aber leicht nachweisen. Denn betrachten wir nur zwei sich
<br/>
entgegenkommende Wellenzüge (vom gleichen Öffnungswinkel),
<br/>
so können wir schreiben:</p>
<center class="par-math-display">
<img src="http://foxridge.mpiwg-berlin.mpg.de/permanent/einstein/annalen/Einst_Stati_de_1910/fulltext/img/Einst_Stati_de_191034x.png" alt=" { ( ) sum 2p-n- ax-+-b-y-+-g-z- Ez = an sin T t- c ( ) + b cos 2p-n- t- a-x-+-b-y-+-g-z n T c ( ) + an'sin 2-pn- t + a-x-+-b-y +-g-z T c 2 p n ( a x + b y + g z )} und + bn'cos ----- t + --------------- { [ T c @-Gz- sum 2p-n-a- 2-p-n 2-pn- @ x = T c - an cos T (...) + bn sin T (...) ]} + a 'cos 2-p-n(...)- b 'sin 2-pn-(...) . n T n T " class="par-math-display"/>
<img src="http://foxridge.mpiwg-berlin.mpg.de/permanent/einstein/annalen/Einst_Stati_de_1910/fulltext/img/Einst_Stati_de_191036x.png" alt=" sum ( t ) Gz = m Bn cos 2p n --- hn , ( T ) @-Gz- sum t- @ x = nCm cos 2p m T - qm . z " class="par-math-display"/>
</center>
<p class="nopar"/>
<p class="noindent">Dann wird: </p>
<center class="par-math-display">
<img src="http://foxridge.mpiwg-berlin.mpg.de/permanent/einstein/annalen/Einst_Stati_de_1910/fulltext/img/Einst_Stati_de_191037x.png" alt=" ( ) 3 c3 3 sum sin gn t f = ----3 T nBn ---3-- cos 2p n -- - hn - gn 16p n T " class="par-math-display"/>
</center>
<p class="nopar"/>
<p class="noindent">und</p>
<center class="par-math-display">
<img src="http://foxridge.mpiwg-berlin.mpg.de/permanent/einstein/annalen/Einst_Stati_de_1910/fulltext/img/Einst_Stati_de_191038x.png" alt=" 3 integral t sum sum -- J = -3c-- T3 d t m n Cm Bn sin-gn 16 p3 n3 [0 { } t- cos 2 p(n + m) T - qm - hn - gn - cos {2p(n - m) t + qm - hn - gn}]. " class="par-math-display"/>