Dimensions of Cosmic Scale Factor ##a(t)##

[AHSAN MUJTABA]

TL;DR

I have confusion regarding the dimensions of the cosmic scale factor, $a(t)$. I have read on the wiki that it is dimensionless, but I wonder about it because it is a function of time, t. I want to use its dimensions to prove the action as non-dimensional.

I know the formula for Hubble's parameter, $\frac{\dot{a}}{a}$, but I cannot infer any dimension of $a(t)$ from it. Please guide me.
Thanks.

 

[Orodruin]

The dimension of $a$ depends on how you define it in the metric.

 

[AHSAN MUJTABA]

The dimension of $a$ depends on how you define it in the metric.

Would you please elaborate a little? Thanks.

 

[AHSAN MUJTABA]

Just check my work to find dimensions of $a(t)$.
I have written the metric as,
$ds^{2}=dt^{2}+a(t)^{2}dx^{2}.$
Now, I am aware of the dimensions of the quantities as:
$ds^{2}=[L]^{2}$, $dt^{2}=[L]^{2}$( I am defining it in terms of length by L=ct, taking c=1.) and $dx^{2}=[L]^{2}$.
I need to define everything in terms of mass dimensions. I have $\lambda=\frac{h}{mc}$ and working in natural units, I can define the dimension of length in terms of mass as, $[M]^{-1}.$
Now, the dimensions of time in terms of mass becomes, $[M]$.
Now, incorporating these dimensions in the metric we get:
$1=1+a(t)^{2},$
$[a(t)]=0.$
So, from this approach, the cosmic scale factor is coming out to be a dimensionless quantity.
Is it a legal approach, please do have a look.

 

[Ibix]

As you have defined it, $a$ is dimensionless, yes. That doesn't stop it being a function of time - for example, if a wave with frequency $f$ passes you then the number of complete cycles that have passed you since time zero is $ft$. That's both dimensionless and a function of time.

 

[AHSAN MUJTABA]

Just one little question, does defining length in terms of mass dimensions by using the relation $L=ct$ is legit or not? In natural units i.e. c=1, the time and length would have the same dimensions.

 

[PeroK]

Just one little question, does defining length in terms of mass dimensions by using the relation $L=ct$ is legit or not? In natural units i.e. c=1, the time and length would have the same dimensions.

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

 

[Ibix]

Just one little question, does defining length in terms of mass dimensions by using the relation $L=ct$ is legit or not? In natural units i.e. c=1, the time and length would have the same dimensions.

That seems to me to be defining time in length units or vice versa. You need an additional choice that $G=1$ to get mass in the same units as length and time. I don't usually do it, and I recall Carroll recommending against it. It isn't wrong, though.

 

[AHSAN MUJTABA]

According to this article, it is seen that $L=ct$ is right because to convert dimensions of time into length, we set c=1. Secondly, we know Compton's wavelength relation, and from that, we can have $(mass)^{-1}$ dimensions of both time and length.

 

[AHSAN MUJTABA]

As you have defined it, $a$ is dimensionless, yes. That doesn't stop it being a function of time - for example, if a wave with frequency $f$ passes you then the number of complete cycles that have passed you since time zero is $ft$. That's both dimensionless and a function of time.

Yes, now I am clear about that. Thanks.

 

[kimbyd]

To elaborate a little bit, there are two common conventions for the scale factor.

One, which usually uses $a(t)$ as its notation, has the scale factor dimensionless and set so that $a(now) = 1$. With this notation, the curvature parameter $k$ is a real number with units of inverse length squared.

The second one, which usually uses $R(t)$ as its notation, has a scale factor with units of length. In this notation the curvature parameter $k$ is an integer value equal to either -1, 0, or 1, representing whether the universe is negatively-curved, flat, or positively-curved. This scale factor is the radius of curvature of the universe.

Please note the "usually" in the above: just because a source has one or the other doesn't mean that's the notation they're using.