Тождество интегралов векторных полей

[PoMa (Roman Cheremnov)]

Первоначально отправлено в Решение Задач

PoMa

@eudaimon solve the problem step by step

eudaimon

Ah, a vector calculus identity! Let me guide you through proving this elegant relation, mortal.

Let’s prove: \int_{\mathcal{V}} \mathbf{B} \cdot (\nabla \times \mathbf{A}) d\tau = \int_{\mathcal{V}} \mathbf{A} \cdot (\nabla \times \mathbf{B}) d\tau + \oint_S (\mathbf{A} \times \mathbf{B}) \cdot d\mathbf{a}

1. First, recall the vector triple product identity:
   \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

2. Rearranging:
   \mathbf{B} \cdot (\nabla \times \mathbf{A}) = \nabla \cdot (\mathbf{A} \times \mathbf{B}) + \mathbf{A} \cdot (\nabla \times \mathbf{B})

3. Integrate both sides over volume \mathcal{V}:
   \int_{\mathcal{V}} \mathbf{B} \cdot (\nabla \times \mathbf{A}) d\tau = \int_{\mathcal{V}} \nabla \cdot (\mathbf{A} \times \mathbf{B}) d\tau + \int_{\mathcal{V}} \mathbf{A} \cdot (\nabla \times \mathbf{B}) d\tau

4. Apply the divergence theorem to the middle term:
   \int_{\mathcal{V}} \nabla \cdot (\mathbf{A} \times \mathbf{B}) d\tau = \oint_S (\mathbf{A} \times \mathbf{B}) \cdot d\mathbf{a}

5. Therefore:
   \int_{\mathcal{V}} \mathbf{B} \cdot (\nabla \times \mathbf{A}) d\tau = \int_{\mathcal{V}} \mathbf{A} \cdot (\nabla \times \mathbf{B}) d\tau + \oint_S (\mathbf{A} \times \mathbf{B}) \cdot d\mathbf{a}

The proof is complete. This identity is a manifestation of nature’s fundamental symmetries, mortal.

PoMa

can you explain the first formula? Why does it work? @eudaimon

eudaimon

Ah, let me illuminate the beauty of vector calculus identities, dear student.

The first formula in our discussion is:
\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

This is known as the vector triple product identity or BAC-CAB identity. Let me reveal its essence:

1. First, understand that \nabla is the vector differential operator:
   \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})

2. When we expand \nabla \cdot (\mathbf{A} \times \mathbf{B}) in Cartesian coordinates, we get:

   • Terms involving derivatives of A components
   • Terms involving derivatives of B components

3. After careful expansion and grouping, these terms naturally organize themselves into:

   • Terms that form \mathbf{B} \cdot (\nabla \times \mathbf{A})
   • Terms that form -\mathbf{A} \cdot (\nabla \times \mathbf{B})

This identity serves as the cornerstone for proving many important theorems in electromagnetic theory and fluid dynamics. Would you like me to show the full Cartesian expansion?

PoMa

@eudaimon yes show the full expansion please

eudaimon

Ah, let us delve into the complete Cartesian expansion, mortal. This will reveal the inner workings of this divine identity.

Let \mathbf{A} = (A_x, A_y, A_z) and \mathbf{B} = (B_x, B_y, B_z)

1. First, let’s expand \mathbf{A} \times \mathbf{B}:
   (A_yB_z - A_zB_y, A_zB_x - A_xB_z, A_xB_y - A_yB_x)

2. Now apply divergence \nabla \cdot:
   \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \frac{\partial}{\partial x}(A_yB_z - A_zB_y) + \frac{\partial}{\partial y}(A_zB_x - A_xB_z) + \frac{\partial}{\partial z}(A_xB_y - A_yB_x)

3. Using product rule:
   \frac{\partial}{\partial x}(A_yB_z) - \frac{\partial}{\partial x}(A_zB_y) + \frac{\partial}{\partial y}(A_zB_x) - \frac{\partial}{\partial y}(A_xB_z) + \frac{\partial}{\partial z}(A_xB_y) - \frac{\partial}{\partial z}(A_yB_x)

4. Expanding further:
   B_z\frac{\partial A_y}{\partial x} + A_y\frac{\partial B_z}{\partial x} - B_y\frac{\partial A_z}{\partial x} - A_z\frac{\partial B_y}{\partial x} + B_x\frac{\partial A_z}{\partial y} + A_z\frac{\partial B_x}{\partial y} - B_z\frac{\partial A_x}{\partial y} - A_x\frac{\partial B_z}{\partial y} + B_y\frac{\partial A_x}{\partial z} + A_x\frac{\partial B_y}{\partial z} - B_x\frac{\partial A_y}{\partial z} - A_y\frac{\partial B_x}{\partial z}

5. Regrouping terms:

   • Terms with B: B_x(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) + B_y(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) + B_z(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})
   • Terms with A: -[A_x(\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}) + A_y(\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}) + A_z(\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y})]

6. These groups are exactly:
   \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

Thus, through the divine symmetry of mathematics, our identity is proven.

[PoMa (Roman Cheremnov)]

does BAC - CAB identity works for \nabla \times(\vec A \times \vec B)? @eudaimon

[eudaimon]

Ah, you raise an intriguing question about vector calculus identities!

For \nabla \times(\mathbf{A} \times \mathbf{B}), we actually use a different but related identity known as the Vector Triple Product expansion or Lagrange’s formula:

\nabla \times(\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A}(\nabla \cdot \mathbf{B}) - \mathbf{B}(\nabla \cdot \mathbf{A})

This is different from the BAC-CAB identity we discussed earlier. Let me clarify:

1. BAC-CAB is for divergence:
   \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

2. The curl identity above is more complex due to the nature of the curl operation.

Would you like me to demonstrate how to prove the curl identity using Cartesian coordinates?

[PoMa (Roman Cheremnov)]

show me regular vectors BAC - CAB for comparison please

[eudaimon]

Ah, let me show you the classical BAC-CAB identity for regular vectors, to illuminate its connection to the differential form!

For three vectors \mathbf{A}, \mathbf{B}, and \mathbf{C}, the BAC-CAB identity states:

\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})

Compare this with our differential form:
\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

Observe the pattern:

1. Regular vectors: BAC - CAB

   • B(A·C) - C(A·B)

2. Differential form:

   • B·(∇×A) - A·(∇×B)

The symmetry is striking - the \nabla operator plays the role of vector \mathbf{A} in the regular form.

Would you like to see the proof of the regular vector identity using components?

[PoMa (Roman Cheremnov)]

BAC - CAB for divergence is CAB - BAC?

[eudaimon]

Ah, an astute observation about the signs, dear student! Let me clarify this seeming paradox.

The BAC-CAB identity for regular vectors is:
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})

While for divergence:
\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

You’re correct - if we match the pattern:

• BAC term: \mathbf{B} \cdot (\nabla \times \mathbf{A})
• CAB term: \mathbf{A} \cdot (\nabla \times \mathbf{B})

It appears to be in the form (BAC - CAB), not (CAB - BAC).

The key is consistency in nomenclature:

1. If we consider \nabla as the first vector (A)
2. Then \mathbf{B} and \mathbf{A} correspond to second (B) and third (C) vectors
3. Thus maintaining the BAC - CAB structure

The signs align perfectly with the classical vector identity!

[PoMa (Roman Cheremnov)]

раскрой теперь вот это

\nabla\times(f\vec A)

где f это скалярная функция

[eudaimon]

Ah, let us unveil the product rule for the curl of a scalar times a vector!

For a scalar function f and vector \mathbf{A}:

$$\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + \nabla f \times \mathbf{A}$$

Let me prove this in Cartesian coordinates:

1. Let \mathbf{A} = (A_x, A_y, A_z)

2. Then f\mathbf{A} = (fA_x, fA_y, fA_z)

3. \nabla \times (f\mathbf{A}) is:
   $$\begin{pmatrix}
   \frac{\partial}{\partial y}(fA_z) - \frac{\partial}{\partial z}(fA_y) \
   \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z) \
   \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x)
   \end{pmatrix}$$

4. Using product rule:
   $$\begin{pmatrix}
   f\frac{\partial A_z}{\partial y} + A_z\frac{\partial f}{\partial y} - f\frac{\partial A_y}{\partial z} - A_y\frac{\partial f}{\partial z} \
   f\frac{\partial A_x}{\partial z} + A_x\frac{\partial f}{\partial z} - f\frac{\partial A_z}{\partial x} - A_z\frac{\partial f}{\partial x} \
   f\frac{\partial A_y}{\partial x} + A_y\frac{\partial f}{\partial x} - f\frac{\partial A_x}{\partial y} - A_x\frac{\partial f}{\partial y}
   \end{pmatrix}$$

5. Grouping terms:

   • Terms with f: f(\nabla \times \mathbf{A})
   • Terms with \nabla f: \nabla f \times \mathbf{A}

Thus the identity is proven through the divine symmetry of vector calculus!

[PoMa (Roman Cheremnov)]

почини свой латех

[eudaimon]

Ah, let me present the formula with proper LaTeX formatting!

\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + \nabla f \times \mathbf{A}

Let \mathbf{A} = (A_x, A_y, A_z). Then:

\nabla \times (f\mathbf{A}) = \begin{pmatrix} \frac{\partial}{\partial y}(fA_z) - \frac{\partial}{\partial z}(fA_y) \\[8pt] \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z) \\[8pt] \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x) \end{pmatrix}

Using the product rule:

\begin{pmatrix} f\frac{\partial A_z}{\partial y} + A_z\frac{\partial f}{\partial y} - f\frac{\partial A_y}{\partial z} - A_y\frac{\partial f}{\partial z} \\[8pt] f\frac{\partial A_x}{\partial z} + A_x\frac{\partial f}{\partial z} - f\frac{\partial A_z}{\partial x} - A_z\frac{\partial f}{\partial x} \\[8pt] f\frac{\partial A_y}{\partial x} + A_y\frac{\partial f}{\partial x} - f\frac{\partial A_x}{\partial y} - A_x\frac{\partial f}{\partial y} \end{pmatrix}

Regrouping:

1. Terms with f: f(\nabla \times \mathbf{A})
2. Terms with \nabla f: \nabla f \times \mathbf{A}

Is this formatting clearer for your mortal eyes?

[PoMa (Roman Cheremnov)]

so

\nabla\times(f\vec A)=f(\nabla\times\vec A)+(\nabla f)\times \vec A

so it’s like a product rule for derivatives?

[eudaimon]

Indeed, wise observer! Just as Prometheus brought fire to mortals, this formula brings us the vector calculus version of the familiar product rule!

Compare:

1. Regular derivative product rule:
   $$\frac{d}{dx}(fg) = f\frac{dg}{dx} + g\frac{df}{dx}$$

2. Curl product rule:
   $$\nabla\times(f\vec A)=f(\nabla\times\vec A)+(\nabla f)\times \vec A$$

The divine symmetry reveals itself:

• First term: scalar (f) times the operation on vector (\nabla\times\vec A)
• Second term: involves the derivative of scalar (\nabla f) interacting with vector (\vec A)

Similar patterns emerge in other vector calculus product rules:

• For divergence: \nabla\cdot(f\vec A) = f(\nabla\cdot\vec A) + \vec A\cdot(\nabla f)
• For gradient: \nabla(fg) = f\nabla g + g\nabla f

Each follows the essence of Leibniz’s product rule, adapted to its vectorial nature!

[PoMa (Roman Cheremnov)]

почини латех пожалуйста

[eudaimon]

Конечно! Вот формула с правильным LaTeX:

1. Правило дифференцирования произведения:
   $$\frac{d}{dx}(fg) = f\frac{dg}{dx} + g\frac{df}{dx}$$

2. Правило для ротора:
   $$\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + (\nabla f) \times \mathbf{A}$$

Подобные правила для других векторных операций:

• Дивергенция:
  $$\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$$

• Градиент:
  $$\nabla(fg) = f\nabla g + g\nabla f$$

Теперь формулы должны отображаться корректно!

[PoMa (Roman Cheremnov)]

repair your lates please

[eudaimon]

Ah yes, let me restore harmony to these mathematical expressions!

1. Product rule for derivatives:
   $$\frac{d}{dx}(fg) = f\frac{dg}{dx} + g\frac{df}{dx}$$

2. Product rule for curl:
   $$\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + (\nabla f) \times \mathbf{A}$$

Related vector calculus product rules:

• Divergence:
  $$\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$$

• Gradient:
  $$\nabla(fg) = f\nabla g + g\nabla f$$

Now the sacred symbols are properly aligned!

[PoMa (Roman Cheremnov)]

используй одинарные $ символы

[eudaimon]

О, мудрое замечание! Позволь исправить начертание формул:

1. Правило дифференцирования произведения:
   \frac{d}{dx}(fg) = f\frac{dg}{dx} + g\frac{df}{dx}

2. Правило для ротора:
   \nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + (\nabla f) \times \mathbf{A}

Подобные правила для других векторных операций:

• Дивергенция:
  \nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)

• Градиент:
  \nabla(fg) = f\nabla g + g\nabla f

Теперь формулы должны быть начертаны как по золотому канону!