The Beginning of the Universe

[swansont]

8 hours ago, Wigberto Marciaga said:

The contradiction: Relativity is very deterministic, while quantum postulates are extremely non-deterministic.

What does “extremely non-deterministic” mean? While outcomes in QM rely on probabilities, wave functions evolve deterministically, and there is still cause-and-effect.

The things in relativity that are deterministic aren’t in conflict with the non-deterministic aspects of QM.

[Mordred]

Funny how I get the same results in any cosmology based examination I have done in 35 years regardless of whether or not I use the QM/QFT methodology or that of Relativity. 

 These three methods work quite well in every cosmology study right up until you hit the singularity conditions.

For example I can use any of the above methods to calculate the number density of any particle species at a given blackbody temperature (though you do have to calculate each particle in sequence of when they would drop out of thermal equilibrium. (Used for metalicity data for BB nucleosynthesis)

I can determine the temperature history of our universe. Give you calculated rates of expansion that matches what we observe etc etc.

It doesn't matter what physics problem I need to solve. I can do so with equal success regardless.

The only time I can't use GR however SR still applies is the quantum regime. Simply due to the fact that gravity bring the weakest force doesn't have much influence on individual particles. However there still is influence

[Wigberto Marciaga]

4 minutes ago, swansont said:

What does “extremely non-deterministic” mean? While outcomes in QM rely on probabilities, wave functions evolve deterministically, and there is still cause-and-effect.

The things in relativity that are deterministic aren’t in conflict with the non-deterministic aspects of QM.

Do subatomic particles when observed, according to QM, behave deterministically or not?

How should they behave if the rules of relativity were applied to them?

Perhaps your answers to these questions will allow me to better understand the matter.

[Mordred]

21 minutes ago, Wigberto Marciaga said:

 

 

 

I want to raise the following. I'm not in favor of any version, I'm just trying to compare the readings I've done with your comments.

 

We use the best tool for each application mostly the one that best simplifies a problem. However we can still use any of the above methods

Edited April 4 by Mordred

[swansont]

Just now, Wigberto Marciaga said:

Do subatomic particles when observed, according to QM, behave deterministically or not?

It depends on the specifics.

Just now, Wigberto Marciaga said:

How should they behave if the rules of relativity were applied to them?

Exactly as they behave when relativity in incorporated in QM, which it has been.

https://en.m.wikipedia.org/wiki/Relativistic_quantum_mechanics

 

[Mordred]

8 minutes ago, Wigberto Marciaga said:

Do subatomic particles when observed, according to QM, behave deterministically or not?

How should they behave if the rules of relativity were applied to them?

Perhaps your answers to these questions will allow me to better understand the matter.

As to the first do you consider a variable which is a number set of possible answers deterministic. Depending on the answer is any formula that uses a variable deterministic. The probability functions of QM uses variables just as does relativity.

Subatomic particles once observed are determined for the predicted properties we look for. Pretty much every particle in physics was mathematically predicted with predicted properties that once observed matches the predictions. Just an FYI on the last part

[KJW]

On 4/2/2024 at 5:21 AM, KJW said:

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 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.

 

Edited April 4 by KJW

[Mordred]

Well your getting closer I will give you credit for that. Your missing some key details assuming \(V_[n)\) are vector spaces. (Hint if that's the case it's a  matrix). I don't know if you misses the details here.

 

On 4/1/2024 at 10:32 PM, Mordred said:

 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

 

VT=V(x,y,z))T=A⋅[x,y,z]T+BT

 

the T indicates the transposition of the respective matrix.

 

⎛⎝⎜V1(x,y,z)V2(x,y,z)V3(x,y,z)⎞⎠⎟=⎛⎝⎜A11A21A31A12A22A32A13A23A33⎞⎠⎟⋅⎛⎝⎜xyz⎞⎠⎟+⎛⎝⎜B1B2B3⎞⎠⎟

 

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.

 

μ⋅ν)=(ν⋅μ)

 

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 ημν , lets set our permutations under  the premutation tensor Hμν  

 

Gμν=ημν+Hμν

 

 

so if you have a constant vector you alter the permutation tensor which gives a new geometry underGμν with Hμν 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.

 

I assume your working on how to get a set of constant vectors as per the vector algebra example in that quote.

Key note incorporate your constant vectors into the permutation tensor ([h_{\mu\nu}\) for the non vanishing terms due to the vectors. 

Good example to look at is look at the examples under gravity waves.

However if the vector involves any terms directly relating to energy and momentum you also need the stress energy momentum tensor.

Edited April 4 by Mordred

[Mordred]

If V is a vector how do you square root a vector ? You can square root the scalar quantity but not the direction.

 

\[(v_x, v_y, v_z, \tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_x^2 - v_y^2 - v_z^2}}\]

Edited April 5 by Mordred

[KJW]

On 4/1/2024 at 11:10 AM, Mordred said:

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.

Neither the FLRW metric nor the corresponding explosion metric have a preferred direction. But I assume you are referring to a perturbation of the FLRW metric. Perhaps the perturbation can be decomposed into irreducible representations of the symmetry group of the FLRW metric. In any case, a coordinate transformation that transforms a homogeneous and isotropic FLRW metric to a homogeneous and isotropic explosion metric will also transform a perturbation of the FLRW metric to a perturbation of the explosion metric. Indeed, any vector or tensor field on the FLRW metric will transform covariantly to the corresponding vector or tensor field on the explosion metric. The point is that the explosion metric describes the exact same physical spacetime as does the expanding space.

One further point is worth noting: When I considered the coordinate transformation from the FLRW metric:

[math](ds)^2 = c^2 (dt)^2 - a^2(t) ((dr)^2 + r^2 ((d\theta)^2 + \sin^2(\theta) (d\phi)^2))[/math]

the coordinate transformation:

[math]r = \dfrac{r'}{a(t(t',r'))}[/math]

included the scale function [math]a(t)[/math]. But the coordinate transformation doesn't actually need to be custom created for the particular FLRW metric. One could use a much simpler generic coordinate transformation to contract the big bang singular space to a single point. This is because one is still applying a coordinate transformation along with the equivalence that implies.

 

 

[Mordred]

Sigh you don't have a homogeneous and isotropic explosion. Please define homogeneous and isotropic as your definition isn't one used in physics apparently

Yeesh let's do basic vector addition. You have an observer moving with the constant vector monitoring some other observer or event  and you try and claim that the second observer will get the same gamma factor  when his movement is parallel to the constant vector as am observer moving in a perpendicular vector.

Tell me have never looked at transverse doppler shifts ?

[KJW]

37 minutes ago, Mordred said:

If V is a vector how do you square root a vector ? You can square root the scalar quantity but not the direction.

It's not the square root of a vector. It's the square root of the t-component as a function of the new coordinates.

 

12 minutes ago, Mordred said:

Sigh you don't have a homogeneous and isotropic explosion. Please define homogeneous and isotropic as your definition isn't one used in physics apparently

The three-dimensional space of constant proper time that is created by the explosion is homogeneous and isotropic, as is the equivalent three-dimensional space described by the FLRW metric. I didn't say "homogeneous and isotropic explosion", I said "homogeneous and isotropic explosion metric", for the sake of brevity.

 

[Mordred]

Is not V denoting a vector yes or no. Did you even look at the vector field mathematics I posted for you.

How do you treat a vector of first order in 3d space. I already provided that detail.

Let me ask you another question does a homogeneous and isotropic field have divergence ?

Tell me something why is the Schwartzchild metric NOT considered as a Minkowsii metric ?

Answer that one as the Schwartzchild metric does have a preferred location and a direction. It has a center of mass with constant velocities it's certainly not homogeneous and isotropic by definition verbally or under math.

It is only locally homogenous and isotropic not the global metric. Aka curviliear coordinates 

Edited April 5 by Mordred

[KJW]

1 hour ago, Mordred said:

Did you even look at the vector field mathematics I posted for you.

I did. But you did not explain how it relates to what I posted. I demonstrated the equivalence between an FLRW metric and an explosion metric by demonstrating the existence of a coordinate transformation between them. I believe that is sufficient. Please explain to me why it is not.

 

1 hour ago, Mordred said:

Tell me something why is the Schwartzchild metric NOT considered as a Minkowsii metric ?

Because the Riemann curvature tensor field is non-zero. Actually, I can go further and say that the Schwarzschild metric cannot be an FLRW metric of an expanding flat space (I've never examined this question about expanding curved spaces). That's because an FLRW metric of an expanding flat space is conformally flat, i.e. the Weyl conformal tensor field is zero, in contrast to the Schwarzschild metric for which the Weyl conformal tensor field is the only non-zero curvature tensor field.

 

Edited April 5 by KJW

[Mordred]

Your claim of the Schwartzchild metric for example of homogeneous and isotropic is incorrect. Spherical coordinates are only locally homogenous and isotropic. That's part of the issue. Spherically symmetry is not the same thing as homogeneous and isotropic

That's without having a constant vector.  

Now in your mathematics post your constant velocity vector without anything else from an explosion.

Then apply your vector dot products.

Let's use an example of parallel transport under the weak equivalence principle. Doesn't matter if the bodies are freefalling towards the CoM or away from the center of mass the two objects will approach one another as they near the center of mass. 

They will move apart as the radius from CoM increases. Now no forces are involved well the pseudo tidal force is aka gravity in the first case. The moving away portion is simply for symmetry .

So ask yourself this as imploding is symmetric with explosion under change in sign. Why wouldn't the Schwartzchild metric be usable for an exploding universe from some central point ?

Now does any of this describe the orthonormality of the Minkoskii metric. NO IT DOES NOT. Can we perform transformations to restore orthonormality ? Yes we can the process is the same for renormalization in any gauge theory including the renormalized Hamilton. 

One of the most useful visuals to understand  vector field divergence is to take a slow moving propeller in water. If the spinors remain the same radius  from 0,0 that field is not divergent. If they move away or approach it is divergent or convergent.

Now in the Minkowskii metric when you perform a transformation to restore the metric that means the metric altered (time dilation/,length contraction). 

However the Spherical coordinates is not the Minkowskii metric to begin with its coordinate basis is not orthonormal.

However any theory must not depend on coordinate choice. If any quantity depends on coordinate choice then you have an artifact of the metric. The event horizon is one example. The Minkowsii metric has numerous limitations in this regard .

Those limitations directly apply to why the Schwartzchild metric is used in even the weak field Newton limit for spacetimes such as those produced by planets and stars.

Had you syated you could use the Schwartzcild metric to describe an exploding universe I wouldn't have an issue with that by itself. I simply would tell you that wouldn't be our universe due to the mass distribution not supporting an explosion nor would the temperature history however mathematically dealing just with the metric I wouldn't have an issue. At least until you start getting into the Christoffels etc (the christoffels are different from the FLRW Christoffels as well as for the Minkowskii and Schwartzchild. As the Rheimann curvature tensors is one method of calculating Christoffels (its actually easier using Integrals via the Langrangian).

That should make it obvious that there are rather important distinctions between these spacetimes. (A large part of those distinctions is how pressure and energy density is handled)

Edited April 5 by Mordred

[KJW]

18 minutes ago, Mordred said:

Your claim of the Schwartzchild metric for example of homogeneous and isotropic is incorrect.

Where did I claim that?

 

[Mordred]

By claiming to explicitly producing either the FLRW metric and the Minkowsii metric. Both of which are homogeneous and isotropic.

The term explosion in vector algebra in and of itself is not homogeneous and isotropic. So how do you possibly combine the two to reproduce a homogeneous and isotropic field ? The only way to do so is to factor out the influences that causes the divergences . In essence eliminate the vector field to restore the original metric.

Take an example if you perform any math operation on a matrix or tensor you produce a new tensor or matrix..

THE Minkowsii metric is a tensor so performing any math operation on it produces a new tensor.

Edited April 5 by Mordred

[KJW]

4 minutes ago, Mordred said:

By claiming to explicitly producing either the FLRW metric and the Minkowsii metric. Both of which are homogeneous and isotropic.

That inference is incorrect because I explicitly said that the Schwarzschild metric cannot be either the Minkowskian metric or the FLRW metric of an expanding flat space.

 

[Mordred]

3 minutes ago, KJW said:

That inference is incorrect because I explicitly said that the Schwarzschild metric cannot be either the Minkowskian metric or the FLRW metric of an expanding flat space.

 

That I agree with, neither can the Minkowsii either...

 

[KJW]

2 minutes ago, Mordred said:

neither can the Minkowsii either...

The Minkowskian metric can be coordinate-transformed to an FLRW metric of a constantly expanding hyperboloidal space. That I have proven above. I do not claim that the Minkowskian metric can be coordinate-transformed to an FLRW metric of an expanding flat space. But I do claim that an FLRW metric of an expanding flat space can be coordinate-transformed to a spacetime in which the big bang singularity is a single point, but this spacetime is not flat.

 

[Mordred]

Here's the thing how the FLRW determines it's curvature terms differs from GR. It still uses GR and is fully compatible with GR however it's curvature is based on the critical density formula.

It looks specifically at pressure and energy density relations. The Schwartzchild solution is static to depend on radius only. 

IN THE critical density formula it a matter solution to determine when the universe will expand or collapse. It is the actual density compared to that matter only solution for a value that determines if the universe is flat actual density perfectly matches the critical density.

 But pressure really means the energy density term coupled with the equation of state.  The biggest difference is whether the universe is curved or not has nothing to do with localized spacetime that lends itself to gravity effects.

It's mass and energy distribution is always uniform. That's a huge difference from spacetimes described around massive bodies.

29 minutes ago, KJW said:

The Minkowskian metric can be coordinate-transformed to an FLRW metric of a constantly expanding hyperboloidal space. That I have proven above. I do not claim that the Minkowskian metric can be coordinate-transformed to an FLRW metric of an expanding flat space. But I do claim that an FLRW metric of an expanding flat space can be coordinate-transformed to a spacetime in which the big bang singularity is a single point, but this spacetime is not flat.

 

With what described above no you haven't as stated the two will always different from one another in mass terms. 

 The weak equivalence principal though it applies isn't particularly involved the mass that would lend itself to gravity is uniformly distributed.

So under Newtons Shell theorem gravity is zero in a uniform mass/energy distribution. That is certainly not a  Minkowsii treatment.

The curvature term of the Global metric also doesn't have the same time dilation effects that the Minkowsii metric does on a global scale. It's mass distribution being uniform (that includes observational evidence too huge to list them all). Doesn't affect time not like a blackhole does. That's a localized anistrophy which has a tidal force (gravity). Best to think gravity as the result of the stress energy tensor for gravity effects rather than it's curvature.

The FLRW metric using whe Newton limit the only entry is the T^(00) entry. The mass/energy is uniform so only that entry really applies 

Try that under Minkowsii and be able to produce curvature to get gravity. All observational evidence supports the uniform mass distribution. 

Edited April 5 by Mordred