Recent content by Alain De Vos

The speed of star on the outer of are solar system is not according to the visible mass. Change the law of gravitation F = 1/r^2 to fix this. Galaxies far away are moving away with increasing speed. Change the law of gravitation F = 1/r^2 to fix this, include a repulsive force to fix. Would this...

Hey, I'm interested in temperature recordings for the last 10 years but also for an estimate in the last billion years, this for hobby and fun. There are a lot of public sources but problem is many sites are torn down and others are created, so I feel very in the wild, it's like finding a tree...

What is the covariant derivative of the position vector $\vec R$ in a general coordinate system? In which cases it is the same as the partial derivative ?

Most processors have 4 rings. Why do operating systems only use 2. Historical reason ? Overhead ? There is no need for more than two rings ?

The basis in this tangent space is then : e(μ)=∂P/∂xμ

So all points in curved spacetime form a manifold but not a vectorspace. It is the collection of all tangents to all curves going through a specific point on this manifold which form a vectorspace. (I must have been thinking 3-D where points and vectors can be interchanged)

Currently reading Carroll spacetime and geometry page 16 and 17. Still unclear for me why tangent vector to a curve does not include change of basis-vectors as lambda changes. I.e. the connection coefficients. Anyone who can shed a light on this very basic issue ?

I wander, which four vectors do you have to choose for the tensors to operate upon in order for the resulting scalar to have an physical meaning. Does it work on any two covariant vectors, only two and the same displacement vectors, the 4-speed vector ?

Exactly Peter, this was my question, I'm currently reading Schutz, afterwards i'll raid Carroll.

Can these tensor be seen as operators on two elements. So given two elements of something they produce something, for instance a scalar ?

I was reading a book by John Dirk Walecka, but his notation confused me, I switched to Lambourne. I have a comparable question in the formula for geodesics. Why is the formula for geodesics a covariant derivative of a standard derivative (being the tangent)? Why is it not a covariant derivative...

With coordinates q en basis e ,textbooks use as line element : ds=∑ ei*dqi But ei is a function of place, as one can see in deriving formulas for covariant derivative. Why don't they use as line element the correct: ds=∑ (ei*dqi+dei*qi) Same question in deriving covariant derivative,

I found out it was just because they dried out.

I learned gradient in 3D space. And gradients where always vectors, pointing in the direction of steepest ... and normal to the surface where the functions is constant. But reading one-forms , a gradient of a function is not always a vector and it has something to do with metric... Can you proof...

Is it because the bloodcells are dried out or because they are attacked by viruses?