Condition for this equality? $int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0$












0












$begingroup$


This is from a derivation in Electrodynamics, however I don't follow the math.



I don't understand why equation $(2)$ and $(3)$ are equivalent.



If $tau =0$ in $(2)$ we only have $chi(t,t')=chi(t,t')$ and not $chi(t-t',0)$ ...?




$$
int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0 tag 1
$$

Here $E$ is an arbitrary function so
$$
chi(t-tau,t')=chi(t,t'+tau) tag 2
$$

for all $t$, $tau$ and $t'$. Or equivalent
$$
chi(t,t')=chi(t-t',0) tag 3
$$

Therefore we can write
$$
int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
$$











share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    This is from a derivation in Electrodynamics, however I don't follow the math.



    I don't understand why equation $(2)$ and $(3)$ are equivalent.



    If $tau =0$ in $(2)$ we only have $chi(t,t')=chi(t,t')$ and not $chi(t-t',0)$ ...?




    $$
    int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0 tag 1
    $$

    Here $E$ is an arbitrary function so
    $$
    chi(t-tau,t')=chi(t,t'+tau) tag 2
    $$

    for all $t$, $tau$ and $t'$. Or equivalent
    $$
    chi(t,t')=chi(t-t',0) tag 3
    $$

    Therefore we can write
    $$
    int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
    $$











    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This is from a derivation in Electrodynamics, however I don't follow the math.



      I don't understand why equation $(2)$ and $(3)$ are equivalent.



      If $tau =0$ in $(2)$ we only have $chi(t,t')=chi(t,t')$ and not $chi(t-t',0)$ ...?




      $$
      int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0 tag 1
      $$

      Here $E$ is an arbitrary function so
      $$
      chi(t-tau,t')=chi(t,t'+tau) tag 2
      $$

      for all $t$, $tau$ and $t'$. Or equivalent
      $$
      chi(t,t')=chi(t-t',0) tag 3
      $$

      Therefore we can write
      $$
      int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
      $$











      share|cite|improve this question









      $endgroup$




      This is from a derivation in Electrodynamics, however I don't follow the math.



      I don't understand why equation $(2)$ and $(3)$ are equivalent.



      If $tau =0$ in $(2)$ we only have $chi(t,t')=chi(t,t')$ and not $chi(t-t',0)$ ...?




      $$
      int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0 tag 1
      $$

      Here $E$ is an arbitrary function so
      $$
      chi(t-tau,t')=chi(t,t'+tau) tag 2
      $$

      for all $t$, $tau$ and $t'$. Or equivalent
      $$
      chi(t,t')=chi(t-t',0) tag 3
      $$

      Therefore we can write
      $$
      int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
      $$








      real-analysis integration multivariable-calculus vector-analysis physics






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      asked Jan 8 at 20:00









      JDoeDoeJDoeDoe

      7701614




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          $begingroup$

          The equality $chi(t-tau,t')=chi(t,t'+tau)$ holds for every value of all three variables.



          Set the first argument of $chi$ to $t-t'$ and the second to $0$, $chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $chi(t,0+t')$. They are equal.



          Other way, make the change in the equality $tauto t'$, $t'=0$ and no change for $t$. We get $chi(t-t',0)=chi(t,t')$






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            The equality $chi(t-tau,t')=chi(t,t'+tau)$ holds for every value of all three variables.



            Set the first argument of $chi$ to $t-t'$ and the second to $0$, $chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $chi(t,0+t')$. They are equal.



            Other way, make the change in the equality $tauto t'$, $t'=0$ and no change for $t$. We get $chi(t-t',0)=chi(t,t')$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The equality $chi(t-tau,t')=chi(t,t'+tau)$ holds for every value of all three variables.



              Set the first argument of $chi$ to $t-t'$ and the second to $0$, $chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $chi(t,0+t')$. They are equal.



              Other way, make the change in the equality $tauto t'$, $t'=0$ and no change for $t$. We get $chi(t-t',0)=chi(t,t')$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The equality $chi(t-tau,t')=chi(t,t'+tau)$ holds for every value of all three variables.



                Set the first argument of $chi$ to $t-t'$ and the second to $0$, $chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $chi(t,0+t')$. They are equal.



                Other way, make the change in the equality $tauto t'$, $t'=0$ and no change for $t$. We get $chi(t-t',0)=chi(t,t')$






                share|cite|improve this answer









                $endgroup$



                The equality $chi(t-tau,t')=chi(t,t'+tau)$ holds for every value of all three variables.



                Set the first argument of $chi$ to $t-t'$ and the second to $0$, $chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $chi(t,0+t')$. They are equal.



                Other way, make the change in the equality $tauto t'$, $t'=0$ and no change for $t$. We get $chi(t-t',0)=chi(t,t')$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 23:21









                Rafa BudríaRafa Budría

                5,7801825




                5,7801825






























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