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I am following Carroll on general relativity. In section 7.3 he introduces the notation $\vec{a}$. I know perfectly well what a 4-vector $a^\mu$ is in GR. $a^\mu$ are the components of the 4-vector $a$ in the tangent space at the coordinate $x^\mu$ on the manifold. But what is meant by $\vec{a}$? Is this just what an observer at the point $x^\mu$ if he tried to measure $\vec{a}$ at this point?

Also, is it true that $g_{ij}a_i a_j = |\vec{a}|^2$ would be the square magnitude an observer at $x^\mu$ would measure? Or perhaps $\eta_{ij}a_i a_j = |\vec{a}|^2$?

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  • $\begingroup$ A 3-vector is the spatial part of a 4-vector. $\endgroup$
    – knzhou
    May 1 '17 at 19:32
  • $\begingroup$ What's $|\vec{a}|$? $\endgroup$
    – Mikkel Rev
    May 1 '17 at 19:34
  • $\begingroup$ It's $\sqrt{g_{ij} a_i a_j}$, just like in freshman calculus. $\endgroup$
    – knzhou
    May 1 '17 at 19:47
  • $\begingroup$ If you write these two sentences into a reply, I will accept it as the answer $\endgroup$
    – Mikkel Rev
    May 1 '17 at 19:52
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As you have correctly realized, ${\vec a}$ is the three-vector formed out of the spatial part of a four-vector. This means that the magnitude of this three-vector, i.e., the scalar product of the three-vector with itself, should be defined as $g_{ij}a^ia^j$ and NOT as $\eta_{ij}a^ia^j$.

A scalar quantity in the sense of a three-vector means that the quantity should be invariant under all the transformations that do not have a boost component to it. This means that $a^2$ is the value of the "length" of the spatial part of a four-vector quantity and remains the same as long as it is measured from the frames related to each other through transformations that do not involve a boost.

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