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Partial differentiation of the transformation law $$ \bar{T}_i = T_r\frac{\partial x^r}{\partial\bar{x}^i} $$

of a covariant vector yields $$ \frac{\partial\bar{T}_i}{\partial \bar{x}^k} = \frac{\partial{T_r}}{\partial x^s} \frac{\partial x^r}{\partial{\bar{x}}^i} \frac{\partial{x}^s}{\partial\bar{x}^k} + T_r \frac{\partial^2 x^r}{\partial{\bar{x}^k}\partial{\bar{x}^i}}. $$

Because of the second term on the right the partial derivative is not a tensor.

Why is this intuitively expected? I thought a tensor was something intrinsic independent of coordinates and thus invariant under coordinate changes. It is not clear to me why a partial derivative fails this.

EditPiAf
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user782220
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    Slightly unrelated, but the failure of the derivative of a tensor at staying invariant under a change of coordinates is one of the prime motivations behind introducing the covariant derivative. – SystematicDisintegration Jun 10 '18 at 08:16

3 Answers3

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Yes, the tensor itself is independent of the coordinate system, but the operation of taking a partial derivative is highly dependent on what coordinate system you're using: you vary one of the coordinates while keeping all the other coordinates (in that coordinate system) constant. And this dependence turns out not to be “tensorial”, when you check what happens if you express the derivative in another coordinate system, as you did in the question.

Hans Lundmark
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  • If I am reading correctly what @SystematicDisintegration mentioned the covariant derivative is the derivative projected to the local tangent plane. So if, for example, there is a map $T: \S \to \mathbb{R}^3$ where $S\subset \mathbb{R}^3$ is a 2 dimensional surface then projecting to the local tangent plane changes the derivative. But what about $T:\mathbb{R}^3 \to \mathbb{R}^3$ where the "tangent plane" is all of $\mathbb{R}^3$ it would seem in this case the partial derivative is no different from the covariant derivative. Does that mean the partial derivative is an invariant in that case. – user782220 Jun 10 '18 at 19:19
  • Well, not quite. If you use a curvilinear coordinate system you will still get nonzero Christoffel symbols. The reason for this is that in spherical coordinates (for example) the basis vector fields $\partial_r$, $\partial_\theta$, $\partial_\phi$ will depend on $(x,y,z)$ so when you compute the derivative that you are going to project to the tangent space (i.e., in this case, do nothing with), you have to take into account the partial derivatives of the components of the vector field and of those basis vectors. So only differentiating the components won't give the right answer. – Hans Lundmark Jun 10 '18 at 20:49
  • See this answer for an explicit example in two dimensions. – Hans Lundmark Jun 10 '18 at 20:51
  • https://youtu.be/BHKd6-IJgVI?t=8m10s gives $\frac{\partial V}{\partial x^k}=[\frac{\partial V^i}{\partial x^k}+V^j\Gamma^i_{kj}]e^i+V^iN^j_{ki}n_j$. In the case of $T:\mathbb{R}^3\to\mathbb{R}^3$ there is no normal to the tangent plane so $n_j$ does not exist. So why doesn't the show the partial derivative reduces to the covariant derivative? – user782220 Jun 10 '18 at 21:31
  • The Christoffel symbols $\Gamma^i_{kj}$ are not zero. – Hans Lundmark Jun 10 '18 at 21:43
  • Why is the Christoffel symbols being not zero relevant? The partial derivative has them appearing. – user782220 Jun 10 '18 at 22:15
  • Yes, the actual partial derivative of the whole vector in the ambient space has Christoffel symbols, but that is not what you get if you just take the partial derivatives of the components of the vector. Which is the whole point: you can't just do “$\partial V^i/\partial x^j$” and get something which is a tensor. – Hans Lundmark Jun 11 '18 at 06:44
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The derivative is independent of coordinates if the basis elements are constant (you’re in Euclidean space), but on a general manifold, you have to use the product rule, but the derivative of a tensor only differentiates the components and thus ignores half of the answer.

Bible Bot
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The second term is due to the fact that in a general coordinate system the components of a vector change under a parallel transport. The covariant derivative of a tensor field is what you get when you subtract the contribution due to the parallel transport from the ordinary derivative.

Count Iblis
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