KJW

I can do better than that. Using the description of a continuum of trajectories given earlier, I can provide an explicit FLRW metric corresponding to an explosion in Minkowskian spacetime. The description of a continuum of trajectories: [math]x(v_x, v_y, v_z, \tau) = v_x t(v_x, v_y, v_z, \tau)[/math] [math]y(v_x, v_y, v_z, \tau) = v_y t(v_x, v_y, v_z, \tau)[/math] [math]z(v_x, v_y, v_z, \tau) = v_z t(v_x, v_y, v_z, \tau)[/math] [math]t(v_x, v_y, v_z, \tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_x^2 - v_y^2 - v_z^2}}[/math] Note that the description of a continuum of trajectories is actually a coordinate transformation in disguise. Let's change to spherical coordinates: [math](ds)^2 = c^2 (dt)^2 - (dr)^2 - r^2 ((d\theta)^2 + \sin^2(\theta) (d\phi)^2)[/math] Then the description of a continuum of trajectories becomes (noting that [math]\beta = \dfrac{1}{c} \sqrt{v_x^2 + v_y^2 + v_z^2}[/math] ) : [math]t(\tau, \beta, \theta, \phi) = \tau\ (1 - \beta^2)^{-\frac{1}{2}}[/math] [math]r(\tau, \beta, \theta, \phi) = \beta\ c\ t(\tau, \beta, \theta, \phi) = \beta\ c\ \tau\ (1 - \beta^2)^{-\frac{1}{2}}[/math] [math]\theta(\tau, \beta, \theta, \phi) = \theta[/math] [math]\phi(\tau, \beta, \theta, \phi) = \phi[/math] [math]dt(\tau, \beta, \theta, \phi) = (1 - \beta^2)^{-\frac{1}{2}}\ d\tau + \tau\ \beta\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta[/math] [math]dr(\tau, \beta, \theta, \phi) = \beta\ c\ dt(\tau, \beta, \theta, \phi) + c\ t(\tau, \beta, \theta, \phi)\ d\beta[/math] [math]= \beta\ (1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ \beta^2\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta + c\ \tau\ (1 - \beta^2)^{-\frac{1}{2}}\ d\beta[/math] [math]= \beta\ (1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta[/math] [math]d\theta(\tau, \beta, \theta, \phi) = d\theta[/math] [math]d\phi(\tau, \beta, \theta, \phi) = d\phi[/math] [math](c\ dt(\tau, \beta, \theta, \phi))^2 - (dr(\tau, \beta, \theta, \phi))^2[/math] [math]= (c\ dt(\tau, \beta, \theta, \phi) + dr(\tau, \beta, \theta, \phi))\ (c\ dt(\tau, \beta, \theta, \phi) - dr(\tau, \beta, \theta, \phi))[/math] [math]= (1 + \beta)((1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta)\ (1 -\beta)((1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau - c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta)[/math] [math]= (1 - \beta^2)((1 - \beta^2)^{-1}\ c^2 (d\tau)^2 - c^2 \tau^2\ (1 - \beta^2)^{-3}\ (d\beta)^2)[/math] [math]= c^2 (d\tau)^2 - c^2 \tau^2\ (1 - \beta^2)^{-2}\ (d\beta)^2[/math] Substituting this expression and the expression for [math]r[/math] into the spherical coordinates Minkowskian metric gives the following intermediate metric: [math](ds)^2 = c^2 (d\tau)^2 - c^2 \tau^2\ ((1 - \beta^2)^{-2}\ (d\beta)^2 - \beta^2\ (1 - \beta^2)^{-1}\ ((d\theta)^2 + \sin^2(\theta) (d\phi)^2))[/math] A coordinate transformation to a more appropriate radial coordinate (note that [math]\beta[/math] is dimensionless, and that [math]t_{\text{ref}}[/math] is a constant reference time corresponding to the time at which the expansion scale is unity): [math]\dfrac{dr'}{c\ t_{\text{ref}}} = (1 - \beta^2)^{-1}\ d\beta[/math] [math]c\ t_{\text{ref}}\ \dfrac{d\beta}{dr'} = 1 - \beta^2 \ \ \ \ \ ;\ \ \ \ \ \beta = 0 \ \ \text{at} \ \ r' = 0[/math] [math]\beta = \tanh\left(\dfrac{r'}{c\ t_{\text{ref}}}\right) \ \ \ \ \ ;\ \ \ \ \ \dfrac{\beta^2}{1 - \beta^2} = \sinh^2\left(\dfrac{r'}{c\ t_{\text{ref}}}\right)[/math] [math]\tau = t' \ \ \ \ \ ;\ \ \ \ \ \theta = \theta' \ \ \ \ \ ;\ \ \ \ \ \phi = \phi'[/math] [math](ds)^2 = c^2 (dt')^2 - \dfrac{t'^2}{t_{\text{ref}}^2} \left((dr')^2 + c^2\ t_{\text{ref}}^2\ \sinh^2\left(\dfrac{r'}{c\ t_{\text{ref}}}\right)\ \Big((d\theta')^2 + \sin^2(\theta') (d\phi')^2\Big)\right)[/math] Thus, a description of an explosion from a single point in Minkowskian spacetime has been coordinate-transformed to an FLRW metric of a constantly expanding hyperboloidal three-dimensional space. Q.E.D.

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.

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.

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. It's relative speed, not just speed. This entire discussion is from the perspective of one of the fragments.

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.

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.

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. That's rather presumptuous.

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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.

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.

It would appear that where we disagree is what an explosion is, and that we do agree that the three-dimensional hyperboloid that I described is indeed homogeneous and isotropic. Therefore, it suffices for me to prove that what I described is an explosion. If I succeed in doing that then this: is irrelevant because you have already indicated that the three-dimensional hyperboloid is homogeneous and isotropic. To me, an explosion is simply a set of straight trajectories in spacetime that originate from a single point in spacetime. The trajectory of the [math]i[/math]-th particle, originating from the origin, and parametised by proper time [math]\tau[/math] is: [math]x_i(\tau) = v_{xi} t_i(\tau)[/math] [math]y_i(\tau) = v_{yi} t_i(\tau)[/math] [math]z_i(\tau) = v_{zi} t_i(\tau)[/math] [math]t_i(\tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_{xi}^2 - v_{yi}^2 - v_{zi}^2}}[/math] where [math]v_{xi}, v_{yi}, v_{zi}[/math] are constant for the [math]i[/math]-th particle. But instead of considering discrete trajectories, one can consider a continuum of trajectories, parametised by [math](v_x, v_y, v_z)[/math], [math]v_x^2 + v_y^2 + v_z^2 \lt c^2[/math]: [math]x(v_x, v_y, v_z, \tau) = v_x t(v_x, v_y, v_z, \tau)[/math] [math]y(v_x, v_y, v_z, \tau) = v_y t(v_x, v_y, v_z, \tau)[/math] [math]z(v_x, v_y, v_z, \tau) = v_z t(v_x, v_y, v_z, \tau)[/math] [math]t(v_x, v_y, v_z, \tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_x^2 - v_y^2 - v_z^2}}[/math] At a fixed value of [math]\tau[/math], [math](x(v_x, v_y, v_z, \tau), y(v_x, v_y, v_z, \tau), z(v_x, v_y, v_z, \tau), t(v_x, v_y, v_z, \tau))[/math] define the three-dimensional hyperboloid: [math]c^2 t^2 - x^2 - y^2 - z^2 = c^2 \tau^2[/math] Or alternatively, a single parameter [math]\tau[/math] family of three-dimensional hyperboloids are defined. Note that [math](v_x, v_y, v_z)[/math] covers all values satisfying [math]v_x^2 + v_y^2 + v_z^2 \lt c^2[/math]. This means that a complete three-dimensional space is defined for all values of [math]\tau \gt 0[/math]. Also, only geometry is being considered, there is no consideration of matter distribution with respect to [math](v_x, v_y, v_z)[/math]. Finally, because the explosion being considered is cosmological, notions of realism that apply to real-world explosions do not apply.

I think I need to point out that the mathematics I wrote says that in terms of geometry an explosion is not anisotropic and inhomogeneous from the perspective of any one of the particles from the explosion. Why hasn't the mathematics been considered during this discussion?

I'm familiar with Dirac notation, though I'm not especially comfortable with it. And I am familiar with the triangle inequality. But I'm not seeing the relevance to this discussion. Do you see any error in the example I gave?

I figure that because you are trying to correct my error that it would be better to use my example.

Consider on the x,y-graph the following three points: (1,0), (0,2), and (3,4) at some time. At double the time, (1,0) will become (2,0) because its velocity is 1 along the x-axis; (0,2) will become (0,4) because its velocity is 2 along the y-axis; and (3,4) will become (6,8) because its velocity is 5 along the (3,4)-direction. The triangle is double the size and all the angles are the same.

I'm not sure I'm visualising your description correctly, but in my assessment of the explosion, triangles formed from any three particles will expand with unchanged angles over time. The above mathematics of the non-relativistic case makes this clear. And I'm quite confident, without explicitly checking, that the same is true for the relativistic case also, in spite of the curvature of the three-dimensional space.

But if one considers any three points forming a triangle, then if the length of each side changes by the same relative amount, the result is a similar triangle with unchanged angles.

In the non-relativistic case, it is clear that angles do not change. I proved that in my first post in this thread. In the relativistic case, the curvature of the three-dimensional space led me to consider the invariance to Lorentz transformations as proof of homogeneous and isotropic expansion. I should point out that I'm not suggesting that an expanding hyperboloidal space is indistinguishable from an expanding flat space. But to distinguish between them is more advanced than the simple geometric arguments often presented against the explosion model of the big bang. I'm not advocating that the big bang was an explosion, btw. But I will suggest that arguments against it are not necessarily valid.

I'm not sure what you are referring to here. What I showed was the coordinates of each particle after a given proper time. Over a continuum of particles, this defines a three-dimensional space. I could've supplied a metric for this space, but did not. Exactly. It was specifically my intention to prove that the three-dimensional space defined by the particles from the explosion is homogeneous and isotropic. What angles are you referring to? The spacetime trajectories of all the particles are orthogonal to the three-dimensional space.

[math]\underline{\text{Non-relativistic:}}[/math] [math]\text{For an explosion at time } t=0 \text{ at the origin of a rectangular coordinate system:}[/math] [math]\text{Let } \textbf{x}_{0i} \text{ be the position vector of the } i \text{-th particle at time } t=T[/math] [math]\text{Let } \textbf{v}_{0i} \text{ be the velocity vector of the } i \text{-th particle at time } t=T[/math] [math]\text{Then: } \textbf{v}_{0i} = \dfrac{1}{T} \textbf{x}_{0i}[/math] [math]\text{Similarly for the } j \text{-th particle at time } t=T \text{ : } \textbf{v}_{0j} = \dfrac{1}{T} \textbf{x}_{0j}[/math] [math]\text{Then: } \textbf{v}_{0j} - \textbf{v}_{0i} = \dfrac{1}{T} (\textbf{x}_{0j} - \textbf{x}_{0i})[/math] [math]\text{Let: } \textbf{x}_{ij} = \textbf{x}_{0j} - \textbf{x}_{0i} \ \ \text{and} \ \ \textbf{v}_{ij} = \textbf{v}_{0j} - \textbf{v}_{0i}[/math] [math]\text{Then: } \textbf{v}_{ij} = \dfrac{1}{T} \textbf{x}_{ij}[/math] [math]\text{From the perspective of the } i \text{-th particle, dropping the } i \text{ :}[/math] [math]\textbf{v}_{j} = \dfrac{1}{T} \textbf{x}_{j}[/math] Thus, all particles are on equal footing with regards to the observation of the other particles of the explosion. [math]\underline{\text{Relativistic:}}[/math] [math]\text{For an explosion at the origin of a Minkowskian coordinate system:}[/math] [math]\text{Let } (x_i, y_i, z_i, t_i) \text{ be the coordinates of the } i \text{-th particle at proper time } \tau[/math] [math]\text{Then: }c^2 t_i^2 - x_i^2 - y_i^2 - z_i^2 = c^2 \tau^2[/math] [math]\text{The set of all particles at proper time } \tau \text{, taken as a continuum, defines a three-dimensional hyperboloid:}[/math] [math]c^2 t^2 - x^2 - y^2 - z^2 = c^2 \tau^2[/math] [math]\text{This is invariant to Lorentz transformations:}[/math] [math]ct = \dfrac{c t' + \beta x'}{\sqrt{1 - \beta^2}}[/math] [math]x = \dfrac{x' + \beta c t'}{\sqrt{1 - \beta^2}}[/math] [math]y = y'[/math] [math]z = z'[/math] [math]\dfrac{(c t' + \beta x')^2}{1 - \beta^2} - \dfrac{(x' + \beta c t')^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math] [math]\dfrac{c^2 t'^2 + 2 c t' \beta x' + \beta^2 x'^2}{1 - \beta^2} - \dfrac{x'^2 + 2 x' \beta c t' + \beta^2 c^2 t'^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math] [math]\dfrac{(1 - \beta^2) c^2 t'^2 - (1 - \beta^2) x'^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math] [math]c^2 t'^2 - x'^2 - y'^2 - z'^2 = c^2 \tau^2[/math] Although Lorentz transformations transform points of the three-dimensional hyperboloid to different points of the three-dimensional hyperboloid, the three-dimensional hyperboloid itself is invariant. Thus, all particles are on equal footing with regards to the observation of the other particles of the explosion. Note that Lorentz transformations take the role of relative velocities. By moving a selected particle to the spatial origin by the application of a Lorentz transformation, the velocities of all the other particles are now relative to the selected particle. Similar to an expanding space, and contrary to what one might expect for an explosion, the space defined by constant age of the particles has no centre. Thus, each particle is able to regard itself as the centre of the space. In particular, the explosion itself has no identifiable location in the space. Because all the particles originated from the explosion, the explosion actually occurred everywhere in the space. One thing that is notable from the above is that for a flat spacetime, the three-dimensional space of constant proper time from a single point is curved like a hyperboloid. For a flat three-dimensional space of constant proper time from a single point, the spacetime will be curved. And although I never dealt with a spherical three-dimensional space of constant proper time from a single point, I think one can extrapolate that the spacetime curvature will be larger than for the flat three-dimensional space.

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[math](ds)^2 = (c^2 - \dfrac{r^2}{t^2}) (dt)^2 + \color{red} {\dfrac{2r}{t} (dt)(dr)} - (dr)^2 - r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2)[/math] For me to complete this, I need to apply a coordinate transformation to remove the part in red. 😉

T is arbitrary. Therefore, the proportionality works for all constant values of T (different proportionality for different T, but proportionality nevertheless). Not necessarily. My concern was about a snapshot in time, not time evolution. My original statement was not incorrect. I was only concerned with the relationship between relative velocity and relative displacement for pairs of particles at a single time. It is you who was fixated on time evolution. When you pointed out that proportionality didn't apply over different times, I clarified what I meant. The entire point of what I wrote was to show that at any given instant in time, the relative velocity between any pair of particles is directly proportional to the relative displacement between that pair of particles, the same as for an expanding space. Your focus on time evolution was a distraction away from my consideration of the pairs of particles at a given time.

What is your point? Are you concerned with the structure of spacetime or how we measure it?