Testing Symbol

  1. [symbol]Ñ[/symbol] x H = [symbol]e[/symbol][sub]0[/sub] [symbol]d[/symbol]E/[symbol]d[/symbol]t + j
  2. [symbol]Ñ[/symbol] x E = - [symbol]m[/symbol][sub]0[/sub] [symbol]d[/symbol]H/[symbol]d[/symbol]t
  3. [symbol]Ñ[/symbol] . H = 0
  4. [symbol]Ñ[/symbol] . E = [symbol]r[/symbol]/[symbol]e[/symbol][sub]0[/sub]

:slight_smile:

God said…

and there was light…

I can only assume that’s Maxwell’s Equations. Pretty cool, but FYI, it won’t look like that to everyone. I use Netscape, and here’s how it looks to me:

Try this:

  1. [symbol]Ñ[/symbol] x H = ε[sub]0[/sub] ∂E/∂t + j
  2. [symbol]Ñ[/symbol] x E = - μ[sub]0[/sub] ∂H/∂t
  3. [symbol]Ñ[/symbol] . H = 0
  4. [symbol]Ñ[/symbol] . E = ρ/ε[sub]0[/sub]

How does that look in Netscape? And anyone know another way to do the del operator?

ƒ

:frowning:

I’d thought Symbol was pretty much universal…

Ah well
Grim

To do a del operator in Netscape you can use:

∇ = ∇

But I don’t think is universal, either.

That shows up as a square in IE.

How was the rest of my post? Did rho, epsilon, mu and the partial differential show up correctly?

Yes they did. Although the dot product looked like a period.

test

tests:
:wink:
:slight_smile:
:stuck_out_tongue:
:B
:E
:smiley:

:stuck_out_tongue:
:open_mouth:
:o

:mad:
:smack:

And for some reason, your B for the magnetic field came out looking like an H, and we all know that’s not right. After all, it’s all vacuum anyway.

<d&r>

You might also try, by the way, just using a superscript period for a dot product. It won’t copy and paste well, but it will at least look right on (almost) all browsers.

I think that just using · will work too. Let’s see:

A · B

D = ε[sub]o[/sub]E + P
H = B/μ[sub]o[/sub] − M

[symbol]Ñ[/symbol] x H = ∂D/∂t + J
[symbol]Ñ[/symbol] x E = − ∂B/∂t
[symbol]Ñ[/symbol] · B = 0
[symbol]Ñ[/symbol] · D = ρ

Happy with that version, Chronos?

And if it was free space, I could write:

[symbol]Ñ[/symbol][sup]2[/sup] x B = μ[sub]o[/sub]ε[sub]o[/sub] ∂[sup]2[/sup]B/∂t[sup]2[/sup]
[symbol]Ñ[/symbol][sup]2[/sup] x E = μ[sub]o[/sub]ε[sub]o[/sub] ∂[sup]2[/sup]E/∂t[sup]2[/sup]

Alternatively, in Heaviside-Lorentz:

D = E + P
H = BM

[symbol]Ñ[/symbol] x H = (∂D/∂t + J)/c
[symbol]Ñ[/symbol] x E = (-∂B/∂t)/c
[symbol]Ñ[/symbol] · B = 0
[symbol]Ñ[/symbol] · D = ρ

May as well do the others, too.

Gaussian:

D = E + 4[symbol]p[/symbol]P
H = B − 4[symbol]p[/symbol]M

[symbol]Ñ[/symbol] x H = (∂D/∂t + 4[symbol]p[/symbol]J)/c
[symbol]Ñ[/symbol] x E = (-∂B/∂t)/c
[symbol]Ñ[/symbol] · B = 0
[symbol]Ñ[/symbol] · D = 4[symbol]p[/symbol]ρ

cgs esu:

D = E + 4[symbol]p[/symbol]P
H = c[sup]2[/sup]B − 4[symbol]p[/symbol]M

[symbol]Ñ[/symbol] x H = ∂D/∂t + 4[symbol]p[/symbol]J
[symbol]Ñ[/symbol] x E = -∂B/∂t
[symbol]Ñ[/symbol] · B = 0
[symbol]Ñ[/symbol] · D = 4[symbol]p[/symbol]ρ

And, lastly, cgs emu:

D = E/c[sup]2[/sup] + 4[symbol]p[/symbol]P
H = B − 4[symbol]p[/symbol]M

[symbol]Ñ[/symbol] x H = ∂D/∂t + 4[symbol]p[/symbol]J
[symbol]Ñ[/symbol] x E = -∂B/∂t
[symbol]Ñ[/symbol] · B = 0
[symbol]Ñ[/symbol] · D = 4[symbol]p[/symbol]ρ

Buncha fuggin’ geeks.