Recent content by Jianbing_Shao

In calculus. the path integral of a vector field with non-zero curl along different paths between two points will get different results. it just means that the movement of a scalar field is path dependent. A vector only contains more components than a scalar. So we can define a path dependent...

Yes, I agree with your view. I think metric and connection are two different types conceptions. they can be defined independently. even in a space with a flat metric we can define a connection, then the movement of a vector is path dependent. so curvature is determined by the property of...

Then the structure of spacetime geometry can't be described using the language of 'infinitesimal action'？ But the connection field can describe the local infinitesimal change of a components of a vector when it parallel transport along a particular path? So can it be regarded as a definition of...

If we define an infinitesimal action at each point in flat spacetime. then a vector moves from a point to another along different paths, then the movement is possibly path dependent. If we observe the movement of a vector in a space and we are not sure if there exists a gravity field, then...

The presence of gravity will result in curvature of spacetime. and the effect of spacetime curvature is that the parallel transportation of a vector is path dependent. Then can I ask such a question: In a flat spacetime is it possible to define an action on a vector which guarantee the movement...

Thanks! Of course a connection includes monmetricity is a general connection, then I agree on your first conclusion. It is just what I mean, perhaps the difference between us is that you think the path dependence of the metric comes from nonmetricity. I think it perhaps a conbination of...

Thanks a lot! logically if we want to prove a property of path independence, we can prove that if we move along different paths we can get the same result, that is just what I am doing. Now if we start from your point, when you say the connection satisfy metric compatibility. compatibility...

So I can express my idea in such a way: From metric compatibility equation##\nabla_\mu g(x)=0##, if choose a particular path ##l##, we can construct a metric confined on path ##l## ##g(x)=P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)##, here ##g(x)## and ##\gamma## are compatible with each other...

Thanks! I just start from an arbitrary connection which includes ##Q\neq 0##. Formula (1)~(7) just the procedure in which I'm trying to prove the metric generated with parallel prapagator can satify the metric compatibility equation. and we write the equation(7) in the form of components ...

Thanks a lot! If we only take the rotation into account, it may be helpful, but it perhaps can not solve the problem. For example, from the result I mentioned above: there exist infinite metric field different from flat metric ##\eta## compatible with a zero curvature connection. then here are...

Thanks! Just because the parallel propagator is also called path ordered product, So when we talked about the parallel propagator, a particular path ##l## have been chosen at first. when I say "So ##\gamma_\mu## is compatible with metric ##g(x)##." I mean that metric ##g(x)## only well defined...

Thanks! So at first we all acknowledge that just as Sean Carroll pointed the parallel propagator ##P(x,x_0)## can solve the parallel transport equation. then if the curvature of the connection is not zero, the parallel propagator ##P(x,x_0)## is path dependent, it is just because this that...

Thanks! Here the logic is we can not get a field just result from path dependence, So we should express ##g## with ##\gamma## and prove that ##g## is path indenpendent for an arbitrary ##\gamma##.

Thanks a lot! But I can not agree with your conclusion. Sean Carroll using parallel propagator $$P(x,x_0;l;\gamma)= P\exp\left(\int_x^{x_0}-\gamma^\nu_{\mu\rho}(x) dx^\rho\right)$$ to describe the parallel trsport of a vector along a particular path ##l##. Simillarly we also can use parallel...

Thanks! Yes it is just what I want to say: Just because I from the discussion about partial differential equations, I find that in most cases, a global solution does not exist, we only can get a solution confined on a curve. So I doubt that to metric compatibility equation, we will get a similar...