The Dirac-Born-Infeld Action, D-Branes and Spacetime Symmetries

By George Shiber Wednesday, December 23, 2015Monday, April 8, 2019 In Cosmology, Grand Unification Physics, Loop Quantum Gravity, M Theory, Mathematical Physics, Philosophy Of Physics, Physics, Super String Theory, Supersymmetry Branes, calabi yau manifold

Ex nihilo nihil fit ~ Parmenides

In this post, in order to solve the Wheeler-deWitt problem-of-time I discussed in my last post, and since string/M-theoretic braneworld cosmology offers the only solution in a unified C*-Heisenberg-algebraic sense, I must first analyse a field theory on a D-brane sigma world-volume and study the Dirac-Born-Infeld (DBI) action. Let me start by giving a proof that the DBI action has a symmetry under the transformation

$\left\{ {\begin{array}{*{20}{c}}{\delta {A_a} = {\Lambda _a} - {\varepsilon ^c}{{\not F}_{ca}} + {\Lambda _k}{{\not \partial }_a}{\Phi ^k}}\\{\delta \,{\Phi ^i} = {\varepsilon ^i} - {\varepsilon ^c}{{\not \partial }_c}{\Phi ^i}}\end{array}} \right.$

of full

$Diff\left( M \right){\dot \propto ^\dagger }\Omega _{closed}^2\left( M \right)$

on a generalized Riemannian structure ${C_ + } \subset \Im M$

First, I must prove the following relation

${\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}$

with $g$ the Riemannian metric on $\Im M$

$g \in {\Im ^ * }M \otimes {\Im ^ * }M$

and $\not S\widehat {\not F}$ the metric on ${\not L_{\widehat {\not F}}}$

$\not S\widehat {\not F} \in {\not L^ * }_{\widehat {\not F}} \otimes {\not L^ * }_{\widehat {\not F}}$

Now, since $\varphi _\Phi ^ *$ is the pullback of the embedding map defined by the field $\Phi$ , we have

${\varphi _\Phi }:\underbrace \to _{pullback}M$

Thus, the determinants on the l. h. s. are those of the $D \times D$ matrices, while the determinant on the r.h.s. is that of the $\left( {p + 1} \right) \times \left( {p + 1} \right)$ matrix distinguished by the index $ab$ . Now, the r. h. s. of

${\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}$

is the Lagrangian ${L_{DBI}}$ of the Dirac-Born-Infeld action and can be proven by combining the following relations of various determinants

1) Let $s$ be the symmetric part of $t$ given by

$\left\{ {\begin{array}{*{20}{c}}{{t^{ij}} = {E^{ij}}}\\{t_a^j = - {E_{ak}}{E^{kj}}}\end{array}} \right.$

and

$\left\{ {\begin{array}{*{20}{c}}{t_b^i = {E^{ik}}{E_{kb}}}\\{{t_{ab}} = {E_{ab}} - {E_{ak}}{E^{kl}}{E_{lb}}}\end{array}} \right.$

and ${t^{ij}} = {E^{ij}}$ the inverse matrix of ${E_{ij}}$ , then we have

$\det s = \det g{\left( {\det {t^{ij}}} \right)^2}$

2) From

$\begin{array}{l}\not S\widehat {\not F}{ = _{{\rm{def}}}}s - \left( {a - \widehat {\not F}} \right){s^{ - 1}}\left( {a - \widehat {\not F}} \right)\\ \in \Gamma \left( {{L^ * } \otimes {L^ * }} \right)\end{array}$

and the definition of $\not S\widehat {\not F}$ , we have

$\det s\,{\rm{det}}\,\not S\widehat {\not F} = {\left( {\det {t_{\widehat {\not F}}}} \right)^2}$

3) Now, by using the explicit expression for ${t_{\widehat {\not F}}}$ , it follows that

${\rm{det}}\,{t_{\widehat {\not F}}} = \det {t^{ij}}{\rm{det}}{\left( {\varphi _\Phi ^ * \left( {g + B} \right) - F} \right)_{ab}}$

Using these relations in 1), 2), 3), one can prove the representation of the DBI action given in

${\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}$

which is

$\begin{array}{l}{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widehat {\not F} = \frac{1}{{\sqrt {\det {t^{ij}}} }}{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}s\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widehat {\not F}\\ = \frac{1}{{\sqrt {\det {t^{ij}}} }}\sqrt {\det {t_{\widehat {\not F}}}} = \sqrt {{\rm{det}}{{\left( {\varphi _\Phi ^ * (E) - F} \right)}_{ab}}} \end{array}$

The integral of the scalar density agrees with the DBI action

${S_{DBI}} = \int_{\varphi \Phi \left( \Sigma \right)} {\sqrt {{\rm{det}}\left( {\varphi _\Phi ^ * {{\left( {g + B - F} \right)}_{ab}}} \right)} } d{x^0} \wedge ... \wedge d{x^p}$

when evaluated on the leaf $\varphi \,\Phi \left( \Sigma \right)$ of ${L_{\widehat {\not F}}}$ at ${x^i} = {\Phi ^i}(x)$ , hence, the DBI action is invariant under the world-volume diffeomorphism on the D-brane.

Now I am in a position to prove that the DBI action is not only invariant under the world-volume diffeomorphism but also under the full target space diffeomorphism and the B-field gauge transformation. Towards this end,

${S_{DBI}} = \int_{\varphi \Phi \left( \Sigma \right)} {\sqrt {{\rm{det}}\left( {\varphi _\Phi ^ * {{\left( {g + B - F} \right)}_{ab}}} \right)} } d{x^0} \wedge ... \wedge d{x^p}$

can be re-written as an integral over the target space $M$ as

${S_{DBI}} = {\int_M {\widehat {\not L}} _{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)d{x^0} \wedge ... \wedge d{x^{D - 1}}$

where

${\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)$

is a Dirac delta function interpreted as a distribution along ${x^i}$ directions. The infinitesimal transformation of the full diffeomorphism and the B-field gauge transformation are parametrized by $\varepsilon + \Lambda$ . So the transformation of the integrand ${\widehat {\not L}_{DBI}}$ is gotten from that of $\det g$

$\begin{array}{l}\delta \det g = - \,{\varepsilon ^M}{{\not \partial }_M}\det g + \det g\\ \cdot \left\{ { - 2{{\not \partial }_M}{\varepsilon ^M}} \right\}\end{array}$

and that of $\not S\widehat {\not F}$

$\begin{array}{l}\delta \det \not S\widehat {\not F} = - \,{\varepsilon ^M}{{\not \partial }_M}{\rm{det}}\,\not S\widehat {\not F} + {\rm{det}}\,\not S\widehat {\not F}\\ \cdot \left\{ { - 2{{\not \partial }_c}{\varepsilon ^c} + 2{{\not \partial }_k}{\varepsilon ^k} - 4{{\widehat {\not F}}^k}_c{{\not \partial }_k}{\varepsilon ^c}} \right\}\end{array}$

And the glorious result is

$\delta {\widehat {\not L}_{DBI}} = - \,{\varepsilon ^M}{\not \partial _M}{\widehat {\not L}_{DBI}} - \left( {{{\not \partial }_c}{\varepsilon ^c} + {{\not \partial }_c}{\Phi ^k}{{\not \partial }_k}{\varepsilon ^c}} \right){\widehat {\not L}_{DBI}}$

with

${{{\not \partial }_c}{\varepsilon ^c} + {{\not \partial }_c}{\Phi ^k}{{\not \partial }_k}{\varepsilon ^c}}$

being central to full diffeomorphism. This is the expected result since ${\widehat {\not L}_{DBI}}d{x^0} \wedge ... \wedge d{x^p}$ is a section of ${\rm{det}}\left( {{\Delta ^ * }} \right)$

On the other hand, the delta function transforms as

$\begin{array}{l}\delta \left[ {{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] = - \,{\varepsilon ^M}{{\not \partial }_M}\\ \cdot \left[ {{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] - \\\left( {\not \partial _k^M{\varepsilon ^k} - {{\not \partial }_k}{\varepsilon ^c}{{\not \partial }_c}{\Phi ^k}} \right){\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)\end{array}$

Now, combining, we get

$\begin{array}{l}\delta \left[ {{{\widehat {\not L}}_{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] = \\ - {{\not \partial }_M}\left[ {{\varepsilon ^M}{{\widehat {\not L}}_{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right]\end{array}$

as desired, hence, the transformation of the integrand in the DBI action

${S_{DBI}} = {\int_M {\widehat {\not L}} _{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)d{x^0} \wedge ... \wedge d{x^{D - 1}}$

is a total derivative and the DBI action is invariant under full target space diffeomorphisms and B-field gauge transformations and the invariance within the static gauge is what will allow us to further analyse the Wheeler-deWitt problem-of-time.