The h Bar

Ryder Rude

07:59

@ClaudioMenchinelli it is correct

Bml

10:43

Hello everyone. In the argument through which invariance in Galilean principle of relativity is proved, it is implicitly being stated that the properties of the position vectors do not change when passing from a system of fixed to one in uniform rectilinear motion and vice versa. It can be shown that the transformations that are part of Galileo's group are simple displacements of the metric.

They therefore preserve the distances between two points and the time intervals. The vectors should therefore not undergo mutation in their length and orientation. However, I have a question to ask with respect to this group. It includes all affine transformations that preserve distances between simultaneous events and time intervals.

I think it is possible to prove that even in the transition from an inertial to a non-inertial system these characteristics are preserved. Do you know a demonstration on this? Where can it be found?

Ryder Rude

@Semiclassical it seems like Bohm himself gave up on Bohmian mechanics later in life. in wholeness and the implicate order, bohm is focusing on idealist viewpoints like the oneness of everything

@Bml one formulation of Newtonian mechanics uses two metrics on spacetime, one for space and another for time. then distances and times wud b scalar becuz of they r contractions of these metrics wrt tangent vectors. not too different from how spacetime interval is a scalar in SR

@Bml this is the theory with the two metrics en.m.wikipedia.org/wiki/Newton%E2%80%93Cartan_theory

Bml

11:09

@RyderRude Thanks. Is there a specific demonstration that it was performed for the specific problem I have made explicit?

naturallyInconsistent

@Bml I just think your logic is backwards. It is difficult to understand what meaning we would be able to ascribe to measurements in non-inertial systems beforehand. Instead, we usually use the fact that we know how things make sense in inertial systems to figure out what kinds of information interpretation would have made sense in a non-inertial one, and finally, after checking a few cases, worked backwards to figure out which non-inertial frame and parameters apply to us.

Bml

11:26

Let's say I understand and don't understand. When the invariance of the Galilean principle of relativity is proved, take two frames of reference, $K$ and $K'$, with the latter moving in uniform rectilinear motion with respect to the former. It is possible to express the position of a material point with respect to the system $K$ ($\vec{r}$) by knowing the position of the point with respect to the system $K'$ ($\vec{r'}$) and knowing the position of the origin of the coordinates of $K'$ with respect to $K$ ($\vec{r_o}$). Thus:

But, "in this argument it is implicitly being stated that the properties of the position vectors do not change when passing from a system of fixed to one in uniform rectilinear motion and vice versa."

My doubt and question come from this initial presentation.

naturallyInconsistent

I very much want to point out that, as long as you are still thinking of position as possibly considered as position vectors, you are already very much tied to a Euclidean flat space, and that is going to very much hinder understanding of curved spacetimes. However, it is clear that you are not at the position to begin learning about these yet.

@Bml Yes, this is true.

@Bml And so, what is it that you are particularly asking for?

Instead, I want to ask you to consider what do you mean by "measuring a length" and "measuring a time", and "comparing length and/or time measurements between different frames of reference"

Because in inertial Galilean relativity these answers are trivial. They stop being trivial the moment you let go of them.

Bml

@naturallyInconsistent What I am asking for is: Is it possible to prove that even in the transition from an inertial to a non-inertial system these characteristics of a Galilean group are preserved? If yes, what is a specific proof of that?

@naturallyInconsistent Why did you specify "inertial" in reference to "Galilean relativity"? Don't Galilean transformations only apply to inertial frames of reference?

naturallyInconsistent

11:41

@Bml And I read it the first time you posted this, and told you that I think your logic is backwards. How are you even able to measure lengths and times in non-inertial frames? If you think lengths, angles, times are easy to measure in non-inertial frames, then you can start arguing about whether the properties you wanted to preserve are preserved. Instead, the actually used logic is that we know the correct answers in inertial frame, and then use them to deduce what the correct procedures

are in non-inertial frames

@Bml Well, I'm using it that way because we get taught how to deal with non-inertial frames in the non-relativistic setting. As long as we are not doing either of SR or GR, then the default conception of transformations is the Galilean one.

So, just to be able to talk to the essence of your problem, the non-inertialness, I needed to have some verbiage distinction to clarify what particular things we are trying to talk about.

Bml

@naturallyInconsistent So, are you implicitly telling me that we actually have to bother GR to deal with my problem, or am I wrong?

naturallyInconsistent

@Bml Very wrong. We will be able to get extremely meaningful understandings just from considering SR, which is still flat spacetime, but even that is not what I am asking you to go into right now either. I'm just asking you to pay attention to the deduction logical flow, and to consider what kinds of headaches you would have to confront, if you decided to want to start with measuring stuff in non-inertial frames.

gtg dinner so hungry

Bml

@naturallyInconsistent Have a nice dinner...

Ryder Rude

12:01

@Bml the transformation to an arbitrary frame can be defined as $x'=x-o(t)$, $t'=t$ where $o(t)$ is the worldline of the observer u r transforming to. note that the differences in x are preserved for simultaneous events at t=t_1 becuz u r subtracting the same amoung o(t_1) from them

and differences in time r preserved trivially.

in multiple space dimensions, u can also include rotations to the above group

ACuriousMind

@Bml You keep talking about a "proof" but you haven't stated what your assumptions are

Obviously, in reality, we know Galilean relativity is false, so what are the idealized assumptions you want to prove it from?

Ryder Rude

Bml doesnt want to prove Galilean relativity is false. they want a proof that transformations to arbitrary frames preserve euclidean distance between simultaneous events and temporal distance, where the distances r defined using the Galilean physics metrics @ACuriousMind

the isometries of these Galilean metrics r larger than the Galilean transforms becuz they also include non inertial transforms

Bml

12:18

@RyderRude Thank you very much for explaining it better than I did.

Ryder Rude

but note that "arbitrary frames" doesnt mean "arbitrary transforms". i gave u the general transform that preserves these two metrics

ACuriousMind

@Bml unfortunately neither of these explanations is clear enough to me

what are the assumptions here and what is the precise statement you want to prove?

Ryder Rude

take a space with two separate metrics: one for time and one for simultaneous events in space (both r euclidean). bml wants to know the group of transforms that preserves these two metrics

Galilean transforms preserve these, but transforms to non inertial frames do too

ACuriousMind

I would prefer to hear an explanation from the person who asked the question, not your mangled interpretation of it

Bml

@ACuriousMind This was the precise statement. I said that the Galilean group includes all affine transformations that preserve distances between contemporary events and time intervals (it is right, isn't it?). The transformations that are part of the Galilean group are simple displacements of the metric (no?). Is it possible to prove that even in the transition from an inertial to a non-inertial system, these characteristics are preserved? I think yes, but what is a solid proof?

(The terms in which @RyderRude put it are correct, it is very close to what I meant)

ACuriousMind

12:27

@Bml so what exactly does "preserve distances" mean here?

let's perhaps not look at the somewhat confusing setting of Galilean relativity, but ordinary 3-dimensional space and the Euclidean distance in it

the default statement here is that the orthogonal group O(3) (the rotations and reflections) preserve Euclidean distance, and if we're allowing affine transformations then the translations do so, too

no other transformations preserve this

Bml

@ACuriousMind You're right, I considered it because my question came to mind when I thought of Galilean relativity.

ACuriousMind

it sounds to me like you want to argue that any transformation (e.g. the one that stretches all coordinate asks by 2) "preserves" the distance even in this case

but what, technically, do you mean by "preserve" here?

Bml

@ACuriousMind By definition, the Galilean Group is the group of all transformations of a Galilean space which preserve its structure (preserves intervals of time and distance between simultaneous events).

ACuriousMind

yes, just like the group of rotations and reflections and translations is the group that preserves the structure of Euclidean space. But apparently you want to say that even transformations outside of the usual 10-parameter Galilean group "preserve distances" when you say they should even be preserved when going to "non-inertial systems", so I'm asking you to make clear what you mean in the simpler example of Euclidean space where we don't have two separate distances to deal with

but either way, I'm asking you to write down some kind of mathematical statement instead of using the ambiguous word "preserve"

Ryder Rude

note that arbitrary transformations do not preserve the galilean metrics, but transformations to non inertial frames do. these r specific transforms of the form : x'=x-o(t), t'=t. o(t) can be any worldline

ACuriousMind

12:33

(I'm not trying to be difficult: I can see at least one interpretation of "preserve" and "transformation" under which it is trivial that all transformations "preserve" any kind of metric on any kind of space, and I suspect you've fallen prey to this linguistic confusion)

Ryder Rude

"preserve" means x'_2-x'_1=x_2-x_1 for simultaneous events

the above is for 1D space

raf

Hi, can anyone please help me find a better strategy to get the expression of $[\nabla_a,\nabla_b]T^{{mn}}_p$ in terms of fundamental tensors?

Bml

@ACuriousMind No, I don't want to say that "even transformations outside of the usual 10-parameter Galilean group preserve distances"... At least, I don't think this statement could be true

Ryder Rude

here is a mathematical formulation. we r looking for the group of transforms $x'(x,t)$ and $t'(x,t)$ such that $\sum (x' _i - y'_i)^2 = \sum (x_i - y_i)^2$, and $t'_2-t'_1=t_2-t_1$. $i$ labels space dimensions

do u agree with this formulation @Bml

ACuriousMind

@Bml So what, exactly, do you mean by your question about "non-inertial" frames "preserving" the distances?

Ryder Rude

12:40

x_i and y_i are space co ordjnates of any two simultaneous events

ACuriousMind means the notion of co ordinate transforms where the components of the metric transform too. in this notion, any transform preserves distances

it is just the invariance of the scalar $g_{\mu \nu} v^{\mu} v^{\nu}$

Bml

The Galilean structure by definition extends the affine structure of the space, so any transformations preserving the Galilean structure ought to preserve the affine structure, too. Let $\phi:\mathbb{R}^3\times\mathbb{R}\rightarrow\mathbb{R}^3\times\mathbb{R}$ be a Galilean transformation. Let $R\in\mathrm{O}(3)$, $\tau\in\mathbb{R}$ and $\mathbf{v},\mathbf{y}\in\mathbb{R}^3$. We can show that
$$\phi: \begin{pmatrix}
\mathbf{x} \\
t
\end{pmatrix}\mapsto \begin{pmatrix}
R & \mathbf{v} \\
0 & 1

@RyderRude Yes, thank you very much. I would also like a more physical proof.

Ryder Rude

just take the transform x' = x-o(t), t'=t, o(t) can be any injective function. this transform preserves spatial distance for simultaneous events and temporal distances for all events

in 1D, the spatial distance is x_2-x_1

for simultaneous events at t_1, x'_1= x_1 - o(t_1), x'_2=x_2-o(t_1). so x'_1 - x'_2=x_1-x_2

in multi dimensions, this transform becomes x'_i = x_i - o_i (t)

Bml

12:56

@RyderRude @ACuriousMind If you like and agree, I can open a question on Physics S.E. so that you can give a more articulate answer. Please let me know.

Ryder Rude

can u pls tell me what u find confusing about this transfirm

the physical interpretation is switching to the frame of an arbitrary observer whose worldline is o(t)

@Bml u can ask this anyway. maybe u will get other PoVs too

Bml

13:16

@RyderRude Nothing, it's a good demonstration, it's just to give the opportunity to make it available to everyone and give a way to confront each other.

ACuriousMind

@Bml What metric? You didn't write down any metric, and you still didn't write down a definition of "preserving"

If you asked this as a question on the main site, I would leave the same annoying nitpicks as comments :P

Bml

@ACuriousMind You are right to do so. I obviously don't have your level of preparation (I wouldn't dream of having it), so I really appreciate your rigour and precision.

Ryder Rude

@ACuriousMind pls see the mathematical formulation i wrotr

here is a mathematical formulation. we r looking for the group of transforms $x'(x,t)$ and $t'(x,t)$ such that $\sum (x' _i - y'_i)^2 = \sum (x_i - y_i)^2$ for events $x,y$ at equal $t$, and $t'_2-t'_1=t_2-t_1$ for all pairs of events. $i$ labels space dimensions.

@ACuriousMind

bml said they agree with this

is there any beef between us currently?

ACuriousMind

@RyderRude They said they are not claiming that transformations outside the usual Galilean group preserve the metric, but what you're formalising is precisely this

I find it unhelpful to throw half-baked formalizations around before we even have clarified what precisely the question is

Ryder Rude

13:39

sorry i mightve misunderstood the question then

so u r not interested in other transforms that preserve this

Bml

@RyderRude I am really very confused. Let's do this: 1) Let's try to examine the transformations belonging to the Galilean group and see if and how they 'preserve' the metric; 2) Let's compare the demonstration in 1) with the one you provided for the transformations outside the Galilean group and see what the substantial differences are. OK? Are you up for it?

@ACuriousMind Regarding metrics: aren't the transformations of the Galilean Group simple displacements of Euclidean metrics? Regarding the meaning of 'preserving', I do not know how to convey it, I cannot find the right words to express it. I mean what is commonly understood by 'preserving the distances between two points and time intervals' because they are simple displacements of the metric.

If you can understand what I am saying and come up with a stricter and more rigorous formulation, it would be much appreciated, as well as more correct.

ACuriousMind

13:55

@Bml I'm not sure what "displacements of Euclidean metric(s)" are

The Euclidean metric on $\mathbb{R}^3$ is just a function of two points $d(x,y)$

you can ask what maps $f : \mathbb{R}^3\to \mathbb{R}^3$ preserve this function in the sense that $d(x,y) = d(f(x),f(y))$. The answer is the "Euclidean group", i.e. reflections, rotations and translations, or equivalently the spatial part of the Galilean group

I have not yet understood what exactly the question about this is

Bml

@ACuriousMind Sorry, then isn't it correct to say that "transformations belonging to the Galilean Group are simple displacements of metrics?"

ACuriousMind

I have no idea what the phrase "displacements of metrics" means

Ryder Rude

@Bml the reason Galilean transforms preserve the space metric is that they reduce to a combination of translation and rotation on each space-like slice

naturallyInconsistent

Note that ACM is specifically pointing out that there is no necessity to state that he is talking about only inertial frames. You, on the other hand, have to put in the effort to explain where the non-inertial part comes into the picture, enough for it to become an answerable question.

Ryder Rude

the reason my more general transform preserves this is the same

and the reason the temporal metric is preserved is trivial as the transform is just a translation

note that my transform is $x'=x-o(t)$. at each space-like slice this is just a translation by a constant vector

so it preserves the space metric

it's just that different slices get translated by different vectors. Galilean boost is a special case of this general transform i gave

when o(t)=vt

Bml

14:08

@ACuriousMind Could you explain to me the relationship between Galilean Group transformations and metrics? This should be related to what I say when I point out that the Galilean Group transformations "preserve temporal intervals and spatial distances".

Ryder Rude

@Bml one important thing. while these transforms to non inertial frames preserve the space and time distances, they do not preserve newtons second law

for the latter, u need galilean transforms

Bml

@RyderRude Which of the many answers are you referring to?

ACuriousMind

14:36

@Bml Galilean spacetime is $\mathbb{R}\times \mathbb{R}^3$ with $d(x,y) = \lvert x^0 - y^0\rvert$ if $x^0 \neq y^0$ and $\lvert \lvert \vec x - \vec y\rvert\rvert$ otherwise, where $\vec x$ are the spatial parts of the $x = (x^0,\vec x)$ and $\lvert \lvert \cdot \rvert\rvert$ is the usual Euclidean metric on $\mathbb{R}^3$. The group of transformations that preserve this function $d$ in the sense above is precisely the Galilean group.

The proof of this is essentially the same as that for $E(3)$ (reflections, rotations and translations) being the group that preserves the ordinary Euclidean metric in $\mathbb{R}^3$, which is why I keep coming back to the simpler Euclidean case

Bml

@ACuriousMind This is the strict formalisation of what I wanted to say. Now, do you think the problem I posed can be addressed, or is there still something missing?

Ryder Rude

@ACuriousMind but $x'=x-o(t), t'=t$ preserves this $d$ but is not Galilean in general

@Bml to preserve Newton's second law, u need galilean group

becuz it's a second order eqn

Bml

@RyderRude Can you elaborate on this? I would like you to show me that metrics and Newton'ssecond law in a Galilean Group are preserved, in a rigorous manner, because while I understand it conceptually, I do not know how to express it in rigorous terms.

Ryder Rude

14:51

@Bml preservance of metric : from t'=t, it follows that $\delta t'=\delta t$ for any pair of events. is it good so far?

i mean t'1-t'_2=t1-t2 for any pair of events

from t'_1=t1, t'_2=t_2

if it's good, ill move on to preservance of spatial metric

Bml

@RyderRude Yes

Ryder Rude

for rotations, each slice of space at a particular moment of time gets rotated. so it should be obvious that the euclidean distance is preserved on each of the slices?

for boosts, x'=x-vt. the slice at time $t$ gets translated by $vt$. so euclidean distance is preserved at each slice as it's just a translation

im sorry. id hav to type a lot for the rigorous proof. u shud see this part in a book

im just translating the math into english

Bml

@RyderRude Excuse me, but aren't these the classic Galilean transformations? How would they show that the characteristics of the transformations of a Galilean group are preserved in the transition from an inertial to a non-inertial frame?

Ryder Rude

then idk what ur question is. im just showing that the classic galilean transforms leave the temporal and the spatial distances invariant

u dont want to use x'=x-vt?

Bml

15:07

@RyderRude OK, sorry, I misinterpreted

Ryder Rude

yes. im not talking about non inertial preservance rn becuz u asked about galilean preservance of metric

@Bml in a Galilean group

about Newton's second law, the lhs is $\frac{d^2x}{dt^2}$. under the transform x'=x-vt, t'=t, this transforms to $\frac{d^2x'}{dt'^2}$

for a non linear transform x'=x+vt+at^2+...., this wudnt hold as the second derivative wudnt be able to kill terms of order t^2 or higher

Bml

@RyderRude I am afraid we are not understanding each other. You said that you agree with me when I say that 'the transformations that are part of the Galilean group preserve the distances between two points and time intervals, so the vectors should not undergo mutation in their length and orientation'. The Galilean Group includes all affine transformations that preserve distances between simultaneous events and time intervals.

You yourself said that it is possible to prove that even when passing from an inertial frame to a non-inertial one, these characteristics are conserved. Now, however, you say that since we are talking about a Galilean group, you are not talking about non-inertial frames. What am I missing?

Ryder Rude

@Bml u asked me here how they r preserved under the Galilean group so i stopped talking about non inertial transforms

Bml

@RyderRude Then I misspoke. I meant the inclusion of non-inertial frames. Sorry again.

Ryder Rude

15:22

here is the proof for non inertial transforms. $(\vec{x}, t)$ transforms to $(\vec{x} - \vec{o(t)}, t)$ under this transform. is this good so far?

Bml

@RyderRude OK

Ryder Rude

now take any time moment, say t1. and take two points on it $\vec{a},t_1$ and $(\vec{b},t_1)$

the original distance between these was $||\vec{a}-\vec{b}||$

good so far?

Bml

@RyderRude OK

Ryder Rude

these two points are taken in the original frame before transform

ACuriousMind

@Bml What is the "problem [you] posed" in this formulation?

Ryder Rude

15:26

after transform, they become $(\vec{a} - \vec{O(t_1)}, t_1)$ and $(\vec{b}-\vec{O(t_1)}, t_1)$

the new distance is $||\vec{a}-\vec{O(t_1)}-\vec{b}+\vec{O(t_1)}||$

this is equal to the previous distance because the vector O(t_1)$ term cancels out

intuitively, the non inertial transform is also just performing a translation by $\vec{O(t_1)}$ at the time slice at $t_1$. being a translation, this preserves the euclidean distance on the slice

Bml

Given your formulation, the Galilean structure by definition extends the affine structure of the space, so any transformations preserving the Galilean structure ought to preserve the affine structure, too. Let $\phi:\mathbb{R}^3\times\mathbb{R}\rightarrow\mathbb{R}^3\times\mathbb{R}$ be a Galilean transformation. Let $R\in\mathrm{O}(3)$, $\tau\in\mathbb{R}$ and $\mathbf{v},\mathbf{y}\in\mathbb{R}^3$. We can show that
$$\phi: \begin{pmatrix}
\mathbf{x} \\
t
\end{pmatrix}\mapsto \begin{pmatrix}
R & \mathbf{v} \\

Mr. Feynman

How broad is String theory? How much stuff does one need to learn to be ready for cosmic strings?

Ryder Rude

@Mr.Feynman do u want to go to the theoretical side?

ACuriousMind

@Bml What does the part where you say "we can show that..." mean? You didn't even specify what conditions $\phi$ has you fulfill, you just said it should "be a Galilean transformation". What is a "transformation to a non-inertial frame of reference"? Is it not also just a map from $\mathbb{R}^3\times \mathbb{R}$ to itself?

Ryder Rude

as opposed to experimentalist

ACuriousMind

15:34

I feel you're getting stuck in a weirdly circular situation here where you simultaneously state you know that Galilean transformations are the unique transformations that preserve the metric and you simultaneously ask whether there are any others :P

Slereah

Any subgroup of the galilean group, presumably :p

ACuriousMind

@Mr.Feynman you do not actually need string theory for cosmic strings :P

Slereah

they do not share much outside of being strings

ACuriousMind

while some string-theoretical models produce cosmic strings "naturally", you can study 1-dimensional defects like cosmic "strings" without ever doing any actual "string theory"

Slereah

Cosmic strings aren't even really strings rly

They're not like one dimensional objects

you can just model them as such

ACuriousMind

15:38

are they like the Dirac string, where the stringy nature is a symptom of trying to force everything into one coordinate chart?

Slereah

They're more like topological defects where there's some phase transition around some defect

The defect is one dimensional but as an object it's not really so

ACuriousMind

oh, that's what you mean

Slereah

you just have a pretty abrupt transition from one to the other

ACuriousMind

yeah, the "object nature" of such defects is generally debatable

Slereah

but typically in GR they are modelled as such because why work harder

Bml

15:40

I mean the most general $\mathbb{R}$-affine map, i.e.
$\phi(\mathbf{x},t)=A(\mathbf{x},t)+(\mathbf{y},\tau)$

Mr. Feynman

@RyderRude is there any non-theoretical string theory...?

ACuriousMind

lol

Mr. Feynman

@ACuriousMind oh, so can I just read a paper about cosmic strings?

Slereah

@Mr.Feynman guitars

ACuriousMind

@Mr.Feynman depends on the paper :P

obviously some paper that talks about how to get cosmic strings from string theory might expect you to know string theory, but as far as I understand it, there should be cosmologists talking about them in non-string-theoretic terms, too

Slereah

15:43

The only book I have on the topic is "Kinks and domain walls" so I guess I'll be recommending that one

but there are presumably better books on the topic

Claudio Menchinelli

Guys, I have some questions: I'm studying the time development of the HO. I have shown that the Hamiltonian operator is time-independent even in the Heisenberg picture, while the position and momentum operators oscillate with time instead

ACuriousMind

@ClaudioMenchinelli The Hamiltonian operator is always time-independent unless it's already explicitly time-dependent in the Schrödinger picture

there is nothing to show, every operator commutes with its exponential so $[H,U(t)] = 0$ always and so also $UHU^\dagger = H$ always (when $H$ is not explicitly time-dependent), this is not specific to the HO at all

Claudio Menchinelli

wait in my case the x and p operators are time dependent

so $H = H(t)$

no wait

ACuriousMind

no, they are not time-dependent in the Schrödinger picture

Claudio Menchinelli

yeah but I'm working in the HP

ACuriousMind

15:52

the time evolution operator is $U(t) = \mathrm{e}^{\mathrm{i}Ht}$ regardless of picture when $H$ is time-independent in the Schrödinger picture.

and so $H$ is also time-independent in the Heisenberg picture, because its Heisenberg equation of motion is just $\partial_t H = [H,H] = 0$

Claudio Menchinelli

Oh yeah now I see

but I have a question

ACuriousMind

you are correct that $x$ and $p$ evolve in time in the Heisenberg picture, but $H$ does not - it never does, unless it is explicitly time-dependent already in the Schrödinger picture

Claudio Menchinelli

yeah as it should be in fact

ACuriousMind

my point is that it's important to realize that this argument is fully general and not specific to the HO at all

Claudio Menchinelli

Yeah I see what you meant to say

thanks

Bml

15:54

@Slereah Yes, you are right. It was a matter of semantic.

Claudio Menchinelli

but I have another question

If I plug in the mean values of the position and momentum operators inside the hamiltonian

am I computing the classical hamiltonian?

ACuriousMind

no, you're computing something without any particular physical meaning in general :P

Claudio Menchinelli

$$H(\langle x(t)\rangle, \langle p(t)\rangle) $$

ACuriousMind

why would you plug the mean values in there?

there's no physical law that would indicate this quantity is of any interest, is there?

Claudio Menchinelli

My professor said to do this

and to show that this is different than $$\langle H\rangle_t$$

ACuriousMind

15:57

note that $\langle p^2\rangle \neq \langle p\rangle^2$, so this is not the same as $\langle H\rangle$

Claudio Menchinelli

and why would he tell me to do this

ACuriousMind

@ClaudioMenchinelli ah, yes, so the whole point of this exercise is to demonstrate to you that this doesn't actually yield anything meaningful

Claudio Menchinelli

I got different results in fact

ACuriousMind

@ClaudioMenchinelli because there are many people who do not realise these two things are different and will try to compute $\langle H\rangle$ by plugging in $\langle x\rangle$ and $\langle p\rangle$

the point of this exercise is to show that it doesn't work like that

if you weren't under this confusion to begin with, good for you!

Claudio Menchinelli

Oh I see hahah

no I thought that plugging the mean values into the hamiltonian amounts to computing the classical hamiltonian

ACuriousMind

16:00

what is "the classical Hamiltonian" and what would computing it mean?

note that classically, the Hamiltonian of the HO is the energy and also a number constant in time on all trajectories that are solutions to the equations of motion

Claudio Menchinelli

yeah in fact I remembered well

my notes are super messy

that's why I was confused

I will correct them. Thanks @ACuriousMind for the clarifications

and today I need to do coherent states

I might come for help later hahah

Bml

@RyderRude OK, thanks

raf

4 hours ago, by raf

Hi, can anyone please help me find a better strategy to get the expression of $[\nabla_a,\nabla_b]T^{{mn}}_p$ in terms of fundamental tensors?

Ryder Rude

@ACuriousMind hello. just to confirm, u agree that distances and times are preserved in non inertial frames using the transformation i gave, right?

raf

I have written about the query and my approach here: physics.stackexchange.com/q/795119/173639

ACuriousMind

16:41

@RyderRude had @Bml indicated that this particular transformation was what they're worried about, I'd have explained that our definition of "Galilean spacetime" is still missing a crucial ingredient: The requirement that the Euclidean distance does not "vary over time", i.e. the urmeter at $t=0$ stays an urmeter at any $t$. Once you properly phrase this it excludes any non-affine transformations like your arbitrarily time-dependent translation.

Bml

16:51

@ACuriousMind "The requirement that the Euclidean distance does not "vary over time"...": that's what I wanted to say, I just didn't know how to express it properly.

ACuriousMind

17:07

@Bml So, geometrically the formalization of that is that, in addition to preserving $d$, the Galilean transformations $f$ need to preserve the following property: If for any two trajectories (spatial functions of time) $q_1,q_2 : \mathbb{R}\to\mathbb{R}\times \mathbb{R}^3, t\mapsto (t,q_i(t))$ the relative velocity $\partial_t(q_1(t) - q_2(t))$ is constant, then it should remain constant after transformation, $\partial_t(f(q_1(t)) - q_2(t)) = \partial_t(q_1(t) - q_2(t))$.

Ryder Rude

this is y Lorentz transforms r so good. for them, the metric alone fixes this property

there is something bad about this c tends to infinity limit

nature had to be Lorentz becuz galilean stuff is just awkward

Semiclassical

17:22

@RyderRude i think this isn't quite right. for one, his final work before he died was this book with Hiley: The Undivided Universe: An ontological interpretation of quantum theory

so he evidently did retain some regard for his earlier work

that said, his views definitely shifted more from physics as such to philosophy more generally

Bml

@ACuriousMind OK, so how to continue?

Semiclassical

and when he introduced his pilot-wave theory, i don't think he regarded that as a "final" theory---he very much wanted to leave open the possibilty of subquantum processes

Ryder Rude

@Semiclassical yes. in wholeness and the implicate order, he says that the hidden variable theory is only for a convenient way to think about QM

in terms of ontology, he seems idealist

it's really weird that he can support both because the consciousness interprrtations r like the polar opposite of hidden variable interpretations in terms of philosophy

Semiclassical

eh, depends on what you mean by that. for instance, there's idealist interpretations of QM which are very explicitly "consciousness causes collapse" and Bohm is not that

Ryder Rude

oh

Bohm says that the implicate order is like a oneness of everything and physics is mind dependent explicate order

this is like Kant's viewpoint

i'm not sure how he links this to quantum physics tho.ive not read that part yet

Semiclassical

17:29

as an example of the type i'm thinking of, i always remember the undergrad quantum textbook i had

which as a mathematical text was perfectly satisfactory

but in terms of philosophy took on positions like "hey, maybe the way to resolve Wigner's friend is that there's not many individual consciousnesses but one universal consciousness!"

which...no, just no

but my own views of "Bohmian mechanics" are much more in the spirit of Bell than Bohm himself at this point

(even if by now my position on interpretations is that of conservation of misery, i.e., each interpretation compels you to be miserable in some way)

Ryder Rude

oh

why does Bohm talk about consciousness and stuff if it's unrelated to the measuremenr problem for him?

Semiclassical

for reference, this is the guy i'm talking about above: philpapers.org/rec/GOSCIQ

Ryder Rude

he says everything is one and there r no boundaries

is this view of his unrelated to quantum physics, or does he link it to quantum physics

@Semiclassical oh

Semiclassical

i'm pretty allergic to consciousness references in QM at this point, so that may be distorting my recollection of Bohm

Ryder Rude

oh. i just thought he had given up on hidden variables since he started talking about idealism. hidden variables are as materialist as is possible.

Semiclassical

17:37

you might find this article interesting then: philarchive.org/archive/WILQMM

Ryder Rude

@Semiclassical how do u re-concile hidden variables with relativity tho

Semiclassical

@RyderRude yeah, this is where the conservation of misery applies to pilot-wave theory :P

if all we're worried about is reproducing the phenomenology of special relativity, that's not beyond the reach of pilot-wave theory. the trouble is that while this is technically doable, it seems to do a lot of violence to the spirit of relativity

Ryder Rude

@Semiclassical thanks

@Semiclassical yeah. the theory becomes really awkward there. especially in QFT

Semiclassical

right

one of the more amusingly perverse suggestions i've seen is that it's actually easier to do Bohmian string theory than it is to do Bohmian QFT :P

i have no way to assess this, to be clear

Ryder Rude

i can see that being true. string theory has a particle interpretation

Semiclassical

17:42

but i like the idea that it's a suggestion that both the philosophers and anyone into string theory are liable to be repelled by for quite different reasons

Ryder Rude

but qft also has a similar interpretation in the worldline formalism

Semiclassical

the main person i know of who's worked on this is Nikolic

e.g. arxiv.org/abs/hep-th/0610138

Ryder Rude

wow it's based on the De Donder formalism

Semiclassical

that's a pretty old paper at this point

Ryder Rude

which is weird becuz the formalism is for field theory

Semiclassical

17:44

for a more recent stab at this: arxiv.org/abs/2205.05986

Ryder Rude

pilot wave theory makes the most sense with particles

i also learned a stochaistic interpretation recently. it's almost identical to Bohmian

Semiclassical

that's another common way to try to extend a trajectory interpretation into QFT, yeah

Ryder Rude

position basis doesnt exist in qft tho. we will have to make the momentum basis the beables

ACuriousMind

@Semiclassical of course it's easier - fewer possibilities to make contact with actual experiments! :P

Ryder Rude

i mean the Fock space

Semiclassical

17:47

@ACuriousMind pffffft

the reason why it's "easier" as i understand it is that you don't have literal particle creation/annihilation in string theory, which otherwise are a big problem for a trajectory interpretation

ACuriousMind

well, you don't really have dynamics at all in string theory, you just have a perturbative definition of scattering amplitudes :P

(even though string theorists like to sell it as if we had something more)

but particles are "created" or "destroyed" in the asymptotic state space just like in QFT, the c/a operators just look a little bit different (they're "vertex operators" on the worldsheet instead of free field modes)

Semiclassical

i'll take your word on that. my understanding of this is pretty much just what little i gleaned from this paper: arxiv.org/abs/hep-th/0702060v4

but mostly i just like how perverse "Bohmian string theory" sounds

(same as rooting for a bad team in NFL to upset a good one. not b/c of how it'd affect the standings, but just for how funny it would be)

Ryder Rude

@Semiclassical i think objective collapse is more promising than hidden variables among materialist interpretations, dont u think?

Semiclassical

meh

Ryder Rude

whyy

Semiclassical

17:55

just doesn't appeal to me? not sure how to put my finger on it

Ryder Rude

instead of having a hidden variable around all the time and piloting it, u just produce it thru collapse sometimes

@Semiclassical okay i guess. do u hav a particular reason to dislike many worlds or relational intepretation?

Semiclassical

i don't know relational interpretation well enough to comment

i'm not fond of having to assume the existence of many worlds to explain our experience of this one

the many-minds version of this is more defensible imo

Ryder Rude

how is that different

Semiclassical

the slogan for this is: "The many-minds interpretation of quantum theory is many-worlds with the distinction between worlds constructed at the level of the individual observer. Rather than the worlds that branch, it is the observer's mind that branches."

Ryder Rude

oh

Semiclassical

18:01

honestly i've mostly given up on finding the "correct" interpretation of QM tho

i just embrace conservation of misery :P

Ryder Rude

@Semiclassical rightly so becuz general relativity is unaccounted for in any QM theory

so it is useless to ask for a consistent interpretation of the world in terms of QM alone

Semiclassical

in the sense of not having a theory of quantum gravity, yes

Ryder Rude

r u sure that there is a quantum gravity, as in, a quantum theory of gravity?

i think it's neither quantum nor classical, whatever it is

Semiclassical

eh, all i mean when i say "quantum gravity" is something that reconciles the two in a satisfactory way

Ryder Rude

yes. this is y i added the "as in" :P

Semiclassical

18:04

what that reconcilation means i have no idea

Ryder Rude

consider that all quantum theories require a background time for the parameter of unitary evolution

this is sufficient to conclude that that theory isnt a quantum theory

Semiclassical

there have been Bohmian attempts at GR, mind: arxiv.org/abs/1801.03353

but that's again beyond my ability to critically assess

Ryder Rude

ya. it's really weird if Bohmian mechanics wud b the first to give a theory of quantum theory :P

it is the most hostile to relativity theory out there

Semiclassical

to the spirit of relativity at least

losing relativity of simultaneity is a hell of a price to pay :P

in the context of non-relativistic QM it doesn't bother me too much, b/c non-relativistic

Ryder Rude

even there, i dont see any reason to think in terms of hidden variables. part of it is becuz i like spookiness

Semiclassical

18:10

but honestly the non-relativistic context is the only one where i feel prepared to defend pilot-wave theory. in that context i think it's adequete, but the moment you want to move past that it becomes a millstone

Ryder Rude

agree

youtube.com/watch?v=sshJyD0aWXg this is the stochaistic interpretation i meant

instead of having a deterministic hidden variable, it has a stochaistic one. rest is like Bohmian i guess @Semiclassical

arxiv.org/pdf/2302.10778.pdf this is their paper

i didnt like it personally

stochaistic processes cannot explain interference due to phase, but they have a way around that

they r defining a more general kind of stocahistic process, which doesnt obey the usual properties

it's basically taking $|\psi (x)|^2$ and calling it a stochaistic evolution of a hidden variable. since the evolution is ultimately derived from Schrodinger eqn, it fulfills interference

Bml

20:34

Hi everyone. Talking about Galilean Group, a question about Galilean relativity came to mind. Is it correct to say that the initially presented proof applies to the statement that: "The laws of mechanics are invariant in the transition from a K
system to a K' system moving in uniform rectilinear motion with respect to K"? (It Is stated in all Physics books). I believe that assuming that the flow of time is the same in all systems is wrong, since the decay time of a particle in motion is longer than that of a particle at rest. How can this be justified?

ACuriousMind

20:48

@Bml The phenomenon regarding particle decay you're talking about is a relativistic effect

i.e. it is not described by Newtonian physics or Galilean relativity

Bml

21:22

@ACuriousMind I understand, but to say that acceleration in the two reference systems K and K' is the same and subsequently assert that "all laws of mechanics are the same in every inertial reference system" is not a logical leap?

ACuriousMind

@Bml What's your point? Yes, strictly speaking Galilean/Newtonian mechanics is false. It's still useful enough in many situations.

Relativisticcucumber

23:08

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