Physics in the Multiverse (alpha)

zephir---the author was using that as an example of where a multi-verse has appeared in the past.

If you cannot see this then you shouldn't be commenting in this thread.

 

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This is dual view to the stance, the author has explained, how the multiverse concept has arrised in string theory historically. You're simply trying to ignore historical conotations. And this is mistake - because just these conotations can help you, what the multiverse concept really means for string theory, the quantum mechanics and the rest of physics. Without it it's just one of many ad-hoced concepts, worth to become memorized in troll's lectures.

Furthemore, you're trying to talk about multiverse in context of some particular idea, although the forum is named Physics in the Multiverse. And I'm not interested to analyze some particular ideas without holistic connection to the context. This is typical problem of specialists: they're trying to understand their theory as in detail, as possible - but because their theory is ad-hoced and separated from context, they'll lose the reason, why such theory was originally developed. The theories are developed to connect the other less advanced theories. My job is to understand and connect all viable concepts of physics, not to fragmentize the scope even more.

If the people would adhere on these trivial rules, they would understood the physics before many years already - because here's nothing really special about physics, it's just curvature game. But the problem of specialists is, they're separatists by their very nature. As the result, the development of physics is more slower, inconsistent and of course expensive, then it could be. Please consider, if you don't learn the mutual respect to others physicists opinions (including those of mine and Everett's), you can never understand, what's the physics is all about.

The point is, not only what the multiverse means in context of some string theory, but other theories as well, including the relativity and Everett "many words" concept. If the string theory wouldn't propose such general explanation, it will be refuted due the correspondence principle violation - so it's just in your interest to play well with others.

 

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zephir---

I have made it abundantly clear that I am dealing with the concept of a multi-verse vis a vis the current situation in string theory.

Note to all: Any further posts on Everett's ``Many Worlds'' interpretation of quantum mechanics will be considered off topic. This interpretation is a completely different subject than the topic of this thread.

Thank you.

 

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Ben, I wasn't ignoring this, just I had to do a lot of reading to ask even half-sensible questions. Anyway, thanks.

Ok, this makes sense, as this ratio (it is a ratio, right?) is a number, and I guess I can see that it must be constant

This caused me a problem at first. I had naively thought that the Big Bang was such a humongous event, that matter was still moving away from us. But it would not account for the acceleration, I see that. OK so far, but then...

I have trouble seeing how constant, a number, a ratio (if I was right about that) can "push" anything. Or is it thought to be a field in the same sense that gravitation is a field? If so, is it a scalar field? Surely not.

Well, accepting the foregoing at face value, this makes sense.

OK, that addresses part of my question, or rather, it underlines my puzzlement!

OK, this is where I start to have real problems (by the way, in the inequality above, I assume you meant n is integer?). First, what is the connection between the CC, "negative gravity", and the vacuum energy? Well, OK, if I'm right, and the CC is some sort of field which exists in the absence of matter, I suppose I can see this, but I don't think I am right!


Anyway, it's challenging stuff. Oh by the way, I figured out quantum foam. What a nice idea!

 

QuarkHead---

Humm... You can look at this and see it from Einstein's equations:

\(G_{\mu\nu} = \kappa T_{\mu\nu} + \Lambda g_{\mu\nu}\)

So gravity is described by G, and energy is described by T (the stress tensor). So gravity sees the cosmological constant as a constant energy density in space---and Einstein's equations say that energy density is related to gravity. (They only say this, they don't explain it... This really bugged Einstein.)

yes, n is an integer.

I wouldn't use the word field, but essentially you are right. The cosmological constant is some energy density which is associated with space. Reiku was more or less right when he said that the vacuum energy (the 1/2 h w) could be the cosmological constant---this is some energy associated with space-time itself, irrespective of there being any matter present. The problem is that treating the cosmological constant this way gives you an error that is 120 orders of magnitude wrong.

Another way to see this is from an action principle. Bear with me

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We can take the Einstein's equatins above and make an action:

\(\mathcal{S}_{\rm EH} = \int d^Dx \sqrt{-g}\left\{\mathcal{L}_{\rm EH}\right\}\).

What is an action? Well, it's a function(al) over field space which describes the dynamics of a physical system. You minimize it to get equations of motion for the fields. (EH stands for Einstein Hilbert.) g is the determinant of the metric.

Now, the action for a scalar field can be written like this:

\(\mathcal{S}_{\rm scalar} =\int d^Dx \sqrt{-g} \left\{\mathcal{L}_{\rm scalar} + a\right\}\).

Suggestively, I have defined the constant term as

\(a = \sum_{k = -\infty}^{\infty} \frac{1}{2}\hbar \omega_k\)

Formally, this sum is divergent, but practically, the largest scale in our problem is the Planck mass. So we can evaluate this sum if we wanted to---it just comes out to be a really big and constant number.

Now, if we couple our scalar field to gravity, naively we would get something like

\(\mathcal{S} = \int d^Dx \sqrt{-g}\left\{\mathcal{L}_{\rm EH}+\mathcal{L}_{\rm scalar} + a\right\}\).

According to gravity, this sum over harmonic oscillators looks just like a constant energy density---this is what I mean when I say that the cosmological constant couples to gravity. When you do the math and find the equations of motion for gravity, you will find something like

\(G_{\mu\nu} = \kappa T_{\mu\nu} + a g_{\mu\nu}\)

up to (overall) constants, perhaps. (I'm being kind of symbolic here, and trying to remember equations from the top of my head, but this is the general idea.)

Now, we could explain this number if we measured it and it was large. (But if it was large we probably wouldn't be around to measure it, because the universe would have blown apart by the time intelligent life had the chance to form.) We could even understand why it could possibly be zero---we are always free to add and subtract a constant term from the action because the equations of motion are defined in terms of derivatives. But what we can't understand is why it should be small and positive.

 

Jeez! I don't how you guys do this, it's so hard! Quickly, as I'm short of time, and just to guide my thinking: what is the Einstein-Hilbert Lagrangian?

Also, in your action integral, what does the notation \(^{d^{^D} x}\) mean?

I suspect I'll regret asking both these questions!

 

It's all just semantics, I think. First of all, note the Einstein Summation convention:

\( A_{\mu\nu}g^{\mu\nu} \Rightarrow\sum_{\mu,\nu}A_{\mu\nu}g^{\mu\nu}\)

A repeated index is always summed over.

The Einstein-Hilbert lagrangian is this (assuming some idiot hasn't edited Wikipedia recently)

\(\mathcal{S} = \kappa \int d^Dx \sqrt{-g} \mathcal{R}\)

The \(\mathcal{R}\) is the Ricci scalar, which is called the curvature scalar by mathematicians. Basically, you make it out of the metric, which is a tensor of rank 2.

\(\mathcal{R} = R_{\mu\nu}g^{\mu\nu}\).

The new R is called the Ricci tensor, or the curvature tensor, and is defined as

\(R_{\mu\nu} = R^{\rho}_{\mu\rho\nu}\)

where THIS R (with four indices) is the Riemann tensor or curvature tensor:

\({{R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\)

The gamma are the Christoffel symbols

\(\Gamma^i {}_{k\ell}=\frac{1}{2}g^{im} \left(\frac{\partial g_{mk}}{\partial x^\ell} + \frac{\partial g_{m\ell}}{\partial x^k} - \frac{\partial g_{k\ell}}{\partial x^m} \right).\)

Anyway, this is all a bunch of differential geometry, and basically amounts to a huge pain in the ass. The imprtant thing to see, though, is that the Ricci scalar \(\mathcal{R}\) can be written purely in terms of the metric.

This is just a D-dimensional differential. I could have written \(d^4x\) if it makes you more comfortable

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