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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 20:00









      JDoeDoeJDoeDoe

      7741613




      7741613






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066651%2fcondition-for-this-equality-int-inftyt-tau-bigg-chit-tau-t-chi%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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

                5,6351825




                5,6351825






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066651%2fcondition-for-this-equality-int-inftyt-tau-bigg-chit-tau-t-chi%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Mario Kart Wii

                    The Binding of Isaac: Rebirth/Afterbirth

                    What does “Dominus providebit” mean?