Differentiating Tensors

  • #1
Hello!

I am very VERY confused!
Would anyone please be kind enough to point me in the right direction.

I read that, in general, the derivative of a tensor is not a tensor.

What do you find when you differentiate a tensor, then?

I thought that you wanted to find the acceleration, to find the curvature. Isn't the acceleration a vector and so a tensor?!

And if a vector is a rank 2 tensor, what is a 4-vector? Because when you differentiate those you get 4-vectors? 4-velocity and 4-acceleration?

I hope that makes sense, I'm getting more and more confused the more I read into it all.

Hannah
 

Answers and Replies

  • #2
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4-vectors are rank-1 tensors. If the components are differentiated, the result is not always a tensor because it does not in general transform in the correct way (unless the Jacobean is independent of the coordinates like in SR).

I can't point you to a good source for this because I only have a rather old book on GR to hand. This might be of some use

http://coastal.udel.edu/~fyshi/notes/tensor_basic.pdf
 
  • #3
WannabeNewton
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Differentiation of tensor components will not result in a valid tensor, in general, because the chart that the tensor's components are defined on will change relative to the bases unlike minkowski space where the bases are orthonormal. When you differentiate the components of the tensor defined relative to a certain bases you are also differentiating the bases which will result in "extra" terms. You must employ the covariant derivative on the tensor components being defined on the chart.
 
  • #4
Mentz114: Thanks, that pdf is great :-)

WannabeNewton: I've been reading about the covariant derivative, I *think* it's starting to fall into place....So when working with non orthogonal bases, you end up with extra terms when you differentiate the tensor...so in order to account for these extra terms and have something which IS covariant (and so a tensor) you introduce those affine connection terms?

Secondly.......I also read that this covariant derivative reduces down to normal differentiation when there is no gravity involved ( when the affine connection = 0 ) ...this is because with no gravity it becomes Minkowski space with orthogonal bases?

And sorry one final thing(!), I've lost the context of it all under all of this maths! When you now have this covariant derivative, this is to find... things like acceleration? Curvature? Or to transform between frames?


Hannah
 
  • #5
WannabeNewton
Science Advisor
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Mentz114: Thanks, that pdf is great :-)

WannabeNewton: I've been reading about the covariant derivative, I *think* it's starting to fall into place....So when working with non orthogonal bases, you end up with extra terms when you differentiate the tensor...so in order to account for these extra terms and have something which IS covariant (and so a tensor) you introduce those affine connection terms?

Secondly.......I also read that this covariant derivative reduces down to normal differentiation when there is no gravity involved ( when the affine connection = 0 ) ...this is because with no gravity it becomes Minkowski space with orthogonal bases?

And sorry one final thing(!), I've lost the context of it all under all of this maths! When you now have this covariant derivative, this is to find... things like acceleration? Curvature? Or to transform between frames?


Hannah
Yes the christoffel symbols do act like correction terms. Covariant derivatives do reduce to ordinary derivatives in a local inertial frame not exactly because there is no gravity but because you are working with a small open subset of the manifold that is diffeomorphic with(to?) an open subset of euclidean space which is flat. So locally the covariant derivative reduces to ordinary derivatives because the metric describing the space - time reduces to the minkowski metric like you said but on global scales the covariant derivative will not generally reduce to ordinary partial derivatives.
 
  • #6
Thank you!
 

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