Talk:Planck temperature

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Why does this particular formula reveal the Planck Temperature? Is there some sort of universal formula for temperature of which this is a special case? The Planck temperature is extremely hot -- appearantly, hotter than anything that currently exists in our universe. What are the implications of this (if any)? Ravenswood 20:09, 17 May 2005 (UTC)

Question on the final value[edit]

Does it make any sense to include two digits on parentheses?

AFAIK, it's a scientific convention denoting that there is some uncertainty in the actual value. Druiffic (talk) 03:18, 17 January 2009 (UTC)

Question about Planck Temperature[edit]

If absolute zero represents the lowest possible temperature because particles are not moving relative to each other, what is happening to the particles at the Planck Temperature? Are they moving relative to each other at the speed of light? If so, it may be appropriate to include this information in the article. Kmorford 22:24, 4 June 2006 (UTC)

They're moving slower than light, but they're more or less at the highest kinetic energy that it makes sense to consider for particles, based on our current understanding of relativity and quantum mechanics. Above the Planck temperature, you'd treat them as black holes, instead. In practice, we expect strange things to happen as this regime is approached, probably resulting either in it being impossible by any means to accelerate particles to higher energies, or in the inevitable decay of the particle into multiple particles of lower energy. There is presently no good description of what happens at these energies. At minimum we'd need a good description of quantum gravity, and for a truly complete description we'd need a theory of everything (as the Planck temperature corresponds to the expected energy scale for unification of all four forces). --Christopher Thomas 01:38, 5 June 2006 (UTC)

I think that it would be extremely useful to provide some form of explanation as to why matter cannot exceed the Planck temperature. This article has me, and I'm sure many other people, quite confused. In fact, additions of this sort to other Planck units and derivatives that do not have them would be desirable. CharonM72 03:23, 15 May 2007 (UTC)

I agree. This article (and the related "Planck constants" articles) explain how Planck's units are such that certain fundamental constants (like c and G) become 1, but this doesn't say anything about the planck constants being any sort of physical limits. For example, one "planck momentum" is just 6.5 kg m/s - obviously not a special momentum at all (just a normal momentum you can encounter in a day-to-day situation). Why is "Planck's temperature" a theoretical limit? Why does it require quantum gravity? This is certainly far from obvious, and not adequately explained in the article. Nyh (talk) 13:52, 14 October 2009 (UTC)

Categorisation[edit]

Why is this a unit of temperature? Surely it's just a constant rather than a way in which we can express temperatuares? hitman012 21:55, 1 September 2006 (UTC)

Planck units were developed to be units, and even if I guess nobody actually measures temperatures in TP it is possible to do that. --Army1987 11:05, 11 November 2006 (UTC)

well, this is dumb[edit]

Planck didn't bother to put in the proper degrees of freedom; this "unitary" expression has 2/2 as the factor, or two dimensions, when it should be 3/2. They should normalise the constant at (2/3)TP. -lysdexia 05:19, 3 October 2007 (UTC)

Conversion table[edit]

The table "Planck temperature temperature conversion formulae" is a conversion table from and to Kelvin, not Planck temperature. Gulliveig (talk) 04:51, 23 August 2011 (UTC)

Discrepency between this and "Orders_of_magnitude_(temperature)" page[edit]

I may be wrong here but on this page Planck temp is said to be about 1.42*10^32 K, but on the http://en.wikipedia.org/wiki/Orders_of_magnitude_(temperature) page, it is said to be 14.2 million YK which unless I'm wrong is 14.2 *10^6 *10^24 which is 1.42*10^31, one order of magnitude off. I haven't done the calc myself so I'm not sure which to believe. I will likely post this on that talk page also.Washyleopard (talk) 12:59, 17 July 2012 (UTC)