Insights Answering Mermin’s Challenge with the Relativity Principle

  • #91
I answered your question many times, the cutoff is not at all arbitrary, you keep whatever you can argue generates conceivable empirical results. I didn't say you can't keep inflationary models. I personally think they're not interesting as articulated by Paul Steinhardt, but others may want to pursue them.
 
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  • #92
Again, the only difference between what I'm claiming and what exists in the common textbook explanations of GR cosmology is that the textbooks say GR cosmology models are problematic because they're "singular" in the sense that they entail ##\rho = \infty##. And I'm saying adynamical explanation does not force those solutions to entail that region. Adynamical explanation allows you to simply omit the problematic region if it is beyond empirical confirmation. So far that's true, the model is working great despite the fact that a purely mathematical extrapolation produces ##\rho = \infty##. So why worry about that region?
 
  • #93
RUTA said:
the cutoff is not at all arbitrary, you keep whatever you can argue generates conceivable empirical results

Then the Einstein-de Sitter model would be valid for any ##t > 0##, since it predicts finite and positive density and scale factor. So are you saying I could use any value of ##B## I like in your modified version of the model, putting the cutoff wherever I want, as long as it doesn't include the ##\rho = \infty## point?

RUTA said:
I didn't say you can't keep inflationary models.

Ok, that helps to clarify your viewpoint.

RUTA said:
Adynamical explanation allows you to simply omit the problematic region if it is beyond empirical confirmation

But ##\rho = \infty## isn't a "region", it's a point. And that point is not even included in the manifold; as I said, it's a limit point that's approached but never reached. So again, I don't see what is wrong with the standard Einstein-de Sitter model, where ##B = 0## in your modified formula, if any finite value of ##\rho## is ok.
 
  • #94
RUTA said:
the textbooks say GR cosmology models are problematic because they're "singular" in the sense that they entail ##\rho = \infty##.

No, that's not what they say. What they say (Wald, for example) is that spacetime curvatures which are finite but larger than the Planck scale are problematic for a classical theory of gravity like GR, because we expect quantum gravity effects to become important at that scale. In the Einstein-de Sitter model, for example, the curvature becomes infinite at the point you have been labeling ##\rho = \infty##, not just the density. And in Schwarzschild spacetime, the curvature is the only thing that becomes infinite at the singularity at ##r = 0##, because it's a vacuum solution and the stress-energy tensor is zero everywhere. But that singularity is just as problematic on a viewpoint like Wald's.
 
  • #95
PeterDonis said:
No, that's not what they say. What they say (Wald, for example) is that spacetime curvatures which are finite but larger than the Planck scale are problematic for a classical theory of gravity like GR, because we expect quantum gravity effects to become important at that scale. In the Einstein-de Sitter model, for example, the curvature becomes infinite at the point you have been labeling ##\rho = \infty##, not just the density. And in Schwarzschild spacetime, the curvature is the only thing that becomes infinite at the singularity at ##r = 0##, because it's a vacuum solution and the stress-energy tensor is zero everywhere. But that singularity is just as problematic on a viewpoint like Wald's.
Yes, the curvature is also problematic as Wald points out in Chapter 9. The ##\rho = \infty## is also a problem for Schwarzschild at ##r = 0## because that's where M is. Am I missing something there?
 
  • #96
PeterDonis said:
But ##\rho = \infty## isn't a "region", it's a point. And that point is not even included in the manifold; as I said, it's a limit point that's approached but never reached. So again, I don't see what is wrong with the standard Einstein-de Sitter model, where ##B = 0## in your modified formula, if any finite value of ##\rho## is ok.
Well, it's difficult to say how "big" ##a = 0## is because the spatial hyper surfaces to that point are ##\infty##. Its "size" is undefined so I was being careful with my language.
 
  • #97
PeterDonis said:
Then the Einstein-de Sitter model would be valid for any ##t > 0##, since it predicts finite and positive density and scale factor. So are you saying I could use any value of ##B## I like in your modified version of the model, putting the cutoff wherever I want, as long as it doesn't include the ##\rho = \infty## point?
Yes, and you could even use ##\rho = \infty## if you could produce empirical verification. Use whatever you need, just don't dismiss the model because you believe an empirically unverifiable mathematical extrapolation leads to "pathologies."
 
  • #98
RUTA said:
The ρ=∞ρ=∞ is also a problem for Schwarzschild at r=0r=0 because that's where M is.

There is no "where M is" in the Schwarzschild solution; it's a vacuum solution with zero stress-energy everywhere and no ##\rho = \infty## (and for that matter no ##\rho \neq 0##) anywhere. Also ##r=0## is not even part of the manifold; it's a limit point that is approached but never reached, so it can't be "where" anything is.

M in the Schwarzschild solution is a global property of the spacetime; there is no place "where it is".
 
  • #99
RUTA said:
don't dismiss the model because you believe an empirically unverifiable mathematical extrapolation leads to "pathologies."

Are you claiming that Wald is "dismissing" the Einstein-de Sitter model or similar models on these grounds? I don't see him dismissing it at all. I just see him (and MTW, and every other GR textbook I've read that discusses this issue) saying that any such model will have a limited domain of validity; we should expect it to break down in regions where spacetime curvature at or greater than the Planck scale is predicted.
 
  • #100
RUTA said:
it's difficult to say how "big" ##a = 0## is

But however "big" it is, it does not extend to any value of ##t## greater than zero in the standard FRW models. That's the point I was making.
 
  • #101
PeterDonis said:
There is no "where M is" in the Schwarzschild solution; it's a vacuum solution with zero stress-energy everywhere. Also ##r = 0## is not even part of the manifold; it's a limit point that is approached but never reached, so it can't be "where" anything is.

M in the Schwarzschild solution is a global property of the spacetime; there is no place "where it is".
You can model Schwarzschild surrounding matter solutions and the Schwarzschild solution holds all the way inside the horizon to that matter. So, as you go smaller and smaller you approach ##\rho = \infty##.
 
  • #102
RUTA said:
You can model Schwarzschild surrounding matter solutions and the Schwarzschild solution holds all the way inside the horizon to that matter. So, as you go smaller and smaller you approach ##\rho = \infty##.

You can do this in a model such as the 1939 Oppenheimer-Snyder model, sure. That's not what I was using "the Schwarzschild solution" to refer to, but yes, it's a valid model, and is subject to the same limitation that Wald describes, assuming Wald's viewpoint that classical GR will break down at Planck scale curvatures is correct.
 
  • #103
bhobba said:
For those interested here is Mermin's original paper:
https://pdfs.semanticscholar.org/76f3/9c8a412b47b839ba764d379f88adde5bccfd.pdf
Feynman in a letter to Mermin said 'One of the most beautiful papers in physics that I know of is yours in the American Journal of Physics.'

I personally am finding my view of QM evolving a bit. Feynman said the essential mystery of QM was in the double slit experiment. I never actually thought so myself, but was impressed with it as an introduction to the mysteries of QM at the beginning level. I am now starting to think entanglement may be the essential mystery.

Thanks
Bill
I ordered the book ILL and copied the letter from Feynman to Mermin and Mermin's response. I'll attach those copies here.
 

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