Steven Charlton
Universal Time, Mass and Length
Imperial Values
From the problem page the following conversion factors were given:
1 ft  30.48 cm
2.2 lb  1 kg
1 cal  4.184 J
1.8 F  1 K
Putting these in the form 1 SI unit = … gives:
1
ft  1 cm
30.48
2.2 lb  1 kg  1000 g , so
2.2
lb  1 g
1000
1
cal  1 J
4.184
1.8 F  1 K
Substituting these in place of the units in the values of the constants gives:
For Newton's Gravitational Constant
G  6.674  10 8 cm3 g1 s 2
3
 1
  2.2

 6.674  10  8 
ft  
lb 
 30.48   1000 
1
s 2
6.674  10  8  1000 3 1  2
ft lb s
30.48 3  2.2
 1.0713  10  9 ft 3 lb 1 s  2

 1.071  10  9 ft 3 lb 1 s  2
For the Speed of Light
c  2.998  1010 cm s -1
 1

 2.998  1010 
ft  s 1
30
.
48


2.998  1010
ft s 1
30.48
 9.83595  108 ft s 1

 9.836  108 ft s 1
Page 1 of 6
Steven Charlton
For Planck's Constant
h  1.054  10 27 g cm2 s 1
2
 2.2
 1

 1.054  10  27 
lb 
ft  s 1
1000
30
.
48



1.054  10  27  2.2
lb ft 2 s 1
1000  30.48 2
 2.49593  10  33 lb ft 2 s 1

 2.496  10  33 lb ft 2 s 1
For Boltzmann's Constant
k  1.38  10 23 J K 1
 1

1
 1.38  10  23 
cal 1.8 F 
4
.
184


1.38  10  23
cal F -1
4.184  1.8
 1.83237  10  24 cal F -1

 1.83  10  24 cal F -1
So the imperial values of the constants are
G  6.674  108 cm3 g1 s 2  1.071 109 ft 3 lb 1 s 2
c  2.998  1010 cm s -1  9.836  108 ft s 1
h  1.054  1027 g cm2 s 1  2.496  1033 lb ft 2 s 1
k  1.38  1023 J K 1  1.83  1024 cal F-1
Page 2 of 6
Steven Charlton
Natural Units
In a system of units where Newton's Constant, the Speed of Light and Planck's Constant are
numerically equal to one, let L be the units of length, M of mass, and T time, then we have:
G  1 L3 M1 T 2
c  1 L T -1
h  1 M L2 T 1
Now we need to solve for L, M, and T in terms of G, c and h.
Note that:
c
L T -1
 3 1  2
G L M T
Cancel an L from the top and bottom, along with a T-1, giving:
c
1
 2 1 1
G L M T
Now multiply both sides through by h, giving:
ch
1
M L2 T 1
 2 1 1  M L2 T 1  2 1 1
G L M T
L M T
Cancel a factor of T-1 and L2 from the top and bottom, and move the M-1 from the
denominator into the numerator as a factor of M, giving:
ch
 M2
G
This can be solved to find M:
ch
M
G
Note that:
h M L2 T 1

c
L T -1
Cancel from the top and bottom a factor of L, and of T-1, giving:
h
 ML
c
Divide both sides of this by the expression found for M above, giving:
h
ch

 ML M
c
G
Simplifying this:
2
L
h
ch h G
h G


  

c
G
c ch
 c  ch
h2 G

c  ch
2
Gh
c3
Page 3 of 6
Steven Charlton
So:
Gh
c3
L
Now all that remains is to find the value of T. Divide both sides of the value for L above by c,
giving:
Gh
L  L T -1 
c
c3


The left hand side simplifies to T, and simplifying the right hand side gives:
Gh 1
Gh 1
Gh
T
 
 2 
3
3
c
c
c
c
c5
So:
T
Gh
c5
Now we have the values for the units in terms of the three constants, all the remains is to
evaluate the results to find the value of the units in SI units:
M

ch
G
2.998  10

cm s -1 1.054  10  27 g cm 2 s 1
6.674  10  8 cm3 g1 s  2
10


2.998  1010  1.054  10  27 cm s -1 g cm 2 s 1
6.674  10  8 cm 3 g1 s  2

2.998  1010  1.054  10  27 cm 3 g s  2
6.674  10  8 cm 3 g1 s  2

2.998  1010  1.054  10  27 2
g
6.674  10  8

2.998  1010  1.054  10  27
g
6.674  10  8
 4.73462990 7   10 10 g
 2.17592   10  5 g
 2.176  10  5 g
And for L:
Gh
L
c3

6.674  10  8  1.054  10  27
2.998  10 
10 3
cm
 1.61571949   10  33 cm
 1.616  10  33 cm
Page 4 of 6
Steven Charlton
And for T:
Gh
T
c5

6.674  10  8  1.054  10  27
2.998  10 
10 5
s
 5.38932451 7   10 - 44 s
 5.389  10 - 44 s
In fact, the Boltzmann constant can also be normalised, and used to find a natural
temperature scale for the universe.
If θ is the unit of temperature in this system of units.
From the equation Work = Force × Distance, then it can be seen that J = Nm, and from the
equation Force = Mass × Acceleration we have that N = kg m s-2. So the units of the
Boltzmann constant are kg m2 s-2 K-1.
So as before:
k  M L2 T 2 θ1
So solving for θ:
M L2 T 2
θ
k
Substituting in what we know from above about the values of M, L and T in terms of G, c,
and h, gives:
2
θ

ch  Gh   Gh 
G  c 3   c 5 
k
2
ch Gh c 5
 3 
G
Gh
c
k

ch c 2

G
k

ch  c 2 
 
G  k 

c 5h
Gk 2
2
The Boltzmann constant must be converted to have the same units as the other constants
before evaluating the result.
k  1.38  10 23 kg m2 s 2 K 1
 1.38  10  23 1000 g100 cm  s  2 K 1
2
 100 2  1000  1.38  10  23 g cm 2 s  2 K 1
 1.38  10 16 g cm 2 s  2 K 1
Page 5 of 6
Steven Charlton
Evaluating this gives:
θ

c 5h
Gk 2
2.998  10 
10 5
 1.054  10  27

6.674  10  8  1.38  10 16

2
K
 1.417187096   10 32 K
 1.42  10 32 K
And so we have the natural mass scale of the universe as 2.176  105 g
The natural length scale of the universe as 1.616 1033 cm
The natural time scale of the universe as 5.389 10-44 s
And the natural temperature scale of the universe as 1.42  1032 K
Page 6 of 6