Thread: double sum designation question..

the telegan law basically states that the total sum of power is zero.
my prof proved lik this:

we choose a node (a point where more then one currents come together)

and decide that the voltage on that node to be zero.
we designate the voltages on the nodes to be $\displaystyle e_k$
$\displaystyle J_k$ is the current.

$\displaystyle v_kJ_k=(e_a-e_b)J_{ab}$
$\displaystyle v_kJ_k=\frac{1}{2}[(e_b-e_a)J_{ab}+(e_a-e_b)J_{ab}]$
$\displaystyle n_t$ is the number of nodes.[/tex]
$\displaystyle B$ is the number of branches.[/tex]
$\displaystyle \sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}\sum _{b=1}^{n_t}(e_a-e_b)J_{ab}$
each J that does not exist in the graph will be zero.
$\displaystyle \sum_{k}^{B}v_kJ_k=\frac{1}{2}\sum_{a=1}^{n_t}e_a\ sum_{b=1}^{n_t}J_{ab}-\frac{1}{2}\sum_{a=1}^{n_t}e_b\sum_{b=1}^{n_t}J_{a b}=0$

because by kcl
$\displaystyle \sum_{b=1}^{n_t}J_{ab}=0$

my problem iswhen he sums for all nodes
he uses
$\displaystyle \sum\sum$ sign which by me represents multiplication
of the sums

why not $\displaystyle \sum+\sum$,thus we can know that ist the sum of many similar equations.

but how he did it doesnt represent a sum

The double sum in this case represents a nested sum (if you're familiar with programming, it's like a nested for loop).
In your problem:
$\displaystyle \frac{1}{2}\sum_{a=1}^{n_t}\sum_{b=1}^{n_t}(e_a-e_b)J_{ab}$
means for every iteration of a, we go through all $\displaystyle n_t$ iterations of b. I like to think of it as sort of a series exponentiation.

Formally if we define: $\displaystyle F = \sum_{i=0}^{n} a_{i}$ and $\displaystyle G = \sum_{j=0}^{n} b_{j}$

$\displaystyle F * G = \sum_{i=0}^{n} a_{i}b_{n-i}$

$\displaystyle F º G = \sum_{i=0}^{n}\sum_{j=0}^{n} a_{i}b_{j}
$
Will yield different answers. I'm not sure about the actual concepts, I'm not very well versed in physics, but your prof is using the nested case.