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Consider the following "classical" example: ^{}_{}
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'=\frac{q}{4πε<sub>ο</sub>r'<sup>3</sup>}(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=\frac{q}{4πε<sub>0</sub>}(0,\frac{γy<sub>ο</sub>}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>}, \frac{-γvt}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>}) and B=-\frac{qv}{4πε<sub>0</sub>c<sup>2</sup>}(\frac{γy<sub>ο</sub>}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>},0,0) or B=\frac{μ<sub>0</sub>}{4π}\frac{qγ}{r'<sub>3</sub>}(vxr') which for γ\rightarrow1 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'=\frac{q}{4πε<sub>ο</sub>r'<sup>3</sup>}(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=\frac{q}{4πε<sub>0</sub>}(0,\frac{γy<sub>ο</sub>}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>}, \frac{-γvt}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>}) and B=-\frac{qv}{4πε<sub>0</sub>c<sup>2</sup>}(\frac{γy<sub>ο</sub>}{(y^{2}_{ο}+γ<sup>2</sup>v<sup>2</sup>t<sup>2</sup>)<sup>3/2</sup>},0,0) or B=\frac{μ<sub>0</sub>}{4π}\frac{qγ}{r'<sub>3</sub>}(vxr') which for γ\rightarrow1 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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