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.


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 $$ n_b = \gamma_{ab}n^a \to e^{2\omega} \gamma_{ab} e^{-\omega} n^a = e^{+\omega} \gamma_{ab}n^a = e^\omega n^b$$

  • $\begingroup$ 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? $\endgroup$ – LorentzNoether Aug 31 '14 at 16:04
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 1 '14 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.