Or is the only difference that a magnetic field is an oscillating electric field?
Asked
Active
Viewed 137 times
-3
-
Magnetism can be explained as an electric effect on charges due to their relative velocity. That's different from a non-relativisticly (slowly) oscillating charge, which will not experience significant magnetic effect. – JMLCarter Nov 21 '17 at 20:28
-
2"Or is the only difference that a magnetic field is an oscillating electric field?" - a magnetic field is not an oscillating electric field. – Alfred Centauri Nov 21 '17 at 21:22
-
Dupe of https://physics.stackexchange.com/q/95815/25301, https://physics.stackexchange.com/q/53916/25301 and all the linked therein. Please try looking for previous questions of the same nature before asking. – Kyle Kanos Nov 22 '17 at 10:56
-
Possible duplicate of What is the difference between an electric and a magnetic field? – Kyle Kanos Nov 22 '17 at 10:57
3 Answers
2
There are a couple of ways of answering this, but at the most obvious level, the force laws in the presence of an electric and magnetic field are very different. In the presence of an electric field, a charge of magnitude $q$ feels a force:
$${\vec F} = q {\vec E}$$
while the same charge feels a force:
$${\vec F} = q {\vec v} \times {\vec B}$$
in the presence of a magnetic field. ($\vec v$ is the velocity of the charge in the relevant reference frame). So, these force equations are very different, so electric and magentic forces manifest quite differently.
More fundamentally, Einstein was able to show that the electric and magnetic fields are not independent entities, but rather belong to an object called the electromagnetic tensor:
$${\bf F} = {\vec E}\cdot dt \wedge d{\vec x} + \epsilon_{abc}B^{a}dx^{b}dx^{c}$$
which is capable of explaining the weird appearance of the velocity in the magnetic force equation, and gives the electromagnetic force the same form in every reference frame, amongst other things.
Zo the Relativist
- 41,373
- 2
- 74
- 143
-
1You didn't give as a yes or no. If you say "no", then is momentum density not intrinsically different from shear-stress? – JEB Nov 21 '17 at 23:30
-
@JEB: it's not answerable as a yes/no, because what version of that you take depends on how you interpret the question -- classically i'ts almost certainly yes, relativistically, they are different components of the same tensor, and "Electric Field" and "Magnetic Field" don't exist as entities, and the question doesn't make sense. – Zo the Relativist Nov 22 '17 at 14:50
1
Is there an intrinsic difference between an electric and a magnetic field?
Yes, the force (on an electric charge) due to a magnetic field can do no work while the force due to an electric field can.
As Jerry Schirmer points out, the magnetic force on a charge $q$ in a magnetic field $\vec B$ is
$$\vec F_m = q\vec v \times \vec B$$
The power is then
$$P_m = \vec F_m \cdot \vec v = q\, \vec v \cdot (\vec v \times \vec B) \equiv 0 $$
Alfred Centauri
- 59,560
-
I'm not sure if this answered my question. You describe the work that can be done by the different fields, but the question is is that due to an intrinsic difference? You also said in the comments above that a magnetic field is not an oscillating electric field. But if you oscillate an electric field what do you get than? – Marijn Nov 22 '17 at 10:34
-
@Marijn, it isn't at all clear to me what your question is if this isn't an answer to it. Perhaps you can start by explaining what you understand intrinsic to mean. – Alfred Centauri Nov 22 '17 at 12:29
0
According to one of Maxwell's equations, $\vec \nabla \cdot \vec B=0$.
As others have suggested, in an $(n+1)$-dimensional spacetime, the electric field is a [polar] vector with dimensionality $n$, whereas the magnetic field is a bivector with dimensionality $n(n-1)/2$ [which, in three spatial dimensions, can be thought of as a pseudovector (with 3 components)].
robphy
- 11,748