How Can Vector Calculus Help Solve a Complex Maxwell Equations Problem?

AI Thread Summary
The discussion revolves around solving a complex problem involving Maxwell's equations and vector calculus. The original poster has made progress but is missing several terms in their solution for the force per unit volume, F_v. They express the force using the equations related to electric fields, magnetic fields, and charge density, and seek assistance in deriving the missing components. A user suggests that the missing terms may cancel out to zero based on Maxwell's equations, which the original poster later confirms they have resolved. The thread highlights the interplay between vector calculus and electromagnetic theory in solving advanced physics problems.
gjfelix2006
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Hi, i am solving a problem about Maxwell Equation that invoves a lot of Vector Calculus, i have a partial solution for it but i have a few terms missing :cry: , i appreciate any help in this. Thanks

The problem is the following

Starting with the expression of the force by volume unit over a free space region with charges and currents:
<br /> F_{v}=\rho E+J\timesB<br />
and using Maxwell's Equations:
<br /> <br /> \begin{array}{l}<br /> \nabla \cdot \mathop E\limits^ \to = \frac{\rho }{{ \in _0 }} \\ <br /> \nabla \times \mathop E\limits^ \to = - \frac{{\partial \mathop B\limits^ \to }}{{\partial t}} \\ <br /> \nabla \cdot \mathop B\limits^ \to = 0 \\ <br /> \nabla \times \mathop B\limits^ \to = \mu _0 \mathop J\limits^ \to + \in _0 \mu _0 \frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \\ <br /> \end{array}<br /> <br /> <br />
and the following vectorial identity:
<br /> <br /> \mathop B\limits^ \to \times \nabla \times \mathop B\limits^ \to = \nabla ({\textstyle{1 \over 2}}B^2 ) - (\mathop B\limits^ \to \cdot \nabla )\mathop B\limits^ \to <br />,

Show that:
<br /> <br /> \begin{array}{l}<br /> \mathop {F_v }\limits^ \to = - \in _0 \frac{\partial }{{\partial t}}(\mathop E\limits^ \to \times \mathop B\limits^ \to ) + \in _0 \mathop E\limits^ \to \nabla \cdot \mathop E\limits^ \to - \frac{1}{2} \in _0 \nabla (E^2 ) + \in _0 (\mathop E\limits^ \to \cdot \nabla )\mathop E\limits^ \to \\ <br /> {\rm{ + }}\frac{1}{{\mu _0 }}\mathop B\limits^ \to \nabla \cdot \mathop B\limits^ \to - \frac{1}{{2\mu _0 }}\nabla \mathop {(B^2 ) + }\limits^{} \frac{1}{{\mu _0 }}(\mathop B\limits^ \to \cdot \nabla )\mathop B\limits^ \to \\ <br /> \end{array}<br />

Now, let me show you my partial solution:

First, by Maxwell Equations, i get J:
<br /> <br /> \mathop J\limits^ \to = \frac{1}{{\mu _0 }}(\nabla \times \mathop B\limits^ \to ) - \in _0 \frac{{\partial \mathop E\limits^ \to }}{{\partial t}}<br />

And i replace it in the first equation for F_{v} to get:
<br /> \[<br /> \begin{array}{l}<br /> \mathop F\limits^ \to _v = \rho \mathop E\limits^ \to + \left( {\frac{1}{{\mu _0 }}(\nabla \times \mathop B\limits^ \to ) - \in _0 \frac{{\partial \mathop E\limits^ \to }}{{\partial t}}} \right) \times \mathop B\limits^ \to \\ <br /> {\rm{ }} = \rho \mathop E\limits^ \to + \frac{1}{{\mu _0 }}(\nabla \times \mathop B\limits^ \to ) \times \mathop B\limits^ \to - \in _0 (\frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \times \mathop B\limits^ \to ) \\ <br /> \end{array}<br /> \]<br /> <br />
Changing the order of the cross product (the sign changes), then
<br /> \[<br /> \mathop F\limits^ \to _v = \rho \mathop E\limits^ \to - \frac{1}{{\mu _0 }}(\mathop B\limits^ \to \times \nabla \times \mathop B\limits^ \to ) - \in _0 (\frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \times \mathop B\limits^ \to {\rm{)}}<br /> \]<br /> <br />
Now i can use the vectorial identity, thus:
<br /> \[<br /> = \rho \mathop E\limits^ \to - \frac{1}{{\mu _0 }}(\nabla ({\textstyle{1 \over 2}}B^2 ) - (\mathop B\limits^ \to \cdot \nabla )\mathop B\limits^ \to ) - \in _0 (\frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \times \mathop B\limits^ \to {\rm{) }}<br /> \]<br /> <br />

Also, from Maxwell equations:
<br /> \rho = (\nabla \cdot \mathop E\limits^ \to ) \in _0 <br />
and replacing it into the last equation for F_{v}, i get:
<br /> \mathop F\limits^ \to _v = \mathop E\limits^ \to (\nabla \cdot \mathop E\limits^ \to ) \in _0 - \frac{1}{{\mu _0 }}(\nabla ({\textstyle{1 \over 2}}B^2 ) - (\mathop B\limits^ \to \cdot \nabla )\mathop B\limits^ \to ) - \in _0 (\frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \times \mathop B\limits^ \to {\rm{) }}<br />
Some terms of what i should get can be seen already, but i have another terms missing. My last step is the following, what you think i should do to get the missing terms?
<br /> \mathop {F_v }\limits^ \to = - \in _0 (\frac{{\partial \mathop E\limits^ \to }}{{\partial t}} \times \mathop B\limits^ \to ) + \in _0 (\mathop E\limits^ \to \cdot \nabla )\mathop E\limits^ \to - \frac{1}{{2\mu _0 }}\nabla \mathop {(B^2 ) + }\limits^{} \frac{1}{{\mu _0 }}(\mathop B\limits^ \to \cdot \nabla )\mathop B\limits^ \to <br />

¿What should I do to get the missing terms?
I think I must develop the first term in the last equation, but I don’t know how, can you help me?

I appreciate any help. Thanks a lot.
 
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Hi gjfelix2006, welcome to PhysicsForums.

First of all, please don't double post. I see you've made a duplicate thread in the other subforum as well. Also, to make it easier for those who wish to help, you might want to read https://www.physicsforums.com/showthread.php?t=8997" thread which explains how to use LaTeX mathematical typesetting. That way you needn't wait till your attachment is approved.
 
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I am sorry for my double post, i am new here and i don't know how to erase one of them, i made a double post because i don't know if my problem is a basic problem or an advanced problem. So sorry, if you can tell me how to erase it, i'll appreciate. Bye
 
I haven't followed your math thru but it looks like that you are missing
3 terms. The one with del dot B is obviously zero from Maxwell's
equations. Try writing a one dimensional version of the other two terms
and I think that they will also cancel out to zero!
 
Thanks a lot for your help. I have solved the problem already. Thanks J Hann.
 
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