Why Do the Terms in the Lagrangian Action Have the Same Dimension?

[shadishacker]

Dear all,
If we consider the lagrangian to have both geometric parts (Ricci scalar) and also a field, the action would take the form below:
\begin{equation}
S=\frac{1}{2\kappa}\int{\sqrt{-g} (\ R + \frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi -V(\phi)\ )}
\end{equation}
which are the Einstein-Hilbert, the kinetic and the potential term respectively.
About the dimensions \begin{equation} [R]=L^{-2}\end{equation}, which is of course the curvature, so the other terms must have the same dimension.
How is this possible?
aren't they energy terms?
don't they have dimension of energy?!

 

[PeterDonis]

In relativity, a system of units is often used which is called "geometric units", because mass and length have the same units (a mass $M$ in conventional units has "mass" $GM / c^2$, which is a length, in geometric units). In these units, energy density and curvature have the same units. These are the units in which the equation you give is written.

In fact, we can see these units explicitly in the second term if we observe that $\phi$ and $g^{\mu \nu}$ are dimensionless in geometric units. The operator $\partial_\mu$ has units of $L^{-1}$, so the kinetic term as a whole will have units of $L^{-2}$.

 

[bcrowell]

@PeterDonis - Are you sure you have that right? Maybe I'm not understanding you, or we're using different conventions.

My analysis would be as follows. Action has units of energy*time, which in geometric units would be L^2. The OP left the $d^4x$ out of the integral, which gives it the wrong units. Putting in the $d^4x$, we get the correct units of $L^2$ on the right.

All of this is in the convention where we check the units of an equation by checking the units of its components in the case where the coordinates are locally Minkowski. This is the system used by Schouten. It's not a universal convention, but if you follow it consistently, it works for checking whether equations make sense dimensionally.

 

[PeterDonis]

Action has units of energy*time, which in geometric units would be L^2.

I wasn't looking at the units of $S$, only the units of the stuff inside the parentheses on the RHS, on the assumption that they all should be the same--i.e., if $R$ has units $L^{-2}$, the other terms should also have units $L^{-2}$. Sorry if that wasn't clear.

As far as the units of $S$ are concerned, if we put the $\mathrm{d}^4 x$ into the integral, which I agree we should, we then get units of $L^2$ for the integral as a whole, yes (assuming that $\kappa$ and $\sqrt{-g}$ are dimensionless). But the units of $R$, and the other two terms inside the parentheses, must be $L^{-2}$ for this to work.

Also, there is another somewhat different convention for units, what might be called "quantum gravity" units, in which the action $S$ has to be dimensionless--in quantum field theory in the path integral formalism, this is because $S$ appears as the argument of an exponential in the path integral. In these units, energy/momentum/mass units and length/time units are the inverses of each other (so the action, energy * time, is dimensionless). In these units, we would have to include an extra coupling constant $G$ in the curvature term, with units $L^{-2}$ (these units are why gravity is not renormalizable as a QFT interaction); and the other terms would have to have units $L^{-4}$ (which can be thought of as energy, or inverse length, over length cubed)--in the kinetic term, for example, the field $\phi$ would have units $L^{-1}$, just like the $\partial_\mu$ operator. These units appear to be used often in the quantum gravity literature.

 

[samalkhaiat]

Dear all,
If we consider the lagrangian to have both geometric parts (Ricci scalar) and also a field, the action would take the form below:
\begin{equation}
S=\frac{1}{2\kappa}\int{\sqrt{-g} (\ R + \frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi -V(\phi)\ )}
\end{equation}
which are the Einstein-Hilbert, the kinetic and the potential term respectively.
About the dimensions \begin{equation} [R]=L^{-2}\end{equation}, which is of course the curvature, so the other terms must have the same dimension.
How is this possible?
aren't they energy terms?
don't they have dimension of energy?!

1) Your expression for the action is not correct. The correct action is of the form
$$S = \frac{1}{k} \int d^{4} x \ \sqrt{-g} R + \int d^{4}x \ \sqrt{-g} \ \mathcal{L}(\phi) .$$
2) In order to ask questions about the units, you need to specify the system of units you are using, and state the dimensions (if any) of the coupling constant.
In the $\mbox{SI}$ units, you have the following: $[ S ] = ML^{2}T^{-1}$, $[ d^{4}x ] = L^{3}T$, $[ R ] = L^{-2}$, $[ \partial ] = L^{-1}$, and $[ g_{\mu\nu} ] = [ g ] = M^{0} L^{0}T^{0}$.

In quantum field theory, we often use the so-called natural units ($\mbox{n.u.}$). This system is obtained by setting $c = \hbar = 1$, i.e., $L = T = M^{-1}$. So, in the n.u. system, the action is dimensionless, $[ S ] = L^{0}$, and $[ d^{4}x ] = L^{4}$.
Okay, let us work out the dimension of the matter field $\phi$. This, of course, depends on the tensorial nature of the field. So, let us assume that $\phi$ is a real scalar field. In this case $\mathcal{L}(\phi) = (1/2) (\partial \phi)^{2}$. In the SI units we have $$[ \mathcal{L}(\phi) ] = [ \partial \phi ] [ \partial \phi ] = [ \partial^{2} ][ \phi^{2} ].$$ So, for the matter part of the action, we get
$$[ S ] = [ d^{4}x ] [ \partial^{2} ][ \phi^{2} ] .$$
Thus,
$$ML^{2}T^{-1} = (L^{3}T)(L^{-2}) [\phi^{2}] \ \Rightarrow \ [ \phi ] = M^{1/2}L^{1/2}T^{-1} .$$
Now, you can work out the dimension of the Lagrangian density
$$[ \mathcal{L} ] = ML^{-1}T^{-2} = \frac{M (L/T)^{2}}{L^{3}}.$$
Thus, the Lagrangian $L(\phi) = \int dV \ \mathcal{L}$ has the units of energy as it should.
In the n.u. system, the scalar field is therefore has one unit of mass dimension $[ \phi ] = M^{1} = L^{-1}$, and $[ \mathcal{L} ] = M^{4} = L^{-4}$.
Okay, let us now work on the Hilbert-Einstein action and find the dimension of the coupling constant $k$. Again, we have
$$[ S ] = [ k^{-1} ][ d^{4}x ][ R ]$$
Substituting the units of $S, R$ and the space-time volume $dVdt$, we find
$$[ k ] = M^{-1}L^{-1}T^{2} .$$
Recall the units of Newton’s constant $[ G_{N} ] = M^{-1}L^{3}T^{-2}$. So,
$$\frac{[ k ]}{[ G_{N} ]} = (L/T)^{-4} .$$
This suggests that $G_{N} \propto c^{4}k$, where $c$ is the speed of light.
Again, as in the $\mathcal{L}(\phi)$ case, the Lagrangian density $R/k$ has the units of energy density
$$\frac{[ R ]}{[ k ]} = ML^{-1}T^{-2} = \frac{M(L/T)^{2}}{L^{3}} .$$
Okay, I hope that will be clear enough for you.

 

[shadishacker]

Thanks a lot for the complete reply.