Logical consistence in Neumann BC in the Nambu-Goto action

Background: In Zwiebach's A First Course in String Theory, 2nd ed. equation $$(6.50)$$ define a kind of momentum along the $$\sigma$$ direction on the worldsheet (Here $$\dot{X} = \partial_\tau X, \, X' = \partial_\sigma X$$):

$$P^\sigma_\mu = - \frac{T_0}{c} \frac{(\dot{X} \cdot X') \dot{X}_\mu - (\dot{X})^2 X'_\mu}{\sqrt{(\dot{X} \cdot X')^2 - (\dot{X})^2(X')^2 } } \tag{6.50}$$

In equation $$(6.56)$$ he states the Neumann BCs for a free open string, which are

$$P^\sigma_\mu(\tau, \sigma^*) = 0,\tag{6.56}$$ where $$\sigma^* = 0, \pi$$.

The problem: Equation $$(6.50)$$ is defined only in points $$(\tau, \sigma)$$ which correspond to a non-zero denominator. However, if equation $$(6.56)$$ is satisfied, then we have

$$(\dot{X} \cdot X') \dot{X}_\mu - (\dot{X})^2 X'_\mu=0. \tag1$$ The LFT of the above equation consists of one-form components. If we apply these to the vector field $$X' = X'^\mu \partial_\mu$$ we get

$$(\dot{X} \cdot X')^2 -(\dot{X})^2 (X')^2 = 0. \tag2$$ But this essentially the argument inside the denominator square root in $$(6.50)$$, hence we have a constradiction. What is the problem here? Am I missing some point?

edit: As @ɪdɪət strəʊlə has mentioned I've tried to see more precisely the denominator behavior. To say that the denominator is zero is to say that

$$\left[(\dot{X} \cdot X') \dot{X}_\mu - (\dot{X})^2 X'_\mu \right]X'^\mu (\tau, \sigma) = 0 \hspace{0,5cm} \text{at} \, \sigma=0,\pi.\tag4$$ If I take the limit $$\sigma \to 0$$ or $$\sigma \to \pi$$, this tell me only the behavior of the numerator when acting on $$X'^\mu$$, not its behavior in general. How am I suppose to get the hole behavior of the determinant at the boundaries without getting the same contradiction?

Of course, I also know that

$$P^\sigma_\mu X'^\mu = \sqrt{(\dot{X} \cdot X')^2 - (\dot{X})^2 (X')^2} \tag5$$

goes to $$0$$ as the expression inside quare root goes to zero, so I could substitute the value of the map by its limit. But this is a specific situation of $$P^\sigma$$ applied to the vector $$X'$$. The boundary condition should be applied for the hole $$P^\sigma$$.

Eq. (6.50) is also defined at points where the denominator is zero, so long as the limits exist. In this case they always exist. You can easily show, using your observation that the denominator is the square root of the numerator contracted with $$X'$$, that the numerator always goes to zero at least as fast as the denominator.
• But I can guarantee that the numerator decrease faster than denominator only if I apply the numerator do $X'$. How do I prove for any admissible vector field? Commented Jun 5, 2023 at 15:52
• @Geni you don't need to apply anything, $P=F/\sqrt{F\cdot G}$, see when the denominator vanishes and convince yourself that the numerator must vanish at least as fast Commented Jun 5, 2023 at 16:45
• Sorry but I only can convince myself of this when the numerator is applied to $X'$. If you could provide some more detailed sketch of a solution it would be great. Commented Jun 5, 2023 at 18:29
• ANother solution I though: I could state that the boundary conditions are $P^\sigma _\mu \to 0$ as $\sigma \to 0^+$ (same for $\sigma \to \pi⁻$). Since the vector field components $X'^\mu$ are continuous, the product of the limits is the limit of products and then we'd have $\lim_{\sigma \to 0⁺} P^\sigma _\mu X'^\mu = 0$, which removes the aparent contradiction of my question. Besides that I could not get anything for your hints. Commented Jun 6, 2023 at 18:17