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4 answers

Not at all. Scalar or vector addition is not affected by order.

2006-11-05 10:41:03 · answer #1 · answered by kirchwey 7 · 0 0

if the vector further in distinctive orders then the resultent rigidity is affected, because of the fact the required field is a vector field with countless variety. the % for countless field implies the sphere quantum would desire to be massless. those are residences of and electromagnetic field, whose quantum is the photon. hence via requiring the community gauge invariance of electron field, existence of electromagnetic field would be inferred. different than for the electromagnetic rigidity and gravitational rigidity there are 2 different forces - the good and the susceptible forces. The good forces are those in charge for containing the nucleus mutually. The susceptible forces can't be categorized as being beautiful or repulsive like all different 3 forces. they alter one particle into yet another. For quantumelectrodynamics, the two fields required interior the thought, electron and photon, the place already undemanding on the time it replaced into being developed. yet, the area replaced into very distinctive for the good and the susceptible interplay. It took an prolonged time in the previous the good and the susceptible forces ought to initiate making sense. the reason replaced into that the fundamental community gauge concept replaced into no longer understood. the form of those gauge theories required advent of latest debris and rigidity, which conformed to the fundamental gauge symmetry and likewise made the theories sensible in terms of coming up them finite

2016-10-21 07:59:00 · answer #2 · answered by ? 4 · 0 0

It isn't. But vector products (dot or cross) are. So be careful ☺


Doug

2006-11-05 10:49:28 · answer #3 · answered by doug_donaghue 7 · 0 0

Not at all. It doesn't matter what order you add them in.

2006-11-05 10:41:17 · answer #4 · answered by Anonymous · 0 0

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