Recent content by Peterdevis

Just like Einstein in 1905 you don't know anything about GR. But in difference with you Einstein understands that when you only can deal with inertial frames (where there is no good definition for) you' ve got to deal with a lot of paradoxes. He bypassed this problem by inventing GR. So here is...

No, at the end K and L run at the same rate but clock K(pe 11:55) runs after on L (12:00). J will run slower then K and L not (only) because of its velocity but because the clock is constant accelerating (making a orbit). So there is no paradox!

May be you don't know but Einsteins Paper from 1905 was only the beginning of the formulation of a whole new theorie. Einstein realized that there where fundamental problems with SR. One of the fundament of SR are the inertial frame and the Lorentztransformation between this sort of frames...

No, because you can't decide which clock is in motion. Motion is always relative to someting. So for clock A clock B moves with a velocity B and for clock B clock A moves with velocity v When you deal with accelleration, your clocks are not longer connected with an inertial frame and so you...

Maybe it's usefull to explain why scientist has choosen differential geometry to describe GR before running to deep into the math. 1) Einstein (and others) recognised that time was not a parameter (Newtonian view) but a (mathematical) dimension, just like the three spatial one's. So the...

The main reason I don't like relativistic mass (and yes it's a personal flavour if you like the term or not) is that nature is coordinate independent. So a physicist has to explain nature in a coordinate independent way and try to avoid coordinate dependent properties. A good text about the...

to set things right: In fact it does! The energy-momentum tensor contains al energy (except gravitational) en influence the curvature.

I agree that a manifold is a topological space (by introducing open sets) that locally lokes like the euclidian space. In my opinion you introduce (if you want it ar not) the standard metric with this last restriction(lokes like the euclidian space) in the topological space. I agree that at...

So if i understand it correct: When a ferromagnetic object comes in a magnetic field, the lowest energy state is the stat where all magnetic spins point out the same direction. So you don't need energy, in fact the object gives energie away. So where do tehe energy go to? Do you know...

In a dutch forum I'm in discussion with a crackpot about extracting energy out of a magnet. He claims that when a iron object enter the range of a permanent magnet it need energie to become ferromagnetic. Is this really so and if it is where comes the energy from?

This is correct. So you can define a large number of connections on a manifold and each of them implies a distinct notion of covariant differentiation. When there is a metric you can define one unique connection that is torsion free and metric compatibele: [p q, r] =...

Maybe, I must reformulate my point of view. As physisist I'm interested in a description of reality independent of a reference system. So i organise "events" (x,y,z,t) as a manifold and now i put one every event a container (I called this a tensor, but this confuse everybody so i call it now a...

I think the curvature tensor let us see that covariant en contravariant components refers to the same object. You can describe the curvature by a (1,3)tensor but the (0,4) tensor contains the same information about the curvation of the manifold

I don't think you have read the link http://www.mathpages.com/rr/s5-02/5-02.htm. When you look at the drawing you see the object P and his covariant en contravariant expression. So one object two expressions!

The question is: G = g_{ij} dx_i \otimes dx_j =g^{ij} \partial_{i} \otimes \partial_{j} ?