User:Quondum/sandbox

template:val links

2 m/s × 3 m/s × 4 m/s (ideally, only the first occurrence should be linked)

candela

cd

dacd hcd kcd Mcd Gcd Tcd Pcd Ecd Zcd Ycd Rcd Qcd

dcd ccd mcd ucd μcd ncd pcd fcd acd zcd ycd rcd qcd

lumen

lm

dalm hlm klm Mlm Glm Tlm Plm Elm Zlm Ylm Rlm Qlm

dlm clm mlm ulm μlm nlm plm flm alm zlm ylm rlm qlm

lux

lx

dalx hlx klx Mlx Glx Tlx Plx Elx Zlx Ylx Rlx Qlx

dlx clx mlx ulx μlx nlx plx flx alx zlx ylx rlx qlx

Breakable gaps template

transclude

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaa|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaa|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

subst

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaa|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaa|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

|aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccccccccc|

GEM

Define

$\mu ={\frac {1}{c^{2}\varepsilon _{0}}}$ (EM)

$\mu _{\text{g}}=-{\frac {4\pi G}{c^{2}}}$ (GEM)

Lorenz gauge used for vector potentials

EM	Proposed	Mashhoon
$(\phi ,\mathbf {A} )$	$(\phi _{\text{g}},\mathbf {A} _{\text{g}})=(\phi _{\text{g}},4\mathbf {A} _{\text{g}}')$	$(\Phi _{\text{m}},\mathbf {A} _{\text{m}})=(-\phi _{\text{g}},-{\tfrac {c}{2}}\mathbf {A} _{\text{g}})$
$(\mathbf {E} ,\mathbf {B} )$	$(\mathbf {E} _{\text{g}},\mathbf {B} _{\text{g}})=(\mathbf {E} _{\text{g}},4\mathbf {B} _{\text{g}}')$	$(\mathbf {E} _{\text{m}},\mathbf {B} _{\text{m}})=(-\mathbf {E} _{\text{g}},-{\tfrac {c}{2}}\mathbf {B} _{\text{g}})$
$(\rho ,\mathbf {j} )$	$(\rho _{\text{g}},\mathbf {j} _{\text{g}})$	$(\rho _{\text{m}},\mathbf {j} _{\text{m}})=(\rho _{\text{g}},\mathbf {j} _{\text{g}})$
${\tfrac {1}{c^{2}}}\partial _{t}\phi +\nabla \cdot \mathbf {A} =0$	${\tfrac {1}{c^{2}}}\partial _{t}\phi _{\text{g}}+\nabla \cdot \mathbf {A} _{\text{g}}'=0$	(1.6) ${\tfrac {1}{c}}\partial _{t}\Phi _{\text{m}}+\nabla \cdot ({\tfrac {1}{2}}\mathbf {A} _{\text{m}})=0$
$\mathbf {E} =-\nabla \phi -\partial _{t}\mathbf {A}$	$\mathbf {E} _{\text{g}}=-\nabla \phi _{\text{g}}-\partial _{t}\mathbf {A} _{\text{g}}'$	(1.7) $\mathbf {E} _{\text{m}}=-\nabla \Phi _{\text{m}}-{\tfrac {1}{c}}\partial _{t}({\tfrac {1}{2}}\mathbf {A} _{\text{m}})$
$\mathbf {B} =\nabla \times \mathbf {A}$	$\mathbf {B} _{\text{g}}'=\nabla \times \mathbf {A} _{\text{g}}'$	(1.7) $\mathbf {B} _{\text{m}}=\nabla \times \mathbf {A} _{\text{m}}$
$\nabla \times \mathbf {E} +\partial _{t}\mathbf {B} =0$	$\nabla \times \mathbf {E} _{\text{g}}+\partial _{t}\mathbf {B} _{\text{g}}'=0$	(1.8) $\nabla \times \mathbf {E} _{\text{m}}+{\tfrac {1}{c}}\partial _{t}({\tfrac {1}{2}}\mathbf {B} _{\text{m}})=0$
$\nabla \cdot \mathbf {B} =0$	$\nabla \cdot \mathbf {B} _{\text{g}}'=0$	(1.8) $\nabla \cdot ({\tfrac {1}{2}}\mathbf {B} _{\text{m}})=0$
$\nabla \cdot \mathbf {E} =c^{2}\mu _{0}\rho$	$\nabla \cdot \mathbf {E} _{\text{g}}=c^{2}\mu _{\text{g}}\rho _{\text{g}}$	(1.9) $\nabla \cdot \mathbf {E} _{\text{m}}=-c^{2}\mu _{\text{g}}\rho _{\text{m}}$
$\nabla \times \mathbf {B} -{\tfrac {1}{c^{2}}}\partial _{t}\mathbf {E} =\mu _{0}\mathbf {j}$	$\nabla \times \mathbf {B} _{\text{g}}'-{\tfrac {1}{c^{2}}}\partial _{t}\mathbf {E} _{\text{g}}=\mu _{\text{g}}\mathbf {j} _{\text{g}}$	(1.9) $\nabla \times ({\tfrac {1}{2}}\mathbf {B} _{\text{m}})-{\tfrac {1}{c}}\partial _{t}\mathbf {E} _{\text{m}}=-c^{2}\mu _{\text{g}}{\tfrac {1}{c}}\mathbf {j} _{\text{m}}$
$\nabla \cdot \mathbf {j} +\partial _{t}\rho =0$	$\nabla \cdot \mathbf {j} _{\text{g}}+\partial _{t}\rho _{\text{g}}=0$	$\nabla \cdot \mathbf {j} _{\text{m}}+\partial _{t}\rho _{\text{m}}=0$
$\mathbf {F} =q(\mathbf {E} +\mathbf {v} \times \mathbf {B} )$	$\mathbf {F} =m(\mathbf {E} _{\text{g}}+\mathbf {v} \times \mathbf {B} _{\text{g}})$	(1.11) $\mathbf {F} =m(-\mathbf {E} _{\text{m}}-{\tfrac {2}{c}}\mathbf {v} \times \mathbf {B} _{\text{m}})$
$\mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B}$	$\mathbf {S} ={\frac {1}{\mu _{\text{g}}}}\mathbf {E} _{\text{g}}\times \mathbf {B} _{\text{g}}$	(1.17) $\mathbf {S} ={\frac {2}{c\mu _{\text{g}}}}\mathbf {E} _{\text{m}}\times \mathbf {B} _{\text{m}}$

rounding

test of template:physconst (trailing zeros)

137.03600

M(¹²C) ≈ 12.00000×10⁻³ kg⋅mol⁻¹

line breaking

test of template:su/sandbox

XXXXXXXXXXXXXXXXXX XXXXXXXXXX⁠^{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}
_{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}⁠XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX⁠^{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}
_{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}⁠XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

XXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX⁠^{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}
_{bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb}⁠XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

test of template:tmath

plainAAAAAAAAAAAAAAAAAAAAAA XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX$zzzzzzzzzzzzz$XXXXXXXXXXXXXXXXXXXXXXXX

NoBreakAAAAAAAAAAAAAAAAAAAA XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX⁠$zzzzzzzzzzzzz$⁠XXXXXXXXXXXXXXXXXXXXXXXX

zwjAAAAAAAAAAAAAAAAAAAAAAAA XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX‍$zzzzzzzzzzzzz$‍XXXXXXXXXXXXXXXXXXXXXXXX

tmathAAAAAAAAAAAAAAAAAAAAAA XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX⁠$zzzzzzzzzzzzz$⁠XXXXXXXXXXXXXXXXXXXXXXXX

reference combining

Variation 1

Body text1. Body text2. Body text3.

Variation 2

Body text1. Body text2. Body text3.

Variation 3

Body text1. Body text2. Body text3.

Variation 4

Body text1. Body text2. Body text3.