# Coordinate independence in spacetime

As far as I know, the $$n$$ coordinates $$(x_1, x_2, ..., x_n)$$ chosen to describe an $$n-$$manifold have to be mutually independent $$\to$$ the mutual derivatives must equal $$0$$ (for example, $$\frac{dx_1}{dx_2} = 0$$).

Anyway, it seems to me that in Minkowski spacetime (coordinates: $$x_1, x_2, x_3, ct$$), there is indeed a dependence between any one of the "spatial" coordinate and the "temporal one" (I simply note that for example, $$\frac{dx_1}{dt}$$ is the "classical" velocity in $$x-$$direction, which is clearly not $$0$$ in general).

So why can this set of coordinates be used to describe a manifold?

• The semi-Riemannian manifold - or Minkowski space - is coordinate independent - in other words, it allows for any arbitrary coordinate system. It's the required physical properties which determined the interval or the metric - or the differential structure on the manifold - snd the treating of time on the same footing as space. – Cinaed Simson Aug 15 '19 at 22:24

Your confusion arises from mixing two notions of a derivative. In particular, a partial derivative and a total derivative along a curve. As you note, as such, there is obviously no codependence between two different coordinates. This means that if you move purely in the direction of one of the coordinates on a manifold, all the other coordinates remain constant. Mathematically, this is expressed in the form of a partial derivative, i.e., we write that $$\frac{\partial x^i}{\partial x^j}=\delta^i_j$$. However, when we talk about motion of a particle in physics, we are talking about a specific curve on the manifold. So, we are not asking as to how much does a spatial coordinate change when we move purely in the direction of the time coordinate. Rather, we are asking as to how much does a spatial coordinate change when we advance the time coordinate by a certain amount along a specific curve. Mathematically, this is expressed in the form of a total derivative, i.e., we are interested in $$\frac{dx^i}{dt}$$ along a specific curve on the manifold. As you note, the value of this quantity depends on the curve we have at hand. In more physical terms, it depends on what kind of a motion we are describing. For example, if the curve (i.e., trajectory) we are talking about is described by $$x^i=2t^2$$ then along that trajectory $$\frac{dx^i}{dt}=4t$$, but, if we had a different trajectory at hand, the value of the total derivative along that trajectory would be something else. Notice that no matter which trajectory we want to analyze, as a general property of our coordinates, $$\frac{\partial x^i}{\partial t}$$ is always zero (where $$x^i$$ are the spatial coordinates).

Just to clarify my possible confusion. Is $$$$\frac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu$$$$ really a necessary condition for a coordinate system of a manifold? This condition looks like an orthogonality condition for me. For example, in two dimensional Euclidean space, let $$x,y$$ be Cartesian coordinates, one can freely choose $$x':=x$$ and $$$$y' := \frac{x+y}{2}$$$$ as the new coordinate system. However, we then have $$$$\frac{\partial y'}{\partial x'} = \frac{\partial }{\partial x}\frac{x+y}{2} = \frac{1}{2} \neq 0.$$$$ It looks like we just need the gradient of the coordinate functions on a manifold to be linearly independent mathematically, isn't it?