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What are the physical dimensions of the $5\times 5$ Kaluza-Klein metric? (the metric should be dimensionless but doesn't look so with the inclusion of the four potential and the scalar field)

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You compensate the mass dimensions with 5d (or whatever dimension your uncompactified spacetime has) Planck mass, so that you metric is dimensionless.

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The metric is not necessarily dimensionless.

The relevant condition is that the line element $$\text{d} s^2 = g_{\mu\nu} \text{d}x^\mu \text{d}^\nu $$ has dimension $(\text{length})^2$, so the dimensions of the metric and the coordinates are related (this is already apparent in 2d Euclidean space in polar coordinates).

Thus, for 5d KK theory, you could either let the extra dimensionl range from 0 to $2\pi$ or some other fixed dimensionless number and have $g_{55}=R^2$ be dimensionful, or have $g_{55}$ be dimensionless and $x^5$ of dimension length.

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If you write the Kaluza Klein metric as

$$\mathrm{d}s^2=g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}+\phi^2(A_{\mu}\mathrm{d}x^{\mu}+\mathrm{d}x^5)^2,$$

then the metric (as well as the scalar and vector field) is dimensionless. The dimensionlessness of these two fields is later compensated by the inclusion of factors of $R$ (the compactificsyjon radius) and the planck mass. That is, you don't get a lagrangian with a canonically normalized kinetic term for these fields until you redefine them with appropriate factors of $R$ and $M_p$.

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