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Consider the following "classical" example: [itex]^{}_{}[/itex]
A system Σ' moves away from a system Σ with relative velocity v in the direction of z'z axis (their origins concur at time t=t'=0). A charge q is located in the origin of Σ' and moves along with it, while an observer sits at (0,yo,0) in Σ. The E-Field as observed in Σ' is according to Coulomb's law: E'=[itex]\frac{q}{4πεοr'3}[/itex](0,yο,-vt) while the B-Field is of course: B'=0.
Now we make use of Lorentz transformation and obtain for Σ the following relations:
E=[itex]\frac{q}{4πε0}[/itex](0,[itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex], [itex]\frac{-γvt}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex]) and B=-[itex]\frac{qv}{4πε0c2}[/itex]([itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex],0,0) or B=[itex]\frac{μ0}{4π}[/itex][itex]\frac{qγ}{r'3}[/itex](vxr') which for γ[itex]\rightarrow[/itex]1 takes actually the form of Biot-Savart law.
We observe that a pure E-Field converts into a mixture of E and B fields. Now arises my question: Is the opposite possible for a single charge? If the answer is no that means that electricity has a more fundamental place in relativity than magnetism and respectively Coulomb's law is more general than Biot-Savart, which by the way is not applicable to single charges. Is it therefore possible that the two fields are not symmetrical and no pure B-Field could be ever detected?
Moreover I would like to know how the leftover E-Field of the above example is compensated for a vast amount of particles (classical example: the straight current-carrying wire).
A system Σ' moves away from a system Σ with relative velocity v in the direction of z'z axis (their origins concur at time t=t'=0). A charge q is located in the origin of Σ' and moves along with it, while an observer sits at (0,yo,0) in Σ. The E-Field as observed in Σ' is according to Coulomb's law: E'=[itex]\frac{q}{4πεοr'3}[/itex](0,yο,-vt) while the B-Field is of course: B'=0.
Now we make use of Lorentz transformation and obtain for Σ the following relations:
E=[itex]\frac{q}{4πε0}[/itex](0,[itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex], [itex]\frac{-γvt}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex]) and B=-[itex]\frac{qv}{4πε0c2}[/itex]([itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex],0,0) or B=[itex]\frac{μ0}{4π}[/itex][itex]\frac{qγ}{r'3}[/itex](vxr') which for γ[itex]\rightarrow[/itex]1 takes actually the form of Biot-Savart law.
We observe that a pure E-Field converts into a mixture of E and B fields. Now arises my question: Is the opposite possible for a single charge? If the answer is no that means that electricity has a more fundamental place in relativity than magnetism and respectively Coulomb's law is more general than Biot-Savart, which by the way is not applicable to single charges. Is it therefore possible that the two fields are not symmetrical and no pure B-Field could be ever detected?
Moreover I would like to know how the leftover E-Field of the above example is compensated for a vast amount of particles (classical example: the straight current-carrying wire).
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