Hierarchy problem: Difference between revisions

Revision as of 14:16, 17 September 2024

Revision as of 06:11, 16 November 2024

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If we live in a 3+1 dimensional world, then we calculate the gravitational force via [[Gauss's law for gravity]]:

: <math display=block>\mathbf{g}(\mathbf{r}) = -Gm\frac{\mathbf{e_r}}{r^2}</math> \qquad (1)</math>

which is simply [[Newton's law of gravitation]]. Note that Newton's constant ''{{mvar|G''}} can be rewritten in terms of the [[Planck mass]].

: <math display=block>G = \frac{\hbar c}{M_{\mathrm{Pl}}^{2}}</math>

If we extend this idea to <math>\{{mvar|&delta</math>;}} extra dimensions, then we get:

: <math display=block>\mathbf{g}(\mathbf{r}) = -m\frac{\mathbf{e_r}}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta}r^{2+\delta}}</math> \qquad (2)</math>

where <math>M_{\mathrm{Pl}_{3+1+\delta}}</math> is the {{nowrapmath|3+1+<math>\delta</math>}} dimensional Planck mass. However, we are assuming that these extra dimensions are the same size as the normal 3+1 dimensions. Let us say that the extra dimensions are of size {{math|''n'' ≪}} than normal dimensions. If we let {{math|''r''&nbsp; ≪&nbsp; ''n''}}, then we get (2). However, if we let {{math|''r''&nbsp; ≫&nbsp; ''n''}}, then we get our usual Newton's law. However, when {{math|''r''&nbsp; ≫&nbsp; ''n''}}, the flux in the extra dimensions becomes a constant, because there is no extra room for gravitational flux to flow through. Thus the flux will be proportional to <math> {{mvar|n^{\{sup|&delta;}}}} </math> because this is the flux in the extra dimensions. The formula is:

<math display=block>\begin{align}

: <math>\mathbf{g}(\mathbf{r}) = -m\frac{\mathbf{e_r}}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta}r^2 n^{\delta}}</math>

: <math>-m\frac{ \mathbf{e_rg}}{M_{(\mathrmmathbf{Pl}}^2 r^2}) &= -m \frac{\mathbf{e_r}}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta} r^2 n^{\delta}}</math> \\[2pt]

: <math> -m \frac{\mathbf{ge_r}(}{M_\mathbfmathrm{Pl}^2 r^2}) &= -m \frac{\mathbf{e_r}}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta}r^2 n^{\delta}}</math>

\end{align}</math>

which gives:

<math display=block>\begin{align}

: <math> \frac{1}{M_{\mathrm{Pl}}^2 r^2} = \frac{1}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta}r^2 n^{\delta}} \Rightarrow </math>

: <math> M_\frac{1}{M_\mathrm{Pl}}^2 r^2} &= \frac{1}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta} r^2 n^{\delta}. </math>\\[2pt]

: <math> \frac{1}{implies \quad M_{\mathrm{Pl}}^2 r^2} &= \frac{1}{M_{\mathrm{Pl}_{3+1+\delta}}^{2+\delta}r^2 n^{\delta}} \Rightarrow </math>

\end{align}</math>

Thus the fundamental Planck mass (the extra-dimensional one) could actually be small, meaning that gravity is actually strong, but this must be compensated by the number of the extra dimensions and their size. Physically, this means that gravity is weak because there is a loss of flux to the extra dimensions.