Integration by parts of a dot product scalar integrand

  • Thread starter bjnartowt
  • Start date
  • #1
284
3

Homework Statement



Is this true or false?

[tex]\int_V {\vec \nabla \Phi \bullet {\bf{E'}} \cdot {d^3}x} = \vec \nabla \Phi \bullet {\bf{E'}} - \int_V {\Phi \cdot \vec \nabla \bullet {\bf{E'}} \cdot {d^3}x} [/tex]
 

Answers and Replies

  • #2
368
12
Well, let's start by looking at what each of these terms actually is. [tex]\Phi[/tex] is a scalar field--a scalar attached to each point in space. Similarly, [tex]E'[/tex] is a vector field--a vector attached to each point in space. So, in the left-hand side, [tex]\nabla\Phi[/tex] is the gradient of a scalar field, which means it's a vector field. [tex]\nabla\Phi\bullet E'[/tex] is the dot product of two vector fields, which means it's a scalar field. Finally, the volume integral over the scalar field sums up the scalars at each point into one big scalar. So the overall type of the expression is a scalar.

Now, if these two expressions are to be equal to each other, they had better be the same type of thing. Using the above logic, what are the types of the middle and rightmost terms?
 
Last edited:
  • #3
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
The identity you want is that for the divergence of the product of a scalar [tex]\Phi[/tex] with a vector [tex]\vec{E}[/tex], which is

[tex] \nabla \cdot ( \Phi \vec{E} ) = ( \nabla \Phi ) \cdot \vec{E} + \Phi (\nabla \cdot \vec{E}). [/tex]

The expression you have is wrong as written, but if you apply the above identity, you'll get the correct result.
 
  • #4
284
3
Hi Chopin, the middle and rightmost terms are scalars. So yes, I have avoided the silliness of adding a vector to a scalar.

fzero, you just wrote down the identity I realized I had to use! I wound up solving the problem (proving Green's reciprocity theorem, if you want to know).

Thank you!
 
Last edited:
  • #5
368
12
Glad to hear you solved it! FYI, though, what I was getting at was that the middle term in your expression isn't a scalar, it's a scalar field. So your expression actually does have some silliness in it--you need some kind of integral there to add up the field into a single scalar. To find it, you can take the volume integral of fzero's identity, and use the divergence theorem to convert the left hand side into a surface integral.
 
Last edited:
  • #6
284
3
>> the middle term in your expression isn't a scalar, it's a scalar field. <<

Oh! good point. that would be illegitimate. Unless, perhaps, my integral on the rightmost part was over a primed variable, leaving unprimed variables as arguments to constitute a scalar field?

Or wait...perhaps not so illegitimate. Wouldn't an integral that is a constant scalar just be a constant additive shift to the middle-term scalar field?

Thinking about these things isn't constructive for my already-solved assignment, but...hey...it may be helpful. What do you think, Chopin, about what I just said in 1st and 2nd paragraph? Is that reasoning sound? That is, I disagree with you in my second paragraph.
 
  • #7
368
12
Not really. First of all, you haven't said anywhere that the integral is a constant scalar. It might be changing throughout the volume, so then your logic doesn't apply. Even if it is constant, though, you still need an integral, because the value of that term is going to be dependent on the size of the volume you're integrating over. (This is also ignoring the fact that that term is just plain wrong--there's a del in it, and there are never any derivative terms in the outside portion of an integration by parts expression.)

Fundamentally, it just doesn't make sense to add a scalar to a scalar field and expect to get out a scalar. Think of it in terms of dimensional analysis--anything inside of a volume integral has a dimension of "stuff per unit volume". Integrating that over a volume gives you a quantity with a dimension of "stuff". So if you were to substitute some kind of physical quantity into your formula, you would find that the units don't work out right when you try to add a "stuff" to a "stuff per unit volume".
 
  • #8
284
3
>> Think of it in terms of dimensional analysis--anything inside of a volume integral has a dimension of "stuff per unit volume". Integrating that over a volume gives you a quantity with a dimension of "stuff". So if you were to substitute some kind of physical quantity into your formula, you would find that the units don't work out right when you try to add a "stuff" to a "stuff per unit volume". <<

Oh...I suppose you could use "dimensional analysis" in this context. Yeah, that kind of does make sense... I have of course never seen [density] + [mass], so that's a good way to check. Well said.
 

Related Threads on Integration by parts of a dot product scalar integrand

Replies
4
Views
1K
  • Last Post
Replies
12
Views
2K
Replies
3
Views
5K
Replies
3
Views
7K
Replies
4
Views
5K
Replies
4
Views
5K
  • Last Post
Replies
9
Views
15K
  • Last Post
Replies
2
Views
1K
Replies
4
Views
667
  • Last Post
Replies
1
Views
3K
Top