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Vector Calculus

  • Thread starter psholtz
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Hello,

I'm working through some problems in the Griffith text on electrodynamics. In one of them, the reader is asked to prove the following identity (which is given in the text), which is a generalization (of sorts) on the divergence theorem:

[tex]\large{ \int_V \left(\nabla T\right) dV = \oint_{\partial V}TdA} [/tex]

where T is a scalar field.

I'm not going to go through the proof here (which is relatively straightforward).

Rather, my question simply is: what meaning (physical or otherwise) can be ascribed to the left-hand side of the equation? I can understand taking the volume integral of a scalar field, but grad-T is a vector. How can you take the volume integral of a vector field?

For instance, in the "traditional" divergence theorem, it is the scalar field "div-F" that is integrated through the volume:

[tex]\large{ \int_V \left(\nabla \cdot F\right) dV = \oint_{\partial V} F \cdot dA }[/tex]

But how can a vector field be integrated through a volume?
 
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Answers and Replies

  • #2
tiny-tim
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Hello psholtz! :smile:
Rather, my question simply is: what meaning (physical or otherwise) can be ascribed to the left-hand side of the equation? I can understand taking the volume integral of a scalar field, but grad-T is a vector. How can you take the volume integral of a vector field?
You just can …

integration is the same as addition …

you can integrate anything you can add …

vectors obey the "vector law of adddition", which you apply, for example, when adding the (vector) forces at different points of a body …

if the force changes continuously (in space), you can integrate it instead of of adding it. :smile:
 
  • #3
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No.. after having thought about is some more, I actually think the trick here is to realize that we are dealing with vectors (on both sides fo the equality), and so what we really have is "three" equations, rather than one.

So working in Cartesian coordinates, the expression on the LHS can be expressed as thus:

[tex]\large{\int_V \left(\nabla T\right) dV =
\int_V \left( \frac{\partial T}{\partial x} \hat{x}
+ \frac{\partial T}{\partial y}\hat{y}
+ \frac{\partial T}{\partial z}\hat{z} \right) dV
}[/tex]

which can be further expressed, component-wise, as:

[tex]\int_V \left(\nabla T\right)dV =
\int_V \frac{\partial T}{\partial x} dV \hat{x}
+ \int_V \frac{\partial T}{\partial y} dV \hat{y}
+ \int_V \frac{\partial T}{\partial z} dV \hat{z}
[/tex]

So we really have three volume integrals, one for each vector component, and each of these three integrals is a "traditional" volume integral, in the sense of being the volume integral of a scalar field w/in that volume.

Similar arguments apply on the RHS of the equality, which likewise is a vector.

That makes sense.. :smile:
 
  • #4
tiny-tim
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yay for vectors!

No.. after having thought about is some more, I actually think the trick here is to realize that we are dealing with vectors (on both sides fo the equality), and so what we really have is "three" equations, rather than one.

So we really have three volume integrals, one for each vector component, and each of these three integrals is a "traditional" volume integral, in the sense of being the volume integral of a scalar field w/in that volume.

That makes sense.. :smile:
Hi psholtz! :smile:

Yes, that's perfectly correct …

if you use coordinates, you can always split a vector integral into three scalar integrals …

and if that makes you happier, by all means continue to do it for the time being. :wink:

However, do remember that the whole beauty of vectors is that a vector can be treated as a single entity, and it's often very helpful to do so.

Vectors are your friends! :biggrin:
 

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