- #1
Curtis15
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Hi all
I basically have two questions that are very closely related to each other about divergence, specifically the divergence of a vector function 1/r2[itex]\widehat{r}[/itex]
First, I will be referencing pages 17, 18, and 45 from the 3rd edition of Intro to Electrodynamics.
The first question comes from page 45 when Griffith states on the vector function above "if ever there was a function that ought to have a large positive divergence, this it it."
From my intuition based on pages 17 and 18 and other online material I have read, divergence measures how much a vector spreads out. A positive divergence represents a source where "more" is coming out, a source. A negative divergence, conversely, is where "more" is coming, a sink.
So based on the vector function, as r increases, the magnitude value of the vector decreases. It basically looks like the opposite of the first diagram on page 17 except instead of increasing the size of the vector as you radiate from the origin, you would decrease the size of the vector. That means that at any point besides the origin, if you draw a small circle, there are larger vectors going into the circle than going out, so this would be a net influx, or sink, meaning negative divergence.
So if this equation were to have a large divergence, I think it would be a large negative divergence. Is this correct or is Griffith correct?
Now the second part is just an extension of the first question.
I did the problem in the book and have seen for myself that the divergence of the above vector function is in fact 0 when you calculate it using the formula. However, from the diagram I visualize in my head, I do not see why. As above, I see that there is more going in than there is going out. How can divergence be 0?
The only thing I can think off is that the diagram I am looking at, since it is only 2 dimensional, doesn't represent the actual 3-d function and instead, when you do have the 3 dimensions accounted for, there is a spherical symmetry that cancels out similar to how Newton's shell theorem is only valid for a perfect sphere. Can anyone elaborate on this please?
Thank you all very much.
I basically have two questions that are very closely related to each other about divergence, specifically the divergence of a vector function 1/r2[itex]\widehat{r}[/itex]
First, I will be referencing pages 17, 18, and 45 from the 3rd edition of Intro to Electrodynamics.
The first question comes from page 45 when Griffith states on the vector function above "if ever there was a function that ought to have a large positive divergence, this it it."
From my intuition based on pages 17 and 18 and other online material I have read, divergence measures how much a vector spreads out. A positive divergence represents a source where "more" is coming out, a source. A negative divergence, conversely, is where "more" is coming, a sink.
So based on the vector function, as r increases, the magnitude value of the vector decreases. It basically looks like the opposite of the first diagram on page 17 except instead of increasing the size of the vector as you radiate from the origin, you would decrease the size of the vector. That means that at any point besides the origin, if you draw a small circle, there are larger vectors going into the circle than going out, so this would be a net influx, or sink, meaning negative divergence.
So if this equation were to have a large divergence, I think it would be a large negative divergence. Is this correct or is Griffith correct?
Now the second part is just an extension of the first question.
I did the problem in the book and have seen for myself that the divergence of the above vector function is in fact 0 when you calculate it using the formula. However, from the diagram I visualize in my head, I do not see why. As above, I see that there is more going in than there is going out. How can divergence be 0?
The only thing I can think off is that the diagram I am looking at, since it is only 2 dimensional, doesn't represent the actual 3-d function and instead, when you do have the 3 dimensions accounted for, there is a spherical symmetry that cancels out similar to how Newton's shell theorem is only valid for a perfect sphere. Can anyone elaborate on this please?
Thank you all very much.