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I'm apologizing for such simple question, I just started to learn string theory. If one end of an open bosonic string is attached to a Dp-brane it means that in directions normal to the brane Dirichlet b.c. are satisfied, and in tangential directions Neumann b.c. are satisfied. Each textbook I checked says that an end point of a string is just pined to the brane, but such descriptions ignores Neumann b.c. in tangential directions. It looks to me that Neumann conditions mean that the string is orthogonal to the D-brane. Is this right?

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The endpoint is pinned in the sense that it can only move in the world volume of the brane. A good visual, I think, is a normal string with a small loop at the end that is tied loosely around a fixed rod. The string moves freely in the direction parallel to the rod, but is fixed in the directions perpendicular to it. This would be like a D1 brand.

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  • $\begingroup$ You saying that there are no Dirichlet conditions in tangental directions, that is just part of the definition. But I'm asking: what second part of the definition (Neumann b.c. in tangential directions) imply? $\endgroup$ – Sasha Jan 24 '15 at 15:27
  • $\begingroup$ Sorry, I misread your question. The other BCs are connected to the fact that no momentum can flow out of the string in that direction, just like for a normal household string. $\endgroup$ – lionelbrits Jan 24 '15 at 15:35
  • $\begingroup$ Yes, I agree with this, but if we don't look at momentum and just look only on coordinates, is not it also true that Neumann's BC's in tangential directions imply orthogonality as I claim? $\endgroup$ – Sasha Jan 24 '15 at 16:43
  • $\begingroup$ Yes, that is correct. Page 129 of zwiebach, if you have it. $\endgroup$ – lionelbrits Jan 24 '15 at 23:29

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