Dimensions of five-dimensional Newton's constant

  • Thread starter papades
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  • #1
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Main Question or Discussion Point

Hello,
I have a question, please let me have your answer, if possible:
In a space with an extra spatial dimension , where the extra coordinate is compact:
[itex]0 \le y \le 2\pi \alpha [/itex]
And the metric of the space is:
[tex]{G_M}_N = \left( {\begin{array}{*{20}{c}}
{{g_\mu }_\nu (x)} & {{A_\mu }(x)} \\
{{A_\mu }(x)} & {\varphi (x)} \\
\end{array}} \right)[/tex]

In the action:


[tex]S = \frac{1}{{8\pi G_N^{(5)}}}\int {{d^5}x\;\sqrt { - G} \;R} [/tex]

Which are the dimensions of
[itex]G_N^{(5)}[/itex]
, that is five-dimensional Newton’s constant?
Many thanks in advance.
 

Answers and Replies

  • #2
649
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Compare your definition to eq. 2.132 here:

http://folk.uio.no/olavau/thesis.pdf

which gives the Einstein-Hilbert action for 4+n dimensional gravity. The prefactor is the 4+n dimensional Planck mass. You'll be able to find the dimension of your 5-dimensional gravitational constant given in units of mass, where c=hbar=1.
 
  • #3
arivero
Gold Member
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Hmm, the Riemann tensor is esentially a bidimensional object, a set of surface intersections so to say, independently of the number of dimensions of the space time. So in principle it does not sound as a good idea to think of Newton's constant depending of the number of dimensions, it could have more sense to put powers of this constant (and c, when needed) to get the exact dimensionality in each case. But perhaps I have done the same mistake in some of my letters, so I am not to launch the first stone.
 
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
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The Ricci scalar has mass dimension 2 for any dimension of spacetime. The Newton constant is defined by the power law at large distances, from which its mass dimension can be determined, or as well as by demanding that the action has mass dimension zero. Therefore, in [itex]n[/itex] dimensions, the Newton constant has mass dimension [itex]2-n[/itex] and is related to the [itex]n[/itex]-dimensional Planck mass by

[itex]G_N^{(n)} \sim \left( M_P^{(n)} \right)^{2-n}.[/itex]
 
  • #5
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-3 :smile:
 

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