钟慢效应和洛仑兹变化t→t'相矛盾到底哪错了?首先来看洛仑兹变换:x'=(x-ut)/√(1-U^2/C^2) y'=y Z'=Z t'=(t-ux/c^2)/√(1-u^2/c^2) 我们都知道钟慢效应t=t’/√(1-u^2/c^2) 也就是以速度u行进的参考系S'里的t
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 17:05:41
![钟慢效应和洛仑兹变化t→t'相矛盾到底哪错了?首先来看洛仑兹变换:x'=(x-ut)/√(1-U^2/C^2) y'=y Z'=Z t'=(t-ux/c^2)/√(1-u^2/c^2) 我们都知道钟慢效应t=t’/√(1-u^2/c^2) 也就是以速度u行进的参考系S'里的t](/uploads/image/z/14950778-50-8.jpg?t=%E9%92%9F%E6%85%A2%E6%95%88%E5%BA%94%E5%92%8C%E6%B4%9B%E4%BB%91%E5%85%B9%E5%8F%98%E5%8C%96t%E2%86%92t%27%E7%9B%B8%E7%9F%9B%E7%9B%BE%E5%88%B0%E5%BA%95%E5%93%AA%E9%94%99%E4%BA%86%3F%E9%A6%96%E5%85%88%E6%9D%A5%E7%9C%8B%E6%B4%9B%E4%BB%91%E5%85%B9%E5%8F%98%E6%8D%A2%EF%BC%9Ax%27%3D%28x-ut%29%2F%E2%88%9A%281-U%5E2%2FC%5E2%29+y%27%3Dy+Z%27%3DZ+t%27%3D%28t-ux%2Fc%5E2%29%2F%E2%88%9A%281-u%5E2%2Fc%5E2%29+%E6%88%91%E4%BB%AC%E9%83%BD%E7%9F%A5%E9%81%93%E9%92%9F%E6%85%A2%E6%95%88%E5%BA%94t%3Dt%E2%80%99%2F%E2%88%9A%281-u%5E2%2Fc%5E2%29+%E4%B9%9F%E5%B0%B1%E6%98%AF%E4%BB%A5%E9%80%9F%E5%BA%A6u%E8%A1%8C%E8%BF%9B%E7%9A%84%E5%8F%82%E8%80%83%E7%B3%BBS%27%E9%87%8C%E7%9A%84t)
xN@_Ń`p0QЪU|ZMĶT}۞|b~4lfg3_O%| AUHO#(PRE-TS.8(]:vv
29iJxv+24/.rK&+l2IVH2(x:íWY:WGqH@ toI/ޙ=TCQ/$na1j#QшoD?J;;9@oR!
[(P>Drz݉l?7=ձuK[;
jX3LTC