- #1
dumbperson
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Hey,
I have not done any proper differential geometry before starting general relativity (from Sean Carroll's book: space time and geometry), so excuse me if this is a stupid question.
The metric tensor can be written as
$$ g = g_{\mu\nu} dx^{\mu} \otimes dx^{\nu}$$
and its also written as
$$ds^2 = g_{\mu\nu} dx^{\mu}dx^{\nu} $$
now let's say you have some metric ##ds^2 = A dt^2 + B dx^2##, in Sean Carroll, and various other places on the internet, they then manipulate this to obtain certain derivatives. For example, if ##ds^2 = 0##, they say that
$$A dt^2 + B dx^2 = 0 , \rightarrow \left(\frac{dx}{dt}\right)^2 = -\frac{A}{B} $$
How is this manipulation of the tensor justified? doesn't ##dx^2## actually mean ##dx \otimes dx## ? why are the elements just treated as numbers?
Thanks
I have not done any proper differential geometry before starting general relativity (from Sean Carroll's book: space time and geometry), so excuse me if this is a stupid question.
The metric tensor can be written as
$$ g = g_{\mu\nu} dx^{\mu} \otimes dx^{\nu}$$
and its also written as
$$ds^2 = g_{\mu\nu} dx^{\mu}dx^{\nu} $$
now let's say you have some metric ##ds^2 = A dt^2 + B dx^2##, in Sean Carroll, and various other places on the internet, they then manipulate this to obtain certain derivatives. For example, if ##ds^2 = 0##, they say that
$$A dt^2 + B dx^2 = 0 , \rightarrow \left(\frac{dx}{dt}\right)^2 = -\frac{A}{B} $$
How is this manipulation of the tensor justified? doesn't ##dx^2## actually mean ##dx \otimes dx## ? why are the elements just treated as numbers?
Thanks
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