How to convert units when calculating a dimensionless quantity?

Click For Summary
SUMMARY

This discussion focuses on converting units when calculating dimensionless quantities in the context of high-energy physics. The original quantity is derived from equations (31-33-34) in the referenced paper, where the relationship between Mpc and GeV is established. The conversion relies on the understanding that in natural units, both length and time are expressed in terms of inverse energy. The dimensions of Mpc (length) and GeV (energy) are clarified, emphasizing their roles in the calculations presented.

PREREQUISITES
  • Understanding of natural units in physics, specifically where ##\hbar=c=1##.
  • Familiarity with dimensional analysis in high-energy physics.
  • Knowledge of the equations and concepts presented in the referenced paper.
  • Basic grasp of the relationship between energy and length dimensions (ML2T-2 vs. L).
NEXT STEPS
  • Research the implications of natural units in high-energy physics.
  • Study dimensional analysis techniques in theoretical physics.
  • Examine the specific equations (31-33-34) in the referenced paper for deeper understanding.
  • Learn about the significance of dimensionless quantities in physical theories.
USEFUL FOR

Physicists, researchers in high-energy physics, and students studying theoretical physics who are looking to understand unit conversions and dimensional analysis in their calculations.

Safinaz
Messages
255
Reaction score
8
Homework Statement
For instance consider calculating this dimensionless quantity:

##
P= \frac{1}{H^4~~ (GeV)^4} \times \left(\frac{1}{k_0 ~~~(Mpc)^{-1}}\right)^{0.2} ~~(1)
##
Relevant Equations
Where H and k are constants. How to convert or choose units to make ##P## dimensionless.
where Mpc##^{-1} = 6.6 \times 10^{-39}## GeV.
The original quantity is given in this paper: [reference][1], equations: (31-33-34), where ##a(\eta)= \frac{1}{H\eta}##, so I considered in (1) only the constants which share by dimensions to ##P##.

Any help is appreciated!


[1]: https://arxiv.org/pdf/hep-th/0703290
 
Physics news on Phys.org
Safinaz said:
where Mpc##^{-1} = 6.6 \times 10^{-39}## GeV.
I am intereted in how do you get this relation ? Mpc has dimension of length L. GeV has dimensitn of energy ML^2T^-2.
 
Last edited:
anuttarasammyak said:
I am intereted in how do you get this relation ? Mpc has dimension of length L. GeV has dimensitn of energy ML^2T^-2.
This is in a system of natural units where ##\hbar=c=1##.
 
  • Like
Likes   Reactions: Orodruin and anuttarasammyak
To add to that, in natural units length and time both have dimensions of inverse energy. It is incredibly common to use such units particularly in high-energy physics, relativity, and related fields.
 
1718061620604.png


(30) and (31) seem to suggest that P(k) has same dimension with k^5. Is it OK? I have not found your (1) in the paper. Where is it ?
 
Last edited:

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K