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High precision tests of Maxwell equations |
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| Jul2-10, 02:29 AM | #1 |
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High precision tests of Maxwell equations
Can anybody point me to some high precision tests of Maxwell's equations. I've tried hard to find some.
Skepticism is a curse, I know. |
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| Jul2-10, 06:33 AM | #2 |
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| Jul2-10, 10:25 AM | #3 |
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| Jul2-10, 11:59 AM | #4 |
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High precision tests of Maxwell equations
Also, the mass of the photon (related to inverse-square law tests) has been measured to be less than 4*10^-48 g (Jackson, p.6).
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| Jul2-10, 02:00 PM | #5 |
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And the force between two conductors -
Exactly 2 × 10–7 newtons when 1 ampere flows between parallel conductors 1 metre apart in vacuum. You can't get any more accurate than that! P.S.
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| Jul2-10, 10:24 PM | #6 |
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The wave nature of light involves all 4 equation to derive. The inverse square law confirms the first equation. The second equation has been probed in monopole experiments. The vacuum part of the third and fourth equations are used to determine the speed of light. I haven't seen a convincing experimental confirmation of them though. Does anybody know of a faraday's or ampere's law experiment that accurate to more than 1 in thousand?.
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| Jul3-10, 12:26 AM | #7 |
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[Additional] Come to think about it, All four laws are just mathematical re-statements of Gauss's law with a bit of relativity thrown in. A test for any of them is a test for the whole construct. |
| Jul3-10, 01:12 PM | #8 |
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Any high precision measurement of the speed of light is also a high precision test of Maxwell's equations.
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| Jul3-10, 03:21 PM | #9 |
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| Jul3-10, 04:00 PM | #10 |
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Hmm, I don't know that I agree with that, but now that I think about it I don't agree with my previous statement either. That c is frame invariant (and the rest of relativity) is predicted by Maxwell's equations, but not its value.
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| Jul3-10, 04:53 PM | #11 |
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| Jul3-10, 06:08 PM | #12 |
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Yes, and with today's SI system it would be more accurate to talk about the precision of measuring the length of a meter since the speed of light has an exact value.
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| Jul3-10, 08:13 PM | #13 |
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This is not as simple a question as it looks.
One has to decide what it means to "test an equation", and usually this means that an alternate form of the equation is posited with some extra parameter(s), such that if this parameter is zero the original equation is recovered. For example, Newton's 2nd law could be expressed as F = ma + x, and experiments undertaken to measure x. However, there are an infinite number of such forms. For example, I could also write down F = (1 + y)(ma), and try and measure y. Where it gets complicated is when you have multiple equations. For example, I can posit a modified Faraday's law: [tex]\nabla \times \mathbf{E} = -(1+k_1) \frac{\partial \mathbf{B}} {\partial t}[/tex] and I will discover there are very stringent limits on k1: it's smaller than 10-10. Likewise, I can instead modify Ampere's Law to get [tex]\nabla \times \mathbf{B} = \mu_0\mathbf{J} + (1 + k_2) \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} [/tex] and I will again discover there are very stringent limits on k2: it's also smaller than 10-10. However, if I made both changes, what I will discover is that [itex]| k_1 + k_2 | < 10^{-10}[/itex], but my actual constraints on k1 and k2 individually are about a thousand times weaker. So by going from a theory with one extra parameter to one with two, I can evade many experimental limits. Put another way, I can always find a (arbitrarily large) set of parameters that will agree with measurements. But that's not very useful. What is more useful is a model with a small number of additional free parameters. One of the most well known is the Proca theory, which has: [tex]\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} - \mu^2 \phi [/itex] and [tex]\nabla \times \mathbf{B} = \mu_0\mathbf{J} +\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} - \mu^2 \mathbf{A}[/tex] where [itex]\phi[/itex] and [itex]\mathbf{A}[/itex] are the potentials of the electric and magnetic fields, and [itex]\mu[/itex] is a new parameter of the theory. It has dimensions*, which is maybe not so nice (a pure number would be easier to interpret), but experimentally it is very small: about 10-30 meters. * It has to, because it links fields and potentials, which have different dimensions. |
| Jul3-10, 09:45 PM | #14 |
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| Jul3-10, 10:09 PM | #15 |
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