Mordred Posted March 31 Share Posted March 31 (edited) Your right you and I do have a different notion of explosion. Particularly in so far as what a geometry would look like if the explosion resulted in what we perceive as spacetime expansion with regards to what the resulting metric would look like. Particularly in what the resulting distribution of mass and its behavior would entail. The primary distinction you keep avoiding us an explosion always results in a preferred direction continously posting math that has no preferred direction will not change my mind. Of course there is a reason why any good cosmology textbook will specify expansion as opposed to explosion. You might want to consider that detail in so far that there is reasons that is the case. I don't know perhaps/perhaps not you might notice the difference if you take a multiparticle system and model each case in vector space would help you see the difference. Expansion after all involves thermodynamics.... Perhaps you can ask yourself how something like the critical density formula even work in the explosion case. Perhaps you can describe the difference between an expanding gas due to temperature change as opposed to one being forced in a particular direction such as popping a balloon might help to understand the difference Edited March 31 by Mordred 1 Link to comment Share on other sites More sharing options...

Mordred Posted March 31 Share Posted March 31 (edited) \[{ds^2} = {d\vec{r^2}} +{r^2}[d\theta^2 + {sin^2} d\phi^2]\] there an exploding spacetime simply use spherical coordinates and designate the radius with a vector. Now all you need is to add your time coordinate for 4d and a scale factor Edited March 31 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted March 31 Share Posted March 31 46 minutes ago, Mordred said: Your right you and I do have a different notion of explosion. Particularly in so far as what a geometry would look like if the explosion resulted in what we perceive as spacetime expansion with regards to what the resulting metric would look like. Are you talking about what an explosion is, or the result of an explosion? I don't think that you can provide a reasonable challenge to my view of what an explosion is, and it would seem that you made no attempt to do so. As for the result of an explosion, the mathematics says that an explosion in flat spacetime produces three-dimensional spaces of constant age that are hyperboloids. Obviously, this does not describe the real universe, and it was never intended to describe the real universe. Its purpose was to demonstrate that an explosion can produce an expanding three-dimensional space that is homogeneous and isotropic. 46 minutes ago, Mordred said: Particularly in what the resulting distribution of mass and its behavior would entail. The spacetime is flat, so there is no energy-momentum involved. The particles are considered to be "test particles", so as not to affect the spacetime geometry. The velocity distribution is simply that required for the resulting three-dimensional spaces to be homogeneous and isotropic. 46 minutes ago, Mordred said: The primary distinction you keep avoiding us an explosion always results in a preferred direction continuously posting math that has no preferred direction will not change my mind. And yet the mathematics proves otherwise. If mathematics doesn't change your mind, then... 46 minutes ago, Mordred said: Expansion after all involves thermodynamics.... Perhaps you can ask yourself how something like the critical density formula even work in the explosion case. Perhaps you can describe the difference between an expanding gas due to temperature change as opposed to one being forced in a particular direction such as popping a balloon might help to understand the difference This is physics. As such, it is outside the scope of what I am discussing, which is pure geometry. So far, I have not even considered the case of a Friedmann-Lemaître-Robertson-Walker (FLRW) metric. Link to comment Share on other sites More sharing options...

Mordred Posted March 31 Share Posted March 31 (edited) sigh honestly do you seriously not know the difference between something that expands with no inherent direction of any particles that volume contains such as Brownian motion as opposed to a inherent direction of gas escaping a previously pressurized container such as a balloon ? Come on man this is seriously getting futile 14 minutes ago, KJW said: Are you talking about what an explosion is, or the result of an explosion? I don't think that you can provide a reasonable challenge to my view of what an explosion is, and it would seem that you made no attempt to do so. As for the result of an explosion, the mathematics says that an explosion in flat spacetime produces three-dimensional spaces of constant age that are hyperboloids. Obviously, this does not describe the real universe, and it was never intended to describe the real universe. Its purpose was to demonstrate that an explosion can produce an expanding three-dimensional space that is homogeneous and isotropic. The spacetime is flat, so there is no energy-momentum involved. The particles are considered to be "test particles", so as not to affect the spacetime geometry. The velocity distribution is simply that required for the resulting three-dimensional spaces to be homogeneous and isotropic. And yet the mathematics proves otherwise. If mathematics doesn't change your mind, then... This is physics. As such, it is outside the scope of what I am discussing, which is pure geometry. So far, I have not even considered the case of a Friedmann-Lemaître-Robertson-Walker (FLRW) metric. ask yourself what causes the universe to expand and tell me why you would ignore the equations of state that cause expansion in your analysis ? The very purpose is to predict how objects such as stars and galaxies move the way they do in regards to how they separate. The very goal is to explain how the distance of stellar objects increase from one another. I 100% guarantee you will not match observational evidence via an exploding universe beginning.... Edited March 31 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted March 31 Share Posted March 31 18 minutes ago, Mordred said: ask yourself what causes the universe to expand and tell me why you would ignore the equations of state that cause expansion in your analysis ? As I said before, this is outside the scope of what I am discussion. You are now arguing against points I did not make. 18 minutes ago, Mordred said: I 100% guarantee you will not match observational evidence via an exploding universe beginning.... And I 100% guarantee that when I consider the appropriate FLRW metric, the exploding universe description will match observation, simply because of the equivalence between them. Link to comment Share on other sites More sharing options...

Mordred Posted March 31 Share Posted March 31 (edited) 1 hour ago, KJW said: As I said before, this is outside the scope of what I am discussion. You are now arguing against points I did not make. And I 100% guarantee that when I consider the appropriate FLRW metric, the exploding universe description will match observation, simply because of the equivalence between them. prove it and in that prove you a preferred direction any mathematics that does not have a preferred direction will be useless to prove your point. Such as the math you have posted so far. While your at it prove you can have a kinetic based exploding volume that matches observational evidence of a cosmological event horizon that exceeds the Hubble Horizon and get recessive velocities in excess of c with your equations above (this occurs beyond the Hubble horizon Don't think using the Homogeneous and isotropic metric of Minkowskii will prove your point. That metric does not describe an inhomogeneous and anisotropic spacetime. By any equivalence principle. For the record its been attempted numerous times by Professors in other cosmology alternative theories. This includes the rotating Godel universe. We were able to even confirm our universe is not rotating. There have been attempts with models such as a universe birthed by a WH or a previous BH. None of these models could properly match observational evidence. so GL I know you likely don't have the skill set for Christoffel connections and Killing vectors so I won't ask you to produce them. Edited March 31 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted April 1 Share Posted April 1 (edited) 3 hours ago, Mordred said: 3 hours ago, KJW said: And I 100% guarantee that when I consider the appropriate FLRW metric, the exploding universe description will match observation, simply because of the equivalence between them. prove it I don't have the appropriate Friedmann-Lemaître-Robertson-Walker (FLRW) metric that describes our universe, so I can't produce anything that matches observation. But given a FLRW metric that has a big bang singularity (a necessary requirement): [math](ds)^2 = c^2 (dt)^2 - a^2(t) ((dr)^2 + r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2))[/math] where [math]a(t)[/math] is the spatial expansion function of time (I've chosen a flat space FLRW for convenience), one can in principle perform the following coordinate transformation: [math]t = t(t',r')[/math] [math]r = \dfrac{r'}{a(t(t',r'))}[/math] [math]\theta = \theta'[/math] [math]\phi = \phi'[/math] where [math]t(t',r')[/math] is a function chosen so that the coefficient of the [math](dt')(dr')[/math] term is zero. The transformation [math]r = \dfrac{r'}{a(t(t',r'))}[/math] transforms a description based on an expanding space to a description based on an explosion (according to the definition I stated earlier). What it does is contract the big bang singularity to a single point. That is, all values of [math]r[/math] at [math]t = 0[/math] in the FLRW metric become [math]r' = 0[/math] at the corresponding value of [math]t'[/math] in the new metric. This is possible because the three-dimensional space at [math]t = 0[/math] is null and therefore metrically indistinguishable from a single point. And because the description based on an explosion is obtained from the FLRW metric by a coordinate transformation, the two descriptions are equivalent in accordance with the principle of general relativity. That is, every invariant property of the FLRW metric is an invariant property of the description based on an explosion. This includes all possible observations. Q.E.D. 3 hours ago, Mordred said: I know you likely don't have the skill set for Christoffel connections and Killing vectors so I won't ask you to produce them. That's rather presumptuous. Edited April 1 by KJW Link to comment Share on other sites More sharing options...

Mordred Posted April 1 Share Posted April 1 (edited) 50 minutes ago, KJW said: That's rather presumptuous. Not really determining a metrics Christoffel and subsequently its killing vectors is a rather tricky process that even many in the field can stumble over. Over the years I've seen even professionally peer reviewed articles overturned due to incorrectly determining either. ok you have the spherical coordinate metric for the FLRW great \[(ds)^2 = c^2 (dt)^2 - a^2(t) ((dr)^2 + r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2))\] OK the equation you have satisfies the Minkowskii metric in so far as Modern cossmology states for a homogeneous and isotropic expansion. You and I both agree on this. We both agree that SR/GR holds. Where we disagree is what happens if you have some preferred direction. A preferred direction can mean many things that direction could be expanding faster or slower than another direction. It could have some variation of flow, it could have some difference in the amount of force or its pressure. it is some feature that requires it to mathematically make it different than any other direction. Same applies for a preferred location. So what an easy example. Well as were also using SR lets simply look at a velocity boost of the metric itself in a given direction. Here is your starting metric [latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex] [latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] Here is how different boosts occur in different directions on that metric. In spherical coordinates as per this scenario. Lorentz group Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[(x^0,x^1,x^2,x^3)=(ct,r,\theta\\phi)\] \[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\] \[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\] generator along z axis \[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\] generator of boost along x axis:: \[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\] boost along y axis\ \[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\] generator of boost along z direction \[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\] the above is the generator of boosts below is the generator of rotations. \[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\] \[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\] \[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\] \[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\] there is the boosts and rotations we will need and they obey commutations \[[A,B]=AB-BA\] You can see you lose the orthogonal condition of the Minkowskii metric regardless of which direction the boost occurs in. Now if you recall an observer moving at relativistic velocity experiences two factors. length contraction and time dilation. Both factors change the metric. Those transforms above restore the changed metric back to the original. That is the very function of the Lorentz transforms. IT IS TO RESTORE orthogonality..... so lets set a preferred direction cause who cares....in the x direction. [latex]\eta=\begin{pmatrix}-c^2&1&0&0\\1&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] now if I had an inertial observer also moving in the x direction those points are already filled. So I must account for them differently Edited April 1 by Mordred 1 Link to comment Share on other sites More sharing options...

KJW Posted April 1 Share Posted April 1 10 minutes ago, Mordred said: Not really determining a metrics Christoffel and subsequently its killing vectors is a rather tricky process that even many in the field can stumble over. Over the years I've seen even professionally peer reviewed articles overturned due to incorrectly determining either. Determining Christoffel symbols can be tedious, but it's not above my skill set if the metric tensor is diagonal (inverting a non-diagonal metric tensor can be problematic, though even that is more tedious than above my skill set). I've never actually dealt with Killing vectors of particular metrics, apart from where it's obvious from the metric itself. But solving the Killing equation doesn't require the Christoffel symbols. However, I have derived the relation between acceleration and time dilation from the Killing equation. Link to comment Share on other sites More sharing options...

Mordred Posted April 1 Share Posted April 1 (edited) 19 minutes ago, KJW said: Determining Christoffel symbols can be tedious, but it's not above my skill set if the metric tensor is diagonal (inverting a non-diagonal metric tensor can be problematic, though even that is more tedious than above my skill set). I've never actually dealt with Killing vectors of particular metrics, apart from where it's obvious from the metric itself. But solving the Killing equation doesn't require the Christoffel symbols. However, I have derived the relation between acceleration and time dilation from the Killing equation. You need to be able to do it for any spacetime curvature term not just the orthogonal case. That's the easiest one. metrics of previous post are added you cross posted even though I stated was still in edit no problem just read my previous post. Please keep in mind I did not include some bulk mass flow in that above. That would require detailing the stress energy momentum tensor which in turned affects the metric. the Bulk flow would be your explosion of energy/mass spewing out of that singularity Edited April 1 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted April 1 Share Posted April 1 (edited) 23 minutes ago, Mordred said: You need to be able to do it for any spacetime curvature term not just the orthogonal case. That's the easiest one. Can Mathematica do it? Actually, I'm far more interested in the general theoretical aspects than dealing with particular metrics. But if I really needed to invert a non-diagonal metric, I could do it. And I do know how to express the inverse of the metric tensor in terms of the metric tensor of arbitrary dimensions in tensor notation. Edited April 1 by KJW Link to comment Share on other sites More sharing options...

Mordred Posted April 1 Share Posted April 1 (edited) 8 minutes ago, KJW said: Can Mathematica do it? Actually, I'm far more interested in the theoretical aspects than dealing with particular metrics. But if I really needed to invert a non-diagonal metric, I could do it. And I do know how to express the inverse of the metric tensor in terms of the metric tensor in tensor notation. Mathematica should be able to do it but if wolfram alpha runs on Mathematica experience has taught me you usually end up doing a large parcel of the work to force it. Wolpram alpha does have tensor packages to do a ot of the tensor operations so its useful in that regard. I don't see why Mathematica doesn't do the same. Inversing the matrix is very useful good skill to have. its one of the common requirements to test specific gauge conditions. I've never tested either though for Christoffels not sure how you would go about it except simple geometry. Edited April 1 by Mordred Link to comment Share on other sites More sharing options...

MigL Posted April 1 Share Posted April 1 I think you guys had better define what an 'explosion' actually is. On 3/30/2024 at 5:51 PM, KJW said: Do you see any error in the example I gave? In no explosion, that I know of, is the speed of the fragments dependent on the separation between them. That only applies to an expansion. 1 Link to comment Share on other sites More sharing options...

Mordred Posted April 1 Share Posted April 1 (edited) 32 minutes ago, MigL said: I think you guys had better define what an 'explosion' actually is. In no explosion, that I know of, is the speed of the fragments dependent on the separation between them. That only applies to an expansion. In any explosion I know of you will always have vectors involved. The vectors of the blast radius as well its fragments. However a simple and very common approach is simply attach to every spacetime coordinate a test particle. Its a very common treatment in GR. The vector field regardless of what's used originating from any central point radiates outward. T The above is a perfect example of a preferred location. ( the origin point) and a preferred direction (the net sum of vectors) a uniform expansion has net no flow of vectors. Nor does it have a center hence the cosmological principle. There is no inherent direction to how the universe expands. nor is there in stellar objects we measure there is no bulk direction flow. Edited April 1 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted April 1 Share Posted April 1 (edited) 32 minutes ago, MigL said: I think you guys had better define what an 'explosion' actually is. I defined it in this post as: An explosion is simply a set of straight trajectories in spacetime that originate from a single point in spacetime. 32 minutes ago, MigL said: In no explosion, that I know of, is the speed of the fragments dependent on the separation between them. That only applies to an expansion. It's relative speed, not just speed. This entire discussion is from the perspective of one of the fragments. Edited April 1 by KJW Link to comment Share on other sites More sharing options...

KJW Posted April 1 Share Posted April 1 15 hours ago, Mordred said: a uniform expansion has net no flow of vectors. Nor does it have a center hence the cosmological principle. There is no inherent direction to how the universe expands. nor is there in stellar objects we measure there is no bulk direction flow. I think it needs to be said that in the case of the explosion in Minkowskian spacetime, although it doesn't appear to be so in a spacetime diagram, all the spacetime trajectories are normal to the three-dimensional hyperboloidal space of constant proper time. In other words, all the particles are at rest in the three-dimensional space. This three-dimensional space really does appear to be expanding. One could even apply to the explosion in Minkowskian spacetime the inverse of the coordinate transformation I wrote above for the FLRW metric to obtain an FLRW metric for an expanding hyperboloidal space. Link to comment Share on other sites More sharing options...

Mordred Posted April 2 Share Posted April 2 (edited) Well unfortunately I still don't agree with that. You may recall I recommended using vector fields ? after all an explosion one common type of vector field that's easily represented. This particular field is a vector field of the first order. a vector field \(V(x,y,z)\) of the first order can be written in the following form of a constant matrix "A" and a constant vector B \[V^T=V(x,y,z))^T=A\cdot [x,y,z]^T+B^T\] the T indicates the transposition of the respective matrix. \[\begin{pmatrix}V_1(x,y,z)\\V_2(x,y,z)\\V_3(x,y,z)\end{pmatrix}=\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\end{pmatrix}+\begin{pmatrix}B_1\\B_2\\B_3\end{pmatrix}\] where B is the constant vector and A the constant matrix. so lets say you have a constant wind in the z direction \(V(x,y,,z),,,B=(0,0,B)\) a rotating vector field would look something like this \(V(x,y,z)=(-y,x,0)\) exploding vector field \(V(x,y,z)=(x,y,z)\) imploding \(V(x,y,z)=(-x,-y,-z)\) Now it doesn't matter the matrix is it could be simply a 3d Euclidean or we can add the 4th dimension to make this 4d or higher dimension. If say we were describing the matter fields or radiation fields we would use the stress energy momentum tensor. However on could have some vector quantity affect geometry as well. Anyways the Minkoskwii metric has no constant vector so there is no B term involved. So claiming it matches any vector space is simply put a mathematically incorrect statement. Recall Minkowskii space is a free fall space with no vectors hence the inner product of two vectors which returns a scalar usage in the Minkowskii tensors. \[\mu \cdot \nu)=(\nu \cdot\mu)\] this statement also tells you its commutative with no preferred direction or reference point. For your observers as per SR observers. and the premises of SR, which a vector field does not satisfy. (the mathematical proof is rather convoluted I'd rather not go into that lmao) but a simplified statement is vector fields fall under SU(N) groups and are not orthogonal where the Minkowskii metric is orthogonal. Under the Poincare group SO(3.1). Now lets skip up to EFE. lets describe our vector as a form of permutations and so we can separate spacetime in any manner of fields including multiple fields. This is regularly done in the EFE its literally one of the most common steps. So lets set our Newtonian limit (MInkowskii) under \(\eta_{\mu\nu}\), lets set our permutations under the premutation tensor \(H_{\mu\nu}\) \[G_{\mu\nu}=\eta_{\mu\nu}+H_{\mu\nu}\] so if you have a constant vector you alter the permutation tensor which gives a new geometry under\(G_{\mu\nu}\) with \( H_{\mu\nu}\) wait a sec that's pretty much the same as what we are doing in the vector spaces. Granted GR loves partial derivatives but an explosion is trivial to describe as divergence of a field using partial derivatives in essence you have some permutation field whether its vectors, scalars etc etc acting upon your baseline metric which forms a new metric statement. Edited April 2 by Mordred Link to comment Share on other sites More sharing options...

KJW Posted April 2 Share Posted April 2 I do not see the point of your previous post. It doesn't matter if you can provide mathematics that shows that an explosion is not homogeneous and isotropic. I have already mathematically proven that under the particular conditions which I described, an explosion that I have already defined will produce a three-dimensional space that is homogeneous and isotropic. Any mathematics that you provide that shows otherwise will not satisfy the conditions I have described and can be dismissed on that basis. But note that I am choosing observers who are equivalent to observers at rest in an FLRW metric. Link to comment Share on other sites More sharing options...

Mordred Posted April 2 Share Posted April 2 (edited) Well whatever you believe if you didn't understand that the Minkowskii tensor which defines the metric would be altered by any vector field. I really can't help you. I even showed you the EFE represention showing that. Edited April 2 by Mordred Link to comment Share on other sites More sharing options...

MigL Posted April 2 Share Posted April 2 This discussion has its roots in Airbrush's claim that spatial expansion could be an explosion. I don't think Airbrush ( or Swansont/Mordred/me ) define explosion as you have. And while the mathematics you guys are posting are very interesting, and remind me of stuff I was once somewhat familiar with, they only match your definition , not the accepted definition of an explosion. After all, if you define a man as a woman, then Bob's your ... aunt ?? Link to comment Share on other sites More sharing options...

geordief Posted April 2 Share Posted April 2 (edited) 1 hour ago, MigL said: This discussion has its roots in Airbrush's claim that spatial expansion could be an explosion. I don't think Airbrush ( or Swansont/Mordred/me ) define explosion as you have. And while the mathematics you guys are posting are very interesting, and remind me of stuff I was once somewhat familiar with, they only match your definition , not the accepted definition of an explosion. After all, if you define a man as a woman, then Bob's your ... aunt ?? If there are different possible definitions of explosions ,might there likewise be different definitions of expansion? Is @KJW basing his definition of an explosion on how things look to a test particle? (as usual I have vanishingly little understanding of this subject and so the question may not be pertinent) Edited April 2 by geordief Link to comment Share on other sites More sharing options...

Mordred Posted April 2 Share Posted April 2 (edited) Well for my definition it's quite simple a center of origin with vectors radiating outward at every angle. That's the mathematics I've shown using V(x,y,z)=(x,y,z) for 3d coordinates for simplicity. It doesn't matter what the vectors they are, they are present so cannot be arbitrarily ignored. Having some observer commoving with the vectors doesn't help eliminate the vectors for other observers. Edited April 2 by Mordred Link to comment Share on other sites More sharing options...

Mordred Posted April 2 Share Posted April 2 (edited) Just to give an applicable example of what sort of influences a vector field can have in the Einstein field equations. I'm going to provide a couple of links of a related theory that has a constant vector field. In this case its a rotating universe that results in torsion. Einstein-Cartan Theory https://en.wikipedia.org/wiki/Einstein–Cartan_theory notice how the stress tensor is affected in that link. It literally doesn't matter how fast or slow the universe is spinning or how minimal of a vector value one has. The mathematics of that model include the entire range of possible values. This is a good example of how the stress tensor gets affected which in turn affects any metric tensor. now one might think if its spinning too slow for any observer to notice then we can ignore it and believe Minkowskii space would work under Einstein-Cartan. This however isn't true. One of the very important aspects of observer is the world line via the geodesic equations. The full geodesic equation is \[\frac{d^2 x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^{\beta}}{ds}=0\] now the following equation will look somewhat different as the article its from has already factored out the terms it requires that and one can replace any symbol for a tensor with any other identifier and the tensor performs precisely the same way. What one uses to symbolize or name a given tensor is simply convenience. anyways in the article the new geodesic is given by equation 10. \[\frac{d^a x^a}{ds^2}+\Gamma^a_{bc}\frac{dx^b}{ds^2}\frac{dx^c}{ds^2}+2S_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds}=0\] https://api.repository.cam.ac.uk/server/api/core/bitstreams/4d357658-b056-45bf-8d29-919db6fac184/content Now to help those not math savvy. What this essentially shows is an antisymmetric universe that resulted from rotation that generates a torsion term. The effect is that the worldlines are in turn subsequently affected as well as a metrics affine connections, Bianchi identities, Christoffels and killing vectors. Its knowing these details that provides me sufficient reason to doubt any claim that an exploding universe can apply the Minkowskii tensor and get 100 percent of the observational evidence we currently have as a perfect match let alone a near match. Now a consequence of a different geodesic equation is that no Observer will get the same results as any observer in a Minkowskii spacetime (Newtonian limit spacetime) let alone in other spacetimes such as the Schwarzschild metric. Now do we look for anistrophic universes absolutely the research in that never stops, here is a 2016 research paper that places the error margin of 121,000 to 1 in disfavor of an anistropic expansion due to explosion/rotation etc. (the paper studies for any form of directional component. How isotropic is the Universe? https://arxiv.org/abs/1605.07178 I have seen papers with higher numbers in disfavor but as I couldn't put my fingers on them this one will suffice. 8 hours ago, geordief said: If there are different possible definitions of explosions ,might there likewise be different definitions of expansion? Is @KJW basing his definition of an explosion on how things look to a test particle? (as usual I have vanishingly little understanding of this subject and so the question may not be pertinent) The question does pertain so I provided my definition with a quick descriptive and the applicable mathematics. I will let @KJW speak for himself on how he thinks an explosion entails and how he is applying his observers. As far as your expansion question what we define as expansion isn't particularly at odds. The debate is what is causing the expansion. In cosmology expansion is literally the results of thermodynamics. Hence the equations of state for cosmology and their incorporation into the FLRW metric acceleration equation. @KJW is attempting an expansion due to an explosion rather than a thermodynamic expansion. Edited April 2 by Mordred Link to comment Share on other sites More sharing options...

Wigberto Marciaga Posted April 2 Share Posted April 2 Hello, blessing, I don't have much control over these topics. But I share a little of what I understand. They say the big bang was like an explosion. But it was not an explosion within a place, but rather the explosion would originate the place. It did not explode within a space, but the explosion would originate the space, they say. Therefore, the explosion would cover all of space. But, he also says, that before and during the explosion (bang), there were no physical laws, at least not like those known today. So applying the limits of the current universe to the Big Bang is not something that seems acceptable from what I understand. Of course, I will tell you honestly, that I read this a while ago on Wikipedia, if I remember correctly. Link to comment Share on other sites More sharing options...

Mordred Posted April 3 Share Posted April 3 58 minutes ago, Wigberto Marciaga said: Hello, blessing, I don't have much control over these topics. But I share a little of what I understand. They say the big bang was like an explosion. But it was not an explosion within a place, but rather the explosion would originate the place. It did not explode within a space, but the explosion would originate the space, they say. Therefore, the explosion would cover all of space. But, he also says, that before and during the explosion (bang), there were no physical laws, at least not like those known today. So applying the limits of the current universe to the Big Bang is not something that seems acceptable from what I understand. Of course, I will tell you honestly, that I read this a while ago on Wikipedia, if I remember correctly. A common descriptive used by pop media, simplified for layman level readers unfamiliar with the BB that is commonly used unfortunately is explosion of spacetime. However the term explosion typically implies a force vector. However a constant vector is not involved in accordance to a huge bulk of observational evidence. A more accurate description is a "rapid expansion of spacetime" due to reducing average /energy mass densities. This is where the ideal gas laws of thermodynamics steps in. this discussion doesn't make any statements of before the BB. The BB model doesn't describe how the universe began but how it evolved. \(10^{-43}\) after the BB Link to comment Share on other sites More sharing options...

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