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The geodesic curvature is given by

$$k=\pm t^a n_b\nabla_a t^b,$$

where $t^a$ is a unit vector tangent to the boundary of the string worldsheet and $n_a$ is an outward vector orthogonal to $t^a$. I don't understand why under the Weyl transformation

$$\gamma_{ab}\rightarrow e^{2\omega}\gamma_{ab},$$

$t^a$ and $n_b$ transform as $$t^a\rightarrow e^{-\omega}t^a,~~~~~n_b\rightarrow e^{\omega}n_a.$$

Is this really as trivial as a normalisation? Also, what do "time-like boundary"$(+)$ and "space-like boundary"$(-)$ mean? I appreciate any discussion related to this. The geodesic curvature was somehow never mentioned in my GR class.

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1 Answer 1

up vote 4 down vote accepted

When we say that they are unit vectors, we mean that the proper length is equal to one. The proper lengths of the two vectors are $$\gamma_{ab} t^a t^b=1,\quad \gamma_{ab}n^a n^b=1$$ and should be equal to one, i.e. $1\to 1$, at all times. (In the Minkowski signature, one of these squared lengths is minus one, but that won't change anything about the text below.) Because $$\gamma_{ab}\to e^{2\omega} \gamma_{ab}$$ but the sum of products has to remain one, it's clear that this extra $\exp(2\omega)$ factor has to be canceled, and we need $t^a\to e^{-\omega}t^a$. We pick two such factors. $n^a$ transforms in the same way, $$n^a \to e^{-\omega} n^a$$ but if we lower the indices, we have $$ n_b = \gamma_{ab}n^a \to e^{2\omega} \gamma_{ab} e^{-\omega} n^a = e^{+\omega} \gamma_{ab}n^a = e^\omega n^b$$

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Thank you. Could you please tell me what they mean by "time-like boundary" and "space-like boundary" which correspond to the $+$ and the $-$ in $k$ for string worldsheets? –  LorentzNoether Aug 31 at 16:04
    
A timelike boundary is a boundary that is a timelike curve i.e. with $ds^2\gt 0$ for each infinitesimal segment, in some conventions, the spacelike one is spacelike. I am totally confident that the terms are self-explanatory. They're not realy new terms. They are combinations of 2 words you should have known since undergrad relativity and basic school geometry, respectively. –  Luboš Motl Sep 1 at 4:09

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