Why is Planck time scaled by c^5?

[apeiron]

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?

 

[phyzguy]

The Planck time is just the Planck length divided by c. This adds two powers of c because they are inside the square root.

 

[dauto]

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

 

[apeiron]

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.

 

[mfb]

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 c^9/2.

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.

 

[Meir Achuz]

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.

 

[ChrisVer]

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 $a,b,d$