in one frame of reference, there is a charge fixed n an observer stationary relative to that charge and in another frame, a person is movin relative to that charge. who will observe magnetic field?
but the answer to the question (it had come in one of my exams), was that neither will. it doesnt make sense, does it? coz for the person who was said to be moving, in his frame, he is at rest and the charged particle and the other observer is moving. its very weird. is your reason the same?
You can see the result by considering the case where the electric charge is stationary, which makes it easy to work out the electric field in spherical coordinates. For a stationary point charge otherwise empty space you get \(\mathbf{E} = \frac{Q}{4\pi \epsilon_{0}}\frac{1}{|| \mathbf{r} ||^{2}}\hat{\mathbf{e}}_{r}\). Writing this in the (x,y,z) coordinate notation but still spherical parameters gives \(\mathbf{E} = \frac{Q}{4\pi \epsilon_{0}}\frac{1}{|| \mathbf{r} ||^{2}}\big( \sin\theta \cos \phi , \sin\theta\sin\phi , \cos\theta \big) = \frac{Q}{4\pi \epsilon_{0}}\frac{1}{|| \mathbf{r} ||^{2}}\big( \sin\theta \cos \phi \hat{\mathbf{e}}_{x} + \sin\theta\sin\phi \hat{\mathbf{e}}_{y} + \cos\theta \hat{\mathbf{e}}_{z} \big) \equiv E_{x} \hat{\mathbf{e}}_{x} + E_{y}\hat{\mathbf{e}}_{y} +E_{z}\hat{\mathbf{e}}_{z}\). The magnetic field is just \(\mathbf{B} = 0 \hat{\mathbf{e}}_{x} + 0\hat{\mathbf{e}}_{y} +0\hat{\mathbf{e}}_{z}= B_{x} \hat{\mathbf{e}}_{x} + B_{y}\hat{\mathbf{e}}_{y} +B_{z}\hat{\mathbf{e}}_{z} \). The reason for this mixing of notation is you can apply a Lorentz boost to these two fields. Maxwell's equations have a Lorentz symmetry which means that if you start with a pair of vectors \((\mathbf{E},\mathbf{B})\) which solve Maxwell's equations then under a Lorentz transform \(\Lambda : (\mathbf{E},\mathbf{B}) \to (\mathbf{E}',\mathbf{B}')\) you'll end up with new electric and magnetic fields \((\mathbf{E}',\mathbf{B}')\) which also satisfy Maxwell's equations. If you crunch the algebra then you can work out the new field components in terms of the old ones (a more elegant and fundamental way is to work with a 4-vector gauge potential but that isn't necessary here). We're interested in the new magnetic field, \(\mathbf{B}'\) and without loss of generality we consider a Lorentz boost in the x direction by speed v (as it is the one given in the wiki link) so \(B_{x}' = B_{x} = 0\), \(B_{y}' = \frac{\gamma v}{c^{2}}E_{z}\) and \(B_{z}' = -\frac{\gamma v}{c^{2}}E_{y}\). Therefore \(\mathbf{B}' = \frac{Q}{4\pi \epsilon_{0}}\frac{1}{|| \mathbf{r} ||^{2}}\frac{\gamma v}{c^{2}} \left( \cos\theta \hat{\mathbf{e}}_{y} - \sin\theta\sin\phi\hat{\mathbf{e}}_{z}\right)\). This can be rephrased in a nice way by defining the velocity vector \(\mathbf{v} = (v,0,0)\) to give \(\mathbf{B}' = -\frac{\gamma}{c^{2}}\mathbf{v} \times \mathbf{E}\).
very interesting. please explain in lay-man's terms Please Register or Log in to view the hidden image! n, according to me, only the moving observer experiences the field and the stationary observer, though sees the moving observer experiencing the effects of a magnetic field, cannot himself observe it... please correct me if i am wrong Please Register or Log in to view the hidden image!
for moving observer there is magnetic field and electric field both.for stationary observer(relative to charge) there is only electric field.
It is easy to write down the electric field for an electron which isn't moving, it just points outwards, while the magnetic field is even easier, it is zero. But we also know how to restructure the electric and magnetic fields if we change our description such that the electron is now moving. When you do that, as I did in my previous post, you find that the magnetic field is no longer zero. The precise form requires crunching the algebra but the basic premise is simple enough.
