# Does Kaluza-Klein Theory Require an Additional Scalar Field?

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.

-

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.

-
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? –  elfmotat Feb 17 at 11:38
@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 ... –  Dilaton Feb 17 at 12:05
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
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. –  elfmotat Feb 17 at 11:24
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. –  TMS Feb 17 at 12:02