How to convert units when calculating a dimensionless quantity?

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Homework Help Overview

The discussion revolves around the conversion of units when calculating a dimensionless quantity, particularly in the context of high-energy physics and natural units. Participants reference specific equations and relationships from a paper to explore the dimensional analysis of quantities such as Mpc and GeV.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the relationship between different units, specifically how Mpc and GeV relate dimensionally. Questions arise regarding the validity of certain dimensional comparisons and the implications of using natural units.

Discussion Status

The discussion is active, with participants seeking clarification on dimensional relationships and the implications of using natural units. Some guidance has been offered regarding the commonality of these units in high-energy physics, but no consensus has been reached on specific interpretations or calculations.

Contextual Notes

There are references to specific equations in a paper that are not universally accessible to all participants, which may affect the clarity of the discussion. Additionally, the use of natural units introduces assumptions that are being examined.

Safinaz
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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
 
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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##.
 
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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 ?
 
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