Question about the metric tensor

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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 lets 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? doesnt ##dx^2## actually mean ##dx \otimes dx## ? why are the elements just treated as numbers?

Thanks
 
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  • #2
Orodruin
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This is a typical example of physicists being a bit sloppy with notation. If you want to be more precise, you can look at a general 4-vector which is light-like and might be the tangent vector of some world line, such as a light-ray. Generally, this tangent vector will have the components
$$
V^\mu = \left(\frac{dt}{ds},\frac{d\vec x}{ds}\right),
$$
where ##s## is the curve parameter. The metric now gives you
$$
g(V,V) = A \left(\frac{dt}{ds}\right)^2 + B \left(\frac{d\vec x}{ds}\right)^2 = 0.
$$
If you use the time coordinate ##t## as the curve parameter, then
$$
\left(\frac{d\vec x}{dt}\right)^2 = -\frac{A}{B}.
$$
 
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This is a typical example of physicists being a bit sloppy with notation. If you want to be more precise, you can look at a general 4-vector which is light-like and might be the tangent vector of some world line, such as a light-ray. Generally, this tangent vector will have the components
$$
V^\mu = \left(\frac{dt}{ds},\frac{d\vec x}{ds}\right),
$$
where ##s## is the curve parameter. The metric now gives you
$$
g(V,V) = A \left(\frac{dt}{ds}\right)^2 + B \left(\frac{d\vec x}{ds}\right)^2 = 0.
$$
If you use the time coordinate ##t## as the curve parameter, then
$$
\left(\frac{d\vec x}{dt}\right)^2 = -\frac{A}{B}.
$$
Ah , thanks!

I don't understand why they don't just do it this way then.
 
  • #4
Orodruin
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I would say because the result is the same and it is intuitively what you would expect if you used dx and dt as small numbers. After all, there is a reason we use dx to denote a one-form. It is also much faster to give plausible arguments which build physical intuition than to dig too deep into the mathematics, at least in the beginning the mathematics of it all is not what you want to emphasise as a teacher - and this is based on my own experience, I do this myself when teaching, most students will not bother too much with it, but I am happy to discuss it in the breaks or after class with more mathematically inclined individuals.
 
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This is a typical example of physicists being a bit sloppy with notation. If you want to be more precise, you can look at a general 4-vector which is light-like and might be the tangent vector of some world line, such as a light-ray. Generally, this tangent vector will have the components
$$
V^\mu = \left(\frac{dt}{ds},\frac{d\vec x}{ds}\right),
$$
where ##s## is the curve parameter. The metric now gives you
$$
g(V,V) = A \left(\frac{dt}{ds}\right)^2 + B \left(\frac{d\vec x}{ds}\right)^2 = 0.
$$
If you use the time coordinate ##t## as the curve parameter, then
$$
\left(\frac{d\vec x}{dt}\right)^2 = -\frac{A}{B}.
$$
Hey again,

I reread what you said, and I don't get it. With respect to what basis is your given V?

$$g = g_{\mu\nu} dx^{\mu}dx^{\nu}$$

and with your vector $$V = (\frac{dt}{ds},\frac{dx}{ds})$$ (I removed the vector sign above the x, so this is just a simple 2d spacetime)

then we have

$$g(V,V) = A (dt\otimes dt)(V,V) + B(dx\otimes dx)(V,V) = A dt(V)dt(V) + B dx(V)dx(V) $$

I don't know how dx and dt act on V? unless V is expressed in the basis of partial derivatives. Am I not getting this at all ? :P

Thanks
 
  • #6
martinbn
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It is given with respect to the basis dual to the basis of one forms ##dx^\mu##.
 
  • #7
Orodruin
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It is given with respect to the basis dual to the basis of one forms ##dx^\mu##.
Which by definition is the basis of coordinate derivatives.
 
  • #8
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It is given with respect to the basis dual to the basis of one forms dx^\mu.
Ah I see, I guess I asked too quickly. I'm sorry. Is what I'm doing here correct?

$$ V= \frac{d\vec{r}}{ds} = \frac{\partial \vec{r}}{\partial t} \frac{dt}{ds} + \frac{\partial \vec{r}}{\partial x} \frac{dx}{ds} $$

(the partial derivatives are the basis vectors)

so

$$g(V,V) = A dt(V)dt(V) + Bdx(V)dx(V) = A V^0 V^0 + B V^1V^1 $$

where $$V^0 = \frac{dt}{ds} , V^1 = \frac{dx}{ds} $$
 
  • #9
bcrowell
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This is a typical example of physicists being a bit sloppy with notation.
No, it's not. It's simply physicists using the notation in the correct way that it was used for 300 years after the invention of calculus. For more on how this applies to differential geometry, with modern standards of rigor, see Nowik and Katz, Differential geometry via infinitesimal displacements, http://arxiv.org/abs/1405.0984 .

doesnt ##dx^2## actually mean ##dx \otimes dx## ?
No, that's Élie Cartan, circa 1930, hijacking the original notation and redefining it. You can use Cartan's redefinition if you like, but it's not mandatory, nor is there any difference in rigor between the two approaches. Using Cartan's approach has some disadvantages, one of which is that it causes the straightforward calculation in your OP appear illegal.

why are the elements just treated as numbers?
Because they are numbers. Infinitesimal numbers.
 
  • #10
andrewkirk
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No, it's not. It's simply physicists using the notation in the correct way that it was used for 300 years after the invention of calculus. For more on how this applies to differential geometry, with modern standards of rigor, see Nowik and Katz, Differential geometry via infinitesimal displacements, http://arxiv.org/abs/1405.0984 .
I'm not convinced that that absolves the author of the OP's text of the sloppiness charge. Infinitesimals is a very niche branch of mathematics, and hardly any students coming into a DG course would have encountered it. What they would have encountered is high-school and first year uni maths teachers telling them things like '##\frac{dy}{dx}## may look like a fraction, but it isn't one, so don't treat it as though it is a fraction unless you are sure that, in the context, that treatment is valid'.

So I feel the onus falls on the physics teacher or author to either make study of infinitesimals a prerequisite for the course (IMO unrealistic and unnecessary), or state that the manipulations they are performing are justified by a theory that the students are not expected to have studied (frustrating for the mathematically-inclined students), or give the justification for their manipulations like Orodruin's explanation above.
 
  • #11
bcrowell
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I'm not convinced that that absolves the author of the OP's text of the sloppiness charge. Infinitesimals is a very niche branch of mathematics, and hardly any students coming into a DG course would have encountered it.
Infinitesimals are the mainstream of mathematics. They have been since Leibniz and Newton. Anyone who took freshman calculus and physics without being exposed to infinitesimals was a victim of a second-rate education.
 
  • #12
andrewkirk
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Infinitesimals are the mainstream of mathematics.
I wonder if we're referring to the same thing. Are you referring to topics like the surreal numbers or the hyperreal numbers, which are only 20th century developments?

So far as I know, neither Newton nor Leibniz provided a rigorous mathematical procedure for treating ##\frac{dy}{dx}## as a fraction rather than a limit, and their approach to infinitesimals was criticised by Cantor, Dedekind and Weierstrass. But I'm happy to be shown to be wrong about that if a link is available.
 

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