# I Problem with Planck units

1. Jul 9, 2016

### facenian

Since $c=1$ we have $[l]=[t]$,
$\hbar=1$ and from Einstein law $E=\hbar\omega\rightarrow [E]=[t^{-1}]$,
$G=1\rightarrow [E]=[m]^2[t^{-1}]$
then we must have $[m]=1$ but since $E=mc^2\rightarrow [E]=[t]^2$
Can someone please point out my mistake

2. Jul 9, 2016

### Staff: Mentor

To my knowledge Planck units are not geometrized, so for example c still has dimensions of L/T. So Energy has the usual dimensions of L^2 M/T^2

3. Jul 10, 2016

### facenian

I should have said "but since $E=mc^2\rightarrow [E]=[1]\neq [t^{-1}]$"

4. Jul 10, 2016

### vanhees71

I think you've got everything right. What Planck intrigued so much about his units was that then physics is in a sense "absolute", i.e., when measuring all quantities in Planck units they are all dimensionless "absolute" numbers.

5. Jul 10, 2016

### Staff: Mentor

They are if you are referring to these Planck units:

https://en.wikipedia.org/wiki/Planck_units

Saying that the Planck units referred to above set $G = 1$ is a little misleading. Actually, what they do is replace $G$ with $1 / m_p^2$, where $m_p$ is the Planck mass (possibly with a factor of $16 \pi$ stuck in there somewhere, depending on the equation). For example, consider the Newtonian equation for gravitational potential energy:

$$U = - \frac{G m_1 m_2}{r}$$

Here $U$ has conventional units of energy. Now suppose we write this same equation in Planck units, in which $[E] = [l^-1]$ (which, as you note, is implied by $c = \hbar = 1$); we will have (if we keep the units of mass the same as the units of energy, which is implied by $c = 1$):

$$U = - \frac{m_1 m_2}{m_p^2 r}$$

The Wikipedia page on Planck units describes this as measuring masses in units of the Planck mass--i.e., it would write $U = - m_1 m_2 / r$, making it look like the units have $G = 1$, but only because $m_1$ and $m_2$ do not have units of mass but are dimensionless numbers. I would prefer to describe it as setting $G = 1 / m_p^2$; this is what many QFT textbooks do, for example, when discussing quantum gravity.

6. Jul 10, 2016

### Staff: Mentor

Hmm, to me tables 1 and 2 both indicate that the Planck units are considered to have SI-like dimensionality.

However, I should be candid that I don't have an authoritative source, just that Wiki page.

7. Jul 10, 2016

### facenian

I think table two only gives the equivalents of plank units in SI units, but Planck units are dimensionless

I read the Wiki page and I believe things have cleared up for me now. I will tell you how I understand it now and ask for you(all) opinion about it.
In Planck units rather than putting physical constants like c,G,etc. equal to one(dimensionless) we rather get rid of them by division making them disappear from the equations so that the physical equations become relations between pure numbers with no dimensions at all.

Last edited: Jul 10, 2016
8. Jul 10, 2016

### facenian

I think I haven't got it right because I tried to assign dimensionality to the magnitudes which must be, as you pointed it out, dimensionless

9. Jul 10, 2016

### facenian

I believe that $[G]=\frac{1}{[m]^2}$ comes from the requirement $c=\hbar=1$ the problem arises when we put $c=\hbar=G=1$ because this leads to a contradiction in dimensionality that caused me the problem that made me post this thread. The Wiki page puts $c=\hbar=G=etc.=1$ and think this is misleading as you said(it migtht be plain wrong)

10. Jul 10, 2016

### Staff: Mentor

Yes, that is my understanding. Basically, any time a "length" appears in an equation, we put in a number representing the ratio of that length to the Planck length; for a "time", we use the ratio of that time to the Planck time; for a "mass", the ratio of that mass to the Planck mass; etc. This makes all numbers appearing in all equations dimensionless.

11. Jul 11, 2016

### Staff: Mentor

That is a good approach. It completely sidesteps the question of dimensionality. It doesn't matter if energy is considered to have dimensions of M or L^2 M/T^2 or even just E. It always shows up as the dimensionless ratio E/Ep

12. Jul 11, 2016

### vanhees71

Well, I think it's pretty simple, at least from the point of view of a theorist. You can define a complete unit system by fixing some purely conventional constants which appear in the SI units, because you have to convert between man-made units that don't fit to the fundamental laws of nature. This is understandable since in Newtonian mechanics there is no fundamental constant. The first time a fundamental constant occured in the history of physics was Maxwell electromagnetism, where the "speed of light" entered the game, which is a universal unit in dimensions of a speed. Nowadays we know it's much more fundamental than that. It's a fundamental property of spacetime as described in the general (and thus also special) theory of relativity. In the SI it's defined as an exact value which was chosen such as to make the relation between the base unit of time (second) compatible with the base unit of length (metre). However, these are conventional units, and thus you can as well set $c=1$, which is very natural, because space and time occur in the theory of relativity in a very "symmetric" way, although there is a distinction, because you need a causality structure to do physics. That's implemented in GR by using a pseudometric with signature (1,3) (or (3,1) which is equivalent) in the description of spacetime as a pseudo-Riemannian continuum.

The next fundamental constant entering the history of physics was Planck's "quantum of action". It's also purely conventional and will probably be fixed to a certain value very soon by the international committee defining the SI. Anyway, for a theoretical physicist it's as inconvenient to keep it as it is inconvenient to keep $c$. So we set it to 1. Now we have only one unit left. That you can choose as some length, time, or energy (momentum or mass). In high-energy particle and nuclear physics we are used to such units and usually choose GeV as units for energies, momenta, masses and fm as units for lengths and times (as well as barns for cross sections. To convert between these choicses you only need one number, namely $\hbar c \simeq 0.197 \mathrm{GeV} \, \mathrm{fm}$ and $10\mathrm{mb}=10^{-30} \mathrm{m}^2=1 \mathrm{fm}^2$.

The SI units are even more inconvenient, because they introduce some (from the point of view of the now established laws of nature superfluous) additional base units. One is the Ampere for electric current. In natural units it's totally superfluous, and you need to remember one more conversion factor between components of the electromagnetic field defined by $\mu_0$ and $\epsilon_0$ with the constraint that $1/(\mu_0 \epsilon_0)=c^2$ (i.e., you measure parts of one and the same quantity in different units; that's a bit like in the US where you measure distances in miles and heights in feet and inches). Another unnecessary base unit is Kelvin for temperature, which needs another conversion factor (the Boltzmann constant $k_{\text{B}}$) between units of energy and temperatures. It's pretty arbitrary using the triple point of water and thus also pretty likely to be substituted by using simply a definition of $k_{\text{B}}$ as with the speed of light $c$.

Anyway, in Planck units all quantities are expressed in terms of dimensionless numbers and in this sense it's the most natural system of units, defining all quantities in terms of fundamental universal constants (as far as we know today in terms of our most fundamental theories, which however can turn out to be wrong in future observations!).

Now in General Relativity there's one more fundamental unit, namely the universal coupling between the gravitational field (represented by the curvature tensor of spacetime) and the energy-momentum-stress tensor of matter and radiation, which conventionally is given by Newton's gravitational constant $G$. Setting also this constant to 0, we are done in the sense that this also fixes the remaining freedom of the choice in the system of natural units described before. Now all quantities are given by dimensionless numbers. To convert to conventional SI units, you only need in addition the corresponding length or energy scale which is Planck length or Planck mass.