Using the wedge product one can pair the generators the Clifford algebra $Cl_{1,3}(\mathbb{R})$ to produce 2-vectors (area elements).

The Nambu-Goto action is a statement on the evolution of invariant area elements of the space-time metric.

Is it appropriate to use these 2-vectors to formulate the Nambu-Goto action.

The possible 2-vectors of $Cl_{1,3}(\mathbb{R})$ are:

$$ \gamma_0\gamma_1,\; \gamma_0\gamma_2,\; \gamma_0\gamma_3,\; \gamma_1\gamma_2,\; \gamma_1\gamma_3,\; \gamma_2\gamma_3 $$

The first three bi-vectors are wedge products involving time $t$ wedge a position $x,y$ or $z$. Intuitively they appear relevant to the Nambu-Goto action.

Can we obtain the Nambu-Goto action using the Lagrangian of an invariant sum of area elements:

$$ S=\int \mathrm dt\, \mathrm dx\, \mathrm dy\,\mathrm dz \sqrt{\left( A(t,x) \gamma_0 \wedge \gamma_1 \right)^2 + \left( B(t,y) \gamma_0 \wedge \gamma_2 \right)^2 + \left( C(t,z) \gamma_0 \wedge \gamma_3 \right)^2} $$

Or, perhaps one should use another combination of 2-vectors?

  • $\begingroup$ What is the square root of a Clifford matrix, what is the motivation for randomly sticking in Clifford elements into a square root of a well-defined quantity, what would happen if you did this in a QFT context for a point particle. $\endgroup$
    – bolbteppa
    Aug 25, 2019 at 19:27

1 Answer 1


OP considers a field theory in 3+1D, while the Nambu-Goto string action is a field theory in 1+1D. OP probably has a $p$-brane action $$ S_p~=~-T_0\int d^{p+1}\sigma ~\sqrt{-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta}} $$ in mind. FWIW, as long as spinor variables doesn't enter the theory, it seems likely that Clifford calculus and gamma matrices are not needed.


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