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<rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Localization of Negative Energy and the Bekenstein Bound · HackerPapers discussions</title><link>https://hackerpaper.kaipfister.de/p/1309.1121/</link><description>Collected HN comments and submissions linking this paper.</description><language>en</language><lastBuildDate>Mon, 02 Nov 2026 11:16:22 GMT</lastBuildDate><atom:link href="https://hackerpaper.kaipfister.de/p/1309.1121/feed.xml" rel="self" type="application/rss+xml" /><item><title>Localization of Negative Energy and the Bekenstein Bound — mentioned by zachf</title><link>https://news.ycombinator.com/item?id=25596947</link><guid isPermaLink="false">urn:hackerpapers:mention:1309.1121:25596947</guid><pubDate>Thu, 31 Dec 2020 20:48:27 GMT</pubDate><description>&lt;p&gt;Well the argument I presented shows that at least one state has a negative energy density at any given point (I suppressed the dependence on position, T = T (x) )— but the argument does not prove that this state has lower total energy than the vacuum because there could be lots of positive energy density somewhere else! In fact this has to be true, because as you correctly pointed out the hamiltonian has to be bounded below, and with some assumptions about the field theory you can show that the vacuum has 0 energy and every other state has higher total energy. You can even put constraints on…&lt;/p&gt;&lt;p&gt;&lt;a href="https://hackerpaper.kaipfister.de/p/1309.1121/"&gt;Paper and discussion history&lt;/a&gt;. A mention is not an endorsement.&lt;/p&gt;</description><category>hep-th</category></item></channel></rss>
