I Converting density unit ##MeV^4## to SI units

[Safinaz]

TL;DR Summary

Converting density unit $MeV^4$ to $kg/m^3$

How to transform density unit in natural units $MeV^4$ to SI units $kg/m^3$,

Here's my trial:

$MeV^4 = (10^6)^4 ~ eV^4 = 10^{24} ~ eV^4$,

$eV = 1.6 * 10^{-19}~ kg~ m^2 / sec^2,$

$MeV^4 = 10^{24} ~ 1.6^4 * 10^{-40} ~ kg^4 m^8 / sec^8$

This is not simply $kg/m^3$!

Any help how to make this conversion!

Thanks!

 

[Orodruin]

One of my favourite limericks is applicable:

To figure the inches you’ve run,
Or to find the slug mass of the sun,
Forget your aversion
To unit conversion.
Just multiply (wisely!) by 1.

https://www.physics.harvard.edu/undergrad/limericks

eV is a unit of energy and is not really a unit of inverse length or mass unless you are using natural units where $c = \hbar = 1$. Combine that information with the suggestions of the limerick.

 

[Orodruin]

Also, 1 eV is 1.6e-19 J, not 1.6e-10 J …

 

[Safinaz]

Also, 1 eV is 1.6e-19 J, not 1.6e-10 J …

Corrected in the post.

 

[Safinaz]

One of my favourite limericks is applicable:

To figure the inches you’ve run,
Or to find the slug mass of the sun,
Forget your aversion
To unit conversion.
Just multiply (wisely!) by 1.

https://www.physics.harvard.edu/undergrad/limericks

eV is a unit of energy and is not really a unit of inverse length or mass unless you are using natural units where $c = \hbar = 1$. Combine that information with the suggestions of the limerick.

I meant conversations like here: https://www.seas.upenn.edu/~amyers/NaturalUnits.pdf

Table(1)

 

[Orodruin]

I meant conversations like here: https://www.seas.upenn.edu/~amyers/NaturalUnits.pdf

Table(1)

Yes? That is the same.

 

[vanhees71]

I guess you have an energy-density here. The dimension is $[E][L]^{-3}$. In natural units, where $\hbar=c=1$, the units are $\text{MeV}^4$. All you need is that $\hbar c \simeq 197 \text{MeV} \; \text{fm}$. To get from the natural units you need just to multiply by the appropriate powers of $\hbar c$. In your case you have
$u =1 \; \text{MeV}^4/(\hbar c)^3.$
You'll get your answer in $\text{MeV}/\text{fm}^3$. Then you only need the conversion factors to Joule and meters: $1 \text{MeV}=10^6 \text{eV} \simeq 1.6 \cdot 10^{-19} \cdot 10^6 \text{J}=1.6 \cdot 10^{-13} \; \text{J}$ and $1 \; \text{fm}=10^{-15} \; \text{m}$.

If you have a mass sensity you have to divide in addition by $c^2$ with $c \simeq 3 \cdot 10^8 \text{m}/\text{s}$.

Of course my conversion factors are just rough approximations. If you really need high precision, look up the precise values in, e.g., the "particle data booklet":

https://pdg.lbl.gov/2022/web/viewer.html?file=../reviews/rpp2022-rev-phys-constants.pdf

 

[Meir Achuz]

I prefer to say 197MeV-fm =1. Introducing hbar c is uneccessary.

 

[Orodruin]

I prefer to say 197MeV-fm =1. Introducing hbar c is uneccessary.

Introducing 197 MeV fm = 1 you just introduced $\hbar c$, you just didn’t call it that.