Recent content by bres gres

thank for correcting my mistake. i should said it is not "ordinary" but "normal-type derivative along the curve"(i mean they don't involve the derivative of basis vector ,therefore there is no connection coefficient involved.

i see. However the problem is not solved yet. it seems the presenter is not correct to set ##\ddot{S^{u_j} _j}= [∇v∇vS]^{uj}## as they are different object.(i am not sure??may be he is correct) the former one is just part of the product: ##\dfrac{dV}{d\lambda} =\epsilon_{ijk}...

Due to my unclear expression in the first post,i might post my question again here.Hope everyone can read it clear sorry about that In this video,the presenter tried to relate the ordinary derivative of the Volume to the Ricci tensor but i get stuck when i try to understand it. My problem is i...

the problem is : in general,"the second derivative of the [component of vector]" itself(which is the double-dot one) is not equal to the "component of the second derivative of the [whole vector]"(which is the double-Nabla derivative) therefore i am not sure why they are equal

this makes sense to me thank you !

In this derivation,i am not sure why the second derivative of the vector ## S_j '' ## is equal to ## R^{u_j}{}_{xyz} s^y_j v^z y^x## could anyone explain this bit to me thank you it seems ## S_j '' ## is just the "ordinary derivative" part but it is not actually equal to ## R^{u_j}{}_{xyz}...

do you mean [V,W] can be specified as it satisfy linearity in some direction? why? i still try to understand what is [V,W] actually that can "link back" to the ##\nabla_V W - \nabla_W V## i just think the [V,W] implied that the basis vectors are commutative but $$\nabla_V W - \nabla_W V $$ are...

i just fix the mistakes in my reply and the question thank for your help :( i am copying and pasting the Latex code from somewhere else and i try to modify them

i see i am watching the video in this link in 4:37 the presenter expanded the vector in this form and i get confused... because he let's $$ V(U)=v^i e_i(U^J e_j)$$ since this is what what we understand $$\nabla_{v} U$$ where $$\nabla_{v} U = v^i e_i(U^j e_j)$$ i cannot see the difference...

therefore what is ##[V,W]## itself ?? i think it is VW-WV actually

in the language of general relativity,we know that we can write $$\nabla_{V}W $$ in this form such that: $$\nabla_{V}W = = w^i d ( V^j e_j)/du^i = w^j e^i (V^j e_j ) = W( V)$$ where $$w^i * d/ (du^i) =W$$ will act on the vector V where $$W = w^i d( ) /du^i $$ and W is a vector as a...

thank you the explanations are clear and it seems i understand

i get trouble when i try to understand a vector which is parallel transport in constant latitude."However, note that the blue vector is not a vector in S^2" why? i think it is "zero vector" in S^2"in R^3 the different tangent spaces at each point in S^2 are oriented differently. So to parallel...

thank for your help the first part is related to the example in 31:32 where u2 = pi/4 (not on great circle) the first question is [ that the vector moving along the latitude in 31:32 leaned toward to left] is unclear i try to understand intuitively what "degree of straight" will the walker on...

question1 : if you draw a small circle around the north pole (it should be the same at every points because of the symmetry of the sphere),then it is approximately a flat space ,then we can translate the vector on sphere just like what we have done in flat space(which translate the vector...