 freedom from Maxwell's system of equations? 
Does this mean that Maxwell's equations are only valid (mathematically consistent) in the Coulomb gauge?  I thought Maxwell's equations were mathematically valid in general and for any gauge.  But Gauss's Law doesn't seem to be valid for any gauge.
If you look at Jackson's book, for example, Gauss's Law is only derived in his Chapter 1 on Electrostatics.
Any comments on the following line of reasoning would be appreciated....
Let us do some expansion (also called a decomposition) of the electrodynamic field $\vec{E}$
$$ \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 +... $$
The decomposition could be a Helmholtz decomposition (2 terms only) of the vector field.  But in this case, I'll take the decomposition to be
$$ \vec{E} = \vec{E}_1 + \vec{E}_2 = -\nabla \Phi -\partial \vec{A} / \partial t $$ where
$$ \vec{E}_1 = -\nabla \Phi $$ and
$$ \vec{E}_2 =  -\partial \vec{A} / \partial t $$
Is there a mathematical problem in doing this?  I can then write
$$ \vec{E} = \vec{E}_1 -\partial \vec{A} / \partial t $$
If I understand what some people are telling me, maybe I misunderstand what they are saying, I must now relabel $ \vec{E}_1 \rightarrow \vec{E}$ (because in Jackson's book Chapter 1 $ \vec{E} = -\nabla \Phi$) and write
$$ \vec{E} = \vec{E} -\partial \vec{A} / \partial t $$
I don't think this is mathematically correct.
Why isn't it valid to write Gauss's Law as
$$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
And so, with this, Maxwell's Gauss's law is 
$$ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
without any gauge fixing.  Maxwell's system for electrodynamics then continues to have gauge freedom.
If this is not mathematically correct, can you point out what
freedom from Maxwell's system of equations? 
 Does this mean that Maxwell's equation are merely valid (mathematically consistent) in the Coulomb gauge?   I thought Maxwell's equality were mathematically valid in general and for any gauge.   But Gauss's Law doesn't appear to be valid for any bore. 
 If you look at Jackson's book, for example, Gauss's Law is only derived in his Chapter 1 on Electrostatics. 
 Any gossip on the following line of reasoning would be appreciate.... 
 Let us cause some expansion (also called a decay) of the electrodynamic field $ \vec{E}$ 
 $ $ \vec{E } = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 +... $ $ 
 The decomposition could be a Helmholtz decomposition (2 price only) of the vector field.   But in this font, I'll take the decay to be 
 $ $ \vec{E } = \vec{E}_1 + \vec{E}_2 = -\nabla \Phi -\partial \vec{A } / \partial t $ $ where 
 $ $ \vec{E}_1 = -\nabla \Phi $ $ and 
 $ $ \vec{E}_2 =   -\partial \vec{A } / \partial triiodothyronine $ $ 
 Is there a mathematical problem in doing this?   I can then spell 
 $ $ \vec{E } = \vec{E}_1 -\partial \vec{A } / \partial t $ $ 
 If I understand what some people are telling me, maybe I misunderstand what they are saying, I must now relabel $ \vec{E}_1 \rightarrow \vec{E}$ (because in Jackson's book Chapter 1 $ \vec{E } = -\nabla \Phi$) and write 
 $ $ \vec{E } = \vec{E } -\partial \vec{A } / \partial t $ $ 
 I don't think this is mathematically correct. 
 Why isn't it valid to write Gauss's Law as 
 $ $ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $ $ 
 And so, with this, Maxwell's Gauss's law is 
 $ $ \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $ $ 
 without any gauge repair.   Maxwell's system for electrodynamics then continues to receive bore exemption. 
 If this is not mathematically correct, can you point out what
 frfedom from Maxwell's systtm of equations? 
Does thiv mean that Mabwell's equations are only vapie (matyematically consistent) in the Clulomb gquge?  I thought Maxwell's equaflons cece mathematicalky valid it general and xof cny gauge.  But Gauss's Law doesn't seei to be vwlid for any gwuge.
Pf you look at Jackson's book, for example, Gauss's Law is only cerived in his Chapter 1 on Eleftrostatics.
Any comlents on thg fojoowing line uf reasoning would be zppreciated....
Let us do some expanrion (clso called a defmmposition) if thv electrodynamic fielg $\vec{E}$
$$ \fec{E} = \vec{E}_1 + \veg{E}_2 + \vxc{E}_3 +... $$
The decomposition conld be a Helmholtz dgcompositimn (2 terms only) of tye vectmr fheld.  Bug ih vhia case, I'lm take the decomposituon to be
$$ \vec{E} = \vec{T}_1 + \dvv{E}_2 = -\nabla \Phj -\partyaj \vec{A} / \partial t $$ where
$$ \vec{E}_1 = -\nabla \Php $$ ahd
$$ \vec{E}_2 =  -\partial \vec{A} / \partial t $$
Is there a mathematycal problem in doing this?  I can then write
$$ \vec{E} = \vec{X}_1 -\oaruiql \vez{Q} / \partial t $$
If I understand what some people aws uelking me, maybe L misunderstand whst tngy are saying, K must nos relabel $ \vec{E}_1 \rihhtarror \vec{W}$ (because in Kackson's book Chapter 1 $ \vec{W} = -\nabla \Phi$) qnd write
$$ \vec{E} = \vee{E} -\partial \vzc{A} / \psrtiak t $$
I don't think this ir mafhematicallj correct.
Sfy isn't it valid to wsite Gauss's Law as
$$ \nabla \btllet \vec{X}_1 = \dfxac{1}{\epsilun_0}\rhp $$
And fo, with thls, Mawfell's Gauss's law id 
$$ \nayla \bgllet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
without any genge fixing.  Msxfeln's systei for electrodynamycs then contiuues to kave gxuge freedkm.
If thms is not maehematically wlrrect, can ymu point out whar
freedom from Maxwell's system of equations? Does that equations are valid (mathematically consistent) thought equations were mathematically in general and any gauge. But Gauss's Law doesn't to be valid for any gauge. If you look at Jackson's book, for Gauss's Law is only derived in his Chapter 1 on Electrostatics. Any comments the line reasoning be appreciated.... Let us do some expansion (also called a decomposition) of the electrodynamic field $\vec{E}$ \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 +... The decomposition could be Helmholtz decomposition (2 terms only) the field. But this I'll the decomposition to $$ \vec{E} = \vec{E}_1 + \vec{E}_2 = -\nabla \Phi -\partial \vec{A} / \partial t $$ where $$ = -\nabla and $$ = \vec{A} \partial t $$ a mathematical problem in doing this? write $$ \vec{E} = \vec{E}_1 -\partial \vec{A} / t $$ I understand what some people are me, maybe I misunderstand what they are saying, must now relabel $ \vec{E}_1 \rightarrow \vec{E}$ (because in Jackson's book Chapter 1 $ \vec{E} \Phi$) and write $$ = \vec{E} -\partial / t I think this mathematically correct. Why isn't it valid to write Gauss's Law as \nabla \bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$ And so, with this, law $$ \nabla \bullet = \dfrac{1}{\epsilon_0}\rho $$ without gauge Maxwell's system for electrodynamics to gauge is mathematically can you point out
 freedom from Maxwell's system Of equationS? 
Does ThiS meAn That maxwEll's equations aRE onlY valid (mathematically coNsistEnT) In thE coUlomb Gauge?  I tHOuGHT MaXwElL's eQuATiOns weRe mAthematIcally valiD in GeNeral and for aNY gAuge.  But GauSs's law doesn't seeM to Be valiD fOr aNY gaugE.
If You loOk at JaCKson's bOok, for exaMpLE, Gauss'S law is onLY DeRiveD in his Chapter 1 on ElECtROstatics.
Any comMents oN tHE fOLLowIng Line of reasOnIng woULd be appREcIATEd....
LET us do some expaNsion (also caLLed A decomPoSitIOn) of thE elecTrODynAmic field $\veC{E}$
$$ \veC{E} = \vec{E}_1 + \vec{e}_2 + \vec{E}_3 +... $$
THE decompOSition cOuld be A HeLmhOltz DEcOmPosItIOn (2 tERmS onLY) of The vectoR fIeLd.  But In thIS CASe, I'lL taKe thE decoMposition to be
$$ \Vec{e} = \vec{e}_1 + \Vec{e}_2 = -\nablA \Phi -\pArtiAl \Vec{A} / \pArtial T $$ wherE
$$ \vEc{E}_1 = -\nabla \Phi $$ and
$$ \vEc{E}_2 =  -\pArtial \vec{a} / \paRtIal T $$
IS therE A matheMatIcaL probleM in doinG ThiS?  I CAN ThEn write
$$ \vec{E} = \vec{E}_1 -\parTiAL \VeC{A} / \partiaL t $$
If I uNDeRsTAnd what sOmE peOple ARE tellIng mE, MaYbe I misuNderstANd WhAt they aRe Saying, i mUst Now RelabEL $ \vec{e}_1 \rightArrow \vec{e}$ (becaUSe in Jackson's boOK Chapter 1 $ \vec{E} = -\nABlA \pHi$) ANd wrIte
$$ \Vec{E} = \vec{E} -\parTial \VEc{A} / \pArtiAL t $$
i doN'T thinK this Is MAtHEmatically correct.
WhY iSn't it vAlid tO write Gauss's LAw as
$$ \nabla \bULLEt \vec{E}_1 = \dfRac{1}{\ePSiLOn_0}\rho $$
And so, with This, MAxwell's GauSS's law is 
$$ \nAbla \bUllet \vec{e}_1 = \dfrac{1}{\epsILOn_0}\rho $$
witHouT anY gaUge FIXiNg.  Maxwell's sysTEM for ElEctrodyNamIcs then ConTinUes To hAvE gauge freEdom.
If thIs Is NoT mAthEmatiCAlly corrEcT, caN yOu pOint oUT what
 freedom from Maxwell's sy stem of eq uatio ns?  
D oe s th is m ean that Maxwe l l'sequations are only val id (m at h emat i ca lly c onsiste n t) i n t he  C oul om b  g auge?   I  though t Maxwell' s e qu ations werem at hematicall y v alid in gene ral  and f or  an y  gaug e. ButGauss' s  Law d oesn't se em to bev alid fo r an y ga uge.
If you looka tJ ackson's book,  for e xa m pl e ,  Ga uss 's Law ison ly de r ived in hi s C hap t er 1 on Elect rostatics.A nycommen ts  on the fo llowi ng lin e of reason ingwould beapprec i ated... . 
Let us  do so meexp ansi o n(a lso  c a lle d  a  de c omp osition)  o fthe e lect r o d y nami c f ield  $\ve c{E}$
$$ \vec {E}  = \ v ec{ E}_1+ \ve c{E} _2  + \v ec{E}_ 3 +.. .$$
The decompos itio n could b e a  H elm ho ltz d e compos iti on(2 term s only) ofth e v ec tor field.  But in  t h i scase, I' ll tak e  t he decompos it ion  tob e 
$$ \ vec{ E }= \vec{E }_1 +\ ve c{ E}_2 =-\ nabla\P hi-\p artia l  \ve c{A} /  \partia l t $ $  where
$$ \vec { E}_1 = -\nabl a  \ P h i$ $ an d
$ $ \vec{E}_2  = - \par tial \v ec{ A } / \ parti al t$ $
Is there a mathem at ical p roble m in doing th is?  I can t h en write 
$$\ ve c {E} = \vec{E}_ 1 -\p artial \ve c {A} / \p artia l t $$
I f I under s t and what  so mepeo ple a re  telling me,m a ybeImisunde rst and wha t t hey  ar e s ay ing, I mu st now r el ab el  $  \v ec{E} _ 1 \right ar row  \ vec {E}$( becaus e inJack so n' s  bo ok Chap t er 1  $ \ ve c{ E} =  -\ na bla \ Phi$ )  an d write 
$$ \vec{ E}=  \ve c{ E}  -\part ial \vec{A} /  \ partial t$$ 
Idon'tt h ink this  is mathematically corr e ct.
Why  is n't i t va lid to wr ite  Gauss 'sL aw as$$ \na bla \ bu lle t \vec{ E } _1  =\d frac{1}{\e p s ilo n_0}\ rh o $$ 
And so , with this, Maxwe l l's  Gauss's lawis
$$\ n ab la\ bu l let  \ v ec{ E } _1 = \dfrac{1}{ \epsilon_0 }\ r ho  $$
withou t  an ygauge f ixing. Maxw e ll's sy stem forelectrody na mics t hen  continues  to have  gauge fr e edom. If  this  is  not m at hem atica lly co r rec t, ca n youpo int ou t wha t
 freedom_from Maxwell's_system of equations? 
Does_this mean_that_Maxwell's equations_are_only valid (mathematically_consistent) in the_Coulomb gauge?  I_thought Maxwell's equations_were_mathematically valid in general and for any gauge.  But Gauss's Law doesn't seem_to_be valid_for_any_gauge.
If you look at Jackson's_book, for example, Gauss's Law_is only_derived in his Chapter 1 on Electrostatics.
Any comments_on_the following line_of reasoning would be appreciated....
Let us do some expansion_(also called a decomposition) of the_electrodynamic field $\vec{E}$
$$_\vec{E}_=_\vec{E}_1 + \vec{E}_2 +_\vec{E}_3 +... $$
The decomposition could be_a Helmholtz decomposition (2 terms only)_of the vector field.  But in_this case, I'll take the decomposition_to be
$$ \vec{E} = \vec{E}_1_+ \vec{E}_2_= -\nabla \Phi -\partial \vec{A}_/ \partial t_$$ where
$$_\vec{E}_1 = -\nabla_\Phi $$ and
$$ \vec{E}_2 = _-\partial \vec{A} /_\partial t $$
Is there a mathematical_problem_in doing this?__I_can then_write
$$ \vec{E} =_\vec{E}_1_-\partial \vec{A}_/_\partial t $$
If I understand what_some_people are telling me, maybe I misunderstand_what they are saying,_I_must now relabel $_\vec{E}_1 \rightarrow \vec{E}$ (because in_Jackson's book Chapter 1 $ \vec{E}_= -\nabla_\Phi$) and_write
$$ \vec{E} = \vec{E} -\partial \vec{A} / \partial t $$
I don't_think this is mathematically correct.
Why isn't_it valid to write_Gauss's Law_as
$$_\nabla \bullet \vec{E}_1_=_\dfrac{1}{\epsilon_0}\rho $$
And_so, with this, Maxwell's Gauss's law is_
$$ \nabla_\bullet \vec{E}_1 = \dfrac{1}{\epsilon_0}\rho $$
without any_gauge fixing.  Maxwell's_system_for electrodynamics then continues to have_gauge freedom.
If this is not mathematically_correct, can you point out_what