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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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he says a 4 vector is defined as the tangent vector to this curve

That is not true in general. A four-vector is not always defined as the tangent vector to a curve. In the book they are computing a tangent vector to a curve in 3+1D spacetime; the tangent vector is just one example of a four-vector.

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 $\lambda = 0$.

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A four vector is always the tangent vector to some curve. Why do you say no? In fact, this is a popular definition of a vector--- a vector at x is an equivalence class of curves through x which do not separate linearly as the parameter increases. –  Ron Maimon Nov 1 '11 at 2:22
    
In the sense that you can always have a curve (or class of curves) to which any given 4-vector is tangent, then yes. But in many cases it's completely useless as an operational definition, because there's nothing interesting about the curves corresponding to the vector in question. –  David Z Nov 1 '11 at 2:35
    
thanks david... i'm wondering why the interval is [0.1]? but this is helpful –  Bozostein Nov 1 '11 at 20:10
    
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
    
this is analgous to choosing a vector with a tail at the origin in 3-d euclidean space? thx –  Bozostein Nov 1 '11 at 20:50
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