What is a metric? [closed]
I was taking a basic course in general relativity. They introduced a concept of a metric which I wasn't able to understand can somebody explain it to me why do we need a metric in curved spaces?
relativity metric-tensor
closed as too broad by user36790, Sebastian Riese, Prahar, Kyle Kanos, user10851 Feb 25 '16 at 2:59
$\begingroup$ Have a look at this question $\endgroup$ – ACuriousMind♦ Feb 24 '16 at 13:04
$\begingroup$ theoreticalminimum.com/courses/general-relativity/2012/fall $\endgroup$ – user73352 Feb 24 '16 at 17:52
why do we need a metric
Clocks don't measure time, they measure the metric applied the the worldline of their path in 4d spacetime. Rulers don't measure distance, they measure the metric along their path in 4d spacetime. Everything you are used to thinking of as a measurement actually measures the metric.
That is why you need it. Because that's what clocks and rulers measure and you want to connect to measurements at some point (physics is an experimental science after all).
So you could ask why you didn't know you needed it. And that's because if you write it as a matrix, then in everyday situations it is very similar (numerically) to the identity matrix.
Which allows you to ignore when you use it, which allows you to confuse lots of things. For instance you could take a directional derivative. Or you could write a vector (the gradient vector) that when dotted (scalar product) with a unit vector tells you the directional derivative in that direction. But that dot product (scalar product) uses the metric. And if you use non cartesian coordinates you'd already have to pay attention. But there is a natural object that takes vectors and gives you directional derivatives and that's the gradient. The gradient vector is the gradient with the metric applied to make it into a vector when really its a function that takes vector and gives directional vectors.
And this is the really issue in relativity. You have to unlearn. You have to find out that thingsvyouvdivide thought were the same (gradient and gradient vector) are actually different. Like forces. Does $\vec F=m\vec a$ or does $\vec F=\mathrm d\vec p/\mathrm d t$? In Newtonian mechanics $m\vec a=\mathrm d\vec p/\mathrm d t$ so it didn't matter which one equally $\vec F$ but in relativity $m\vec a\neq \mathrm d\vec p/\mathrm d t$ so they can't both equal the same thing. And so you have to learn that forces are about changes in momentum, not about mass times acceleration. That's the real difficulty of learning Relativity. And so now you have to learn to distinguish between column vectors like the gradient vector and row vectors like the gradient.
If you write the metric as a matrix $G$ then the relationship between a column vector $c$ and a row vector is naturally $r$ but if you have two column vectors $a$ and $b$ you can compute $(Ga)^Tb.$ And that's a scalar. So the row vector associated with a column vector $a$ is not $a^T$ it is $(Ga)^T.$ But why would you even think it is $a^T.$
The only problem is the matrix $G$ is sometimes an identity matrix, in fact in a Euclidean space with Cartesian coordinates it is the identity matrix. So you might have thought $a^T$ and $(Ga)^T$ are the same since you practiced a situation where they are the same. And since that was the first one you learned, you might not notice using it when it's the identity matrix. In non Cartesian coordinates it isn't the identity matrix. So you can practice with non Cartesian coordinates and get used to it. Learning the difference between a change in coordinates and an actual metric distance.
For instance if you use polar coordinates $r$ and $\theta$ then the $\theta$ doesn't even have units of distance. Clearly a coordinate difference isn't giving you something measured by a ruler. For that you need the metric.
TimaeusTimaeus
A metric is a distance function for a space. It takes two space(time)-coordinates and gives the distance between them. We usually consider points which are infinitesimally close to one another, because that way it applies to every kind of space, curved or not. In a 2D Euclidean space, the metric is given by pythagoras' theorem: $ds^2(dx_\mu)=dx^\mu dx_\mu=dx^2+dy^2$. In the flat 1+1D Minkowski space, $ds^2(dx_\mu)=dx^\mu dx_\mu=c^2dt^2-dx^2$.
In both these cases the absolute values of the coordinates do not matter, only their differences are used to calculate the distance. When a space is curved, this changes. The distance function changes as we move along the space, due to the change in curvature. This dependency is captured in the metric tensor. We have in general $$ds^2(x_\mu,dx_\mu)=g^{\mu\nu}(x_\mu)dx_\mu dx_\nu\equiv dx^\mu dx_\mu.$$
The previous two flat spaces are also described by this general equation. For the Euclidean space, the metric tensor is the 2x2 unit matrix. For the Minkowki spacetime, the metric tensor is a 2x2 diagonal matrix with 1 in the topleft and -1 in the bottomright. In both cases they are constant matrices, independent of the absolute coordinates. The metric tensor is always there, but it is only in a space with non-constant curvature where it changes from one place to another.
vosovvosov
Pythagoras' theorem is an example of a metric. Notice that this theorem is postulated rather than derived (in fact, it defines flat 2D space). Notice also that (except for very short distances) it isn't valid on a globe, where a different rule (to calculate distances) needs to be specified instead.
benjiminbenjimin
$\begingroup$ You're making confusion between the metric and the application of the metric tensor onto two vectors (which gives back Pythagoras' theorem). $\endgroup$ – gented Feb 24 '16 at 13:46
It is a way to quantify the behavior of the space, whether it is flat or curvilinear.
$ ds^{2} = g_{\mu \nu} \triangle x_{\mu} \triangle x_{\nu} $
Where ds is the interval, g is the metric and the x's are the co-ordinates. As it has been said the x's can be drawn analogous to pythagoras in euclidean space, and the metric is a machine that translates that to curved space, if it is so.
EDIT: $ds^{2} $ - corrected by Lewis Miller.
CppgCppg
$\begingroup$ Don't you mean $ds^2$? $\endgroup$ – Lewis Miller Feb 24 '16 at 17:04
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