Answering Mermin’s Challenge with the Relativity Principle

[PeterDonis]

Or because our intuitions are quite fine but theories with causal loops make no sense.

If "make no sense" just means you prefer a physical theory that says they're not physically valid, that's fine. But they make perfect sense from the standpoint of logical and mathematical consistency.

 

[Elias1960]

Only if you wish to ignore the wealth of astronomical data we have.

Really? Without the need to extend our theory beyond 5000 years into the past? Of course, this would lead to a lot of things remaining unexplained - but only in dynamical thinking.

 

[RUTA]

Only if the model violates at least one of the premises of the singularity theorems.

They're looking for past extendability and found it. Why were they looking for that? Because they were thinking dynamically. Here is an analogy.

Set up the differential equations in y(t) and x(t) at the surface of Earth (a = -g, etc.). Then ask for the trajectory of a thrown baseball. You're happy not to past extend the solution beyond the throw or future extend into the ground because you have a causal reason not to do so. But, the solution is nonetheless a solution without those extensions. Same for EEs with no past extension beyond a(0) and a choice of a(0) not equal to zero. Why are you not satisfied with that being the solution describing our universe? There's nothing in the data that would ever force us to choose a(0) = 0 singular. The problem is that the initial condition isn't explained as expected in a dynamical explanation. All we need in 4D is self-consistency, i.e., we only have to set a(0) small enough to account for the data. Maybe someday we'll have gravitational waves from beyond the CMB and we'll be able to push a(0) back to an initial lattice spacing approaching the Planck length. But, we'll never have to go to a singularity.

 

[RUTA]

Really? Without the need to extend our theory beyond 5000 years into the past? Of course, this would lead to a lot of things remaining unexplained - but only in dynamical thinking.

Correct, you're trying to find a 4D model to account for all the data and our models go well beyond 5000 years into the past to do that. But, not all the way to a singularity.

 

[PeterDonis]

They're looking for past extendability and found it.

I'm sorry, but whatever they were "looking for" is irrelevant to what the theorems actually say mathematically. Any claim that you can "choose" to just make the model not have a singularity can only be true if your model violates at least one of the premises of the singularity theorems. That is true regardless of what the intentions of the people who proved the theorems were.

Why are you not satisfied with that being the solution describing our universe?

I have said nothing whatever about what I personally would or would not be "satisfied" with. I am simply pointing out a mathematical fact that it seems to me that any claim about solutions must take into account. Are you disputing this mathematical fact? If not, then you must acknowledge that any solution that has the property you appear to prefer (not having an initial singularity) must violate at least one of the premises of the singularity theorems.

The problem is that the initial condition isn't explained as expected in a dynamical explanation.

If you have a spacetime that meets the conditions of the singularity theorems, and which therefore has an initial singularity, it seems to me that the singularity theorems themselves would provide an adequate nondynamical explanation of the initial singularity, since, as I've said, those theorems are not dynamical, they're geometrical.

 

[RUTA]

If you have a spacetime that meets the conditions of the singularity theorems, and which therefore has an initial singularity, it seems to me that the singularity theorems themselves would provide an adequate nondynamical explanation of the initial singularity, since, as I've said, those theorems are not dynamical, they're geometrical.

So, you tell me, why do I have to past extend beyond $a(0) \neq 0$ to a singularity in a simple Einstein-deSitter model for example?

 

[PeterDonis]

So, you tell me, why do I have to past extend beyond $a(0) \neq 0$ to a singularity in a simple Einstein-deSitter model for example?

I have made no such claim, so I don't see why I should have to justify it.

You don't appear to even be reading what I'm actually saying. I'm just asking which of these two options you are choosing:

(1) You are disputing that the singularity theorems are mathematically correct: you think there can be a model that satisfies all of the premises of the singularity theorems but does not have an initial singularity; or

(2) You are accepting that the models you are interested in, which do not have initial singularities (and I am not in any way disputing that you can have legitimate reasons for being interested in such models), violate at least one of the premises of the singularity theorems (the obvious ones to violate would be the energy conditions, since we already know inflationary models and models with a positive cosmological constant violate them anyway).

I don't see a third option.

 

[PeterDonis]

why do I have to past extend beyond $a(0) \neq 0$ to a singularity in a simple Einstein-deSitter model for example?

Since the Einstein-de Sitter model satisfies the conditions of the singularity theorems (assuming you mean the model described on this Wikipedia page), it must have an initial singularity. Whether you call the point of that initial singularity $a(0) = 0$ or redefine your coordinates so the singularity occurs at some coordinate time before $t = 0$ doesn't make a difference to the global geometry of the solution.

 

[RUTA]

I have made no such claim, so I don't see why I should have to justify it.

You don't appear to even be reading what I'm actually saying. I'm just asking which of these two options you are choosing:

(1) You are disputing that the singularity theorems are mathematically correct: you think there can be a model that satisfies all of the premises of the singularity theorems but does not have an initial singularity; or

(2) You are accepting that the models you are interested in, which do not have initial singularities (and I am not in any way disputing that you can have legitimate reasons for being interested in such models), violate at least one of the premises of the singularity theorems (the obvious ones to violate would be the energy conditions, since we already know inflationary models and models with a positive cosmological constant violate them anyway).

I don't see a third option.

You have to look at what they proved. If you have a copy of Wald, read chapter 9 section 1, "What is a Singularity?" I'm not going to type that entire section here, but you'll find that the notion of a singularity is difficult to define, so the singularity theorems proved "the existence of an incomplete timelike or null geodesic." That such a place is exists is deemed "pathological" because "it is possible for at least one freely falling particle or photon to end its existence within a finite "time" (i.e., affine parameter) or to have begun its existence a finite time ago." So, my choice of $a(0) \neq 0$ would satisfy their definition of a singularity. It's not infinite density or infinite curvature, it's an entirely well-behaved "singularity," so I am not calling it a singularity. Here is the last sentence in that section, "Unfortunately, the singularity theorems give virtually no information about the nature of the singularities of which they prove existence."

 

[DarMM]

Regarding adynamical explanations: I think it would be fine if we said that there is a 4D manifold of events $\mathcal{M}$ with some set of data $\Sigma\left(\mathcal{M}\right)$ at each event but that $\Sigma\left(\mathcal{M}\right)$ must obey some constraint $\delta \Sigma\left(\mathcal{M}\right) = 0$ with $\delta$ intended in some highly symbolic way as I'm not proposing such a constraint.

If we then went out and checked that the physical content of events matched that predicted by the constraint, that would be clearly scientific and sensible. I don't think it is required that the data on each leaf of a foliation of $\mathcal{M}$ must be related to that on any other leaf via some Green's function or other expression of dynamical propagation in order for the theory to be scientific or count as an explanation.

Classical theories were always such that data on later leaves followed from that on earlier leaves via integration against some kernel (or similar), but I don't think this must be true.

 

[PeterDonis]

You have to look at what they proved.

Yes, I know that, strictly speaking, "singularity" means "geodesic incompleteness".

my choice of $a(0) \neq 0$ would satisfy their definition of a singularity.

Why? Since you appear to be saying the spacetime in the model you describe here is extendible past $t = 0$ indefinitely (i.e., to arbitrary negative values of $t$), then geodesics in the model would be similarly extendible. So the spacetime in your model would not be geodesically incomplete, hence would not contain a singularity.

If you think your model would still be geodesically incomplete while still being extendible indefinitely past $t = 0$, then you will have to give quite a bit more detail about your model, because I don't understand how it would be, given what you have said so far and what kind of model you appear to be interested in.

 

[PeterDonis]

you will have to give quite a bit more detail about your model

Or a pointer to a paper or other reference that describes the sort of model you are referring to would be fine.

 

[RUTA]

Why? Since you appear to be saying the spacetime in the model you describe here is extendible past $t = 0$ indefinitely (i.e., to arbitrary negative values of $t$), then geodesics in the model would be similarly extendible. So the spacetime in your model would not be geodesically incomplete, hence would not contain a singularity.

If you think your model would still be geodesically incomplete while still being extendible indefinitely past $t = 0$, then you will have to give quite a bit more detail about your model, because I don't understand how it would be, given what you have said so far and what kind of model you appear to be interested in.

If you choose to have the "beginning" be at $a(0) \neq 0$ in the EdS model, for example, then you have a "singularity" per their definition. It's "pathological" because "it is possible for at least one freely falling particle or photon to ... have begun its existence a finite time ago." That's only pathological from the dynamical perspective, as I explained using the thrown ball example.

 

[PeterDonis]

If you choose to have the "beginning" be at $a(0) \neq 0$ in the EdS model

You can't; the Einstein-de Sitter model is a known exact solution of the EFE and has $a(0) = 0$.

I suppose you could quibble about this by changing coordinates so that the value of $t$ where $a = 0$ is not $t = 0$; but it will have $a = 0$ at some value of $t$. That is a known geometric property of the model.

You seem to be confusing the (true) statement that the singularity theorems by themselves don't tell you very much about the actual properties of the singularities, with the (false) statement that you can just handwave any kind of singularity you want into a specific model. We know a lot more about the EdS model than just what the singularity theorems tell us.

If you want to construct some other model that looks like the EdS model for some range of $t$ (such as, for example, redshifts smaller than $z = 1000$ or so), but then differs at values of $t$ before that, that's fine. If you can show that such a model, while satisfying the premises of the singularity theorems and therefore being geodesically incomplete, nevertheless has $a(0) \neq 0$, that's fine too. But you can't just wave your hands and say "EdS model" to do that; you have to actually construct the other model and show that it has the properties you claim it has.

 

[RUTA]

You can't; the Einstein-de Sitter model is a known exact solution of the EFE and has $a(0) = 0$.

It's a second-order differential equation, of course you can freely choose $a(t)$ at two different times to find a particular solution. There is no quibble, it's a mathematical fact.

 

[PeterDonis]

It's a second-order differential equation

The Einstein Field Equation is a second-order differential equation, for which you can freely choose as you say.

The Einstein-de Sitter model is not; it is a particular solution to that equation, in which there is no more freedom to choose $a(t)$; it's exactly specified for the entire solution.

There is no quibble, it's a mathematical fact.

A mathematical fact about the wrong thing. See above.

 

[RUTA]

The Einstein Field Equation is a second-order differential equation, for which you can freely choose as you say.

The Einstein-de Sitter model is not; it is a particular solution to that equation, in which there is no more freedom to choose $a(t)$; it's exactly specified for the entire solution.

You're arguing semantics now. Call it something else, then. The point is, we have a solution of EE's for the spatially flat, matter-dominated cosmology model without infinities. If this solution bothers you, you need to ask yourself, "Why does this solution bother me?" Wald was clear about why it would bother him, but that is purely dynamical bias.

 

[RUTA]

Again, you're free to have a dynamical bias and ignore adynamical explanation. But, that sentiment does not in any way refute the point I'm making, i.e., adynamical constraint-based thinking avoids the problem caused by dynamical thinking in this case.

 

[PeterDonis]

You're arguing semantics now.

I'm using the same terms you used. I'm just correcting your erroneous usage of them.

The point is, we have a solution of EE's for the spatially flat, matter-dominated cosmology model without infinities.

Then you need to show me one, because the EdS model is not one.

you're free to have a dynamical bias

I have said nothing at all about my personal preferences. If you make statements that are wrong as a pure matter of math, you should expect to have them corrected. Correcting them is not "bias", it's just correcting erroneous statements. Your claim that the EdS model is "without infinities" is wrong as a pure matter of math: the EdS is a specific solution of a specific equation with specific properties, and those properties include $a = 0$ at $t = 0$.

Your claim that there might be some solution of the EFE that is spatially flat, matter-dominated, but without any point at which $a = 0$ might be true; but you can't just wave your hands and claim it. You need to show such a solution, or prove that one exists. You have done neither. My pointing that out is not "bias"; it's just asking you to show your work.

 

[RUTA]

I have said nothing at all about my personal preferences. If you make statements that are wrong as a pure matter of math, you should expect to have them corrected. Correcting them is not "bias", it's just correcting erroneous statements. Your claim that the EdS model is "without infinities" is wrong as a pure matter of math: the EdS is a specific solution of a specific equation with specific properties, and those properties include $a = 0$ at $t = 0$.

Your claim that there might be some solution of the EFE that is spatially flat, matter-dominated, but without any point at which $a = 0$ might be true; but you can't just wave your hands and claim it. You need to show such a solution, or prove that one exists. You have done neither. My pointing that out is not "bias"; it's just asking you to show your work.

Go to this Insight and you'll see the solution I'm talking about. I have not said anything "mathematically incorrect." I assumed you were familiar with the differential equation resulting from the spatially flat, matter-dominated cosmology model called Einstein-deSitter, which is the differential equation I am solving. The only dispute you have raised here is that you claim the EdS solution entails $a(0) = 0$, while I am using the term to mean the spatially flat, matter-dominated model. We can argue semantics if you like, but it doesn't change anything.

 

[PeterDonis]

Go to this Insight and you'll see the solution I'm talking about.

Thank you for the pointer, it's much better to talk about a specific model.

the differential equation I am solving

Do you mean equation (18) in the Insight?

 

[RUTA]

Do you mean equation (18) in the Insight?

Yes

 

[PeterDonis]

Yes

Ok, then yes, I agree you can pick $a(0) \neq 0$ in your solution, and, as far as I can tell, that also makes $\dot{a}$, $\ddot{a}$, and $\rho$ finite at $t = 0$ (basically because you have substituted $t + B$ for $t$, so all of the values at $t = 0$ are proportional to some power of $B$ instead of diverging).

However, this model is obviously extensible to negative values of $t$, and when you reach $t = - B$, your model has $a = 0$ and $\dot{a}$, $\ddot{a}$, and $\rho$ all infinite. So your model is not a different model from the standard one, it's just a shift of the $t$ coordinate by $B$ (strictly speaking there is a rescaling of $t$ as well). Considering the patch $t \ge 0$ in this model is simply equivalent to only considering the patch $t \ge B$ in the standard Einstein-de Sitter model. This is not a model in which the singularity theorems are violated; it's just a model in which you have artificially restricted attention to a particular patch.

 

[RUTA]

Ok, then yes, I agree you can pick $a(0) \neq 0$ in your solution, and, as far as I can tell, that also makes $\dot{a}$, $\ddot{a}$, and $\rho$ finite at $t = 0$ (basically because you have substituted $t + B$ for $t$, so all of the values at $t = 0$ are proportional to some power of $B$ instead of diverging).

However, this model is obviously extensible to negative values of $t$, and when you reach $t = - B$, your model has $a = 0$ and $\dot{a}$, $\ddot{a}$, and $\rho$ all infinite. So your model is not a different model from the standard one, it's just a shift of the $t$ coordinate by $B$ (strictly speaking there is a rescaling of $t$ as well). Considering the patch $t \ge 0$ in this model is simply equivalent to only considering the patch $t \ge B$ in the standard Einstein-de Sitter model. This is not a model in which the singularity theorems are violated; it's just a model in which you have artificially restricted attention to a particular patch.

Right, the singularity theorem is not violated because it is still true that there are timelike and null geodesics with finite affine parameter lengths into the past (finite proper time). But, all the observables and physical parameters are finite (except meaningless ones like the volume of spatial hypersurfaces of homogeneity). It is absolutely "artificial" in that there is no dynamical reason whatsoever for not extending the solution into the past (with negative values of t) all the way to $a = 0$. But, in the 4D global self-consistent view, there is no reason to do that. You only need as much of the spacetime manifold as necessary to account for your observations. I don't foresee a need for $\rho = \infty$, i.e., $a = 0$, but if we ever do need such $\infty$, then you can include it at that point.

 

[PeterDonis]

all the observables and physical parameters are finite

Not at $t = - B$. There the density $\rho$ is infinite.

in the 4D global self-consistent view, there is no reason to do that

Yes, there is, because in the 4D global self-consistent view, the manifold is its maximal analytic extension. Arbitrarily cutting it off at some point prior to that makes no sense on that view. If you think it does because of some "adynamical constraint", what is that constraint? It can't be "because RUTA prefers to cut off the solution at $t = 0$ in his model".

You only need as much of the spacetime manifold as necessary to account for your observations.

Not if you want your model to make testable predictions about observations that haven't been made yet.

 

[RUTA]

Not at $t = - B$. There the density $\rho$ is infinite.

Yes, there is, because in the 4D global self-consistent view, the manifold is its maximal analytic extension. Arbitrarily cutting it off at some point prior to that makes no sense on that view. If you think it does because of some "adynamical constraint", what is that constraint? It can't be "because RUTA prefers to cut off the solution at $t = 0$ in his model".

Not if you want your model to make testable predictions about observations that haven't been made yet.

As I explained in the Insight, EEs of GR constitute the constraint. Any solution of EEs that maps onto what you observe or could conceivably observe is fair game. There is nothing in GR that says you must include extensions of M beyond what maps to empirically verifiable results. But, if you have a prediction based on $a = 0$ and $\rho = \infty$, by all means include that region.

 

[PeterDonis]

As I explained in the Insight, EEs of GR constitute the constraint.

That doesn't explain why you would cut off a solution of the EFE short of its maximal analytic extension.

There is nothing in GR that says you must include extensions of M beyond what maps to empirically verifiable results.

Again, you have to do this if you want your model to make testable predictions about observations that haven't been made yet.

Also, the position you appear to be taking seems highly implausible on your own "blockworld" viewpoint. Why would a "blockworld" just suddenly have an "edge" for no reason? It seems much more reasonable to expect any "blockworld" to extend as far as the math says it can.

 

[RUTA]

That doesn't explain why you would cut off a solution of the EFE short of its maximal analytic extension.

Again, you have to do this if you want your model to make testable predictions about observations that haven't been made yet.

Also, the position you appear to be taking seems highly implausible on your own "blockworld" viewpoint. Why would a "blockworld" just suddenly have an "edge" for no reason? It seems much more reasonable to expect any "blockworld" to extend as far as the math says it can.

Look again at the partial parabola for the trajectory of a ball with $y(0) = 3$. We don’t include the mathematical extension into negative times demanding therefore we must include $y = 0$. Why? Because we don’t believe there can be any empirical evidence of that fact. So, in adynamical thinking the onus is on you to produce a prediction with empirical evidence showing you need to include $a = 0$ with $\rho = \infty$. We can then do the experiment and see if your prediction is verified. If so, according to your theory, we need to include that region. There is no reason to include mathematics in physics unless that mathematics leads to empirically verifiable predictions. So, what is your prediction?

 

[martinbn]

They're looking for past extendability and found it. Why were they looking for that? Because they were thinking dynamically. Here is an analogy.

Set up the differential equations in y(t) and x(t) at the surface of Earth (a = -g, etc.). Then ask for the trajectory of a thrown baseball. You're happy not to past extend the solution beyond the throw or future extend into the ground because you have a causal reason not to do so. But, the solution is nonetheless a solution without those extensions. Same for EEs with no past extension beyond a(0) and a choice of a(0) not equal to zero. Why are you not satisfied with that being the solution describing our universe? There's nothing in the data that would ever force us to choose a(0) = 0 singular. The problem is that the initial condition isn't explained as expected in a dynamical explanation. All we need in 4D is self-consistency, i.e., we only have to set a(0) small enough to account for the data. Maybe someday we'll have gravitational waves from beyond the CMB and we'll be able to push a(0) back to an initial lattice spacing approaching the Planck length. But, we'll never have to go to a singularity.

I am missing something very basic here. Take for example $y''=2$ on the interval $[0,2]$ with $y(1)=1$ and $y(2)=4$. The only solution is $y(x)=x^2$. How do you make $y(0)$ not equal to zero?

 

[RUTA]

I am missing something very basic here. Take for example $y''=2$ on the interval $[0,2]$ with $y(1)=1$ and $y(2)=4$. The only solution is $y(x)=x^2$. How do you make $y(0)$ not equal to zero?

There are any number of reasons you might want to use $y = 0$, but you have to come up the reason to do so. You don’t use the math to dictate the use of $y = 0$. What if I want to use the math for throwing a ball? I don’t use $y = 0$ because I believe it is not possible to find empirical verification of that fact. Again, the empirically verifiable physics drives what you use of the math, not the converse. So, again, what is your prediction requiring I keep $a = 0$ with $\rho = \infty$? Produce that prediction and its empirical verification and we’ll know we have to keep that region.