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Relation between H and B fields, and D and E fields 
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#1
Dec706, 05:22 PM

P: 129

Hey!
I am having some trouble understanding why the magnetic flux density B and the magnetic field strength H does not have the same units. I mean, as far as I understand, the H field is just the B field with the magnetic properties (magnetization) of the material taken into account? Isn't this correct? So why isn't the H field just the B field multiplied by some dimensionless scalar (or tensor)? The same thing applies to the electric field E and the electric displacement D. If the D field is the electric field with the polarization taken into account, why do they not have the same units, and why is D not just E times some scalar value which depends on the polarizability of the material? 


#2
Dec706, 06:35 PM

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The polarisability however has units, it is not a dimensionless quantity, so when we multiply some quantity by another, nondimensionless (dimensioned?) quantity (i.e. multiplying E by polarisability), we must necessarily obtain a quantity that has different units. So your question really boils down to "why does the polarisability have units?" In which case the answer should be fairly selfevident. Claude. 


#3
Dec806, 07:39 PM

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PF Gold
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Most elementary and intermediate texts use the confusing SI system, but working physicists more often use gaussian for their own calculations. The 3rd ediltion of Jackson triles to go to SI, but even he as to use gaussian to complete the subject. 


#4
Dec906, 02:46 AM

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Relation between H and B fields, and D and E fields
The confusion arises only when one's taught SI units in highschool and follows a CGSbased book in college, or viceversa.
If one's being kept in the same "environment" along his academic studies (HS+ college), there's no confusion. Daniel. 


#5
Dec1006, 05:07 AM

P: 129

Thanks for the replies, good to know my thinking was right!
It seems strange however that E and D have different units in one system of units, and the same units in another one. I mean, one could never define another system of units where a newton had the same units as coulomb, right? 


#6
Dec1006, 08:35 AM

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In gaussian units, charge and current are defined in absolute terms by the forces between two charges and between two wires.
Then there is no arbitrariness about their units. 1 stacoulomb=1 dynecm^2 from Coulomb's law. 1 statamp=1 sc/sec if the 1/c^2 is put into the force between two wires. That connection (1/c^2) was measured by Weber and Kolrausch in 1856. SI was developed 100 years ago because an engineer named Georgi thought that electric charge was a new unit, independent of force. Thus SI was originally called MKSA (A for ampere). Geogi did not know that there were other charges besides electric charge, and that after SR and QM physicists would realize that all charge (in terms of alpha) is dimensionless. Trying to give a dimension to an intrinsicly dimensionless quantity is why SI has a different unit for so many things (E,P,D,B,M,H) that should have the same unit. If SI can give units to empty space, then it would be equally sensible (really nonesensical) to define the unit of charge as one newton if epsilonzero were given different arbitrary units than it has in SI. 


#7
Dec1106, 02:53 AM

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#8
Dec1106, 06:52 AM

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(It would be divided by 4piepsilonzero in SI). Putting numbers in in ANY system gives the dimensionless number 1/137.036. This is called alpha and is the dimensionless charge value. 


#9
Dec1306, 09:24 PM

P: 2,251

from an SI pointofview, D is measuring flux. think of this in terms of the natural meaning of an inversesquare action. we have a point charge Q "emmiting" a total quantity of flux that is the same as the amount of charge Q. these lines of flux are emmitted equally in all directions in 3 dimensional space. the flux density is the density of these lines of flux per unit of area that is perpendicular to the lines of flux (perpendicular to the line connecting Q to that unit area). a sphere has a surface area of [itex] 4 \pi r^2 [/itex] so this total flux of Q is distributed equally among that entire surface area. that means that that the flux density, D, is the total flux, Q, divided by the total surface area [itex] 4 \pi r^2 [/itex] or [tex] D = \frac{Q}{4 \pi r^2} [/tex] or in vector form (assuming Q is at the origin) [tex] \mathbf{D} = \frac{Q}{4 \pi  \mathbf{r} ^3} \mathbf{r} [/tex] so flux density, D, is the physical quantity with dimension of charge per area or charge per square of length [Q][L]^{2}. now the electrostatic field, E, is the physical quantity that represents how much force per unit charge that there is applied to a test charge, q, placed at position r from the main charge Q (which is at the origin). from the Coulomb Force equation we know that [tex] F = \frac{1}{4 \pi \epsilon_0} \frac{Q q}{r^2} [/tex] or in vector form (assuming Q is at the origin) [tex] \mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{Q q}{ \mathbf{r} ^3} \mathbf{r} [/tex] and the E field is [tex] E = \frac{F}{q} = \frac{Q}{4 \pi \epsilon_0 r^2} [/tex] or in vector form (assuming Q is at the origin) [tex] \mathbf{E} = \frac{\mathbf{F}}{q} = \frac{Q}{4 \pi \epsilon_0  \mathbf{r} ^3} \mathbf{r} [/tex] electrostatic field, E, is the physical quantity with dimension of force per charge or masslength per timesquared per charge: [M][L][T]^{2}[Q]^{1}. so comparing the equations, what is the relationship between flux density, D, and electric field strength, E? 


#10
Dec1306, 09:51 PM

P: 2,251

[tex] \alpha = \frac{e^2}{\hbar c} [/tex] is only dimensionless (and currently believed to be 1/137.03599911) in unit systems that define the unit charge so that the Coulomb Force constant [itex] 1/(4 \pi \epsilon_0) [/itex] is 1 and disappears from the equations of physical law, namely the Coulomb force law (sorta like defining the unit force so that the constant C in [itex] F = C dp/dt [/itex] is 1 and goes away). this is an arbitrary human decision. in any system of units, the general expression for the Finestructure constant is [tex] \alpha = \frac{e^2}{\hbar c (4 \pi \epsilon_0)} [/tex] . (i know you know this, Meir, but i think in introducing this to someone, one should not say simply "e^2/(hbar c)".) one thing also to point out is that this Finestructure constant really is the square of the elementary charge when measured in Planck units: [tex] \alpha = \frac{e^2}{\hbar c (4 \pi \epsilon_0)} = \left( \frac{e}{q_P} \right)^2 [/tex] and can be thought of, in a world of natural units where all these scaling constants in the field equations go away, and given a constellation of charged bodies all with fixed numbers of protons and electrons in these charged bodies, that [itex] \alpha [/itex] represents the strength of the electromagnetic action. 


#11
Dec1306, 10:14 PM

P: 2,251

BTW, Meir, i know that this is an old thread (and a different thread, but one of the issues are the same), i must say that i agree with you and marcusl fully about this. "H" should be called "flux" and "B" should be called "field".



#12
Dec1406, 10:25 AM

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1. The term "fine structure constant" is the historical designation because it was first noted in atomic spectrocopy 100 years ago. Now that we know that alpha is a standard ratio between many numbers in physics, it is probably time to not limit it to fine structure, and just call it alpha, but that is not an important point. 2. I did mean to imply that alpha had somewhat different algebraic forms in different systems of units. That is why I explicity mentioned the division by fourpiepsilonzero in SI. I did not mention, but should have that it is simplest in the form of natural units used today by most HE theorists, where alpha=e^2. Fortunately, I know of no one (you and I included) who redefines alpha to be anything other than 1/137. e^2, on the other hand, has many different values in different unit systems. 3. I am glad we both agree that "[itex] \alpha [/itex] represents the strength of the electromagnetic action". I would only add: whatever system of units is used. Thank you for your interest. 


#13
Dec1406, 10:28 AM

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#14
Dec1406, 10:58 AM

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[tex] \alpha = \frac{e^2}{\hbar c (4 \pi \epsilon_0)} [/tex] and then just note that [itex] 4 \pi \epsilon_0 [/itex] gets set to 1 in some systems of units because of the manner that the unit charge is defined in those systems of units. personally, i think it's much more natural to define charge so that [itex] \epsilon_0 = 1 [/itex] (and, for gravitation set [itex] 4 \pi G = 1 [/itex]) and then (with [itex] c = \hbar = 1 [/itex]) you get [itex] \sqrt{4 \pi \alpha} = e [/itex] which i think is the more natural and salient dimensionless number for which \alpha is derived. 


#15
Dec1406, 10:59 AM

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From a more abstract viewpoint, H and B aren't even the same type of geometrical object. H is a [twisted]1form in space, which is associated with a lineintegral, and B is a 2form in space, which is associated with a surfaceintegral. Similarly, E is a 1form and D is a [twisted]2form.
http://arxiv.org/abs/physics/0407022 (see pictures on page 7) 


#16
Dec1406, 11:10 AM

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#17
Dec1406, 01:31 PM

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[tex]\Phi = \int \vec{B} \cdot d\vec{A}[/tex] and has units of Webers (SI) or Maxwells (Gaussian/cgs). It might be better to join authors such as E. Weber and Julius Stratton who call H "magnetic intensity." 


#18
Dec1506, 02:34 AM

P: 2,251

[tex]\Phi = \int \vec{B} \cdot d\vec{A}[/tex] but, if E is "field" (because it is related to intensity of effect) and D is "flux density" (because it is related to the density of how much of the source of the effect is present at some point), to be consistent, shouldn't they have named "H" as "magnetic flux density" and "B" as "magnetic field"? isn't calling "B" a flux density an historical accident? 


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