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 Discussion on question by SergioMarqu…

Imported from a comment discussion on physics.stackexchange.co...
Dvij D.C.
May 1, 2022 19:11
[...] An example of the latter kind of apparent (i.e., standpoint-dependent) would be the Doppler effect and the even more trivial example of the latter kind of apparent (i.e., purely a human illusion) would be a mirage or something like that.
Dvij D.C.
May 1, 2022 19:11
@m4r35n357 I get the spirit in which you are making the point about length contraction being apparent, but for someone who is new to the subject, I would like to point out that there is apparent (i.e., observer-dependent where observer means a choice of frame of reference) and then there is apparent (i.e., not only observer-dependent but also standpoint-dependent or something even more trivial, e.g., a purely a human illusion). Length contraction is the former kind of apparent. [...]
 

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Dvij D.C.
Apr 29, 2022 10:21
@Slereah I remember reading in Weinberg that the metric has 4 superfluous degrees of freedom due to diffeomorphisms and this is accounted for by the 4 Bianchi identities leading to the Einstein equation not fixing the 4 superfluous degrees of freedom of the metric. Which did make sense, whereas counting Bianchi identities as additional removal of dof seems weird :/
Dvij D.C.
Apr 28, 2022 21:31
I love these arXiv memoir submissions. Hartle has been posting a bunch recently: arxiv.org/search/….
Dvij D.C.
Apr 6, 2022 17:15
Since $f_{ijk}$ are subject to several constraints, I'd have to do actual counting to figure out the number of independent equations I'm getting.
Dvij D.C.
Apr 6, 2022 17:15
Hmm, OK, I think it's something I can write but is largely useless. Let's say a complete set of basis is $\{O_i\}$ with $i=1,\dots,n^2$ and the non-degenerate operator I have in mind is $O_k$ with $k\in\{1,\dots,n^2\}$. Given that the basis is complete, I can write $[O_i,O_j]=\sum_k f_{ijk}O_k$ with $f_{ijk}\in\mathbb{R}$. Now, for a generic observable $G=\sum_i g_iO_i$ to commute with $O_k$, I need $\sum_i a_i[O_k, O_i]=\sum_{il} a_if_{kil}O_l=0\implies \sum_i a_if_{kil}=0,~\forall l$.
Dvij D.C.
Apr 6, 2022 16:59
and thus, all observables that commute with $S_z$ can be parameterized by two real numbers, the coefficients of $S_z$ and $I$. Obviously, this is a singularly simple system but perhaps similar comprehensible parametrizations can be obtained for any finite-dimensional Hilbert space via writing out the most general observable as a linear combination of a complete set of observables and imposing commuting requirements?
 

 Discussion on question by Libertarian…

Imported from a comment discussion on physics.stackexchange.co...
Dvij D.C.
Apr 20, 2022 14:48
Let us continue this discussion in chat.
Dvij D.C.
Apr 20, 2022 14:48
The non-homogeneous part is just the addition of a constant vector everywhere. :/
Dvij D.C.
Apr 20, 2022 14:48
Furthermore, you're talking in terms of trajectories which is weird to me. The transformation maps a point in spacetime to another point in spacetime. It will obviously induce a transformation on trajectories, but it is not defined as a transformation of trajectories. For example, if I have a trajectory $x=\sqrt{t}$, then the transformation induced on this trajectory by the Galilean transformation won't even be linear. The transformed trajectory will be $x'=vt'+\sqrt{t'}$. But that has nothing to do with anything.
Dvij D.C.
Apr 20, 2022 14:48
It's obvious that the non-homogeneous part of the transformation is rigid. Now, the homogeneous part of the transformation is given by the matrix $G = \begin{pmatrix} R_{3\times 3} & v_{3\times 1}\\0_{1\times 3} & 1_{1\times 1}\end{pmatrix}$, i.e., $(x'~t')^\mathrm{T}=G(x~t)^\mathrm{T}$. So the transformation doesn't depend on the point at which you're applying the transformation (i.e., $G$ doesn't depend on $x,t$). By your standard, rotation is also not rigid because when I write the transformed coordinates $x'=Rx$, the $x'$ depends on $x$.
Dvij D.C.
Apr 20, 2022 14:48
I can measure velocity. I refute it thus.
 

 Discussion between Dvij D.C. and Libe…

Imported from a comment discussion on physics.stackexchange.co...
Dvij D.C.
Apr 20, 2022 06:51
Dude, I already told you without using matrices! It preserves the Euclidean norm. Anyway, I agree this is not going anywhere, although I carried it longer than any other conversation that I've had with a libertarian :) Bye.
Dvij D.C.
Apr 20, 2022 06:48
I agree tho that I have said pretty much everything I can say here. I will summarize my main two points: If you are allergic to using the matrix notation where I write $(x~t)^\mathrm{T}$ in a single vector then you can see the rigidity in purely Euclidean geometry and notice that the distances remain the same. The second point is that the more insightful way to look at the matter is look at the transformation matrix, and see that the matrix itself doesn't depend on time or space.
Dvij D.C.
Apr 20, 2022 06:46
No, SO(3) gauge theory does not. SO(3) gauge theory in classical mechanics would mean applying position-dependent rotations. Say I apply a rotation of 0 degrees on (1,0,0) and a rotation of 45 degrees clockwise around z axis on (0,1,0). Would the distance between the two points I mentioned remain the same?
Dvij D.C.
Apr 20, 2022 06:41
Lol, I have been responding precisely to your messages. Tell me why you disagree with the accepted definition of rigidity?
Dvij D.C.
Apr 20, 2022 06:39
Rigid transformation
In mathematics, a rigid transformation (also called Euclidean transformation or Euclidean isometry) is a geometric transformation of a Euclidean space that preserves the Euclidean distance between every pair of points.The rigid transformations include rotations, translations, reflections, or any sequence of these. Reflections are sometimes excluded from the definition of a rigid transformation by requiring that the transformation also preserve the handedness of objects in the Euclidean space. (A reflection would not preserve handedness; for instance, it would transform a left hand into a right...
Dvij D.C.
Apr 20, 2022 06:39
I just did: chat.stackexchange.com/transcript/message/60931099#60931099. And it is the definition of rigidity.
Dvij D.C.
Apr 20, 2022 06:38
That's one way to look at it
Dvij D.C.
Apr 20, 2022 06:38
It preserves the Euclidean norm
Dvij D.C.
Apr 20, 2022 06:38
@LibertarianMonarchistBot I told you why it's rigid!
Dvij D.C.
Apr 20, 2022 06:38
which involves (x,t)^T with which you're uncomfortable because you think it's the same as accepting that space and time are on an equal footing, they are not
Dvij D.C.
Apr 20, 2022 06:37
In order to do that, you have to write a matrix relation like the one I wrote
Dvij D.C.
Apr 20, 2022 06:37
I am not assuming anything about your knowledge of linear algebra, I'm just pointing out that you can't avoid representation theory of galilean group if you want to represent its action on coordinates in the way you wrote
Dvij D.C.
Apr 20, 2022 06:36
which would involve acting on (x,t)^T
Dvij D.C.
Apr 20, 2022 06:36
In other words, you have to represent the boosts as matrices
Dvij D.C.
Apr 20, 2022 06:36
If you want an element of galilean group to act on coordinates in the way you are writing, you have to use linear algebra
Dvij D.C.
Apr 20, 2022 06:35
Representation theory of the Galilean group
In nonrelativistic quantum mechanics, an account can be given of the existence of mass and spin (normally explained in Wigner's classification of relativistic mechanics) in terms of the representation theory of the Galilean group, which is the spacetime symmetry group of nonrelativistic quantum mechanics. In 3 + 1 dimensions, this is the subgroup of the affine group on (t, x, y, z), whose linear part leaves invariant both the metric (gμν = diag(1, 0, 0, 0)) and the (independent) dual metric (gμν = diag(0, 1, 1, 1)). A similar definition applies for n + 1 dimensions. We are interested in projective...
Dvij D.C.
Apr 20, 2022 06:35
So it is a matrix?
Dvij D.C.
Apr 20, 2022 06:33
No, but what is B(t) mathematically? You're multiplying it with x, so you have to define its properties.
Dvij D.C.
Apr 20, 2022 06:32
I'm not sure what is B(t) then
Dvij D.C.
Apr 20, 2022 06:32
Is B(t) supposed to be a matrix here?
Dvij D.C.
Apr 20, 2022 06:28
Just to reconfirm, I'm not smuggling in any relativity. These are pure Euclidean distances.
Dvij D.C.
Apr 20, 2022 06:27
Now, the reason x \to x + vt is rigid is that by the definition of a rigid transformation, it is a transformation that does not change the distance between points. Before the transformation, distance between $x,y$ is $\vert x - y\vert$, after the transformation, it's $\vert x - vt - y + vt\vert$ which is the same.
Dvij D.C.
Apr 20, 2022 06:27
LOL, it's not irrelevant because it points out the underlying principle, but OK. Regarding spacetime, what's your point? Space and time are obviously not on an equal footing in Newtonian mechanics (neither are they in relativity for that matter, but OK).
Dvij D.C.
Apr 20, 2022 06:21
OK, please go ahead.
Dvij D.C.
Apr 20, 2022 06:21
I keep mentioning rotations because it goes like x \to Rx which is "dependent" on x by your standards.
Dvij D.C.
Apr 20, 2022 06:20
No, it's not. The transformation parameter is v, not vt.
Dvij D.C.
Apr 20, 2022 06:20
I just want you to spell out for me why you think rotation is rigid but boost is not.
Dvij D.C.
Apr 20, 2022 06:19
You agree that the rotation part of the Galilean transformation is rigid, right?
Dvij D.C.
Apr 20, 2022 06:19
Tell me why rotation is rigid but boost is not
Dvij D.C.
Apr 20, 2022 06:18
because it's clearly not x(t)\to x(t) + a
Dvij D.C.
Apr 20, 2022 06:18
Wait, so rotation is also not rigid now?
Dvij D.C.
Apr 20, 2022 06:16
I'm not talking about a field theory either! A geometric transformation is a geometric transformation. My point is that Lagrangian is irrelevant (either in field theory or in particle mechanics) to determine whether a transformation is local or global. Obviously, it is relevant to determine the trajectory of a particle.
Dvij D.C.
Apr 20, 2022 06:15
About the second comment, the one with "This does not look rigid to me", I point to my earlier comment with the big matrix. Can you tell me why rotation is rigid even if there is a x on the RHS?
Dvij D.C.
Apr 20, 2022 06:12
Regarding your Lagrangian comment, why is it relevant to the discussion of whether the Galilean transformation is local or not? The transformation has nothing to do with the Lagrangian, the Lagrangian will tell you if the transformation is a symmetry of the system or not -- not whether it is global or local. In any case, for precisely these reasons, I'm not sure what it means to "start from a Lagrangian". Feel free to elaborate more on that.
Dvij D.C.
Apr 20, 2022 06:10
Furthermore, you're talking in terms of trajectories which is weird to me. The transformation maps a point in spacetime to another point in spacetime. It will obviously induce a transformation on trajectories, but it is not defined as a transformation of trajectories. For example, if I have a trajectory $x=\sqrt{t}$, then the transformation induced on this trajectory by the Galilean transformation won't even be linear. The transformed trajectory will be $x'=vt'+\sqrt{t'}$. But that has nothing to do with anything.
Dvij D.C.
Apr 20, 2022 06:10
The non-homogeneous part is just the addition of a constant vector everywhere. :/
Dvij D.C.
Apr 20, 2022 06:10
It's obvious that the non-homogeneous part of the transformation is rigid. Now, the homogeneous part of the transformation is given by the matrix $G = \begin{pmatrix} R_{3\times 3} & v_{3\times 1}\\0_{1\times 3} & 1_{1\times 1}\end{pmatrix}$, i.e., $(x'~t')^\mathrm{T}=G(x~t)^\mathrm{T}$. So the transformation doesn't depend on the point at which you're applying the transformation (i.e., $G$ doesn't depend on $x,t$). By your standard, rotation is also not rigid because when I write the transformed coordinates $x'=Rx$, the $x'$ depends on $x$.
Dvij D.C.
Apr 20, 2022 06:10
I can measure velocity. I refute it thus.