tag:blogger.com,1999:blog-13869903.post4079495048376268324..comments2018-01-21T11:31:19.843-06:00Comments on nanoscale views: A thermoelectric surprise in metalsDouglas Natelsonhttps://plus.google.com/101165937354831985246noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-13869903.post-10411941768631304412017-07-23T10:41:25.906-05:002017-07-23T10:41:25.906-05:00Anon, yes, it seems so. Variations of around 10% ...Anon, yes, it seems so. Variations of around 10% or so along the length of the device would be about right.Douglas Natelsonhttps://www.blogger.com/profile/13340091255404229559noreply@blogger.comtag:blogger.com,1999:blog-13869903.post-69699614994930912882017-07-22T00:11:21.940-05:002017-07-22T00:11:21.940-05:00I see, thank you for the explanation! And do you t...I see, thank you for the explanation! And do you think that the Seebeck coefficient can vary that much along the wire? I.e., does this all work out quantitatively?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13869903.post-45994880717034664912017-07-19T16:33:32.476-05:002017-07-19T16:33:32.476-05:00(somehow blogger ate the rest of my explanation.) ...(somehow blogger ate the rest of my explanation.) For each position of the laser, we get some profile \(T(x)\), and we can compute that integral. If \(T(x)\) happens to be largest right where \(dS/dx\) is positive, you'd likely end up with a positive number. Conversely, if \(T(x)\) is largest where \(dS/dx\) happens to be negative, then you could get a negative net voltage. Depending on how the Seebeck coefficient varies along the sample, you could then get sign flips of \(V\) as you scan the laser spot along the sample.Douglas Natelsonhttps://www.blogger.com/profile/13340091255404229559noreply@blogger.comtag:blogger.com,1999:blog-13869903.post-54087847104137700372017-07-19T16:22:30.544-05:002017-07-19T16:22:30.544-05:00Anon, if there is a spatially varying Seebeck coef...Anon, if there is a spatially varying Seebeck coefficient \(S(x)\), and the sample is heated inhomogeneously via the spot from the laser, it's not that hard to have voltage vs. laser position be non-monotonic. One way to see this is to start from the definition of \(S\), where locally \(\nabla V \equiv -S \nabla T\). Then \(V = -\int S(x)~dT(x)/dx~dx\) where the integral runs from one end of the sample to the other. If you integrate by parts, you get \(V = -[S(1)T(1)-S(2)T(2)] + \int T(x)~dS/dx~dx\), where 1 and 2 denote evaluation at the ends of the sample. That second integral is the key, because \(dS/dx\) can take on either sign as \(S\) varies along the sample.Douglas Natelsonhttps://www.blogger.com/profile/13340091255404229559noreply@blogger.comtag:blogger.com,1999:blog-13869903.post-72774067679311427182017-07-19T14:31:16.171-05:002017-07-19T14:31:16.171-05:00Hi Doug, why do you think the sign of photovoltage...Hi Doug, why do you think the sign of photovoltage changes along the wire?<br /><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13869903.post-59055917299441205752017-07-16T16:46:00.855-05:002017-07-16T16:46:00.855-05:00Nifty. Thank you for writing this up.Nifty. Thank you for writing this up.Ted Sandershttps://www.blogger.com/profile/15001183656827732917noreply@blogger.com