an orthonormal basis from it

By the Gram-Schmidt construction

Ryder Rude

@Slereah thanks. I had forgotten about this

So, for non-inertial observerse, we can choose a locally minkowski basis

Slereah

Gram-Schmidt I remember for SR is for a flat metric otoh*

lemme check if it works to diagonalize it

I vaguely remember Fermi coordinates having two different timelike vectors involved

philsci-archive.pitt.edu/21311/1/Local%20flat.pdf

it does indeed vanish on the curve

see 1.7

Ryder Rude

12:58

Yeah. I just saw

But they also say they're doing it for geodesic worldlines

@Slereah I guess u can also do this for non-geodesics using Gram Schmidt. But what u cant do for non geodesics is to make the first-derivative of the metric vanish...( Perhaps something like this)

But you can still diagonalise it locally

Slereah

Oh wait

roma1.infn.it/teongrav/gr20_21/fermi_coordinates.pdf

looks like the metric is indeed changed by acceleration

if you consider the basis to be constructed from the tangent

I guess that way you locally obtain that the observer is non-inertial

Ryder Rude

so this means that non-inertial observers dont experience local minkowski spacetime?

But i thinl Gram schmditt should work. I'm confused now

I will have to read in detail. Thanks

Slereah

Gram-Schmidt works at establishing an orthogonal basis in a given metric, and you can diagonalize the metric given an appropriate basis

The hard part is making sure you can make the two coincide

Although idk if what I'm saying is accurate, this is just on top of my head

Slereah