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Does Kaluza-Klein Theory Require an Additional Scalar Field?

+ 6 like - 0 dislike
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I've seen the Kaluza-Klein metric presented in two different ways., cf. Refs. 1 and 2.

  1. In one, there is a constant as well as an additional scalar field introduced: $$\tilde{g}_{AB}=\begin{pmatrix} g_{\mu \nu}+k_1^2\phi^2 A_\mu A_\nu & k_1\phi^2 A_\mu \\ k_1\phi^2 A_\nu & \phi^2 \end{pmatrix}.$$

  2. In the other, only a constant is introduced:

$$\tilde{g}_{AB}=\begin{pmatrix} g_{\mu \nu}+k_2A_\mu A_\nu & k_2A_\mu \\ k_2A_\nu & k_2 \end{pmatrix}.$$

Doesn't the second take care of any problems associated with an unobserved scalar field? Or is there some reason why the first is preferred?

References:

  1. William O. Straub, Kaluza-Klein Theory, Lecture notes, Pasadena, California, 2008. The pdf file is available here.

  2. Victor I. Piercey, Kaluza-Klein Gravity, Lecture notes for PHYS 569, University of Arizona, 2008. The pdf file is available here.

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user elfmotat
asked Feb 17, 2013 in Theoretical Physics by elfmotat (30 points) [ no revision ]

2 Answers

+ 8 like - 0 dislike

When you write the five dimensional Kaluza-Klein metric tensor as

$$ g_{mn} = \left( \begin{array}{cc} g_{\mu\nu} & g_{\mu 5} \\ g_{5\nu} & g_{55}\\ \end{array} \right) $$

where $g_{\mu\nu}$ corresponds to the ordinary four dimensional metric and $ g_{\mu 5}$ is the ordinary four dimensional vector potetial, $g_{55}$ appears as an additional scalar field. This new scalar field, called a dilaton field, IS physically meaningful, since it defines the size of the 5th additional dimension in Kaluza-Klein theory. They are natural in every theory that hase compactified dimensions. Even though such fields have up to now not been experimentally confirmed it is wrong to call such a field "unphysical".

"Unphysical" are in some cases fields introduced to rewrite the transformation determinant in calculations of certain generating functionals, or the additional fields needed to make an action local, which may have conversely to such dilaton field, no well defined physical meaning.

answered Feb 17, 2013 by Dilaton (4,175 points) [ revision history ]
Why is $g_{55}$ necessarily a scalar field? What's the motivation to promote it from being a constant to a field? Is it just because a 5th dimension with variable size is more general?

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user elfmotat
@elfmotat yes, as the whole metric is a field, it is more general and consistent to consider its components, such as the dilaton, to be fields too. However, the dilaton field should better not vary too much at scales smaler than cosmilogical ones, since in ST it determines not only the size of compactified dimensions but the value of the string coupling constant too. And this again determins the fundamental laws of nature which should be about constant in our universe ...

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user Dilaton
+ 3 like - 0 dislike

It will be better if you were mentioned the sources.

However, as I remember, that in some old works on this theory, they used to assume that $\phi=const$ because the main purpose of the theory was looking for a geometrical unification of Gravity and electromagnetism, and no physical scalar fields was known at that time

Also it will be not very accurate to say "Requires additional" scalar field, because this field in addition to the Electromagnetic field raises naturally (almost, considering cylindrical condition) in the theory after applying the least action principle on five dimensional scalar curvature, and this "natural way" is the whole point of the theory.

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user TMS
answered Feb 17, 2013 by TMS (40 points) [ no revision ]
My sources are: weylmann.com/kaluza.pdf math.arizona.edu/~vpiercey/KaluzaKlein.pdf I'm not sure why you say applying the the action principle on the 5D Ricci scalar naturally implies an additional field. When considering the second form of the metric in my OP, isn't the Ricci scalar just: $\tilde{R}=R+\frac{k}{4}F^{\mu \nu}F_{\mu \nu }$ If $k$ is a constant I don't see how that implies a scalar field.

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user elfmotat
Yes as you mentioned, there is no reason to put $k=const$, the first paper you mentioned is the original way it was done, including the scalar field just makes it more general, the approach that taken in the second paper.

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user TMS
I disagree with your last paragraph. Since to unify gravity and EM you need five dimensions, an additional parameter is needed in front of the new fifth component when writing proper time (or distance) by the five dimensional KK metric. And this parameter is exactly the additional field (which my have a fairly constant value) that appears naturally as 55 component in the KK metric, it IS needed.

This post imported from StackExchange Physics at 2014-03-17 03:58 (UCT), posted by SE-user Dilaton

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