The h Bar

Prof. Legolasov

03:30

Hi all! Can anyone take a look at this Wikipedia page: https://en.wikipedia.org/wiki/Geodesics_in_general_relativity. Specifically, the section titled "Deriving the geodesic equation via an action".

I think that derivation is just completely wrong.
Specifically, on the third full-line formula:

I think it is incorrect to apply the chain rule here.
As a simple counterexample, consider

Moreover, that derivation completely ignores the fact that $\lambda$ is an unphysical parameter in the action that can be chosen arbitrarily, but the form of the equation of motion that they derive is valid only when $\lambda$ is the proper time along the geodesic. Hence the e.o.m. can't follow from the action without special care (i.e. fixing the gauge) in the derivation, which is completely absent from theirs.

I think they got the correct answer by accident.
I tried arguing there, but they seem to not believe me :)
Could anyone here check my reasoning?

abhas_RewCie

04:28

@Prof.Legolasov The formula is correct, I guess....

@Prof.Legolasov $\delta \sqrt{\dot x^2 } \neq \delta \dot x$

But, instead, $\delta \sqrt{\dot x^2} = \frac{\delta \dot x}{2\sqrt {\dot x}}$

@Prof.Legolasov $\lambda$ is some arbitrary coordinate (in their derivation) not only time, time is denoted by $\tau$ there. (I guess)

$\lambda \in \{x^n\}_{n=0}^3$

Prof. Legolasov

04:46

@abhas_RewCie thanks for your reply. I'm a bit confused – do you mean that their formula is incorrect or mine is?

@abhas_RewCie if $\lambda$ is an arbitrary coordinate, they should not have gotten the geodesic equation in its usual form. $\ddot{x}^{\sigma} = - \Gamma^{\sigma}_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$ is only valid for when $\lambda$ is the proper time, the general form of the geodesic equation for an arbitrary $\lambda$ is different and more involved.

abhas_RewCie

@Prof.Legolasov Theirs...

Prof. Legolasov

Ok, thanks for the sanity check!

abhas_RewCie

@Prof.Legolasov Give me a time to check references, I've to check what they are doing with $\lambda$ and $\tau$

Prof. Legolasov

I'd like to fix that on Wikipedia. I feel uncomfortable with this incorrect (assuming I'm right) derivation being there for students to read

I wonder if anyone here has any experience arguing on Wikipedia :)

abhas_RewCie

@Prof.Legolasov They replace $\lambda$ with $\tau$ later, to get the required form

Note $g_{\mu \nu}$ is independent of $\tau$ but not $\lambda$

@Prof.Legolasov Honestly, the way they are changing parameters doesn't look correct to me either.

@Prof.Legolasov It should be changed...

Prof. Legolasov

05:04

@abhas_RewCie they replace $\lambda$ with $\tau$ but only in a sense of replacing one Greek letter with another

They don't impose the gauge fixing condition $\sqrt{-g \dot{x} \dot{x}} = 1$

Which is the correct way of passing from an arbitrary parameter to proper time parameter

abhas_RewCie

@Prof.Legolasov I doubt that....

Prof. Legolasov

Ok, why?

abhas_RewCie

@Prof.Legolasov I think they are trying to do something else with different notations...

@Prof.Legolasov agree...

@Prof.Legolasov page - 36 if.nu.ac.th/sites/default/files/bin/BS_chakkrit.pdf

that's the correct way^

Yuvraj Singh...

05:53

@abhas_RewCie hi

Do you have Antivirus? Other than avasti, actually I have Bitdefender Antivirus and my subscription ended last day

Guru Vishnu

@YuvrajSingh... A small correction - It's Avast not Avasti; I hope you spelled the author's name Narendra Avasthi (Physical Chemistry) :-)

user434058

07:47

@GuruVishnu BTW Neeraj Kumar's book is better :P

user434058

Enjoy before it gets nuked :D

John Rennie

08:02

Boom!

I love Anna's comment insanity does not reside in the mountains though I'm not entirely sure what it means :-)

bolbteppa

08:43

@Prof.Legolasov doesn't look like there's any error on the wiki

user434058

09:45

Oops! I have become so habitual of flagging HW questions that sometimes I often mistakenly flag a bad off topic question with the same HW reason... Ahhh.... Muscle memory ruins it all!!

user434058

0

Q: Objects viewed across a body of water

I'm sitting on the back patio of my lake house, (looking across the lawn and across the lake) and viewing the mansion across the lake... But when I go next door (flat lot), the house looks larger/closer. It's very freaky. I didn't believe my neighbor till she physically walked me over to see. I w...

user434058

BTW, There's a slight chance that the OP just wants to show how rich they are :P

user434058

And anyways this ☝️ question would be closed, because it lacks info, right?

user434058

11:25

2

user434058

Seems like there are many "curious" users on PSE...

user434058

BTW, what would be the most used account name on PSE?

user434058

Is this, by any chance, flag-worthy?

user434058

This chatroom feels so lonely whenever I am around -_-

Yuvraj Singh...

11:56

@JohnRennie Anna is the only one to explain it:-P

Prof. Legolasov

12:07

@bolbteppa I can definitely be wrong, but do you mind explaining why my arguments don't hold?

bolbteppa

Two immediate errors are $\sqrt{\dot{x}^2} \neq \dot{x}$ and $\delta \dot{x} \neq 0$,

Also $\delta \dot{x}^2 \neq - \ddot{x}$

Slereah

quite a lot of errors

abhas_RewCie

12:22

I thought that it was a different notation...

it should be $g_{\mu \nu} \frac{d \dot x^{\tau}}{d\lambda} \frac{\dot x^{\nu}}{d \tau}$

Yuvraj Singh...

@abhas_RewCie hi

abhas_RewCie

hello

Yuvraj Singh...

Have you read my comment

?

abhas_RewCie

@YuvrajSingh... no... which one? Antivirus?

Yuvraj Singh...

Yup

My subscription ended yesterday

abhas_RewCie

12:26

@YuvrajSingh... buddy, I use Linux, we don't have that here

Yuvraj Singh...

Even I am using linux

abhas_RewCie

@YuvrajSingh... XFCE Desktop, cool, fitted with a Vim Editor.

@YuvrajSingh... You don't need AV for Linux.

just don't install programs as root user.

Yuvraj Singh...

@abhas_RewCie bro I have two pc one is old and and one is new

I use old for surfing.

abhas_RewCie

@YuvrajSingh... Cool... Don't use AV on older ones, it slows things.

k...

Yuvraj Singh...

Bro it is windows 8 and i5

abhas_RewCie

12:29

@YuvrajSingh... cool

@FakeMod LAMO

Yuvraj Singh...

Ah!

@abhas_RewCie lmao?

user434058

He meant LMAO

user434058

Or LMFAO

abhas_RewCie

@YuvrajSingh... LMAO`

in a bit fancy way

Yuvraj Singh...

Why you refer it here?

abhas_RewCie

12:31

@YuvrajSingh... no, nothing, it was for fakemod

28

Q: Are these UFOs on the videos released by the Pentagon still unidentified?

The Pentagon recently released videos of UFOs taken by Navy pilots. According to this Guardian article, The Pentagon on Monday [27 April 2020] released three declassified videos that show US Navy pilots encountering what appear to be unidentified flying objects. (...) The videos ...

ufo

user434058

Nice

Yuvraj Singh...

13:00

Rubbish.

abhas_RewCie

13:15

Rubbice

Sha Vuklia

13:28

How does a change of observer (in Minkowski spacetime) induce a "change of observer" map for some quantum mechanical system?

here, $M$ is Minkowski spacetime, and $f\colon M\to M$ is defined as the bijection which sends one spacetime event (with coordinates (x,t)) to the spacetime event which has the same coordinates (x,t) for the new observer

oh maybe a better question; can this induced map be just the identity? (when we changed observers in Minkowski spacetime)

Slereah

Sure

It's a group even

Sha Vuklia

because, if we were to think of spin for instance, then we wouldn't care about a change of observers right

but if we were to measure position or sth, then it might change

Slereah

Changing observer would change the spin, certainly

since the observer (with a tetrad) defines an orientation for the spin

Sha Vuklia

oh right

had forgotten that spin is measured along a direction

abhas_RewCie

Well there was a time, we all were crazy to solve Riemann Hypothesis.

Sha Vuklia

13:59

Second question: why do we get this homomorphism? The thing is, by Zeeman we know that the semidirect product of $M$ and $\mathcal L_t\times\mathbb R^*$ is the group of transformations of $M$ which correspond to changes of observer such that causality is preserved.

However, if we restrict ourselves to observers which have the same idea of transition probability, shouldn’t we then take a subgroup of $M\rtimes(\mathcal L_T\times\mathbb R^*$)? This would then yield a homomorhism from this subgroup of $\operatorname{Aut}(\hat H)$.

or would this $T$ then send such a "good" change of observer (i.e., causality and transition probability are preserved) to the corresponding automorphism of $\hat H$, and any other one just to $0$?

Slereah

Errr isn't that by the wigner theorem

Sha Vuklia

Slereah

If the probability is the same then it's a whatever symmetry group

Sha Vuklia

I guess it doesn't matter, because later on they say that they are going to focus on $\mathcal L_0$ anyways, and then we do have a homomorphism (since then causality and transition probabiliy are both preserved)

@Slereah you mean this?

Slereah

I mean the theorem that symmetries on the Hilbert space that preserve the probability can be represented as a unitary operator

Sha Vuklia

14:07

yea, that's basically wat that tells us

Slereah

I'm afraid I can't read sequences :p

Sha Vuklia

it says that any automorphism of $\hat H$ is just the projection of a unitary or anti-unitary operator on $H$

since we have $\operatorname{Aut}(\hat H)\cong \tilde U(H)/U(1)$, where $\tilde U(H)$ consists of all unitary and anti-unitary operators on $H$

Slereah

I see

bolbteppa

Yeah that's Wigner's theorem (unitary or anti-unitary)

abhas_RewCie

I found a cool website.

epdf.pub

geocalc33

14:28

0

Q: How do you show that the Lorentz metric is preserved for $\zeta^{3,1}?$

The following is a visualisation of a Lorentz transformation: https://www.desmos.com/calculator/u5qpd135uc. The lines of constant time in Minkowski space, $\Bbb R^{1,1},$ are hyperbolas: $$ xy=t$$ The lines of constant time in $\zeta^{1,1}$ are: $$\varphi_S(x)=e^{\frac{S}{\ln(x)}}.$$ Where ...

special-relativity

reopen?

abhas_RewCie

@geocalc33 I still don't understand your question... :(

geocalc33

:(

abhas_RewCie

@geocalc33 I don't know what do you want by explicitly mentioning that and lorentz metric is preserved

bolbteppa

$\phi_S(x) = e^{\frac{S}{\ln x}} = e^{\ln(x)^2/\ln x} = e^{\ln x} = x$?

abhas_RewCie

@bolbteppa Good Evening sir...

geocalc33

14:35

@bolbteppa!

bolbteppa

Non-linear Minkowski space seems to make no sense as well

geocalc33

hi

abhas_RewCie

What is non linear Minkowski space? I've only heard linear ones....

geocalc33

@bolbteppa $S=\{\log^2(x):x\in \Bbb Q \cap (0,1)\}$

bolbteppa

In the post you used $\mathbb{R}$ not $\mathbb{Q}$

geocalc33

14:37

okay it doesn't really matter

$S$ is a generator and generates the curves right? So $x$ can be R or Q it's not that important really

bolbteppa

Are you actually saying $\phi_S(x) = e^{ \frac{\{\ln(x)^2 : x \in \mathbb{R} \cap (0,1) \}}{\ln x}}$ or are you saying $\phi_s(x) = e^{\frac{S}{\ln(x)}} = e^{\frac{\ln(x)^2}{\ln(x)}} = e^{\ln x} = x$

geocalc33

first one is probably better notation!

bolbteppa

The first one makes no sense

geocalc33

oh sorry I guess I didn't realize that

I've been using that notation for a while

ACuriousMind

15:24

@Slereah Better get back to learning homological algebra, then! ;)

Apr 26 at 7:23, by ACuriousMind

@dumbasarock Please don't post links to your questions here directly after you've posted them; if everyone did that the chat would consist of little else.

Slereah

@ACuriousMind Alas!

I'd like to learn the BRST complex

it is homology rich

Yuvraj Singh...

15:57

We have plenty of theories starting from the atomic to genreal relativity, so a question arises in my mind that out of this theories, what are the numbers of theories which are complete in itself and has proven correctly since it's discovery?

I asked this question because I see that either theory is upgraded to some new version or degraded as false, I was looking for a theory which last long till the time I am writing this text!

Aaron Stevens

19:33

2

Q: What is wrong with this calculation of work done by an agent bringing a unit mass from infinity into a gravitational field?

Let us assume that a gravitational field is created by a mass $M$. An agent is bringing a unit mass from $\infty$ to distance $r < \infty$, both measured from mass $M$. The agent is always forcing the unit mass with a continuously changing force $\vec F(\vec x)$, $\vec{x}$ being the distance poi...