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Physically speaking, "pseudo-vectors" are vectors $v\in \mathbb{R}^3$ which transform as $ v'= (\det {R})v$ if the "system were to transform as $R\in O(3)$". However, what does this mean mathematically? And in particular, why is the magnetic field $B$ a pseudo-vector?

I would imagine that by "vectors", we actually mean smooth differential forms with the isomorphism $v\mapsto \sum v_i dx_i$, and by "transforming the system as $R\in O(3)$", I would imagine that it means we are applying a pullback on 1-forms corresponding to the map $x\mapsto Rx$. Assuming that $B =*dA$ where $*$ is the hodge-star operator, how would $\det R$ be factored into this transform?

Qmechanic
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Andrew Yuan
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    Related: https://physics.stackexchange.com/q/130098/50583, https://physics.stackexchange.com/q/313091/50583 – ACuriousMind Feb 14 '21 at 23:28
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    Since the source of the B field is current that is a line-like creature when properly thought of B should be some kind of surface creature whose sheets "radiate out" of the current line. This is in analogy with a point charge and its E field whose lines radiate out of the point. An elementary surface creature is a small parallelogram, ie., a bivector that in 3D can also be represented as a (pseudo-) vector perpendicular to the plane of the bivector while forming a right hand frame with it. – hyportnex Feb 15 '21 at 00:38
  • Not really central to your query, and you may have just made a typo, but $\star dA$ is not the magnetic field, this is just the Hodge dual of the Faraday $2$-form. "The" magnetic field can only be defined once you have a splitting of the manifold into a time and space piece $M\cong \Bbb{R}\times S$. Then, with respect to this splitting we can write $dA =dt\wedge E+B$ for some 1-form $E$ and 2-form $B$ on $M$. Taking the dual gives us $*dA=-dt\wedge \tilde{B} + \mathcal{E}$ for some other 1-form $\tilde{B}$ and 2-form $\mathcal{E}$. – peek-a-boo Feb 16 '21 at 18:56
  • Then by taking an interior product we get $\tilde{B}=-\iota_{\frac{\partial}{\partial t}}(*dA)$. Note that this is a 1-form on the spacetime manifold $M$ (relative to a particular decomposition into time and space). Finally, if you want to "vectorize" everything you need to first, for each time, pullback $\tilde{B}$ to space. This yields a time-dependent family of 1-forms $\beta_t$ on $S$. Finally using a metric you can use musical isomorphisms to make this into a time-dependent family of vector fields on $S$. But anyway, it's $\star$ which causes all the funny "issues" regarding orientation. – peek-a-boo Feb 16 '21 at 19:00
  • @peek-a-boo I haven't quite thought it through, but I do agree that if we were to incorporate the electric field $E$, then we would indeed need to have a space-time manifold. However, that seems to be overcomplicating things since all I care about is $B$ and spacial rotations/reflections. Therefore, if I were to assume the isomorphism between vectors $v$ and differential forms $\sum v_i dx_i$, then I think $B=*dA$ is indeed the right way of writing, since $B=\nabla \times A$. – Andrew Yuan Feb 16 '21 at 19:53
  • Ah I see, your $A$ refers to simply the "spatial" vector potential; i.e the $1$-form on space corresponding to $\mathbf{A}=A_1\mathbf{e}_3+A_2\mathbf{e}_3+A_3\mathbf{e}_3$. In my mind I was thinking of the full potential on spacetime (hence the more elaborate formulae) which (unfortunately?) is also simply just denoted as $A$ (because recently I've been studying some special relativity so my brain reserved $A$ purely for this). Ok then we're in agreement, $B$ is just $*dA$ (up to a musical isomorphism to convert 1-forms to vector fields). – peek-a-boo Feb 16 '21 at 20:13

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I don’t know the math jargon, but the magnetic field is a pseudo vector because it is determined for a given configuration of moving charges by the right hand rule. This means that if you look at a mirror reflection of a physical setup with the B vector drawn in, it will be reversed. Angular momentum is another example of this. This property of reversing direction relative to the rest of the scenery upon mirror reflection is how I recognize pseudo vectors.

Ben51
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  • I guess that's the very definition and there should not be a need to look much into the mathematical details as far as only the identification is concerned. – ultralegend5385 Feb 14 '21 at 23:44
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I thought about this a little further, and I think I can explain that $B$ is a pseudo-vector (without using too much math jargon).

Let $R\in O(3)$ and let $\omega:\mathbb{R^3} \to \mathbb{R^3}$ be a smooth vector function such that $A$ or $B$. Let me define transforming the vector as $R$ as group action on $\omega$ defined by $$ (R\cdot \omega)(x) = R\omega (R^{-1}x) $$ That is, to get the new vector function $\omega'=R\cdot \omega$ at some postion $x$, we need to transform the system back as $R^{-1}x$, find the corresponding vector of $\omega$ at that value, and then rotate/reflect that vector based on $R$. One way to see that this is the canonical way of transforming the system is to think about the differential forms $$ \omega_i(x)dx_i $$ If we take $y=Rx$ (i.e. system transforms as $R$), then the differential form becomes $$ (R\cdot\omega)_i(y) dy_i $$

Then using the fact that $B=\nabla \times A$, we see that $$ R\cdot B = (\det {R}) (\nabla\times(R\cdot A)) $$ This is where the determinant comes from. In particular, if for some special system $R\cdot A =A$ for some reflection $R$, then $R\cdot B = -B$.

Andrew Yuan
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  • Hm, while it is true that the curl of a vector (or the cross product of two vectors) is a pseudovector to me it seems all you have done is pushed the same problem into another place. How do you know $A$ is a vector and not a pseudovector? – Andrew Feb 16 '21 at 02:38
  • The normal logic is what is given in another answer: under parity all currents reverse direction, and so the $B$ field they generate reverses direction, so $B$ must be a pseudovector, and then your argument would then be used to establish that $A$ must be a vector. – Andrew Feb 16 '21 at 02:40
  • There are probably equivalent ways of doing so. For my purpose, I would say that the Maxwell equations (assuming static electric field) would give something like $-\nabla^2 A = J$ assuming that we chose our gauge to be $\nabla\cdot A=0$, so you could reason that since $J$ is a vector, $A$ must also be a vector. I personally wanted to start out with the vector potential $A$ because that what you usually use in minimal substitution or Peierls substitution. – Andrew Yuan Feb 16 '21 at 02:44
  • Ah ok, I see. Indeed, so long as you are using Maxwell's equations to relate the vector potential to the current and using the properties of the current to infer the transformation properties of $A$, then there is nothing circular and your argument is fine (and at least morally equivalent to the other one). – Andrew Feb 16 '21 at 02:47
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In the Lorentz force $\vec F=q\vec E + q\vec v\times\vec B$,
the force $\vec F$, the electric field $\vec E$, and the velocity $\vec v$ are [ordinary, polar] vectors, and the electric charge $q$ is a scalar.

Since the cross-product of two ordinary vectors is a pseudo-vector, the magnetic field $\vec B$ must be a pseudo-vector so that $\vec v\times\vec B$ is an ordinary polar vector.

(Even though $\vec E$ and $\vec B$ are "vectors", since the "magnetic field vector" is a pseudovector, it can never be added [as vectors] to an "electric field vector".)

robphy
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