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What is a tensor?

I think groups representations are the most natural way to understand tensors. The following parts can be read essentially independently. The 2nd is the most relevant. Why tensors in physics can be ...
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Contravariant Vector Component Transformation from Polar to Cartesian

It's because basis are defined with satisfying normalization condition. Basis are covariant, because they are partial derivatives(Consider them as gradient). $$\hat{r} = \frac{\partial}{\partial r} = ...
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How do I expand: $\langle u , \nabla\rangle u$?

The object $\langle \vec{u}, \nabla\rangle$ is an operator defined by $\langle \vec{u}, \nabla\rangle = u_x \frac{\partial}{\partial x} + u_y \frac{\partial}{\partial y} + u_z \frac{\partial}{\partial ...
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Generalizing Fermi-Walker Derivative/Transport to General Vector Bundles

Rather than viewing the Fermi-Walker derivative as a specific differential operator (which is what is causing the issues with generalization and the dependence on the curves $\gamma_1,\gamma_2$), one ...
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4 votes
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Riemann curvature tensor in an inertial frame

The fact that a function's first derivative vanishes at a point does not mean that its second derivative vanishes at that point. Note that for $f(x)=x^2$, $f'(0)=0$ but $f''(0)=2$.
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1 vote

From the point of view of physics, why is it useful to know the irreps of rotation group?

In a nutshell: If we know the tensor/spinor/vector components in one coordinate system, and we know how it transforms, then we know it in every coordinate system, which is pretty useful. This sort of ...
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-1 votes

Are tensors constructed such that one forms "act" on some complex vector field?

If you understand tensors as they ate usually described in GR texts, then understanding differential forms is straightforward. Theyy are exactly the antisymmetric covariant tensors. They do not need ...
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Expressing Maxwell's equations in tensor form using Electromagnetic field strength tensor

Ah, I've been here before(I got help from a professor offsite)! The matrix multiplication representation of electromagnetism in general relativity is $$\begin{matrix}\vec{f}=\widetilde{F}\...
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Are tensors constructed such that one forms "act" on some complex vector field?

Is this supposed to be interpreted as $\omega$ 'acting on' some vector field, such that we get a number? Yes. [...] I am not sure if this object is a vector field at all. It is. I'm not entirely ...
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3 votes
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Expressing Maxwell's equations in tensor form using Electromagnetic field strength tensor

Since $F_{ij} = \epsilon_{ijk}B^k$ one has $\epsilon^{lij} F_{ij} = \epsilon^{lij}\epsilon_{ijk}B^k = 2\delta_{lk} B^k$, hence $B^l = \frac{1}{2} \epsilon^{lij} F_{ij}$. The third Maxwell's equation ...
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Expressing Maxwell's equations in tensor notation

The equation $$ \partial_\mu F^{\nu\mu} = J^\nu $$ is to hold for all $\nu$, i.e. for $\nu=0$ and for all $\nu=i$. For $\nu=0$ this reads $$ \partial_\mu F^{0\mu} = J^0 = \partial_i F^{0 i} $$ ...
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2 votes

What is the idea behind 2-spinor calculus?

May I suggest a somewhat intuitive way to try to answer the question, inspired by the late Sir Michael Atiyah's view that "spinors are the square root of geometry". Atiyah argues in his ...
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1 vote
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A particular contraction of Levi-Civita symbols and tetrads

It is zero. In $\epsilon_{\alpha\beta ij}$, all indices have to be distinct. So one of them must be $0$, and therefore $e^\alpha\wedge e^\beta\wedge e^0\wedge e^k$ vanishes, since $0$ appears twice. ...
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2 votes

Kerr Solution metric ansatz for EFEs

Pedagogical derivations (the ones I found by googling for a few minutes) seem to prefer the use of "oblate spheroidal coordinates" (OSC) as a good starting point to describe static, axially-...
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1 vote

Finding the proportionality constant in $\varepsilon^{\mu\nu}A_\mu^{\ \lambda} A_\nu^{\ \rho}\propto \varepsilon^{\lambda\rho}$

The properties of the epsilon tensor (as exploited by Cosmas Zachos) give a very elegant answer here. But, if you didn't know about those identities, you can still get the answer in a straightforward ...
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2 votes
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Finding the proportionality constant in $\varepsilon^{\mu\nu}A_\mu^{\ \lambda} A_\nu^{\ \rho}\propto \varepsilon^{\lambda\rho}$

Just use the standard properties of the tensor in solving for C, $$ \varepsilon^{\mu\nu}A_\mu^{\ \lambda} A_\nu^{\ \rho} = C \varepsilon^{\lambda\rho} ~~~~~\leadsto \\ \epsilon _{\lambda \rho} \...
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Show that the contraction of a covector and a vector is Lorentz invariant

$s=\omega_\mu V^\mu$ is scalar. It has "no" indices; the $\mu$ that looks like an index is internal to the definition of $s$ (it is summed over), in the same way that $c=\int_0^5x\,dx$ is a ...
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  • 2,297
4 votes

What is the idea behind 2-spinor calculus?

A $2$-spinor $\psi\in V\cong\mathbb{C}^2$ is here a (left) Weyl spinor, so Penrose & Rindler are e.g. exploring the fact that the complexified Minkowski spacetime $\mathbb{C}^{1,3}\cong V\otimes \...
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1 vote

Interpreting stress at the ends of a bar

The statements "stress is based on internal force [only]" and "you cannot associate stress with an external force" are misconceptions. Elasticity theory, among other fields, doesn'...
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3 votes

How to calculate the rank of a tensor?

Just count unpaired indices. You have $\alpha$ down and $\sigma$ up, so you have a (1,1) tensor. (In the same way, you know that $T_{\alpha\beta}$ is (0,2).) (FYI unless $T_\gamma^\sigma$ is symmetric ...
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1 vote
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How to calculate the rank of a tensor?

You can easily find the rank of a tensor by looking at the number of free indices. To find dummy indices, look for where you have the same index on the top (contravariant), and the bottom (covariant), ...
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2 votes
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Components of the fully contravariant Kronecker Delta in Schwarzschild Metric

The index structure of the Kronecker delta is $\delta^\mu_\nu$, not $\delta_{\mu\nu}$ OR $\delta^{\mu\nu}$. In ANY coordinate system, $\delta^\mu_\nu$ has value 1 if both indices are equal and 0 if ...
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4 votes

Ricci Identity with Torsion Proof

Remember that $\nabla_{\nu} Z^{\sigma}$ is a type $(1,1)$ tensor, so $\nabla_{[\mu} \nabla_{\nu]} Z^{\sigma}$ will spit out terms like $\Gamma_{[\mu \nu]}^{\lambda} \nabla_{\lambda} Z^{\sigma}$.
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3 votes

Ricci Identity with Torsion Proof

We have $$\nabla_\mu \nabla_\nu Z^\sigma=\partial_\mu(\nabla_\nu Z^\sigma)+\Gamma_{\lambda\mu}^\sigma \nabla_\nu Z^\lambda - \Gamma_{\nu\mu}^\rho \nabla_\rho Z^\sigma$$ as $\nabla_\mu$ is acting on ...
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  • 360
5 votes
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Ricci Identity with Torsion Proof

Hint: You can, in a first step, expand the outer derivative (write $D_\nu Z^\sigma=A_\mu^\sigma$ if you wish). You will get a partial derivative acting on $A_\mu^\sigma$ and two terms with ...
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The foundations of geometric formulation of Newton's axioms

$\nabla (dt) = 0$, means that given all clocks are synchronized, there's no time flow gradient going from point to point in system, or in other words : $$ \frac {\partial (\Delta t)}{\partial x} + \...
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The foundations of geometric formulation of Newton's axioms

I think I sort of understand the $\nabla (dt)$ condition. Suppose we have a room and consider any point in the room, in the Newtonian case, it must be that if you kept a clock at each point, then , ...
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3 votes

Argument of a scalar function to be invariant under Lorentz transformations

A scalar Lorentz invariant function satisfies $$ f(k) = f(\Lambda k). $$ for all $\Lambda$ satisfying $\Lambda^T \eta \Lambda = \eta$. Let us look at the infinitesimal version of this equation. ...
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2 votes

Preservation of symmetries of Tensors under lowering and raising indices

If $$T^{ij} = T^{ji},$$ Then $$T^{ij}g_{jk} = T^{ji}g_{jk}$$ $$T^{i}_{\ \ k} = T_{k}^{\ i} = (T^{i}_{\ \ k})^T$$ And $$T^{i}_{\ k}g_{li} = T_{k}^{\ i}g_{li}$$ $$T_{lk} = T_{kl},$$
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Metric tensor from hyperbolic PDE

Suppose that we have an equation of the form $$ g^{\rho \sigma} \frac{\partial^2 \psi}{\partial x^\rho \partial x^\sigma} + A^\tau \frac{\partial \psi}{\partial x^\tau} = 0 \tag{1} $$ and we suspect ...
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3 votes

Why is a Lorentzian metric still Lorentzian after a general coordinate transformation?

Under general coordinate transformations $$ g' = M g M^T , \qquad M_\mu{}^\alpha = \frac{\partial x^\alpha}{\partial x'^\mu} \in GL(n,{\mathbb R}) . $$ It's clear that $\det g' = \det g ( \det M )^2$...
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Software recommendation for a tensor calculation

The free Mathematica package xTensor can perform calculations such as this. You will also need to companion package xCoba, ...
2 votes

Maxwell's stress tensor and pressure

It seems like you're curious as to what the three different components of the stress tensor mean. Roughly speaking, $T_{xx}$, $T_{yy}$, and $T_{zz}$ tell you how much force per area is being exerted ...
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Maxwell's stress tensor and pressure

My question is how do we get the answer from those three components of stress tensor ? What you are asking is not clear. Are you asking how to get the off-diagonal components? Or are you asking at ...
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