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Following on from this comment - do uncharged particles radiate under acceleration? (no offense to the commenter, just clarifying as I've not heard of this before)

Why is this? I vaguely recall a satisfying explanation (will link if I find it) for charged particles that rested on them having charge, so am a bit lost imagining how this works for uncharged particles.

Broader question: does this mean that anything that accelerates, necessarily radiates something else?

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Xeren Narcy
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  • I believe leptons would radiate Ws and Zs, but truthfully, I don't think one can actually separate the electroweak force in the standard model like that. – CuriousOne May 28 '15 at 08:06
  • what kind of acceleration... "g?" and is this Larmor formula applicable in this case? – user6760 May 28 '15 at 08:42
  • Please read my answer http://physics.stackexchange.com/q/186361/ – HolgerFiedler May 28 '15 at 18:51
  • https://indico.cern.ch/event/153576/material/slides/1?contribId=45 (Coherent Bremsstrahlung from Fast Neutrons) – HolgerFiedler May 28 '15 at 18:59
  • @HolgerFiedler Thanks, I had a look at your answer already but wasn't sure how it applies to uncharged; I'll have a look at the article and come back to this. – Xeren Narcy May 28 '15 at 23:23
  • Ok, so if I'm understanding it correctly, it's still the same effect as for charged particles, but the issue is that some particles considered uncharged like neutron actually have non-zero electric dipole and magnetic moments that permit them to interact (albeit weakly, however, necessarily) with photons? – Xeren Narcy May 29 '15 at 00:02
  • Xeren Narcy, neutrons don't have electric charge but have magnetic dipole moment and intrinsic spin. – HolgerFiedler May 29 '15 at 03:44
  • @HolgerFiedler I'm not suggesting neutrons have electric charge, but that they may (or do) have non-zero electric dipole moment. I would think a spinning electric dipole would generate a magnetic moment... – Xeren Narcy May 29 '15 at 03:53
  • Magnetic dipol moment is a intrinsic property and it is related to the intrinsic spin only. – HolgerFiedler May 29 '15 at 08:06

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