- #1
Nick R
- 69
- 0
Faraday's law is often stated in SI units as
\nabla x E(x,y,z,t) = \frac{\partial B(x,y,z,t)}{\partial t}
But x, y, z (or some other set of coordinates) are variables not functions and thus have no "t dependence".
So it would seem that the "total derivative" of B is the same as the partial:
\frac{dB}{dt}=\frac{\partial B}{\partial t}+\frac{\partial B}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial B}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial B}{\partial z}\frac{\partial z}{\partial t} = \frac{\partial B}{\partial t} + 0 + 0 + 0
It seems that there isn't any difference between the partial of B wrt to t and the "full derivative" of B wrt t.
The reason I'm concerned is the issue where you CANNOT do the following (or maybe sometimes... like in this case you can?):
\int^a_b\frac{\partial f(x,y)}{\partial x}dx = f(b,y) - f(a,y)
\nabla x E(x,y,z,t) = \frac{\partial B(x,y,z,t)}{\partial t}
But x, y, z (or some other set of coordinates) are variables not functions and thus have no "t dependence".
So it would seem that the "total derivative" of B is the same as the partial:
\frac{dB}{dt}=\frac{\partial B}{\partial t}+\frac{\partial B}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial B}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial B}{\partial z}\frac{\partial z}{\partial t} = \frac{\partial B}{\partial t} + 0 + 0 + 0
It seems that there isn't any difference between the partial of B wrt to t and the "full derivative" of B wrt t.
The reason I'm concerned is the issue where you CANNOT do the following (or maybe sometimes... like in this case you can?):
\int^a_b\frac{\partial f(x,y)}{\partial x}dx = f(b,y) - f(a,y)
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