Condition for this equality? $int_{-infty}^{t-tau} bigg(chi(t-tau,t')-chi(t,t'+tau)bigg)E(t'), dt'=0$
$begingroup$
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
$$
int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
$$
real-analysis integration multivariable-calculus vector-analysis physics
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
$endgroup$
add a comment |
$begingroup$
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
$$
int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
$$
real-analysis integration multivariable-calculus vector-analysis physics
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
$endgroup$
add a comment |
$begingroup$
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
$$
int_{-infty}^{t} chi(t-t') E(t'), dt'=0 tag 4
$$
real-analysis integration multivariable-calculus vector-analysis physics
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
$endgroup$
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
real-analysis integration multivariable-calculus vector-analysis physics
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
asked Jan 8 at 20:00
JDoeDoeJDoeDoe
7741613
7741613
add a comment |
add a comment |
1 Answer
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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 $chi(t-t',0)=chi(t,t')$
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
$endgroup$
add a comment |
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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 $chi(t-t',0)=chi(t,t')$
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
$endgroup$
add a comment |
$begingroup$
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 $chi(t-t',0)=chi(t,t')$
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
$endgroup$
add a comment |
$begingroup$
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 $chi(t-t',0)=chi(t,t')$
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
$endgroup$
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 $chi(t-t',0)=chi(t,t')$
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
answered Jan 8 at 23:21
Rafa BudríaRafa Budría
5,6351825
5,6351825
add a comment |
add a comment |
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