# Swell free

As a source free or free of sources in is physics and potential theory a vector field designates whose field lines do not have a starting point in the area under consideration. Source-free is z. B. the outer space of a force or gravity field if it does not contain any mass points or charges.

This ideal case rarely occurs in nature because almost everywhere - including in interplanetary space - there are residual gas molecules, dust particles and free electrons. For scientific practice and in astronomy , however, there is no source wherever the matter or gas density is below a few particles per cm 3 . For laboratory physics, the best technically producible high vacuum can be the reference value which, with a residual gas pressure of around 10 −11 millibars, is far higher.

## Field lines and divergence

In electrodynamics and fluid mechanics , source-free spaces are characterized by the fact that as many field lines enter the considered space section as they exit. This behavior of the field lines can be described mathematically by the divergence of the vector field.

The mathematics is called the term "source-free" also divergence-free , because the lack of resources is coupled with the disappearance of the divergence: in a source-free vector field applies: ${\ displaystyle {\ vec {a}}}$ ${\ displaystyle \ operatorname {div} \, {\ vec {a}} \, = \, 0}$ Here stands for the divergence operator (see also Nabla operator ). ${\ displaystyle \ operatorname {div}}$ Conversely, the presence of sources is characterized by and the presence of sinks is characterized by${\ displaystyle \ operatorname {div} \, {\ vec {a}}> 0}$ ${\ displaystyle \ operatorname {div} \, {\ vec {a}} <0.}$ ## Interpretation and examples

Physically, the divergence of a vector field can be interpreted as a measure of the source strength, because according to Gauss's theorem , the integral of the divergence over a volume is equal to the flow through the surface of the volume. Accordingly, a vector field whose divergence is zero is called source-free, because here the flux is equal to zero for any closed surfaces, i.e. In other words, no more net flows out than in. So there are neither sources nor sinks inside the volume that is enclosed by the surface .

Important examples of source-free fields in physics are the magnetic field , more precisely: magnetic induction , and the velocity fields of incompressible flows, which are source-free due to the continuity equation .

For all twice continuously differentiable vector fields , according to Schwarz's theorem, applies${\ displaystyle {\ vec {V}} ({\ vec {r}})}$ ${\ displaystyle {\ mbox {div}} \, ({\ mbox {red}} \, {\ vec {V}} ({\ vec {r}})) \, \ equiv \, 0.}$ Thus the rotation of a twice continuously differentiable vector field is always source-free.

(For a vector field gives the source density and the so-called. Vortex density .) ${\ displaystyle {\ vec {V}} ({\ vec {r}})}$ ${\ displaystyle {\ mbox {div}} \, {\ vec {V}}}$ ${\ displaystyle {\ mbox {red}} \, {\ vec {V}}}$ The converse also applies: A once continuously differentiable vector field that is free of sources everywhere, such as B. the magnetic field, thus with for all, has a vector potential that is twice continuously differentiable everywhere, so that the following applies, and can also be calculated from the vortex densities by integration. The proof is not trivial (see Poincaré's lemma ). ${\ displaystyle {\ vec {B}} ({\ vec {r}}),}$ ${\ displaystyle {\ mbox {div}} {\ vec {B}} ({\ vec {r}}) \ equiv 0}$ ${\ displaystyle {\ vec {r}},}$ ${\ displaystyle {\ vec {A}} ({\ vec {r}}),}$ ${\ displaystyle {\ vec {B}} ({\ vec {r}}) = {\ mbox {red}} \, {\ vec {A}} ({\ vec {r}}) \ \ (= { \ vec {\ nabla}} \ times {\ vec {A}} ({\ vec {r}}))}$ ${\ displaystyle {\ vec {A}} ({\ vec {r}})}$ ${\ displaystyle {\ vec {B}} ({\ vec {r}})}$ If, on the other hand, the field is not source-free, but eddy-free, as in the electrical case, it has a scalar potential instead of the vector potential from which it can be derived through gradient formation. The scalar potential can be calculated from the source densities by integration. The proof steps are analogous to the above (see theoretical electrodynamics ). ${\ displaystyle {\ vec {E}} ({\ vec {r}}),}$ ${\ displaystyle {\ vec {A}} ({\ vec {r}})}$ ${\ displaystyle \ Phi ({\ vec {r}}),}$ ${\ displaystyle {\ vec {E}} ({\ vec {r}}) = - {\ mbox {grad}} \, \ Phi ({\ vec {r}}) \ \ (= - {\ vec { \ nabla}} \ Phi ({\ vec {r}})).}$ ${\ displaystyle {\ vec {E}}}$ 