Reporting on an open problem

[00:00:01]Welcome to another live stream on the assumptions of physics. Today we are going to go back to an old open problem because there have been some people in the community that have been discussing and making some progress. So we are going to review what has been done and what hasn't been done. H maybe I'll pull out a document to see what is it that we're going to talk about. I'm slightly not ready, but let me go get it.

[00:00:39]Hopefully, if I just look on the open problem section, there's going to be a link on the things that I want to show.

[00:00:49]And that is if my internet works. And of course, we are live so it's not going to work.

[00:00:59]Ah, and this is why I have to prepare everything before. No, nothing. This is absolutely not working.

[00:01:10]Okay.

[00:01:12]Okay. Let me share the this.

[00:01:19]Okay. So, what we need to talk about is is there a link here? Yes. There we go.

[00:01:29]Uh so, we saw that in classical mechanics if we decide to uh h formulate the simplectic form in terms of position and velocity. We get this form where the mass times the metric tensor appears on the position velocity uh part of the simplectic form and in the position position part appears the uh the electromagnetic force tensor and appears this G alpha beta gamma contracted with the velocity and this G alpha beta gamma is basically some type of curl on the first index on G. And so this raises the question because this is not something that appears typically when people do differential geometry.

[00:02:28]And so the questions are since this appears where the electromagnetic force appears, can I interpret this whole thing as some type of inertial force or not? H since uh we know that the metric tensor uh represents so sorry the simplective form it's something that allows us to um uh define the entropy on phase space. Uh what does what do both of these term mean in terms of that? How do the the g alpha beta gamma influence the appearance of the um uh of the entropy?

[00:03:09]what happens in the entropy is effectively a non-cartisian uh coordinate system that's probably what's going to happen and uh in another question is what for example what happens when these are zero and what happens I don't know if I had it here but one of the thing that happens if these all these terms of zero probably it's there in the open problem uh these basically you can write the metric tens answer as the hessen of a scalar function and this kind of looks like what you have with kalar manifold in quantum mechanics where you have a metric that is generated by some type of hessen even though it's complex and so it's a little bit different and so the question is can we make sense of this pseudo tensor understand what it is and characterize it because since it appears so naturally the instinct should be it it should be important. So Logan has prepared a set of slides for what he has found and other people has found. So I'll give it up to Logan and please go ahead.

[00:04:23]>> All right, I'll come here and share my screen. Hopefully this all works like it was before.

[00:04:35]All right. Is it all all shared and everything? Is it working for you?

[00:04:39]>> It's streaming correctly.

[00:04:42]>> Okay. Um Okay. So, just for conventions, I'll use the um the factor out in front of the symmetric and anti-ymmetric parts. Um so that every every tensor will decompose equally into the symmetric and anti-ymmetric parts and that changes the wedge product. This is I make note of this because this differs from the conventions used in the assumptions of physics document where the wedge product is just directly the difference without the factor of 1/2.

[00:05:11]>> Okay. Um so the the way I started was sort of more from the mathematical side and then reconnecting with the physics.

[00:05:21]Um and tried to make everything as formal as possible. And the way I did that was defining metric one forms where for each value of sigma h subigma is just a one form on the manifold. And you can evaluate the exterior derivative of that. And because it will be anti-ymmetric the term you get is there equation two and this object G appears and the definition is given there in equation three. I also note the relationship to the anti-ymmetric christopho symbols which becomes important with some of Julius's work and Isaac's work with frame transport and how G can appear effectively whenever the christophal symbols appear this object should correspond to something happening physically.

[00:06:09]Um, and then this is just a note of why these metric one forms are natural objects to study in the first place besides just a random mathematical thing we would define. And that's because the sigma metric one form will extract from a vector field its sigma coariant component. And if we sum all of these together, or I shouldn't say sum all of these together, if we combine all of these together, we get the co- vector field associated with the vector field.

[00:06:39]Yeah, let let me let me digest it because the the whole like that notation is really like something that >> Yeah, that's that. Yeah, that's because I I was using a mix of both index and index free notation because for me the index free notation is more natural but I also know that in physics the standard is to use the component notation.

[00:07:04]>> Yeah. also because there is the whole issue that the symbol mean different things. So even if you're in component notation but you write dx that dx is still a one form for you >> and for me it's instead dx is typically something else. So I'm I'm trying to digest what is it that is going on. So basically that's just >> so so this is just right uh for a fixed sig sigma index. Um so in the same way for tetrad when we have latin latin components we just the latin index we generally treat as fixed. Um so this is just for each fixed sigma applying to the vector field itself not necessarily just applied to the vector components but applied to the entire vector field will yield one component.

[00:07:53]>> Yeah. So basically the h sigma is just picking out the component of the vector.

[00:07:59]It's picking out the covariant component of the vector >> and then it says the sigma component it's >> it's yeah it's the of the yeah the so you start with the vector field x and if you lower the vector field down to the co-angent bundle using the metric where once again I'm using the index free notation using the flat symbol. So what doesn't uh Okay, so I'm I'm trying to understand what this is. No, so the x lower sigma is a one form that takes >> it. No, it yeah a h lower sigma is a map from the co-angent sorry a map from the tangent bundle to individual co-ang well yeah I I I completely confusing to me okay so x is a vector field okay and h when you apply >> when you apply h sub sigma you just get a function you get a function x sub sigma is one of the components of the co- vector field which is just a function over the manifold >> right but then >> because h sub sigma is a one form in and of itself so we apply to a vector field and you'll get a function now we we we these functions that we get for each value of sigma naturally assemble together into a cove vector field >> right so the the thing that's absolutely not clear to me is that if I got the component I would get an x upper sigma and not an x lower sigma.

[00:09:48]So I don't understand why if if that's a a a scalar function where where why do I get the the the lower form?

[00:10:03]because we have uh part of the definition of the metric one form we have the metric in there um which will lower it down to the co-angent bundle.

[00:10:16]>> Okay. Okay. So there there in the in in the middle of equation 4 the dx mu and partial new will turn into a delta change the new index to a mu and then the metric will lower down to sigma >> right because the real thing is that I'm contracting x m new with g sigma okay that's what we are doing so I'm getting the component and I'm uh okay and I'm doing the the inverse in terms of units and then uh the the sigma component of x.

[00:10:55]Okay. So what is x flat now? It's one form.

[00:11:01]>> Yeah. or a co- vector field over yeah co- vector field which is a one form >> right which is going to get is basically do going to do the uh the inner product between the uh incoming vector field and the vector field of x. Okay. [clears throat] >> Um this is a lot of math just like formalism that's not particularly that important. Um but I wanted to include it as a way of staying very mathematically rigorous and grounded. Um this is basically just what um was done with for example when in your introduction on capital G you wrote that if all of the capital G components vanish the metric must be generated by some Hessen potential.

[00:11:58]>> Yeah. Um this is basically just making the applications of point P points lema rigorous but instead of considering the case of all components G mu sigma vanishing this is asking what does it mean for a particular combination of G mu newu sigma indices to vanish and it's not particularly that important as far as the mathematics go I'm trying to find here right so what I defined here with u restricted to mu is basically it's coordinate dependent it's the mu uh mu new coordinate sheet that is over a point x kn and what it means for a particular combination of mu new sigma indices on capital g to vanish means that the metric the sigma sector of the metric is generated potential on the munu plane. That's right. So it it has to do basically with the sigma sector of the metric curling in the mu new plane.

[00:13:14]And that gives us an understanding of each particular combination of indices rather than just the condition that all components of G vanish because all components of G vanishing is a very strict condition to be met on neighborhoods because most metrics are not generated by Hessen. So that's why I was wondering okay well what does it mean for a particular combination of indices um and then this carries carries for the point's lema applications and shows the hessen um hessen potential and other people have also noted when looking at this that because g is not tensorial we can change a coordinate system at a given point >> to one in which the metric is hessen at any given point on any given manifold.

[00:14:03]In the same way that remon normal coordinates choose a coordinate system to make all of the christophal symbols vanish or the connection coefficients vanish. Um >> at a point or at a neighborhood you can do it >> at at a point you can do it. The the real geometric restriction is whether or not we can do it for a neighborhood >> and on and on on an open neighborhood.

[00:14:24]We can only do that if the metric is actually generated by a hessene potential.

[00:14:30]Um and then this gets even more coordinate free. This is basically because earlier we had h subigma was one one form.

[00:14:43]What this notation means is we're effectively assembling those together >> and h by itself with no index is just a homeomorphism from the tangent bundle to the cotangent bundle. And in equation 12 I just write that in component form. H applied to a vector field just gives the cove vector field of X. The only reason why I did all of the formalism in equation 11 is to show how the metric one forms appear in this lowering map and to stress the importance of the metric in how we lower vector fields into co- vector fields.

[00:15:23]>> So wait an so how is H different from just G?

[00:15:28]H is different from just G because we have um because G by itself obviously each component G mu new is just a scalar function over the manifold. If we contract it with dx mu we have a one form over the manifold.

[00:15:47]So H is basically the one form version of G each for each index for yeah each metric each each yeah each metric one form is just the one form version of the metric in a particular sector for >> G is a function of two vectors while H is a function of uh >> just one vector right >> one form >> sorry it's just like evaluating g uh half right not plugging in two just plugging in one >> yes yes it's plugging in one um because right yeah because if you apply if you apply h to a vector field you get the co- vector field and then you can apply that >> right >> as a one form to a vector this is a basically yeah you could think of the square root of the metric >> well I I'm just thinking that the metric takes two vectors and gives you a number and this H gives me a vector. It takes a vector and gives me a co vector.

[00:16:53]>> Yes.

[00:16:53]>> Right. That's it. Yeah.

[00:16:54]>> Yeah. It it's just a partially applied metric.

[00:16:58]>> Yeah. Um and then you can evaluate the exterior derivative of this mapping by taking exterior derivative of each individual component. And the reason you can do that is because the only part that will vary along the manifold is going to be the metric dependency.

[00:17:19]Um all of that is to say okay so now now I get into how this connects to classical mechanics is because the lejand uh >> to to bring it home. Sorry I keep interacting because if I don't understand each point sorry I skipped. So the point that you're making I think is that that g alpha beta is going to be that exterior derivative of h right.

[00:17:44]>> Yes. Yes. Um so obviously here down in equation 13 what I have dh sub mu is you can substitute in for capital g and this is going to appear. Yeah.

[00:17:56]>> Yeah. Okay. Um and so we can say that dh without any indices vanishing on a region is precisely the condition under which the metric admits a hessen coordinate description on that neighborhood. The reason why I say hessen coordinate description is because normally the hessen condition is defined in terms of covariant derivatives.

[00:18:21]Um but here we're using uh yeah normally the hessen condition is using coariant derivatives. Here we're using coordinate derivatives. So this hessene condition is going to depend on our coordinate system. Um and then I introduce a bit more notation um that basically just changes how we think about the position dependence of instead of considering this family of H mappings, we just consider the mapping at a point X. And that mapping at a point X is going to be a homeomorphism from the tangent space at X to the cotangent space at X which is just going to take in a tangent vector at X and output the cotangent vector at X. This is exactly what Julius was saying of just it's the metric applied to one vector.

[00:19:12]um leandre map uh behaves in a similar way where we have leandre map is just a way of associating to each vector at a point x a cotangent vector. So we can relate to this to classical mechanics by given a particle of mass m charge q on a manifold. Um we note that the for velocity u at a point x is just tangent vector at x. And if we let a subx be the electromagnetic for potential at the point x we define the fiber wise a genre map. So this is just going to take in this depends on the point x um because it is fundamentally a map from the tangent fiber to the cotangent fiber at point x and is defined as as so in equation 16.

[00:19:58]um which is a it's a well- definfined sum even though we have the velocity in one part and not in the other because both of those pieces will be elements of the co-angent fiber at point x. They'll both be co- vectors. Um and this is defined as script p because this combination is going to be our uh canonical momentum associated to the velocity.

[00:20:25]Um this is not particularly that important but I just note because we're operating from a Hamiltonian Hamiltonian privilege point of view. Um this is not necessarily as important but it can be noted that you can identify the lrangeian that produces this after you transform it with the leandre transform but I I don't necessarily prefer this approach because this treats the genre this treats the lranian as primitive and then you genre transform it. I'm looking at the genre transform more as the fundamental object um that associates velocities to momenta.

[00:21:00]Um this is just the uh the formal way of separating the velocity dependent part and the velocity independent part. Um just splitting up our definition of momenta into the part that depends on the velocity and the part that doesn't.

[00:21:17]This just separating into the kinetic part and the electromagnetic part.

[00:21:21]Now we can ask if we fix a given velocity. So if we imagine just a vector field across the manifold that is completely constant and we ask what is the position dependence of this map.

[00:21:35]Um the way we evaluate that is we'll differentiate both lx and alpha x um with the exterior derivative and we get equation 20.

[00:21:46]Once again we get this dh object. Um and then F is used as the electromagnetic electromagnetic um uh Faraday tensor.

[00:21:57]All of this connects back to um classical mechanics. Sorry, this is not okay. Um the important part is the physical interpretation of this. The mathematics is just for the formalism and making sure everything is rigorous, but that's not actually what we're interested in obviously. Um but MDH measures a scale uh the failure of this morphism which L subx is just how we assoc the kinetic part of the leandre transform the failure of that to have a scalar potential and QF is the failure of the electromagnetic potential to have a scalar generator. both of which >> generator >> um this has to do back with the uh point craze lema on um so right so down in equation six um when we when we noted that if a specific combination of these specific combination of G vanishes that means that we can write some of the metric components in terms of the derivatives of a potential or a generator depending on which word you want to use. Um, if we think about all the possible ways metric components might be generated from these potentials, that's what DH equation 16 encodes.

[00:23:33]>> And so that's what I mean.

[00:23:35]>> I thought there was also an an F, right?

[00:23:39]>> Yeah. So >> um right because F would just be the curvature of the electromagnetic >> right. So form well QF measures the failure of the fine translation alpha X to emit a local generator that would be the fact that uh the vector potential does not come from a potential.

[00:24:05]>> Yes. Yeah. Okay. Um, so now we can ask if we instead of just evaluating this at a point I I'm by evaluating I mean evaluating MDH evaluating this map at a point we can ask okay what if we evaluate it at every tangent vector along a curve.

[00:24:26]Um, and when I say with minimal consideration of how the velocity changes, that basically means as we move along the curve, the tangent vector velocity is going to change. And that will obviously be encoded. But we don't want to do anything extra with it. The only incorporation of that change we want to have is the bare minimum of just how it changes. Really what we're interested in is the position dependence of the velocity to momentum map expressed in local coordinates. That gives you equation 21 which is precisely what appears in the simplectic form when we use tangent space.

[00:25:08]All of that. So basically the punch line is that the simplectic form expressed over the tangent bundle instead of the co-angent bundle naturally incorporates the position dependent variation in the leandre maps at every single point along an evolution along a trajectory. This leandre map is our rule for converting velocities into momenta.

[00:25:31]When we try to express this simple form in terms of the tangent space, we have to include position dependency of this map because this map is not going to be the same at every single point in general.

[00:25:46]The metric dependent variation is encoded in DH while the electromagnetic variation is encoded in QF.

[00:25:58]So f appears also when we don't go to velocities and we just go to kinetic momentum. So we use the the lower indices form of the um of the kinetic momentum.

[00:26:16]>> Yes. Um in that case our uh the mapping we still have we would still have the this aphine translation is the key point of we aren't necessarily mapping from the tangent space to the co-angent space but we're shifting everything by this translation of I write alpha alpha is just defined qa >> yeah so I understand the math it's the problem is that I still I'm still not uh I I don't feel that I have made any progress on the physics like it seems that the math is is put say more better together >> but I still don't understand uh what's happening. So these >> how at at every point along along a manifold whatever we think of the manifold you know at at every point in space >> we we have a rule of associating a velocity to position however that rule isn't going to be the same everywhere >> and the reason it's not going to be the same everywhere is because the metric will vary the electromagnetic potential will vary and all of this um how this rule varies is depend is encoded precisely in G G encodes the kinetic part of the Leandre transform and how that varies, >> right? So in in that sort of nebulous sense, yes, the the thing to me is I need to understand what each components say >> and and then how is it related to a force like like it seems to me that there has to be a geometric way to understand both the the potentials the P and the forces here. And I want to understand both for f that appears in a slightly different way because f appears even if I just say I am I go from position to not conjugate so from velo conjugate momentum not to velocity but to let's say kinetic momentum. So I haven't I haven't moved to the to the tangent. I'm still in the cotangent bundle but f already appears there. So that's picking up let's say a part of of of this uh change of map coordinate whatever you want to call it and then we we we go to the other side. So I in the diagonal for example it's a lot easier to understand for me why when I just use position and kinetic momentum the uh the diagonal the off diagonal terms are still the identities because I'm still in the cotangent bundle and all that I did was a change between p and u that is just dependent on x. So the way that I count state, it basically didn't change. It's it's still the same. But then when I go to velocity, clearly the way that I count state is going to be different. And clearly there is a a position dependent because uh if I have a potential forces, my velocity with relation to to momentum is going to be different. And also if I have a if I'm not in flat space then things are gonna get screwed up. But I still don't understand exactly why and how is that related to forces. Does that make sense?

[00:29:36]>> Yeah. Yeah. Uh I I I still have a few more things. I'll go through those and then we can take it back.

[00:29:41]>> Okay. I thought [laughter] >> Yeah. Sorry. Sorry. It's deceptive.

[00:29:44]Deceptive.

[00:29:45]>> I thought you were done. Okay. Sorry. Go ahead.

[00:29:49]>> Okay. Cuz uh I I also wanted >> Okay. There you go. Okay. I I also wanted to formalize because a few other people have said in the discord like this is you know the curl of the metric.

[00:29:59]I wanted to formalize how exactly this appears when we talk about curl and vorticity.

[00:30:04]>> Um and that's effectively just using if we want to talk about the vorticity of a vector field at a point. We compute the line integral of the vector uh lowered vector field around that area. Um and then just doing the formal stuff of okay if you shrink down the neighborhood on what you're considering down to a point the line integral will stay but the area integral will just return the integrant yada yada yada.

[00:30:30]Effectively what we're doing is we're just evaluating the exterior derivative of the co- vector field at the point. Um so the ve uh the vorticity of a vector field V at point X. Oh I scrolled. Um is given in equation 26 which is just the exterior derivative after you lower it.

[00:30:52]But crucially what I did was I in equation 25 I split the lowered component of the vector field into the metric and co uh contravariant components.

[00:31:06]>> Okay. Okay. because there are so many things I need to unpack because the notation just jumbles everything in. So >> uh when you're doing the line integral of v uh uh flat I don't even know how you say that how you say >> yeah v flat yeah >> so you're effectively doing the line so the one form that you're integrating it's v contracted with g >> and then on the other side you're going around the So you're basically taking the V component along the path. That's what you're doing that you're that's what you're integrating over.

[00:31:53]>> Yeah. And around that loop uh infinite decimal loop around the point.

[00:31:58]>> Okay. And you need so I'm trying to understand so this is something that you can't do if you do not have a metric tensor because if I if you don't have a metric tensor the only thing that you can integrate are one forms.

[00:32:12]>> Correct. And if you start with the vector field, you have no way of identifying a unique one form.

[00:32:19]>> Okay. And then you say Stokes theorem, then you just do that and then you got it. Okay.

[00:32:26]>> Yeah. And yeah, equation 25. I used product rule to split up the met the metric and the contravariant components um while maintaining this anti-ymmetry because the exterior derivative appears in Stokes theorem it's going to be anti-ymmetric over these indices >> um and then I put that in equation 26 um instead of writing the anti-ymmetric brackets um you have the wedge product that appear there because fundamentally what you're going to get from this is a two form and the wedge product will enforce anti-ymmetry.

[00:33:05]On the next slide, I express that in components um each component of the vorticity which is equation 27 which with a little bit of notation there.

[00:33:18]But this what this what equation 27 is is it is the vorticity of the vector field V at point X in the mu new plane.

[00:33:29]Now this will obviously be coordinate dependent because when we talk about different coordinates what we mean by the munu plane will change but in a given coordinate system this is well defined and the reason why this split is helpful [snorts] is because G capital G is encoding the anti-ymmetric partial derivatives of the metric while partial star is evaluating the anti-ymmetric partial derivatives of the contravariant components of the vector field. The effectively what's happening is capital G is encoding how the metric is curling in the mu new plane and the other term is encoding how the vector field itself is curling in the mu new plane and we need both when we talk about the vorticity of the vector field because of the fact that to talk about the line integral in the first place from the previous slide we have to lower the vector field with the metric first.

[00:34:37]In that way, capital G encodes how the sigma sector of the metric curls in the MUN plane, which directly connects to why G appears as an obstruction to the metric being generated by a potential is this uh this notion of the metric curling.

[00:35:06]Yeah, I just need to understand that at some point geometrically what it means because again I understand it mathematically by I >> yeah what is it that is curling about the metric because >> um yeah effectively yeah when you when when you want to talk about how a vector field curves or not how a vector field curves how the circulation and the vorticity of a vector field part of what you're going to get is the actual vector field circulating but part of what you get is going to be a remnant of the fact that the metric itself might be changing in a circular manner and that is encoded in G.

[00:35:44]>> Yeah. So as I said I can understand it mathematically what I would need to understand is okay the metric defines you know length with rods and therefore you're going to make a path and this is the rods that you're taking. Look at this. When you when you do a full loop, your rod does not come in the same place. And that means like I would need something like that, right? Right.

[00:36:08]>> Yes. That that that's what um because back to uh the metric one forms and the exterior derivative of the metric one forms.

[00:36:19]Where do I have that?

[00:36:25]Equation two. um the the failure of H subigma to be closed is encoded in G because it's the exterior derivative of that. the failure of a one form to be closed corresponds to the anhalonomy.

[00:36:45]Um, which like you said, >> look at you're still throwing a math.

[00:36:50]>> Okay. Yeah, I I was just going to bring this back. I was just going to bring this back of exactly what you were talking about with um if you transport rods or rulers around around a loop >> um the ruler may change length when as you get back um and that is encoded in the failure of these metric one forms to be closed which is >> right. So what I need to understand so I I suspect that the uh so imagine it like this I have a a rod I make it go through a square >> and then there's going to be a change in the rod right >> yes >> or in the thing that evaluates the rod because these are one form so I suspect that these are the things that I give the distance like I give it the rod and and it gives me the the distance along some uh trajectory but but that's fine.

[00:37:48]So the point should be I have a rod that starts in a particular direction. I make it move through two direction and I'm going to get something right. And so I suspect that the mu and the new are the two direction in which I'm making the loop.

[00:38:06]>> Yes. Um >> right because that's the plane.

[00:38:09]>> Correct. So if you can think about your ruler as being a vector >> that we that we keep constant around the loop. We we we want to keep the same direction and everything and we're just curious about moving the this vector around. In that case the second term on the right hand side of equation 27 will be zero >> because of the partial derivatives. It will vanish. Um, >> okay. Okay.

[00:38:39]>> So, as you move around, what you're going to get is the vorticity will the vorticity in the mu new plane directly encodes the failure of this vector to come back to exactly where it started. And if the vector itself is constant, in the case of moving a ruler around a loop, the only nonzero term you're going to get is g mu sigma contracted.

[00:39:05]>> Yeah. Okay. Good. So, Mu and and new we agree are the direction in which I make my my little closed loop.

[00:39:12]>> Correct.

[00:39:13]>> Then sigma is going to be the direction in which I oriented my ruler.

[00:39:18]>> Correct.

[00:39:19]>> Okay. Then I come back and junu tells me the difference in that direction.

[00:39:28]>> Yeah. How how that has changed. And I'm not >> if it's positive it's going to increase and if it's negative it's going to decrease or it's a multiplicative factor. If it's positive it's get you know stretched and also if it's greater than one it gets stretched or if it's less than one it gets shrink like what is the coefficient telling me? um be because this came from vorticity and stokes theorem the the coefficient would tell you how much this vector is rotated in the positive orientation so you know counterclockwise when you get back to the starting point >> how much is rotated >> yes I do not believe you would [snorts] have any intrinsic stretching >> okay okay so This is exactly the type of thing that I want to be clear. Look just like just in general, right? When I I have this question. What is this thing mean? I want to understand this at level. Okay.

[00:40:28]So I have I have two direction in which I make the loop. I take my my my rod which is in a separate direction. I I make it go around >> and you say it rotates. But now to say that I have a rotation, you need to give me a plane. So is it the rotation along the plane? Yes, it it will necessarily be in the same um uh in the same plane of rotation as you moved it around and and I and be because of how g was defined with um a two form in the mu new plane. um whi- which this this is still what I'm actively thinking about um because I suspect part of why this rotation will necessarily be in the munu plane will have to do with something with either metric compatibility or something but that also doesn't make entirely sense because we haven't necessarily introduced a connection here we aren't using coariant derivatives we're using the partial derivative um >> but but but I believe just from how g was defined in terms of the exterior derivative of the metric in the mu plane that this rotation will necessarily be constrained to the mu new plane without any stretching unless the vector itself changes then the vorticity can be whatever which way because of the second term. But if you keep the vector constant and move it around the plane, the only possibility of it not returning to where it started will be a rigid rotation in the mu plane that is encoded by G mu contracted over the sigma index, >> which is the same plane that you transported it in. As a test, if the direction of sigma is perpendicular to both mu and u I would expect g to be zero then.

[00:42:20]>> Mhm.

[00:42:21]>> Because it cannot rotate, >> right?

[00:42:29]>> Yeah, that would be that would be a good chest. Yeah. Maybe a small comment is whether this applies to all the planes because uh how many planes can you define in the whole space. So one plane is not enough to set the metric to you know to trivial one. You need to apply this for for what three four planes and then conclude this otherwise in one plane you only have this conclusion that is flat. In other planes you still have an arbitrary you know dependence >> right. Um because yeah you you can you can consider transporting it in all possible planes attached to a point and sort of get a complete picture of how this transport would work and that would the number of planes that you have attached to a point will depend on the dimensionality of the space.

[00:43:15]Yeah I'm not I'm not entirely sure on the the perpendicular point which is something that I'll have to think about more.

[00:43:24]>> Yeah. So look, this is the type of thing that you just or anybody who's watching, these are the type of things that I'm interesting. So I have the math, right?

[00:43:31]Good. And I understand it's or or you know, it's I'm pushing symbols. So I understand it as pushing symbols, but I understand what's going on. And then I want a tight physical interpretation in terms of stupid simple operations and then they give me some insight and then of course I want to see that that insight is back in the math because otherwise I have the completely the wrong intuition.

[00:43:52]which yeah this this this intuition of moving the vector around and not coming back to where it started was exactly what I you know was picturing when I when I was saying that the metric rotates I just didn't make that as explicit which is obviously which is what I should have done instead of just relying on the math. Yeah, but that's fine. I'm here exactly to sort of as long as you accept that I'm gonna stay there and keep going back to it.

[00:44:15]Absolutely fine. I mean, this for me is a lot of fun. So, so good. Okay. So we have a very good candidate in in in what then G is with respect to you know rods movement because you're basically saying for the forticity the important thing is that if the rod remains the same then that second part is zero and therefore [clears throat] the only thing that I'm doing is this transport around which the thing that it's interesting to me and that that's another thing that I want to understand is that when we do the um the curvature right the the Romanian uh uh you know for tensor we are doing a similar thing.

[00:44:55]So how is it different? Because I feel that this is going to pick up some part of that idea, but this is coordinate dependent while the the the R is according independent. And I think there has to be a because my secret goal is to hope that there is some type of like simple in operation that we can do to G to get R, right? because we are kind of doing this thing. We're looping around and doing this something like that >> and uh yeah. So anyway, go ahead.

[00:45:34]>> Yeah, that that pretty much uh that pretty much finishes off what I had. I had I just attached on two sides on uh frame transport today because yesterday we were talking in the Discord server about frame transport. Um, but those slides were mostly for me because they are just a wall of text that I don't think helps anyone. Um, so I will save this for when when we talk about what some other people are thinking about.

[00:46:01]Um, so yeah, that finishes as far as my uh explicit presentation goes.

[00:46:09]>> Yeah. So uh [laughter] yeah, first well congrats for writing up all this maths.

[00:46:15]I would not have guts to even attempt to do it. There's a lot of latte practice.

[00:46:19]I will say that. Yeah. [laughter] >> Yeah. That that's why I'm just you know usually attaching these you know handwritten slides uh instead of latte because I just don't have enough patience to type the formulas which might not might be only you know distracting in the end. But uh going back to the topic uh uh I would uh suggest maybe like giving a like different uh you know view but for that maybe uh could anyone share u the part from the book where we have uh let's say this definition of big G and this um let's say Newton's second law in terms of the Christopher symbol and the Faraday tensor. Um yes I'll do it.

[00:47:07]So you want uh this part >> uh maybe lower like uh at the end of the page do Yes. For example, yes this one for example.

[00:47:23]>> So um this is basically uh Newton's uh second law in disguise [laughter] if you read it correctly. And now uh so we don't have too much time so I want to be you know uh just very quick. Um so uh what we see on the right side uh we have um uh the force which uh from the electromagnetic part it depends linearly on the velocity and on this Christophal part it depends quadratically and um now um what we know about so this is now discussion let's say from physics so let's say this reverse physics approach because we have reverse physics and physical mathematics and I want to you know take them separately because you know when you know otherwise we get into confusion. Um so what we know from electromagnetic forces in uh general relativity what what we know is that uh they only introduce the um they act perpendicular to the velocity meaning that uh the vector uh how to say the energy is preserved. The only thing is that the vector is rotating and that's what we see in the um in the simplectic form that's again it's the same part which is responsible for the rotation of the vector. Um now uh of course uh we should also pay um how to say a certain attention to um to the electromagnetic force as a physical theory and um this electromagnetic force that we derived uh from the vector potential because the vector potential was let's say this you know constant of integration that we introduced a priori this is not necessarily related to any uh you know field theory it can be even an offset by you let's say offset of uh momentum by some constant basic basically different point of you know reference point. So uh in this case let's say we use the same symbol for two different things but uh regardless what we see in the simplectic form and what we see from physics is that the electromagnetic force only introduces the rotation and uh that got me into thinking what the simplectic what information simplectnic form extracts and uh from what and uh here what would be maybe also helpful is if we introduced a force generated by a gradient of scalar. So let's say we had some potential that is not a vector potential not coming from a metric just some scalar potential which which we usually exclude or we say that we can absorb it into parameterization. It's fine. Uh what we would see is that scalar potential would not appear in the simplectic form and uh why because u uh scalar potential uh affects the energy directly. There is no freedom to rotate a vector. it only uh how to say we only see it explicitly in the Hamilton's equations of motion in the in the simplectic form what we usually see is um are the effects which do not change the energy is that it's any how to say possible operations of kinematical variables which preserve the energy or the length or whatever but uh they still allow some freedom so the motion is not completely frozen but it is not uh you So there is no friction there is no like energy exchange. So what I thought is that uh that g big g should also correspond to something like rotation meaning that we are not changing lengths we're not changing uh angles which are also coming from the inner product. So it's the same thing we can use the metric you know to define the length and to define inner products. So uh this operation should not do that. It means that what again remains is rotation of the frame.

[00:51:17]>> Okay.

[00:51:17]>> And I >> but this one I but here there should be a distinction uh of from uh rotation versus uh uh carrying a vector along around a closed loop because the these are different operations.

[00:51:32]So basically so so now the if the interpretation from the physics part is that simplectic form is basically is kinematics. It's not dynamics. We do not see why the force is generated why the energy is increased reduced. It's only it might be relabeling of indices. It can be mixing of the reference frame but it cannot be you know anything that you know accelerates or decelerates the system. So now uh what happens uh if we now go back to the Newton second law uh we have that this uh Faraday tensor is already antiymmetric. So uh if we tried to multiply by for velocity it disappears. So it means that the length is not affected fine. Uh however when we look at the christoal symbol uh with the with respect to the first two indices uh it's well we have anything it's not symmetric it's not anti symmetric anything you sell out. So now what is usually very useful in uh in algebra is this uh decomposition which is called uh trace trace free symmetric and anti-ymmetric and uh symmetric part encompasses the trace trace and trace free while antiymmetric is you know its own part.

[00:52:51]So now symmetric part usually describes this isotropic uh behavior something that does not depend on the angle. Uh so this is like expansion or you know contraction of the whole volume and uh this part uh actually uh this is the part that uh some people might uh remember when they calculate the trace of Christopher symbol which gives this derivative of logarithm of determin of square root of determinant of metric.

[00:53:21]There are these things sometimes appearing and what this does this generates basically when we contract with for velocity this will generate a part which is proportional to for velocity and anyone who knows uh anything about let's say conformal uh let's say conformal rescaling they will already tell that okay you can kill this part by changing parameterization. So this is the part this pure isotropic part is something that you can kill just by you know changing parameterization along your trajectory. So you change the time you kill this part for any you know coordinate system. Then there is the remaining part which is also of the symmetric part. This is uh called sheer or this um yeah basically it's like the the c the part that deforms circle into an ellipse. So this is really the part that comes for uh well this is a different part which let's say not let's not discuss for now for much but it also belongs to the symmetric part and it so what we can agree that neither of these two appear in the simplectic form so this are irrelevant what is remaining is only the antiymmetric part uh and so it is neither uh neither expansion nor the shear so the the the remaining part is usually called verticity and we saw the word here used uh very often which is verticity. So now uh how to understand that uh so uh yeah we so I was thinking about it uh for some time and then uh yeah yesterday I think we had some discussions in the discord server and uh we agreed uh and I we agreed on the notes that Logan made yesterday but yeah didn't add no didn't finish maybe for this talk uh but uh basically we we got reached the same conclusion so to extract if we want a physical operation to extract this anti-ymmetric part uh what we need to do is we need to define an orthonormal frame. So basically four vectors unit length all of them are you know 90° with each other and we ask what what happens when you take uh uh you know derivative on along the curve. Since uh these uh these four vectors span the whole space and they're a unit length, the net operation has to be uh the some rotation matrix. It has to preserve their lengths. And from here uh now of course since we're using ordinary derivatives uh there is always this additional part that comes uh from that derivative of the metric or this lead drag of the metric. Uh if we used covariant derivative then this rotation matrix would be anti-ymmetric and this would uh and this mat rotation matrix can be expressed in terms of uh by vector by vector spann by any two legs of this frame contracted with the christophal symbol and it and due to symmetries it extracts exactly the antiymmetric part of the christoal symbol. So, so you can interpret this that the net rotate that each vector of the orthonormal frame as we propagate along the trajectory it is rotated uh you know or and basically is written with in terms of the rest of the vectors and this rotation matrix is uh covered by the anti-ymmetric part of Christophel. So, so basically if we want to measure this part, we need to measure how the frame is rotating. So now to get back >> we we would need I I completely get lost. So I would need some kind of like slow with with the equations on like I I I'm not able to follow.

[00:57:18]>> Okay. Okay. So then uh so then let's say then the net conclusion is that uh the anti-ymmetric part it what it does is just uh let's say if we start with some initial >> yeah coordinate system I I I need I need equation otherwise I I completely get lost. So the thing that is interesting though >> that I didn't realize uh of something that you said and relates to the others.

[00:57:46]So if we're looking to the for velocity, the norm of the for for velocity is always constant. And so the fact that the only thing that the evolution can do is just rotate the the for velocity that that's basically an obvious it has to be like that. It can't do anything else. So the fact that the electromagnetic force field does that is it's in that. But now it also tells us that the inertial part also have to do that. And so the what we are interested in terms of the forces then it's going to be exactly the ferticity where the the part of the vector remains the same like the the vector remains of the same length because we're actually looking how those things rotate. So I even more uh uh like integrating the the parts that I've understood of what you were saying and what Logan was showing. I even see a a a even more of a physical reason why you are interested in this curl of of the metric tensor because it exactly tells you how the vector is going to rotate therefore how the velocity is going to change by the the action of the fact that you have different coordinate systems. So that that now makes a lot more sense to me.

[00:59:11]>> Yeah. Now of course uh uh even though the velocity is the magnitude of velocity is conserved uh it's it does not say that the Christophel is already antiymmetric because uh we are how to say we have to interpret uh it's um with respect to the notion that we have in flat space uh in flat space and cartisian coordinates because uh if we took this as you know flat space but different coordinates. Uh that's why we have additional terms.

[00:59:45]That's why christophal is not the anti-ymmetric part but the whole term.

[00:59:49]If it was pure rotation then yes we only have anti-ymmetric part but we have the full christophal because this um uh how to say we still have uh well uh we have to differentiate in which uh coordinate system we measure the length because if we measure in cartigian then the coefficients of the metric are you know minus one one on one if we change the basis now they are you know position dependent uh so it's not minus one1 it is something else and these differences then appear as uh as the symmetric part of the christole >> so that's why we have full Christopher but yes it is the conclusion is that uh it's it the anti-ymmetric part should tell us about the rotation of the frame but it is the frame tracked by this single point on the trajectory it is Not it is some it is not the vector taken along a loop that is a different thing which should agree. So that the two things are related and it needs to be clear how they are related but there is still the idea that on one side we're taking something and and making it a loop and then the other thing is that we are transporting the same thing and see how it's that thing itself is rotating.

[01:01:10]Yeah. Yeah. Yeah. Those are two different things are related and we need to understand how they're related.

[01:01:15]>> Yes.

[01:01:16]If Isaac wanted to to say something since he has been quiet or or it's fine.

[01:01:21]>> I mean I mean that's the point that if Isaac could share the slide that he put on discord last day then this would be exactly what I wanted to say. So I don't know why it was not included [laughter] >> because that was exactly the equation.

[01:01:36]>> You you asked me to share from the book so I shared from the book.

[01:01:39]>> Uh yeah yeah yeah but yeah. Oh no I mean Logan Logan had something. Sorry I forgot. I forgot who was whom I was discussing with but there was full page of calculations and it was exactly my formula. [laughter] So uh uh at uh yes this one.

[01:02:06]So, so basically what they said is that here if we we need to take sai and so now there's this kai ab mu mu is the coordinate index which is what we used a are so-called tetrate indices they are labeling these orthonormal legs so we have four vectors spanning the space hence we have you know uh 0 1 2 3 uh going into AB and if we contracted uh SI mu with four velocity the the remaining part is the rotation matrix and yeah I think uh yeah I think there was one outcome that uh we discussed in text but I I it must does not appear in formulas but yes It's uh it's due to rotation of the frame.

[01:03:10]>> Okay. I I couldn't uh like I didn't have enough time to to to digest the map >> I guess.

[01:03:18]>> Okay. So then I so I will uh so I guess uh for now uh also from my side I don't think I can contribute more to this. I I think uh like I my message to take away from here is that this part is that extra part is basically rotation of orthonormal frame that's uh we started initially with some labeling we had you know xyz t and then because of met behavior of the metric uh the labeling is uh this frame is changing you know uh because we always have vector with a matrix >> I have to go slowly because otherwise I don't Okay. So we start with EA and EB. Those are two coordinate the coordinate bases, right? They're the basis vector.

[01:04:06]>> Yes. No. A E A are the orthonormal vectors.

[01:04:12]>> EA and EB are two orthonormal vectors.

[01:04:15]>> Yes.

[01:04:16]>> Okay.

[01:04:17]>> They have their repres they have representation in coordinate basis. But at this point these are just vectors which are pointing in four different directions.

[01:04:25]>> Yeah. Yeah. Okay. Okay. So we have we have ena. Okay. So then you say I make I I move things and they will remain orthogonal. And so I have a relationship between their changes and the changes of G. And then I say well I want to define G on one of the vectors and on the change of the second vector with respect to that direction. And so I define this c a b mu which is just going to be 12 minus 12 the derivative of g uh okay on how do I know that it's the second one it doesn't matter that it's the second one because it's going to be symmetric so who cares and then you make the anti-ymmetric okay of course and that's going to be now the the the the g that we that we have so you have uh the the the anti-ymmetric part is going to be that plus 12 the derivative and then you get uh okay or you are just making the change of one or the other which we can verify is equal to zero and then uh you find that uh you have an omega ab mu which is that thing that we were doing.

[01:06:02]Oh, we are using then the covariant derivative.

[01:06:12]Okay. Yeah, I can't really understand.

[01:06:14]Okay. So, e >> Yeah, I don't know if we're gonna have time for me to to digest a little So I I I think this >> you're muted.

[01:06:26]>> Sorry. So So this last line uh I think we don't have to pay attention. This spin connection is what appears if we used if you study this tetrate field or frame field theory you know rigorously.

[01:06:39]This is not necessary in this case. I think what is important is that uh at some point you can express this uh kai as a contraction of uh this bi vector formed of two normal vectors contracted with christopho >> and [clears throat] what remains is yeah >> and yeah and that's what you were mentioning in the conversation as it progressed that I didn't have time to include in that write up was that that explicit rotation connection but I I I agree with the conclusion you got to Yeah.

[01:07:14]>> Maybe we'll discuss it some other time when we have a better uh narrative so that I can follow better because like that I didn't follow that much.

[01:07:25]>> Right.

[01:07:28]>> Anything else else you guys want to discuss uh when we close it here? Are there a question from the chat? There were far some viewers on on YouTube as well, but I I didn't see any activity on the chat, which I am monitoring. Uh >> I mean I mean yeah, there's there's obviously more to discuss, but I think all of that can be saved until a later date done in the Discord server. All of that, >> right? The point is that if we have something that can be discussed in 10 minutes and not in another hour, [laughter] >> right? Yeah.

[01:08:05]Okay. No, that was very good. like uh I think yeah this is definitely progress and uh as I said I'm I'm more interested in this like silly operational thing of what is it that we're actually doing and uh and yeah so we we can uh uh uh we can proceed uh I don't know if I have any topic for uh next live am I think I'm traveling next week let me check.

[01:08:39]Yes. So, next week I'll be traveling.

[01:08:42]So, I doubt that I'm going to be able to uh to be uh online. I'll I'll double check, but uh tentatively we we can uh and then I'm going to be traveling in August, but I'm going to be here August 1 at least.

[01:09:01]So, yeah. So if there are uh topics that people want to discuss on uh on the next year like we've with the conferences I found a couple of people that I want to invite to uh uh to chat. So that would be good and uh Logan Kaylor relationship and yes all of that.

[01:09:29]I I was just briefly looking through the Kaylor stuff that Nicholas put onto um onto there and it it is directly related to the whether or not the metric is Hessen which comes from the whether or not the metric is itself curling. So there there is a direct relationship there. Yeah. And then there is the so the the the kalor thing is also interesting because uh so it's something that I will want to recover in the space of ensembles because uh in uh so there is a natural metric metric that comes out of the um uh of informationational geometry that it's uh comes from the the entropy which is the the hessen of the entropy.

[01:10:20]So you take the SM of the entropy that gives you a metric on the space of ensemble both in classical mechanics and quantum mechanics and in quantum mechanics you also have a simplectic form on the space of pure states what I need is that simplectic form which is not going to be a simple form I need a a a plus structure extended in the space of ensembles and it should be that uh the space of ensembles is foliated by the entropy. So every uh iso every fiber of the entropy should be in some sense a simple well not manifold because we're going to be infinite dimensional but you know it has to is going to have some type of simple structure and it's going to have some type of uh uh uh the metric that comes from the uh the the entropy itself.

[01:11:19]And so there has to be a link between the two. And uh whether it's exactly the type of link that you would have in a kalor manifold, I don't know. And again, it's not going to be a manifold anyway because we need to be able to do it in infinite dimensional even in classical mechanics because if we look at the space of probability distribution, that's going to be infinite dimensional.

[01:11:40]But again, we're going to have this type of things. It's stuff that I want to generalize, but I do not know whether the the the the Kaylor is too much of a um ad hoc case for let's say finite dimensional uh uh things that then if you go to infinite dimension, it's going to become much more complicated. But it's it's in the realm of things that that I that I want to understand. So >> So okay. So we can >> we we can schedule for the next one. We we we we can already say that we're going to we're talking about this.

[01:12:15]>> Yeah. And because especially if you're gone next week and if it's in two weeks, that's enough time to >> develop things further and connect things to each other and everything.

[01:12:24]Yeah.

[01:12:27]>> Julius, you wanted to say something. I think I interrupted you. uh yeah maybe not about uh the next topic but more just you know the comment that's uh even if the kaler manifold can be useful for quantum mechanics and uh what you mentioned uh in this particular case uh this is actually quite constraints the possible geometries because uh we know that this is standard exercise that we need you know 20 independent components uh to which are then covered by reman tensor and if we have the scaler then it's just you know one component we have still 19 components to describe the rest so uh if there is a need for scalar manifold it should not come probably from here [laughter] okay so I think we're done for today and thank you everybody thank you Logan for putting everything together thank you for the nice discussion and we'll see if not that next week in two weeks by Oh yeah, there we go.