Understanding the Mass-Length Relation: Solving for V in Simple Equation

[Safinaz]

Homework Statement

How to get the length in this eqution in the right unit

Relevant Equations

$M_p^2 = M^3_s V$

Hello,

If i have this relation:

$M_p^2 = M^3_s V$

where $M_p$ and $M_s$ are masses in GeV and V is a length. Let $M_p = 10^{18} ~$ GeV and $M_s = 10^3$ GeV , what is V in meters ?

My solution :

The equation becomes

$V = 10^{30}$ GeV , but $1 m \sim 10^{15} ~ GeV^{-1}$ or $1 m^{-1} \sim 10^{-15} ~ GeV$

which means $V = \frac{10^{30} ~ GeV. m^{-1}}{10^{-15} ~ GeV} = 10^{45} ~ m^{-1}$ ! is there any wrong here?? Cause V is so large , and also i want it in meters not $m^{-1}$Any help is appreciated!
Thanks!

 

[anuttarasammyak]

V has dimension of $E^{-1}$.

 

[Orodruin]

Your math is off. $V$ must have energy units of inverse energy for your initial relation to make sense.

 

[Safinaz]

V has dimension of $E^{-1}$

 

[Safinaz]

If like in this case $M_s = 1000$ GeV and $M_p = 10^{18}$ GeV, ( $M_s$ is much less than $M_p$ ) . This means this equation should be rewritten to add some factor of dimensions $GeV^{-2}$. Then the equation become:

$M_p^2 = \frac{1}{g^2} M^3_s V$ But according to this paper

https://inspirehep.net/files/da3f5a4d325281183a9b6134ad7c7dce

g is dimensionless cause it's a gauge coupling. So I don't get how in this paper after equation : ( 2.1 ) he got $V = 10^8$ km.

 

[Steve4Physics]

Homework Statement:: How to get the length in this equation in the right unit
Relevant Equations:: $M_p^2 = M^3_s V$

Hello,

If i have this relation:

$M_p^2 = M^3_s V$

where $M_p$ and $M_s$ are masses in GeV and V is a length. Let $M_p = 10^{18} ~$ GeV and $M_s = 10^3$ GeV , what is V in meters ?

My solution :

The equation becomes

$V = 10^{30}$ GeV , but $1 m \sim 10^{15} ~ GeV^{-1}$ or $1 m^{-1} \sim 10^{-15} ~ GeV$

which means $V = \frac{10^{30} ~ GeV. m^{-1}}{10^{-15} ~ GeV} = 10^{45} ~ m^{-1}$ ! is there any wrong here?? Cause V is so large , and also i want it in meters not $m^{-1}$Any help is appreciated!
Thanks!

I guess you are using natural units, i.e. taking ℏ = c = 1. (If this is correct, it would have been helpful if you had stated this explicitly.)

$V = \frac {M_p^2 }{M^3_s} = \frac {(10^{18} GeV)^2 }{(10^3 GeV)^3} = \frac {10^{36} GeV^2}{ 10^9 GeV^3} = 10^{27} GeV^{-1}$

A length of $1GeV^{-1}$ in natural units ≈ 0.2 fm (=0.2x10⁻¹⁵ m) in SI units.

So, for the final step, you can now convert $10^{27} GeV^{-1}$ to metres.

(Edited, as original version included the final step - which meant you had nothing to do for yourself!)

 

[Safinaz]

I guess you are using natural units, i.e. taking ℏ = c = 1. (If this is correct, it would have been helpful if you had stated this explicitly.)

$V = \frac {M_p^2 }{M^3_s} = \frac {(10^{18} GeV)^2 }{(10^3 GeV)^3} = \frac {10^{36} GeV^2}{ 10^9 GeV^3} = 10^{27} GeV^{-1}$

A length of $1GeV^{-1}$ in natural units ≈ 0.2 fm (=0.2x10⁻¹⁵ m) in SI units.

So, for the final step, you can now convert $10^{27} GeV^{-1}$ to metres.

(Edited, as original version included the final step - which meant you had nothing to do for yourself!)

Thanks! That's very helpful.. It seems I'm just confused.