# How can one $E$-field formula be rewritten to another version [closed]

I will start with going through a rewriting of one E field to another that I understand which is that the total E field out of a sphere is derived as

$$\pmb{r}=x\pmb{i}+y\pmb{j}+z\pmb{k}=r\cos\theta \sin\varphi\pmb{i}+r\sin\theta \sin\varphi\pmb{j}+r\cos\varphi\pmb{k}$$

$$\dfrac{\dfrac{\partial\pmb{r}}{\partial r}}{\left|\dfrac{\partial\pmb{r}}{\partial r}\right|}=\cos\theta \sin\varphi\pmb{i}+\sin\theta \sin\varphi\pmb{j}+\cos\varphi\pmb{k}$$

we want to find the E flux out of a sphere

$$\dfrac{q}{4\pi\varepsilon_0 |r|^3 }\iint(r\cos\theta \sin\varphi\pmb{i}+r\sin\theta \sin\varphi\pmb{j}+r\cos\varphi\pmb{k})\cdot dA$$

$$=\frac{q}{4\pi \varepsilon_0 |r|^2 }\iint(\cos\theta \sin\varphi\pmb{i}+\sin\theta \sin\varphi\pmb{j}+\cos\varphi\pmb{k})\cdot (cos\theta \sin\varphi\pmb{i}+\sin\theta \sin\varphi\pmb{j}+\cos\varphi\pmb{k})r^2\sin\varphi d\theta d\varphi$$

$$=\frac{q}{4\pi\varepsilon_0} \int_{0}^{\pi}\int_{0}^{2\pi}(\cos^2\theta \sin^2\varphi+\sin^2\theta \sin^2\varphi+\cos^2\varphi)\sin\varphi d\theta d\varphi$$

$$=\frac{q}{4\pi\varepsilon_0} \int_{0}^{\pi}\int_{0}^{2\pi}\sin\varphi d\theta d\varphi=\frac{q}{\varepsilon_0}$$

In the derivation of the electromagnetic wave by Maxwell they introduce displacement current in the following way

If we integrate the current density over a surface S we get the charge in q inside the surface per time. This is given as this integral

$$\iint J(t) \cdot n dA=-\frac{dq}{dt}$$

Now assume we have a surface that encloses a volume V and we have charges flowing through this volume

If the charges that go through this volume then there is no change in charge per unit time and

$$\iint J(t) \cdot n dA=-\frac{dq}{dt}=0$$

But if more enters than leaves we have that

$$\iint J(t) \cdot n dA=-\frac{dq}{dt}\neq 0$$

Gauss law is given as:

$$\iint E \cdot n dA=\frac{q}{\varepsilon_0}$$

so that

$$\varepsilon_0\iint \frac{\partial E}{\partial t} \cdot n dA=\frac{dq}{dt}$$

So we obtain

$$\iint J(t) \cdot n dA+\varepsilon_0\iint \frac{\partial E}{\partial t} \cdot n dA$$

From this we see that we have a new definition of current density. It can be derived that

$$\nabla \times \pmb{B}=\mu_0 J(t)$$

So with our changing current density we obtain in differential version

$$\nabla \times \pmb{B}=\mu_0 J(t)+\mu_0 \varepsilon_0 \frac{\partial E}{\partial t}$$

Above in the part about displacement current they have established a new formula for curl of the magnetic field by using Gauss law for change in current density. Gauss law derived in the upper first part is based on a static environment of charges.

Then as they derive the E field that describes speed of light they do the following. First current is 0 which leads to the following Maxwell's equations:

$$\nabla \cdot \pmb{E}=0$$

$$\nabla \cdot \pmb{B}=0$$

$$\nabla \times \pmb{B}=\mu_0 \varepsilon_0 \frac{\partial E}{\partial t}$$

$$\nabla \times \pmb{E}=- \frac{\partial B}{\partial t}$$

By using the vector identity

$$\nabla \times(\nabla \times \pmb{E})=\nabla(\nabla \cdot \pmb{E})-\nabla^2 \pmb{E}$$

We want to combine the constraints above and

$$\nabla \times(\nabla \times \pmb{E})=- \frac{\partial \nabla \times B}{\partial t}=-\mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2}$$

By using the vector identity

$$\nabla^2 \pmb{E}-\mu_0 \varepsilon_0 \frac{\partial^2 E}{\partial t^2}=0$$

So that

$$(\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2}) (E_x,E_y,E_z)-\mu_0 \varepsilon_0 \frac{\partial^2 (E_x,E_y,E_z)}{\partial t^2}=0$$

If we only look at the x-dim:

$$(\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2}) (E_x)-\mu_0 \varepsilon_0 \frac{\partial^2 E_x}{\partial t^2}=0$$

Which has the solution

$$E_x=E_{x,0}e^{i(kx-\omega t)}$$

From calculus it is obtained that

$$e^{(i\theta)}=\cos \theta+i \sin\theta$$

$$E_{x,0}e^{i(kx-\omega t)}=E_{x,0}[\cos(kx-\omega t) +i \sin(kx-\omega t)]$$

Now I finally can ask my question. How can you start with a static version of the E field and rewrite it to another type of E field by solving maxwell equations? What allows the E field to transform? If you use the static E field how can it at the same time be equal to another type of E field? Can this be illustrated with a clearer example where a function is transformed? Or does anyone have a proof?

• What static field gets transformed into what 'another type' of electric field? I don't see what it is you are asking here. – Kyle Kanos Dec 19 '19 at 12:00
• In the first part they obtain the amount of E field going out of a sphere by adding all going out of a sphere. By doing so thet add all pieces of E field out. That I understand. But in the last part where they use maxwell equations to find the non static E field they start with amongst other the displacement current that uses E from gauss law which is static. Then they equate this static E field and suddenly end up with a new non static E field. cont. – torgny Dec 19 '19 at 12:17
• I dont see how the linear additions or the equality sign work here. Is there another more simpe example where you use one function and equate to something else and transform your function into something else that someone has? Or just a proof for this transformation. – torgny Dec 19 '19 at 12:18
• A static electric field will always be static. But in what you wrote you consider charged particles moving and thus you have an electric current. This electric current induces a magnetic field which induces an electric field and vice versa. Your "new" electric field is time dependant and not the same as your static field, so you did not transform anything. You simply constructed an electro-magnetic wave. – Tera Dec 19 '19 at 13:44