Help understanding proof involving Maxwell equation

In summary, the conversation discusses a demonstration involving Maxwell's equations and the wave equation. It is shown that if \vec{J} = 0 in Maxwell-Ampere's law, it satisfies the wave equation. One small step involves taking the curl of \nabla \times \vec{B}. The conversation also mentions the commutation of time derivatives and the curl, as well as Cairault's theorem on equality of mixed partials.
  • #1
U.Renko
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1
help understanding "proof" involving Maxwell equation

I'm currently taking a course on mathematical methods for physics.
(Like always I'm a bit confused about where exactly I should post these questions, should it go to the homework forum? )


anyway as I was reading the lecture notes I found this demonstration that if [itex] \vec{J} = 0 [/itex] in Maxwell-Ampere's law satisfies the wave equation: [itex] \left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 \right)\vec B = \vec 0 [/itex] :

There is one small step I'm not sure why/how he did it.

he took the curl of [itex] \nabla \times\vec B [/itex], that is, [itex]\nabla\times\left(\nabla\times\vec B \right) = \nabla\left(\nabla\cdot\vec B\right) - \nabla^2\vec B [/itex]

then since [itex] \nabla\cdot\vec B = 0 [/itex] and [itex] \nabla\times\vec B = \mu_0\epsilon_0\frac{\partial\vec E}{\partial t} [/itex] he says [itex] \nabla\times\left(\nabla\times\vec B \right) = \nabla\left(\mu_0\epsilon_0\frac{\partial\vec E}{\partial t}\right) = \mu_0\epsilon_0\left(\nabla\times\frac{\partial\vec E}{\partial t}\right) [/itex]

so far I understand everything, but now he says:
[itex] \mu_0\epsilon_0\left(\nabla\times\frac{\partial\vec E}{\partial t}\right) = \mu_0\epsilon_0\frac{\partial\left(\nabla\times\vec E\right)}{\partial t} [/itex] and it is this very last step that I'm not sure if it's allowed. And if is, why it is?

the rest of the demonstration follows easily, supposing this last step is correct.
 
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  • #2
That step just says that time derivatives commute with the curl. You should try this out for yourself and see that it is so. (Basically it's true simply because partial derivatives commute and the basis vectors are not time-dependent.)

Just write out the curl of E, take the time derivative. Then write out the curl of the time derivative of E and they are identical.
 
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  • #3
You missed one:
$$ \left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2 \right)\vec B = 0\\
\implies \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec B =\nabla^2\vec B\\
\implies \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec B = \nabla(\nabla\cdot\vec B)-\nabla\times (\nabla\times \vec B)$$

Note:
$$\nabla\cdot\vec B = 0\\
\nabla\times \vec B = \mu_0\epsilon_0\frac{\partial}{\partial t}\vec E\\
\nabla\times \vec E = -\frac{\partial}{\partial t}\vec B$$

https://www.google.co.nz/search?q=m...0j5j0j69i60.3357j0j1&sourceid=chrome&ie=UTF-8
 
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  • #4
By Cairault's theorem [see http://calculus.subwiki.org/wiki/Clairaut's_theorem_on_equality_of_mixed_partials] [Broken] continuous partial derivatives commute.
 
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  • #5


Hello,

I am happy to help you understand the proof involving the Maxwell equation. In this proof, we are trying to show that if the current density \vec{J} is equal to zero, then the magnetic field \vec{B} satisfies the wave equation \left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 \right)\vec B = \vec 0.

The step that you are confused about is where the author takes the curl of \nabla \times \vec{B} and then rewrites it as \nabla \times \left(\nabla \times \vec{B}\right) = \nabla \left(\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}\right). This step is allowed because of a mathematical identity known as the vector identity. This identity states that the curl of the curl of a vector field is equal to the gradient of the divergence of that vector field minus the Laplacian of that vector field. In this case, the vector field is \vec{B}, so the identity becomes:

\nabla \times \left(\nabla \times \vec{B}\right) = \nabla \left(\nabla \cdot \vec{B}\right) - \nabla^2 \vec{B}

Now, since we know that \nabla \cdot \vec{B} is equal to zero (as stated in the proof), we can simplify the above equation to:

\nabla \times \left(\nabla \times \vec{B}\right) = -\nabla^2 \vec{B}

Finally, we can use the fact that \nabla \times \vec{B} is equal to \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} (as stated in the proof) to rewrite the above equation as:

\nabla \times \left(\nabla \times \vec{B}\right) = \mu_0 \epsilon_0 \frac{\partial}{\partial t}\left(\nabla \times \vec{E}\right)

I hope this explanation helps you understand the last step of the proof. It is important to note
 

1. What are Maxwell's equations and why are they important in science?

Maxwell's equations are a set of four fundamental equations that describe the behavior of electric and magnetic fields. They are important in science because they provide a mathematical framework for understanding and predicting the behavior of electromagnetic waves, which are essential for technologies such as radio, television, and cell phones.

2. How do Maxwell's equations relate to light and other forms of electromagnetic radiation?

Maxwell's equations explain the properties and behavior of electromagnetic waves, including light. These equations show that light is a form of electromagnetic radiation and can be described by the same principles that govern electric and magnetic fields.

3. Can you explain the significance of Maxwell's equations in the development of modern physics?

Maxwell's equations were crucial in the development of modern physics as they provided a unified theory of electricity and magnetism. They also paved the way for the development of the theory of relativity and quantum mechanics, which have had a major impact on our understanding of the universe.

4. How are Maxwell's equations used in practical applications?

Maxwell's equations have many practical applications, including the design and operation of electronic devices such as computers, televisions, and cell phones. They are also used in fields such as telecommunications, radar technology, and medical imaging.

5. Are there any limitations or challenges associated with understanding and applying Maxwell's equations?

While Maxwell's equations are incredibly useful and accurate, they do have some limitations. For example, they do not account for the effects of quantum mechanics or gravity. Additionally, some of the equations can be complex and difficult to solve, making it challenging to apply them in certain situations.

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