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My Theory Of Everything Is Now Largely Complete


Dubbelosix

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Actually I've just had a look at the Friedmann equations and so now I know what a and k are. I find I was wrongly taking k to be Boltzmann's constant (seeing as Doublesox's equation contains temperature so is implicitly thermodynamic) but it is isn't that. In the context of Friedmann's equations, k is a dimensionless parameter, set to be between -1 and +1, according to whether a 3-spherical, flat or hyperbolic universe is being considered. So it is a sort of "shape" factor. a, on the other hand is the "scale factor", again dimensionless. 

 

So I need to re-do my dimension check on the second term in the LHS bracket. I now get something a lot simpler, viz. L²/T².  So it is still a nonsense, as one cannot add terms with different dimensions, but at least we don't have to worry about daft things like √M.  Maybe if Doublesox can add something to the first term with dimensions L², he can make it look more superficially plausible, at least.  :winknudge:

 

 

Oh is this you in a round-about way saying you may have been wrong... Let me help you articulate it better, you are wrong.

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Here is an example of my proclivity for dimensional analysis. I noticed an error in a wiki article, made a public post on stack exchange (which is full of physicists) and it wasn't answered for a while... until I had to answer my own question. I exposed the wiki article as wrong.

 

https://physics.stackexchange.com/questions/467765/heaviside-vector-units/472807#472807

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