Kaluza Klein theories, dilation field, and dimensional reduction


Let us quickly run through the standard KK compactification. We start with a $d+1$ dimensional theory $$ S = \frac{1}{16\pi G_{d+1}} \int d^{d+1}x \sqrt{G} R_{d+1} $$ More general actions on the $d+1$ dimensional space can be considered, but this will suffice for our purposes. The metric $G_{MN}$ can be decomposed as $$ ds^2 = G_{MN} dx^M dx^N = e^{2\Phi} \left( dt + A_\mu dx^\mu \right)^2 + g_{\mu\nu} dx^\mu dx^\nu $$ We assume now that $t$ is the compactified direction with $t \sim t + 2 \pi R$. The theory has the following symmetries

  1. $d$-dimensional diffeomorphims, $x^\mu \to x'^\mu(x)$ under which $A_\mu$ and $g_{\mu\nu}$ transform as rank 1 and 2 tensors respectively.

  2. Gauge transformations along the compactified directions, $t \to t + \lambda(x)$, $A_\mu \to A_\mu - \partial_\mu \lambda$. This symmetry essentially describes the local choice of origin in the compactified direction.

Now, if the length scales of our problem are large compared to the radius of the compactified circle $R$, we then assume that $\Phi$, $A_\mu$, and $g_{\mu\nu}$ are only functions of $x^\mu$ and not $t$. (This is only done here to make things simpler. One can consider the more general case where the fields are expanded into modes in the $t$ direction. This gives us massive particles in the $d$-dimensional space. We will not consider this here). With this assumption, we find $$ R_{d+1} = R_d - 2 e^{-\Phi} \nabla^2 e^{\Phi} - \frac{1}{4} e^{2\Phi} F_{\mu\nu} F^{\mu\nu},~~ \sqrt{G} = e^{\Phi} \sqrt{g} $$ The action then takes the form $$ S = \frac{2\pi R}{16\pi G_{d+1}} \int d^d x \sqrt{g} e^{\Phi} \left[ R_d - \frac{1}{4} e^{2\Phi} F_{\mu\nu} F^{\mu\nu} + \nabla_\mu \Phi \nabla^\mu \Phi \right] $$ Thus, we note that in the $d$-dimensional space, we have a gauge field and a scalar field. The action is not quite in the Einstein-Hilbert form since the scalar field $\Phi$ couples to $R_d$ and $F_{\mu\nu} F^{\mu\nu}$ non-trivially. Also, the kinetic term for $\Phi$ has the wrong sign. This field is called the dilaton.

To understand why $\Phi$ is called the dilaton (related to dilation, or in other words scale), let us go back to the $d+1$ dimensional metric. Consider the $S^1$ living at a fixed point $x^\mu$. The induced metric on this circle is $$ ds^2_{S^1} = e^{2\Phi(x^\mu)} dt^2 $$ The size of this circle is $$ \int ds_{S^1} = \int_0^{2\pi R} dt e^{\Phi(x^\mu)} = 2\pi R e^{\Phi(x^\mu)} $$ Thus, we see that the effective radius of the circle at $x^\mu$ is $R e^{\Phi(x^\mu)}$. In other words the dilaton field controls the size of the circle.

As an aside, the compactified action above is written in what is called the string frame (name comes from string theory). It possible to go to the more standard Einstein frame (where the action takes the form $\sqrt{g} R$, etc) by doing a field redefinition $g_{\mu\nu} \to e^{2\omega} {\tilde g}_{\mu\nu}$ and appropriately choosing $\omega$. In this frame, the scalar kinetic term has the right sign. However, we still have a non-trivial coupling to $F_{\mu\nu}F^{\mu\nu}$.

Kevin Ye
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Kevin Ye

Pursuing in physics research. Eager to learn more.

Updated on March 06, 2020


  • Kevin Ye
    Kevin Ye over 3 years

    I am reading something about Kaluza Klein theories and compactification. I have some conceptual question:

    (1) Why do we call the fifth scalar field $\Phi$ the dilation field? Is there any scaling property for that?

    (2) What happens to this field after dimensional reduction?

    Many thanks if you can provide me with further reading material! :)

    • John Rennie
      John Rennie over 9 years
      dilation or dilaton?
    • Heterotic
      Heterotic over 9 years
      As John Rennie pointed out it is called the dilaton field. It is related to the size of the manifold we are using for the compactification.
    • Kevin Ye
      Kevin Ye over 9 years
      Thank you! Sorry I guess it is a auto correction of words generated by my Mac. It seems Mac does not know dilaton at all...:)
  • Prahar
    Prahar over 9 years
    $t$ is not a time-like coordinate (take a look at the metric signature). The remaining $x^\mu$ might have a time-like coordinate if you wish. The discussion however, is completely general.
  • Kevin Ye
    Kevin Ye over 9 years
    Thanks a lot! Sorry for delayed reply. Your answer is very informative and helpful. :) Just two quick question. (1) So in the first compactification scheme you eliminate coordinate $t$, and $\mu$ denote the index range from 1 to d. So in the remaining coordinate will there be timelike direction? Namely is Minkowski signature preserved here? (2)Also, what about other compact manifold for $t$, will the statement that the dilaton field control the geometrical size of the manifold still holds?
  • achatrch
    achatrch over 8 years
    Can I ask something about this scalar field (dilaton). I read that originally it was just put constant, but I dont see why this is the case. On the other if it has its own dynamics, than how can we fix the radius of extradimension and also wouldnt it affect the other two terms in the action due to its coupling to them? Thanks