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There are two mathematical concepts which are both called vector. The first one, the vector from linear vector space is the basic "multicomponent object" which you seem to mainly talk about. The second notion of a vector is of a member of the so-called "tangent bundle" of a manifold. The second notion is the one which is defined equivalently with the ...


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Let me start with a tautology: Vectors are geometrical objects living on a vector space XD So far it says nothing, but we always have had the mental image of a vector as an arrow. A bit further into abstraction (still with or idea of an arrow representing a vector), one can find a set of transformations of vectors which preserves the properties of vectors, ...


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Mathematically, the idea of a vector is prior. You could define object that fulfill all properties of a vector space without referring to components or anything. From the notion of a vector one can derive that there exist a maximum number of linearly independent vectors and any vector in your vector space can be represented uniquely by a linear combination ...


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Recall the integral definition of the gradient: $$\nabla \varphi = \lim_{V \to 0} \frac{1}{V} \oint_{\partial V} \varphi \hat n \, dS$$ This should tell you the gradient's components transform the same way as those of the normal vector $\hat n$, which is known to have covariant components. You can verify that the normal vector has covariant components by ...


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Gradient is covariant! Why? The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called $q$'s, they are denoted $q^i$. Therefore, the gradient (or a derivative if you prefer) is $$\partial_i = ...


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This is really a comment, but it got a bit long for the comment field. I'd guess that, like me, your experience in physics is from an area where solving differential equations is a routine part of the job. We're used to analysing a problem, writing down a differential equation that encapsulates the physics and solving it, analytically if we're lucky or in ...


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"Covariant vectors as expressed in a dual basis" is exactly the same thing that "orthogonal projections to achieve covariant components" Choose one generating system $\vec e_1, \vec e_2$, non necessary orthogonal, a vector $\vec v$ may be expressed by $\vec v = v^1 \vec e_1 + v^2 \vec e_2$. The coordinates $v^1, v^2$ are called contravariant coordinates of ...


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The Ricci tensor is built adding up some of the components of the Riemann tensor, as the definition specifies: $$ R_{\mu\nu} = R^{\lambda}_{\ \mu\lambda\nu} $$ The repetition of two indices above and below means that all these components must be summed: $$ R^{\lambda}_{\ \mu\lambda\nu} = R^{0}_{\ \mu 0\nu} + \dots + R^{3}_{\ \mu 3\nu} $$ This operation ...


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In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ...


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You may find this question and especially Emilio Pisanty's answer to be enlighting with regards to what co- and contravariant vectors really are. Now, if I may rephrase your original question a bit: *How do parallelograms and lines in $\mathbb{R}^3$ relate to co- and contravariant vectors in $T_p \mathbb{R}^3$ and $T^*_p\mathbb{R}^3$? The first one is ...


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A covariant vector is commonly a vector whose components are written with ``downstairs" index, like $x_{\mu}$. Now, the gradient is defined as $\partial_\mu := \dfrac{\partial}{\partial x^\mu}$. As you can see the covariant vector $\partial_\mu$ is the derivative with respect to the contravariant vector $x^\mu$. the contravariant form of $\partial_\mu$ is ...


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You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is $$ T^* = Q T Q^T $$ However, if you take material derivatives on both sides, you have, $$ \dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T ...


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Angular displacement is an example of what's generally called a pseudovector. This is a quantity that is similar to a regular vector, except for the fact that it behaves differently under improper rotations such as reflections (it gains an additional sign flip). Any quantity which is the cross-product of two polar vectors will generally be a pseudovector. ...


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It looks like the question boils down (at least in part) to the following: can a fluid have negative ABSOLUTE pressure? This question has been discussed here several times. My take is: it can (although such state is probably metastable in the best case), because the force between two molecules can be attractive. See, e.g., ...


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Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as $\sigma$. You can pick a Cartesian basis $\{ e_1, e_2, e_3 \}$ and express the components of the tensor in that basis $$ \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & ...


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Comment to the question (v2): Apart from the issue of various overall sign conventions found in the literature, note that: On one hand, there is the Levi-Civita symbol with upper (lower) indices, whose entries are only $0$s and $\pm 1$s; it is a contravariant (covariant) pseudotensor density, respectively. On the other hand, there is the Levi-Civita ...


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I just wanted to comment on what I think the question from Zee's book is getting at. I don't think that this question from Zee is a particularly good question, and I think ACuriousMind's answer is excellent and gets at the real points that you should be trying to learn. So this is being written purely for the sake of trying to clarify Zee's book, rather than ...


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Tensors in abstract mathematics are just functions with linear arguments. In abstract index notation, the placement of indices--up vs. down--tells you whether that particular argument should be a vector or covector. For example: $$T^{\mu \nu} \equiv T(\text{covector}, \text{covector})$$ Since both the indices are up, it means both the first and second ...


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Regarding what you quoted: a vector is represented by the sum of a set of basis vectors times the vector components. If the components transform according to $L$, then the bases will transform according to $L^{-1}$, which means that when you multiply the bases with the components (to make the vector), you will get the same result every time (since $L\cdot ...


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Let's do this systematically: (I must admit that the part of your question with "prove that $p'$ is not a vector if $p$ is" did not really make sense to me, so I decided to show you how vectors in GR are really defined and how transformations really work. Feel free to tell me to mind my own business if this is not what you are looking for ;) ) Let ...



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