Questions tagged [covariance]

How a quantity behaves under a change of basis vectors. This tag covers relativistic covariance, as well as contravariant and covariant tensors not necessarily in the context of relativity. DO NOT USE THIS TAG for statistical covariance.

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What is the commutator of a Poisson bracket and the covariant derivative?

Consider a classical vector field $V^\mu$ on a curved background. We make a 3+1 split of coordinates into $t,x^i$, where $x^i$ are coordinates on spatial hypersurfaces $\Sigma$ and $t$ the parameter ...
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Confusion regarding the $\partial_{\mu}$ operator

I've been confused about the $\partial_{\mu}$ operator. Peskin and Schroeder defines it as $\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$ For example, the Euler Lagrange equation of motion is ...
1k views

Do contravariant and covariant partial derivatives commute in GR?

I'm considering something like this: $\partial_{\mu}\partial^{\nu}A$ . I feel like we should be able to commute the derivatives so: $\partial_{\mu}\partial^{\nu}A = \partial^{\nu}\partial_{\mu}A$. ...
3k views

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Which is the difference between the delta tensor $\delta_{a}^{b}$ and the metric tensor $g_{ab}$?

I don't understand which is the difference between the delta of Kronecker $\delta_{a}^{b}$ and the metric tensor $g_{ab}$. They looks to have the same effect when raising and lowering the indices, but ...
81 views

PDG review says “The flavour of a given neutrino is Lorentz invariant.”

This the starting paragraph of section 14.1 PDG review (PDF) asserts: The flavour of a given neutrino is Lorentz invariant. What does this really mean? A neutrino of a given flavour $\alpha$, i.e.,...
112 views

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Why does the analogy between electromagnetism and general relativity differ if you consider them as gauge theories or fiber bundles?

Electromagnetism and general relativity can both be thought of as gauge theories, in which case there is a natural analogy between them: (Strictly speaking, the gauge symmetry of diffeomorphism ...
144 views

Inverting Operators, and Propagators on Curved Spacetime

I am a bit confused about inverting operators, and calculating propagators on a curved spacetime. Consider the following example: If I have a Lagrangian for a charged scalar field on a curved ...
340 views

Does it make sense to ask how the covariant derivative act on the partial derivative $\nabla_\mu ( \partial_\sigma)$? If so, what is the answer?

I want to find out how the covariant derivative acts on terms containing a partial derivative, e.g. $\nabla_\mu(k^\sigma\partial_\sigma l_\nu)$. But I don't know how to evaluate the terms of the form ...
321 views

Is potential energy a scalar operator?

If a scalar operator $\hat{S}$ is defined as an operator that is invariant under rotations, i.e $$U^\dagger S U = S,\,\,\,\,\,\,\, U=e^{-i\theta\hat{\mathbf{J}}\cdot{\mathbf{n}}}$$ which is ...
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Inconsistency with partial derivatives as basis vectors?

I have been trying to convince myself that it is consistent to replace basis vectors $\hat{e}_\mu$ with partial derivatives $\partial_\mu$. After some thought, I came to the conclusion that the basis ...
550 views

Transformation of generalized coordinates

One of the advantages of Lagrangian formulation is that the equations of motion have the same form regardless of the choice of generalized coordinates. Suppose that a system has $s$ degrees of freedom,...
190 views

How is the expression for the Stress-energy tensor in Cosmology a covariant expression?

Consider the energy-momentum tensor $$T_{\mu\nu}=(p+\rho)u_\mu u_\nu+pg_{\mu\nu}$$ used in Cosmology. I have a problem with this equation. Since this a tensor equation the RHS should transform in the ...
614 views

Deriving Maxwell Equations in their covariant form

Mawell Equations, in a particular unit system, are: \begin{eqnarray} \nabla \cdot \vec{E} &=& \rho &(1)\\ \nabla \times \vec{B} &=& \frac{\partial \vec{E}}{\partial t} + \vec{J}&...
131 views

$\Delta x^{\alpha}\equiv x_{2}^{\alpha}-x_{1}^{\alpha}$, the difference between two points, is not a vector

I want to show $\Delta x^{\alpha}\equiv x_{2}^{\alpha}-x_{1}^{\alpha}$, the difference between two points, is not a vector. By definition, if $\Delta x^{\alpha}\equiv x_{2}^{\alpha}-x_{1}^{\alpha}$ ...
338 views