Consider the inhomogeneous Maxwell equations in (3+1) notation (Gaussian CGS units) with electric charge-density and electric 3-current-density on the RHS.GRQFT said:I'm at a level where I I know enough to ask about stuff like the Hodge dual, but not really understand it.
What I meant was that the process of making ##c## dimensionless is to absorb it into another quantity. The reason is that the dimensions/units of that other quantity had to change to accommodate the fact that ##c## became dimensionless.Dale said:No. If ##c## is a dimensionless 1 then you could equally say that it was absorbed into the space coordinate. Or that its cube was absorbed into the time coordinate and its fourth root was absorbed into the space coordinate. Far easier to just say it just doesn’t exist in those units. There is no reason for it to be absorbed anywhere and inherent ambiguity if you try to assert that it is absorbed somewhere.
Indeed. Just find the reason why it does exist in the other. The absence of that reason is the reason that it doesn’t exist in the one.
For instance, ##\epsilon_0## exists in SI units to make Coulomb’s law dimensionally consistent. That is why it does exist in SI. In Gaussian units Coulomb’s law is naturally dimensionally consistent. That is why it doesn’t exist in Gaussian units.
You can do a similar exercise with every dimensionful constant that is present in one system of units and not in another.
So ##k## is dimensionless. It does not affect the units.Dale said:As an exercise for you, trace why the constant ##k## in Newton’s 2nd law, ##\vec F_{net}=km \vec a##, doesn’t exist in SI units.
Also, if you want to assert that ##k## does exist in SI, but is absorbed somewhere, then where is it absorbed and how do you know it was absorbed there and not somewhere else?
Interesting example. I agree that the two ways to see the conversion is effectively equivalent. However, I still prefer to think of it as being absorbed, because it explains the change in the units of the quantity into which it is absorbed.renormalize said:I would say instead that ##c## simply disappears by choosing uniform units, as I tried to explain in post #118.
Please indulge me by pondering an exactly analogous situation in 3D that persists to this day in US sea navigation. American maritime vessels measure distance traveled in nautical miles (NM), but the depth below them is given in fathoms (ftm). The relation between these quantities is ##1\,\text{NM}=\mathfrak{c}\,\text{ftm}## where the "mysterious" conversion factor ##\mathfrak{c}## is ##1012.69##. But the rest of the world, not being wedded to imperial tradition, more rationally measures height, width and depth all in the same uniform units, namely meters. The rest of the world thinks that the "conversion" between transverse-distance and longitudinal-depth is simply unity (if they think about it all). So what happened to the "dimensionful" constant ##\mathfrak{c}##? You can say it was absorbed into the depth coordinate, whereas I claim it disappeared by making the natural choice of uniform units for height, width, and depth. But in either view, the result is consistent with saying ##\mathfrak{c}=1\,##, and is dimensionless, in uniform units.
And that same reasoning applies to ##c## in the unified theory of Minkowski spacetime.
Interesting. Applying the Hodge dual in this way seems specific to EM given that it comprises two different yet equally important fields (E & B).Sagittarius A-Star said:Consider the inhomogeneous Maxwell equations in (3+1) notation (Gaussian CGS units) with electric charge-density and electric 3-current-density on the RHS.
##\nabla \cdot \mathbf{E} = 4\pi\rho_{el}##
##\nabla \times \mathbf{B} - \frac{\partial \mathbf{E}}{\partial {(ct)}}= \frac{4\pi}{c}\mathbf{J_{el}} ##
Take these equations and substitute ##B \mapsto E##, ##E \mapsto -B##.
Then you will get the homogeneous Maxwell equations in (3+1) notation (Gaussian CGS units) with magnetic charge-density and magnetic 3-current-density on the RHS.
##\nabla \cdot \mathbf{B} = 4\pi\rho_m =0##
##\nabla \times \mathbf{E} +\frac{\partial \mathbf{B}}{\partial {(ct)}}= \frac{4\pi}{c}\mathbf{J_m} = 0##
Now consider the EM field tensor, Gaussian CGS units with (+++-) convention.
##F^{\mu\nu} = \begin{pmatrix} 0 & B_z & -B_y & -E_x \\ -B_z & 0 & B_x & -E_y \\ B_y & -B_x & 0 & -E_z \\ E_x & E_y & E_z & 0 \end{pmatrix}##
Do the same substitution ##B \mapsto E##, ##E \mapsto -B##. Then you get the Hodge dual.
##
\tilde F^{\mu\nu}
=
\begin{pmatrix}
0 & E_z & -E_y & B_x\\
-E_z & 0 & E_x & B_y\\
E_y & -E_x & 0 & B_z\\
-B_x & -B_y & -B_z & 0
\end{pmatrix}
##
Therefore, you can convert Maxwells inhomogeneous equation
##\partial_\nu F^{\mu\nu} = \frac{4\pi}{c} {J_{el}}^\mu##
into Maxwells homogeneous equation
##\partial_\nu \tilde{F}^{\mu\nu} = \frac{4\pi}{c} {J_m}^\mu = 0##
by substituting the EM field tensor by it's Hodge dual.
The Hodge dual is quite a beautiful operator defined on the exterior algebra of an orientable manifold with a metric. For such a manifold of dimension ##n##, it defines an isomorphism between ##k##-forms (or ##k##-vectors) and ##(n-k)##-forms (or ##(n-k)##-vectors).GRQFT said:Interesting. Applying the Hodge dual in this way seems specific to EM given that it comprises two different yet equally important fields (E & B).
The stress energy tensor is a proper tensor, it can't both be zero and non-zero at a given point in spacetime. I assume that's what you meant by your correspondence, for both the homogeneous and inhomogeneous Maxwell equations are valid at a given point in spacetime.GRQFT said:Then again, I wonder if this can also be done for gravitational radiation (in the weak-field limit) to get 'homogeneous' and 'inhomogeneous' gravitational wave equations corresponding to zero and non-zero stress-energy tensor, respectively.
Applying the Hedge dual in this way switches the interior derivative of the field between electric and magnetic sources (zero, because magnetic monopoles do not exist).GRQFT said:Interesting. Applying the Hodge dual in this way seems specific to EM given that it comprises two different yet equally important fields (E & B).
Source:Wikipedia said:As a result of quantization, the quantum electrodynamic vacuum can be considered a material medium.[20] It is capable of vacuum polarization.[21][22] In particular, the force law between charged particles is affected.[23][24] The electrical permittivity of the quantum electrodynamic vacuum can be calculated, and it differs slightly from the simple ε0 of the classical vacuum. (Likewise, its permeability can also be calculated and differs slightly from μ0.) This medium is a dielectric with relative dielectric constant greater than 1, and is diamagnetic, with relative magnetic permeability less than 1.[25][26]
Sagittarius A-Star said:They say that the QED vacuum can be considered a material medium with ##\epsilon_r >1## and ##\mu_r <1##.
Is this misleading or false?
Yes, it is dimensionless in SI units, but not in imperial units. In imperial units it is ##0.031 \mathrm{\ lbf \ s^2 \ lbm^{-1} \ ft^{-1}}##. So where did ##k## get absorbed in SI?flippiefanus said:So ##k## is dimensionless. It does not affect the units.
So, in the process of making ##k## dimensionless in SI units, which quantity had to change to accommodate the fact that ##k## became dimensionless in SI?flippiefanus said:What I meant was that the process of making ##c## dimensionless is to absorb it into another quantity. The reason is that the dimensions/units of that other quantity had to change to accommodate the fact that ##c## became dimensionless.
That is a possible alternative. Just make it a personal preference. Then you can use it when and where you find it convenient, and not use it where you don’t.flippiefanus said:I still prefer to think of it as being absorbed, because it explains the change in the units of the quantity into which it is absorbed.