Why is Planck time scaled by c^5?

apeiron
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Just curious.

The Planck length is lpl = (hG/c3)1/2 = 10-33cm

And it seems intuitive that it's c cube because space has three dimensions for the action.

But the Planck time is tpl = (hG/c5)1/2 = 10-43s

So is there some obvious physical reason why c is to the power of five here?
 
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The Planck time is just the Planck length divided by c. This adds two powers of c because they are inside the square root.
 
apeiron said:
Just curious.

The Planck length is lpl = (hG/c3)1/2 = 10-33cm

And it seems intuitive that it's c cube because space has three dimensions for the action.

But the Planck time is tpl = (hG/c5)1/2 = 10-43s

So is there some obvious physical reason why c is to the power of five here?

Yes, that's the only combination of physical constants that gives you a constant with time units
 
phyzguy said:
The Planck time is just the Planck length divided by c. This adds two powers of c because they are inside the square root.

Thanks. Beautifully simple.
 
apeiron said:
Just curious.

The Planck length is lpl = (hG/c3)1/2 = 10-33cm

And it seems intuitive that it's c cube because space has three dimensions for the action.
I don't see how a power of 3/2 looks natural.
The Planck volume has c9/2.

Those odd factors just show how "unnatural" the SI-units (where c, h, G, k are not nice numbers) are in terms of fundamental physics.
 
Newton said Gmm'/r=energy. In natural units, that means G is a (length)^2.
The hbar and c are just put into get G in cm^2. This is always unique.
 
what does "space has 3 dimensions for the action"? I mean that it's kind of weird, we don't know whether at Planck scale you need more than 3 spatial dimensions, so it's not so intuitive...
On the other hand, everything seems normal under what is called dimensional analysis... So you have some constants (G, c, \hbar) and you want to build characteristic quantities out of them ... So for everything, you just write:
[X]= [c]^{a} [\hbar]^{b} [G]^{d}
and you solve for a,b,d
 

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