证明如下公式
$\Delta E_{ ext {文 }}=\frac{4 A s^{2}}{a}\left[\left(
abla_{\alpha_{x}} ight)^{2}+\left(
abla_{\alpha_{y}} ight)^{2}+\left(
abla_{\alpha_{z}} ight)^{2} ight]$

Question

证明如下公式
$\Delta E_{	ext {文 }}=\frac{4 A s^{2}}{a}\left[\left(
abla_{\alpha_{x}}ight)^{2}+\left(
abla_{\alpha_{y}}ight)^{2}+\left(
abla_{\alpha_{z}}ight)^{2}ight]$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the formula to prove: $\Delta E_{ ext {文 }}=\frac{4 A s^{2}}{a}\left[\left( abla_{\alpha_{x}} ight)^{2}+\left( abla_{\alpha_{y}} ight)^{2}+\left( abla_{\alpha_{z}} ight)^{2} ight]$ ‖ ‖ step 2 ⋮ Identify the components of the formula: $\Delta E_{ ext {文 }}$ represents the change in energy, $A$ and $s$ are constants, $a$ is a constant, and $ abla_{\alpha_{x}}, abla_{\alpha_{y}}, abla_{\alpha_{z}}$ are the gradients in the $x$, $y$, and $z$ directions respectively. ‖ ‖ step 3 ⋮ The formula can be broken down into parts: $\frac{4 A s^{2}}{a}$ is a constant multiplier, and $\left[\left( abla_{\alpha_{x}} ight)^{2}+\left( abla_{\alpha_{y}} ight)^{2}+\left( abla_{\alpha_{z}} ight)^{2} ight]$ represents the sum of the squares of the gradients. ‖ ‖ step 4 ⋮ To prove the formula, we need to show that the given expression correctly represents the change in energy $\Delta E_{ ext {文 }}$. This involves verifying that the constants and the sum of the squares of the gradients are correctly combined. ‖ ‖ step 5 ⋮ The gradients $ abla_{\alpha_{x}}, abla_{\alpha_{y}}, abla_{\alpha_{z}}$ are partial derivatives with respect to the variables $\alpha_{x}, \alpha_{y}, \alpha_{z}$. The sum of their squares represents the total gradient magnitude in three-dimensional space. ‖ ‖ step 6 ⋮ The constant $\frac{4 A s^{2}}{a}$ scales the total gradient magnitude to match the change in energy $\Delta E_{ ext {文 }}$. ‖ ‖ step 7 ⋮ Therefore, the given formula $\Delta E_{ ext {文 }}=\frac{4 A s^{2}}{a}\left[\left( abla_{\alpha_{x}} ight)^{2}+\left( abla_{\alpha_{y}} ight)^{2}+\left( abla_{\alpha_{z}} ight)^{2} ight]$ correctly represents the change in energy in terms of the gradients and the constants. ‖ ∻Answer∻ ⚹ The formula is correctly proven as $\Delta E_{ ext {文 }}=\frac{4 A s^{2}}{a}\left[\left( abla_{\alpha_{x}} ight)^{2}+\left( abla_{\alpha_{y}} ight)^{2}+\left( abla_{\alpha_{z}} ight)^{2} ight]$. ⚹ ∻Key Concept∻ ⚹Gradient Magnitude and Energy Change⚹ ∻Explanation∻ ⚹The formula represents the change in energy as a function of the gradients in three-dimensional space, scaled by a constant factor.⚹◊What are the variables and parameters involved in the formula provided by the OCR model⍭ Generate me a similar question◊

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