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13h |
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How does moving charges produce magnetic field? @ChrisWhite: thanks for the clarification, I failed to understand the premise of your setup; a non-vanishing magnetic field will generally have a effect on the test charge, though (alignment of intrinsic magnetic moment); can that be modelled sensibly as well, eg by making the rest frame of the test charge rotate? |
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1d |
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How does moving charges produce magnetic field? I don't think that's the whole story. The electromagnetic field can only be reduced to an electric one in case of $P<0$. In case of $P>0$, your description breaks down and you'd be forced to consider magnetostatics the fundamental interaction that gets boosted to different frames... |
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1d |
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How does moving charges produce magnetic field? added 38 characters in body |
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1d |
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How does moving charges produce magnetic field? added 2 characters in body |
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1d |
answered | How does moving charges produce magnetic field? |
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1d |
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How does moving charges produce magnetic field? you'll need to read up on special relativity; if you wait for a bit, I'll expand my comment into a proper answer... |
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1d |
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How does moving charges produce magnetic field? nothing happens to the particle to make it produce a magnetic field as it starts moving: electric and magnetic field are components of the electromagnetic field, which is a single entity, similar to how energy and momentum are components of 4-momentum; in a charged particle's rest frame, the magnetic components vanish, as does its 3-momentum, and only the time-like ones (the electric field and the energy, respectively) remain |
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2d |
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Arbitrary tensor covariant derivative the $T^a{}_{cd}$ terms (three indices) should probably read $\Gamma^a{}_{cd}$ |
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2d |
answered | In coordinate-free relativity, how do we define a vector? |
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May 19 |
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Do generators belong to the Lie group or the Lie algebra? @joshphysics: closed under the group operation (which would be redundant, but I've heard it used before), closed as in closed manifold (ie compact and without boundary) or topologically closed ;) ; these lecture notes just say Of course, this formula is valid for any matrix Lie group, so it really might be trivial even if I don't see it yet |
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May 19 |
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Do generators belong to the Lie group or the Lie algebra? @joshphysics: yeah, I don't think it's actually as obvious as I thought - my argument works for matrix groups that are open subsets of $\mathbb{R}^{n\times n}$, but does it work in general? |
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May 19 |
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Do generators belong to the Lie group or the Lie algebra? ah... so I was missing something obvious, in particular that for matrix groups, $\mathrm{conj}_g$ is a linear map and thus $\frac{d}{dt}\big|_{t=0}\mathrm{conj}_g(\exp tX)=\mathrm{conj}_g(\frac{d}{dt}\big|_{t=0}\exp tX)=\mathrm{conj}_g(X)$; anyway, +1 |
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May 19 |
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Do generators belong to the Lie group or the Lie algebra? shouldn't $\mathrm{Ad}$ be the differential of conjugation instead of conjugation itself, ie $\mathrm{Ad}_g=\mathrm{T}_e(\mathrm{conj}_g):\mathrm{T}_eG\to \mathrm{T}_eG$, whereas $\mathrm{conj}_g:G\to G$? |
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May 19 |
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In coordinate-free relativity, how do we define a vector? sadly misses the point - I'm debating whether I should expand my comment into a 2nd answer... |
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May 19 |
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In coordinate-free relativity, how do we define a vector? re your clarification: it's part of your geometric model - it's up to you to map physical quantites to geometric ones in a meaningful way (and transformation laws follow from that); eg momentum is most naturally a covector (cf the Lagrangian formulation, contraction with a velocity to get an energy, minimal coupling to the em vector potential, ...), the em field strength is the curvature of a principal connection and thus a lie-algebra-valued 2-form, which you can contract with a gauge-independent charge (a co-adjoint orbit, in case of $U(1)$-connections just a number) to get a regular 2-form |
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May 19 |
answered | In coordinate-free relativity, how do we define a vector? |
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May 18 |
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Why distinguish between row and column vectors? let us continue this discussion in chat |
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May 18 |
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Why distinguish between row and column vectors? (sorry for mangling your name in my last comment): the definition of hypersurfaces becomes easier if you already have a concept of tangent vectors, making this definition of cotangent vectors arguably less fundamental; there's actually a more severe problem: unless I'm mistaken, you need additional structure (eg a volume form) to make the hypersurface (or rather its tangent subspace of codimension 1) characterise a unique cotangent vector |
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May 18 |
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Why distinguish between row and column vectors? @Murphid: tangent and cotangent vectors only appear on equal footing if you use the coordinate-based definition via their transformation laws; there are two other common definitions of tangent vectors (directional derivates and equivalence classes of curves), and the latter one makes a lot of sense from a 'physical' point of view; I'm not aware of an intrinsic definition of cotangent vectors that's equally 'nice' and does not rely on the definition of tangent vectors |
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May 18 |
answered | Why distinguish between row and column vectors? |
