Mathematical trivial double about lorenz gauge $partial_muA^mu$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why is the total time derivative of this partial space derivative zero?What does it mean to square a partial derivative with a one dimensional vector (scalar)?Simplifying to Linear Differential Operator?Laplacians on Kaehler manifoldsA question about notation in derivatives inn $partial_barAL^I$Meaning of partial differentiationMultivariable calculus chain rule and change of variablesWhy is $partial_mupartial^muphi=partial^mupartial_muphi$?Gauge invariance of the Hamiltonian for particle on external electric fieldProduct rule involving partial derivatives
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Mathematical trivial double about lorenz gauge $partial_muA^mu$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why is the total time derivative of this partial space derivative zero?What does it mean to square a partial derivative with a one dimensional vector (scalar)?Simplifying to Linear Differential Operator?Laplacians on Kaehler manifoldsA question about notation in derivatives inn $partial_barAL^I$Meaning of partial differentiationMultivariable calculus chain rule and change of variablesWhy is $partial_mupartial^muphi=partial^mupartial_muphi$?Gauge invariance of the Hamiltonian for particle on external electric fieldProduct rule involving partial derivatives
$begingroup$
$$D_mu.=partial_mu+iqA_mu$$
$(D_muD^mu+m^2)phi(x)=0$
Expliciting:
$(partial^mupartial_mu+iqpartial^muA_mu+iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Setting Lorenz gauge: $partial_muA^mu=0$
I do not understand why I get:
$(square +2iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Instead of:
$(square +iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
partial-derivative quantum-mechanics special-relativity
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
$endgroup$
add a comment |
$begingroup$
$$D_mu.=partial_mu+iqA_mu$$
$(D_muD^mu+m^2)phi(x)=0$
Expliciting:
$(partial^mupartial_mu+iqpartial^muA_mu+iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Setting Lorenz gauge: $partial_muA^mu=0$
I do not understand why I get:
$(square +2iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Instead of:
$(square +iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
partial-derivative quantum-mechanics special-relativity
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
$endgroup$
add a comment |
$begingroup$
$$D_mu.=partial_mu+iqA_mu$$
$(D_muD^mu+m^2)phi(x)=0$
Expliciting:
$(partial^mupartial_mu+iqpartial^muA_mu+iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Setting Lorenz gauge: $partial_muA^mu=0$
I do not understand why I get:
$(square +2iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Instead of:
$(square +iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
partial-derivative quantum-mechanics special-relativity
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
$endgroup$
$$D_mu.=partial_mu+iqA_mu$$
$(D_muD^mu+m^2)phi(x)=0$
Expliciting:
$(partial^mupartial_mu+iqpartial^muA_mu+iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Setting Lorenz gauge: $partial_muA^mu=0$
I do not understand why I get:
$(square +2iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
Instead of:
$(square +iqA^mupartial_mu-q^2A^2+m^2)phi(x)=0$
partial-derivative quantum-mechanics special-relativity
partial-derivative quantum-mechanics special-relativity
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
asked Apr 1 at 8:26
Stefano BaroneStefano Barone
402210
402210
add a comment |
add a comment |
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