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## Hot answers tagged covariance

26 votes

### Lagrange's equation is form invariant under EVERY coordinate transformation. Hamilton's equations are not under EVERY phase space transformation. Why?

You should really think about the variables we use as being like coordinates on some manifold, the configuration space (roughly the same as the phase space, I won't be careful about the distinction). ...
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25 votes
Accepted

### Covariant vs contravariant vectors

We don't talk of covariant and contravariant bases. Start with the basis $\{\mathbf e_i\}$. Then a general vector can be written $$\mathbf v = v^i \mathbf e_i$$ Now if you double the length of a basis ...
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23 votes
Accepted

### Is it "mathematically wrong" to ignore dual spaces, 1-forms, and covariant/contravariant indices in classical mechanics?

As long as you restrict yourself to orthonormal bases, then that's fine. The reason for this is that indices are "raised" or "lowered" via the metric, and in an orthonormal basis ...
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23 votes

### Definition of four-velocity: why define it with proper time of the object?

$dx^\mu$ is covariant and $d\tau$ is invariant. So $dx^\mu/d\tau$ is manifestly covariant while $dx^\mu/dt$ is not. Covariant quantities are of interest because the laws of physics are covariant. So ...
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19 votes
Accepted

### Vectors as functions?

However, I have read that the modern way to learn these concepts is to think of vectors as multilinear functions of covectors This is actually not quite true, though the distinction is subtle. In the ...
• 70.3k
18 votes
Accepted

A Lorentz invariant quantity does not change under a Lorentz transformation. For example, charge is invariant, but energy is not. The components of a vector transform contravariantly, i.e. $A^\mu \to \... • 103k 16 votes Accepted ### What is the reason to believe that the laws of physics are same in all frames of reference? Is there any supportive evidence which suggests so other than the evidence of common sense and intuition? Yes, there is quite a substantial body of experimental evidence supporting the two postulates ... • 104k 15 votes Accepted ### Covariant formulation of electrodynamics It's true that the first equation has the same form in all Lorentz frames, but it's not obvious unless you know how$\rho$and${\bf J}$individually transform. For example, the similar-looking ... • 48.5k 15 votes Accepted ### Usefulness of Curl and Divergence as Multilinear Maps This answer takes into account the elaboration on the question which takes place in the comments. Let's build a tensor space. I won't bother with the definition of a tangent vector as a "... • 70.3k 15 votes Accepted ### Critique on tensor notation You are correct: a tensor can be viewed as a linear function in many different ways. We define a$(p,q)$tensor$T$as a function that takes$p$covectors and$q$vectors and returns a number: $$T: {V^... • 28.4k 14 votes Accepted ### Gradient, divergence and curl with covariant derivatives For the gradient, your mistake is that the components of the gradient vary contravariantly. On top of that, there is a issue with normalisation that I discuss below. I don't know if you are familiar ... • 2,295 14 votes Accepted ### Inconsistency with partial derivatives as basis vectors? Raising and lowering indices in a vector is not a valid operation. Basis vectors are no exception. While x_\mu=g_{\mu\nu}x^\nu is a valid operation, \hat e^\mu=g^{\mu\nu}\hat e_\nu is not. The ... 14 votes ### Why does the analogy between electromagnetism and general relativity differ if you consider them as gauge theories or fiber bundles? There is an additional structure at play here. Depending on which level you're looking at, this is either the fact that the tangent connection \nabla admits the exchange of arguments (if \nabla_XY ... • 11.3k 14 votes ### Momentum operator in QM - scalar or vector? Momentum is a vector operator. Period. When restricted to one-dimensional problems, momentum becomes a one-dimensional vector, which coincides with scalars in that space. • 134k 14 votes ### Lagrange's equation is form invariant under EVERY coordinate transformation. Hamilton's equations are not under EVERY phase space transformation. Why? A more geometric approach to the Hamiltonian formulation is to consider the (2n+1)-dimensional contact manifold {\cal M} with coordinates (q^i,p_j,t). The Hamiltonian action functional is$$S_H[\... • 207k 13 votes Accepted ### Can Einstein-Hilbert action be derived from symmetry considerations? Well, one can reason as follows: One wants a diffeomorphism-invariant action, which must be of the form $$S=\int d^4x\sqrt{-g}L,$$ where$L$is a scalar in terms of transformation properties. ... • 11.3k 12 votes ### How to show the spacetime interval is invariant in general? You cannot derive the invariance of the line element because it is one of the assumptions on which relativity (both flavours) is based. When you say: I understand how to derive the spacetime interval ... • 358k 12 votes ### Double covariant derivative of tensor So you have to compute $$\nabla_a \nabla_b h_{cd}$$ First you have to think of$\nabla_b h_{cd}$as a$(0,3)$tensor so that $$\nabla_a \nabla_b h_{cd} = \partial_a ( \nabla_b h_{cd} ) - \Gamma^e_{... • 26.7k 11 votes ### Co and contravariant: tensors or components? Yes, the contravariant components make reference to a different geometrical object than the covariant components. The covariant components are components of a vector from the dual space to the vector ... • 22.7k 11 votes Accepted ### Origin of \sqrt{-g} in the integral of action S When you're working in general relativity, the coordinates are basically arbitrary and they do not even need to have dimensions of length. This means that the "usual" volume element,$$ dx^{... • 134k 10 votes ### Why do we need a metric to define gradient? The main point of the gradient is to have the following equation: $$df(v) = \langle \nabla f , v \rangle$$ for every vector$v$of the tangent space where$\langle \cdot , \cdot \rangle$is the ... • 101 10 votes Accepted ### What is "a general covariant formulation of newtonian mechanics"? The development of general relativity has led to a lot of misconceptions about the significance of general covariance. It turns out that general covariance is a manifestation of a choice to represent ... • 2,726 10 votes Accepted ### How to show the spacetime interval is invariant in general? Lets look at an arbitrary invertable coordinate transformation: $$x^\mu \rightarrow x'^{\mu}=x'^{\mu}(x^\nu).$$ The corresponding Jacobian$\Lambda$\Lambda^\mu_{~~\rho}=\frac{\partial x'^{\mu}}{... • 3,021 10 votes Accepted ### Confusions about Covariant and Contravariant vectors You're dealing with different geometric objects: Tangent vectors, which can be realized as equivalence classes of curves, and cotangent vectors, which can be realized as equivalence classes of real-... • 13.7k 10 votes Accepted ### Is there differential form notation for Maxwell's equation in curved spacetime? Differential forms are natural objects that depend solely on the smooth structure, therefore they are valid in any spacetime. Note that the Hodge star$\star$does, however, depend on a metric tensor.... • 11.3k 10 votes ### Critique on tensor notation Forget tensors for a moment and just think about our old friend the matrix. Take a square matrix for example. You can multiply it onto a column vector and get back a column vector. Or you can put a ... • 59.8k 10 votes Accepted ### If momentum is a covector, how does$p=mv$? Yes, OP points to the fact that classical mechanics typically relies on the existence of a fiducial/distinguished/background metric structure on the configuration manifold so that we can apply the ... • 207k 10 votes ### Notion of Co- and Contravariance in Dirac-Notation In Dirac notation,$\langle \psi |$is a covariant vector, while$| \psi \rangle$is a contravariant vector. Of course, this nomenclature is hardly ever used, because it is more convenient to call ... • 22.3k 9 votes Accepted ### Einstein Summation Convention: One as Upper, One as Lower? In the 'strict' sense, you should only apply the summation convention to a pair of indices if one is raised and another is lowered. For example, consider a vector$v$and a dual vector$f\$ (i.e. a ...
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