# Tag Info

## New answers tagged covariance

0

It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ...

0

Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...

0

The coordinate invariant volume element/measure on a manifold with metric $g$ is $$d^d x \sqrt{|g|}$$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$d^d x' ... 0 A scalar is, like other scalars, merely just a number. Think about their matrix representation:$$ \psi=(\phi_R\; \phi_L)^T$$and$$\bar{\psi}=\psi^\dagger\gamma^0 =(\phi^*_L \; \phi_R^*). It is clear that $\bar{\psi}\psi$ is a 1x1 matrix (scalar), and of course the operation is legitimate. Those other forms are also 1x1 matrices. However under Lorentz ...

6

To add more details to jimjo's answer, I would like to explain the "at most" in my comment Vector, at most, can be covariant. Three-vectors are only covariant under rotations, but if you include boosts then three-vectors transform in a non-covariant way. Therefore, Newton's second Law is non-covariant under the full Lorentz Group. To get a covariant ...

7

Newton's second law is covariant, as it does not change its form if we switch to another frame of reference. As already explained by @AccidentailFourierTransform in his comment, Newton's second law is a vector law. This means the quantities in the law are vectors, which have different values in different frames of reference. Only scalars, by definition, do ...

-1

I have been thinking about this question for the last few years and my conclusion is: When physicists say natural units means c = 1, they are sloppy What they mean is [v] = c Remarks: [] means "unit of" according to IUPAP convention (see red book) [v] = c means "the unit of speed is the speed of light" Once we realize this, the answer to your ...

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