Show that 1/ρ D_ρ/D_t = □(→┬∇ ) . □(→┬V )
To show that 1/ρ Dρ/Dt = ∇ · V, we can start by expanding the expression on the left-hand side.
1/ρ Dρ/Dt = 1/ρ (∂ρ/∂t + V · ∇ρ)
Next, we can apply the continuity equation, which states that ∂ρ/∂t + ∇·(ρV) = 0. Rearranging this equation, we have ∂ρ/∂t = - ∇·(ρV). We can substitute this expression into the equation above.
1/ρ Dρ/Dt = 1/ρ (- ∇·(ρV) + V · ∇ρ)
Now, let's use a vector identity called the product rule which states that ∇·(A B) = B · ∇A + A · ∇B. We can apply this identity to the second term above.
1/ρ Dρ/Dt = 1/ρ (- ∇·(ρV) + V · ∇ρ) = 1/ρ (- ∇·(ρV) + (∇ρ) · V)
Notice that ∇·(ρV) is equal to (∇ρ) · V + ρ ∇·V by the product rule. We can substitute this into the equation.
1/ρ Dρ/Dt = 1/ρ (-((∇ρ) · V + ρ ∇·V) + (∇ρ) · V)
The (∇ρ) · V term cancels out.
1/ρ Dρ/Dt = 1/ρ (-ρ ∇·V)
Finally, we can simplify the expression.
1/ρ Dρ/Dt = - ∇·V
Therefore, we have shown that 1/ρ Dρ/Dt = ∇ · V.