# why evaluate at lambda = 0

I am trying to understand Herbert Goldstein's introduction to 4-vectors. He describes a 1-D curve in spacetime $P_(\lambda)$ then he says a 4 vector is defined as the tangent vector to this curve $$v = \biggr ( \frac {dP} {d\lambda}\biggr)_{\lambda =0}$$

why is $\lambda$=0? what does that have to do with anything? I have been staring at this for like 20 minutes I still don't understand what he is talking about... it's giving me problems because i need to understand this part later because it is relevant to how tensors transform also he says $\lambda$ is a measure of a length along the curve... i don't really follow that point either... i though $\lambda$ could be any parameter like proper time etc.

any help on this??

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In particular, the formula given tells you how to compute the tangent vector at a specific point $\mathcal{A}$. Since the curve runs from $\mathcal{A}$ to $\mathcal{B}$ and is parametrized by $\lambda \in [0,1]$, $\lambda = 0$ is the value which corresponds to the point $\mathcal{A}$. So if you're going to define the tangent vector at $\mathcal{A}$, you need to set $\lambda = 0$.
You have to pick some interval for the parameter, and $[0,1]$ is just easy to work with. – David Z Nov 1 '11 at 20:21