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fyc802

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lsloneil

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ÒýÓûØÌû:
3Â¥: Originally posted by fyc802 at 2013-04-22 17:01:04
At t=infinity, <B(0)*B(t)> = <B>^2£¬ÎªÊ²Ã´£¿ÎÒ¸Õ¸Õ¸úÀÏʦÌÖÂÛ¹ýÁË£¬¿ÉÄÜÊÇÎÒÃǶԡ°Ïµ×ÛÆ½¾ù¡±µÄ¸ÅÄîÀí½âÓÐÎó¡£ÎÒÏëͨ¹ý±à³Ì¼ÆËã×ÔÏà¹Øº¯Êý¡£ÇëÎÊÄúÓÐûÓйØÓÚ¡°Ïµ×ÛÆ½¾ù¡±µÄÏà¹Ø×ÊÁÏ»ò¹«Ê½£¿×îºÃÊÇ ...

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5Â¥2013-04-23 08:58:32
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lsloneil

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fyc802: ½ð±Ò+20, ¡ï¡ï¡ïºÜÓаïÖú 2013-04-26 09:22:43
Not always zero. It depends on the ensemble average . At t=infinity, = ^2, so if is not equal to zero, then A(t) won't go to zero at infinite time.

For velocity-velocity auto TCF or dipole-dipole TCF in uniform liquid, it should decay to zero since =0. However, at some interfaces or in confinement, the dipole-dipole TCF might not decay to zero since molecule's orientations are not uniformly distributed.
2Â¥2013-04-22 13:32:16
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fyc802

½ð³æ (ÖøÃûдÊÖ)


ÒýÓûØÌû:
2Â¥: Originally posted by lsloneil at 2013-04-22 13:32:16
Not always zero. It depends on the ensemble average <B>. At t=infinity, <B(0)*B(t)> = <B>^2, so if <B> is not equal to zero, then A(t) won't go to zero at infinite time.

F ...

At t=infinity, = ^2£¬ÎªÊ²Ã´£¿ÎÒ¸Õ¸Õ¸úÀÏʦÌÖÂÛ¹ýÁË£¬¿ÉÄÜÊÇÎÒÃǶԡ°Ïµ×ÛÆ½¾ù¡±µÄ¸ÅÄîÀí½âÓÐÎó¡£ÎÒÏëͨ¹ý±à³Ì¼ÆËã×ÔÏà¹Øº¯Êý¡£ÇëÎÊÄúÓÐûÓйØÓÚ¡°Ïµ×ÛÆ½¾ù¡±µÄÏà¹Ø×ÊÁÏ»ò¹«Ê½£¿×îºÃÊǹØÓÚż¼«¾ØµÄ¡£
3Â¥2013-04-22 17:01:04
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luqing6879

ľ³æ (ÖøÃûдÊÖ)

ÒýÓûØÌû:
3Â¥: Originally posted by fyc802 at 2013-04-22 17:01:04
At t=infinity, <B(0)*B(t)> = <B>^2£¬ÎªÊ²Ã´£¿ÎÒ¸Õ¸Õ¸úÀÏʦÌÖÂÛ¹ýÁË£¬¿ÉÄÜÊÇÎÒÃǶԡ°Ïµ×ÛÆ½¾ù¡±µÄ¸ÅÄîÀí½âÓÐÎó¡£ÎÒÏëͨ¹ý±à³Ì¼ÆËã×ÔÏà¹Øº¯Êý¡£ÇëÎÊÄúÓÐûÓйØÓÚ¡°Ïµ×ÛÆ½¾ù¡±µÄÏà¹Ø×ÊÁÏ»ò¹«Ê½£¿×îºÃÊÇ ...

Personally, I don't think "At t=infinity, = ^2". The integral should approximate to 0 at t=infinity.
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4Â¥2013-04-23 05:29:25
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