Tensor Calculus 5a: The Tensor Property

What people said

Recommends user56834 · 2017-02 · Mathematics Stack Exchange · on “How to reconcile these two definitions of tensors”

In this [this video] lecture series a definition of a tensor is given roughly as follows: (Definition 1) : [we] define tensors as variants that transform from one coordinate system to another by a very special rule. A variant $T_i$ is called a covariant tensor if its values $T_i$ and $T_{i'}$ in the coordinate systems $Z^i$ and $Z^{i'}$ are related by $$T_{i'}= T_i J^i_{i'}$$ [then a similar definition for a contravariant tensor] How is this reconciled with the definition that seems to me to be much more general: (Definition 2): Given vector space $V$, a tensor is a multilinear map $V^* \times ... \times V^*\times V \times ... \times V \to \mathbf{R}$ It seems to me that the first definition is e

Recommends Clemens Bartholdy · 2021-05 · Mathematics Stack Exchange · on “Of what coordinates is a transformed contravariant tensor a function of?”

I found the answer from this lecture [this video] at 1:28:16 by Pavel Grinfeld on this topic. It goes as follows: $$S_j^i = \frac{\partial T^i}{\partial Z^j}$$ Now, in primed coordinates: $$ S_{j'}^{i'} = \frac{\partial T^{i'}}{\partial Z^{j'} }$$ So, the derivative must be taken in primed coordinates. Basically the object in the new coordinate system by the new derivative operator.

Cites Chill2Macht · 2016-08 · Mathematics Stack Exchange · on “What is the most general setting in which the Einstein convention is relevant?”

Question: To what extent do the terminology and concepts surrounding the term "tensor" (particularly as used in engineering and more applied physics, e.g. ignoring Fock spaces in QM) rely on or assume the existence of a background Riemannian metric? Do these concepts only extend to the level of generality of Riemannian geometry, or do they use the full level of generality in which the term "tensor" can be used in abstract algebra? Background/Motivation: When one refers to "vectors", one means either (1) the concept of "arrows" in $\mathbb{R}^3$ or perhaps more generally some Euclidean space, or (2) elements of an arbitrary abstract vector space (or even an arbitrary module), "arrows" being a

Cites Antoni Parellada · 2022-04 · Mathematics Stack Exchange · on “Step in the prove that the covariant basis is a covariant tensor”

From this step in this video on YouTube [this video] : $$\frac{\partial \vec R(Z')}{\partial Z^{i'}}\overset{?}=\frac{\partial \vec R(Z(Z'))}{\partial Z^{i'}}$$ where $\vec R$ is a position vector differentiated with respect to the different coordinates $Z^{i'}$ in the prime coordinate system $Z'.$ I think the denominator on the RHS should be in the un-prime coordinate system: $$\frac{\partial \vec R(Z')}{\partial Z^{i'}}\overset{?}=\frac{\partial \vec R(Z(Z'))}{\partial Z^{i}}$$ Is this correct? Was there a slight mistake on the presentation? He explains it in his book step-by-step right after equation $(6,10),$ and would be the answer to the question from another user that has been linked to th

Critical Clemens Bartholdy · 2021-05 · Mathematics Stack Exchange · on “Raising and lowering indices on the derivative operator”

So I've learnt, we can 'raise' and 'lower' indices of the derivative operator in the following way: $$ \frac{\partial }{\partial x^{\alpha} } = g^{\alpha \beta} \frac{\partial }{\partial x_{\beta}}$$ Alternately: $$ g_{\alpha \beta} \frac{\partial}{\partial x^{\alpha } } = \frac{\partial}{\partial x^{\beta} }$$ Source: Eqtn 16.24 of this site [LINK] . To contradict the above, this point is discussed at 1:20:39 in lecture 5a [this video] on Tensor Calculus by Pavel Grinfeld at 1:20:39, where he states that only $ \frac{\partial}{\partial x^{\alpha}}$ exists (as he says, you can only take derivative with respect to coordinates) My question is does such a thing as $\frac{\partial}{\partial x_{\beta}