[問題]griffiths電磁第五章,"A"的部分
我用3rd edtion
→ →
第235頁,有一式是▽^2 A = -μ0 J
而底下的註解寫道,在卡式坐標中可將上式寫為
▽^2 Ax = -μ0 Jx 、 ▽^2 Ay = -μ0 Jy 、 ▽^2 Az = -μ0 Jz
但在curvilinear coordinates中,卻不能寫成
▽^2 Ar = -μ0 Jr
而要以
→ → →
▽^2 A = ▽(▽.A)-▽×(▽×A)
來處理
原文是:
In curvilinear coordinates the unit vectors themselves are functions of
position, and must be differentiated, so it is not the case, for example,
▽^2 Ar = -μ0 Jr
我看不懂這段的意思,也不知道為何▽^2 Ar = -μ0 Jr不一定正確,煩請高手解答,謝^2
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◆ From: 218.162.104.193
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不能寫成▽^2 A = (▽^2 Ar)r + (▽^2 Aθ)θ + (▽^2 Aψ)ψ=-μ0(Jrr+Jθθ+Jψψ)
然後透過r分量相等,得到▽^2 Ar = -μ0 Jr嗎?
好像沒有多出什麼東西
※ 編輯: ed78617 來自: 218.162.104.193 (07/11 19:02)
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