Electrodynamics: Derivatives involving Retarded-Time

  • #1
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Hi all,

I have ran into some mathematical confusion when studying the aforementioned topic. The expression for retarded time is given as
$$t_R = t - R/c$$
##R = | \vec{r} - \vec{r'} |##, where ##\vec{r}## represents the point of evaluation and ##\vec{r'}## represents the source position. I rewrite the above equation into
$$x_i = x'_i + \text{terms independent of x' and x}$$
where ##x_i## is a cartesian coordinate. If I had the following derivative ##\partial _{x_i}##, could I then say this?
$$\partial _{x_i} = \frac{ \partial }{ \partial x'_i} \frac{ \partial x'_i}{ \partial x_i} = \partial _{x'_i}$$

If so, there is the following identity
$$\nabla(1/R) = - \nabla ' (1/R)$$
where the prime means that the derivative is with respect to source coordinates ##x' _i##

Suppose I start from the RHS:
$$\nabla (1/R) = \sum_i \partial_ {x_i} (\frac{1}{R}) \hat{x_i} = \sum_i \partial_ {x' _i} (\frac{1}{R}) \hat{x_i} = \nabla' (1/R)$$

which is clearly wrong. What have I done wrongly?

Thanks in advance!
 

Answers and Replies

  • #2
Delta2
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Your mistake is that you consider the dependency equation ##x_i=x'_i+C## between the field coordinates and the source coordinates but there can be no such dependency, we treat all field variables ##x_i## and all source variable ##x'_i## as independent from each other when proving the identity ##\nabla (1/R)=-\nabla'(1/R)##.
 
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