Is this "derivation" for negative black hole entropy correct?

+ 1 like - 0 dislike
201 views

This "derivation" will use relativity-corrected Planck units to show black hole entropy should be negative. Do you see an error in this line of reasoning?

From Einstein's "Relativity" appendix 2 $meters=i*c*seconds$ where $i=sqrt(-1)$. I'll use this to convert seconds to meters/i/c.  It forces c = i*3E8 unitless but not dimensionless due to i, but not "dimensional"  if you use that word to mean "has units". In relativistic Planck units c=i instead of 1.

This gives negative energy, E= - mc^2 ,in accordance with gravitational cosmology. It also gives momentum "i" which causes it to be the negative of force, in accordance with their opposite directions from the point of view of the mass being pushed. Force does not carry the "i" which tells the user it is not translational like momentum is. Planck's constant carries 1/i. It might be better to express -i and -1/i as 1/i and i.

For black holes, $S=kAc^3/(4Gh)$

I get the relativity-corrected Planck units to have $G=m/kg/-1$ and $h=-kg m/i$, $c^3=-i$, and $k=-1/-1$. k is units of heat energy divided by temperature which is a kinetic energy constrained to a strict distribution profile, so it is also unitless. Temperature is a measure of random kinetic energy, period.  All this gives the entropy of a black hole to be

$S = -A/4$  (in unitless, relativity-corrected planck units)

That you can also substitute mass for area is important which I'll mention later.

It is really a description of the number of states lost, so you have to have a starting number of states and an ending number of states. In other words, using the negative entropy idea from Schrodinger's "What is life?" I get

$S = ln(0) - ln(e^(A/4)) = 1-A/4$

So there might be a very very slight error in the black entropy equation, i.e. 1.

What's inside of a black hole is not observable, so it is not part of physics. It can't be entropy or negative entropy. It is the amount of entropy lost from the observable universe when it collapsed, so where did it go?

My suggestion below is not part of the question, but I feel the need to express a possible solution since I view information conservation as rigid. In short, entropy and a comoving volume are interchangeable and negative of each other like mass and energy. The comoving area integrated from a center point outward for a volume plus the entropy inside could be conserved like mass and energy, and if you include the negative signs described above, both sums over cosmological scales are zero. On smaller scales it might be something like for a given volume mass+V=constant and E+S=constant where V is the integrated surface area that includes the effects on space from nuclear particles and E might be strictly gravitational and mass is all other E.

BTW, The acceleration of the Hubble constant has relativity-corrected Planck units of -1/m^2.

The OBSERVABLE Universe decreasing in surface area as a result of the accelerated expansion might be emitting entropy (via greater and greater red shift) like a black hole. A larger comoving surface area minus our observable surface area (1-A related?) would be proportional to the entropy we have emitted so far. This would mean our observable plus non observable entropy per comoving volume is constant but local proper volumes like our solar system could decrease in entropy.

Maybe a black hole collapse is a sudden decrease in proper surface area that offsets the entropy lost, so it is conserved in the form of an area contraction instead of entropy, and emitted again as it returns the lost space. A comoving volume much larger than the event would be constant and from the reasoning above the entropy + surface area summed over the radius of the volume would also be constant.

E=-m so maybe S=-A. If S=1-A then maybe E=1-m from some quantum effect.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.