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I have a question about brane polarisation. In the most famous cases I know of--the original giant D2 brane of Myers, the Polchinski-Strassler resolution of the $\mathcal{N}=1^*$ IR singularity, and the KPV metastable state corresponding to an NS5 brane carrying anti-D3 charge, the lower dimensional brane charge carried by the blown up brane is described by a magnetic field strength. For example, in the original Myers calculation, the D0 charge of the D2 brane is given by $\int_{S^2} F \propto N$.

I'm confused because in 1006.3587 Klebanov and Pufu essentially repeat the KPV analysis for an M-theory version of the Klebanov-Strassler solution. Here they add an electric field strength, $F_{01} \propto \mathcal{E}$. They calculate the brane potential in terms of the electric field $\mathcal{E}$, and then preform a Legendre transformation to get the potential in terms of the anti-F1 charge (which they call $p$).

Why do they take a different approach from the other examples of the Myers effect? Are the two equivalent, or different for an important reason that I don't understand?

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The electric field, rather than the associated (string) charge enters into the action. This means that the action is not the Hamiltonian, and does not give you the potential energy for a static configuration.

The electric field is like a velocity $\dot q$. To get the Hamiltonian, you have to consider the Legendre transform of the Lagrangian $$ H = p \dot q- L(q,\dot q) , $$ with $p = dL/d\dot q$ the conjugate momentum.

Same goes for a Lagrangian depending on $\cal E$ $$ H = \Pi {\cal E}- L({\cal E}) $$ with $\Pi = dL/d{\cal E}$.

In the earlier examples of the Myers effect you are referring to, the action, after plugging in the static ansatz, does not depend on any velocity or electric field and one finds simply that $H = -L$.

For more information, see for instance the original supertubes paper http://arxiv.org/abs/hep-th/0103030, where they also need such a Legendre transform for the (somewhat related) supertube effect.

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