Figuring out the units for the cosmological constant

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Discussion Overview

The discussion revolves around determining the units of the cosmological constant (Λ) within the context of the Einstein Field Equation. Participants explore the relationships between various components of the equation, including the metric tensor (gab) and the energy-momentum tensor (Tab), and how these relate to the units of Λ.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that Λ has the same units as the Ricci scalar curvature R, which is (1/length)^2, due to the derivatives involved in the Riemann tensor.
  • There is a claim that the metric tensor gab has no units, which is supported by multiple participants.
  • One participant notes that when G and c are not set to one, the term 8π in the equation has units of m/J, leading to a discussion about consistency in units across the equation.
  • Another participant questions whether g00 must have units of c² to ensure dimensional consistency when multiplied by dt², suggesting that this is necessary for the metric to align with the units of meters².
  • There is a discussion about the nature of the metric as a matrix that multiplies dx-dx terms, resulting in a diagonal matrix with no units, while dx terms themselves have units of meters.
  • One participant introduces the concept of proper time and its relationship to the metric, defining proper time² as -ds²/c² and discussing its implications for the units involved.

Areas of Agreement / Disagreement

Participants generally agree that the metric tensor has no units and that Λ shares units with the Ricci scalar curvature. However, there are differing views on the specific units of g00 and its relationship to proper time, indicating that the discussion remains unresolved in these areas.

Contextual Notes

There are unresolved aspects regarding the definitions and assumptions about the metric tensor and its components, particularly in relation to the units of g00 and the implications for proper time.

Mike2
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How does one figure out the units of measure for the cosmological constant? In the Einstein Field Equation:

\[Rab - \frac{1}{2}Rgab + \Lambda gab = 8\pi Tab\]


Lambda is a constant but the units of measure for gab and Tab differ for each combination of indices. For example T00 is an energy density. So what is the measure for g00 so that Lambda comes out to a measure of one over meters squared?

Thanks.
 
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Dimensionally, \Lambda has the same units as the Ricci scalar curvature R, which has units of (1/length)^2, each (1/length) factor introduced by a derivative operator in the Riemann tensor.
 
Mike2 said:
How does one figure out the units of measure for the cosmological constant? In the Einstein Field Equation:

\[Rab - \frac{1}{2}Rgab + \Lambda gab = 8\pi Tab\]


Lambda is a constant but the units of measure for gab and Tab differ for each combination of indices. For example T00 is an energy density. So what is the measure for g00 so that Lambda comes out to a measure of one over meters squared?

Thanks.

So there are no units of measure for the metric gab?
 
Mike2 said:
So there are no units of measure for the metric gab?
Indeed, there are no units for the metric tensor.
One has also to be careful with the 8 Pi. When G and c are not set to one, the term 8 Pi becomes 8 Pi G /^(c^4) which has units of m /J. So, if we move this term to the left side, the left side is multiplied by J /m. Since the Einstein tensor has units of 1 /m^2, the product of both thus becomes J/m^3, thus an energy density, which has thus the same units as the energy stress tensor. So everything is consistent. Note also that pressure and energy density have the same units. So, even when terms look different in the energy stress tensor, they have the same units.
 
notknowing said:
Indeed, there are no units for the metric tensor.
One has also to be careful with the 8 Pi. When G and c are not set to one, the term 8 Pi becomes 8 Pi G /^(c^4) which has units of m /J. So, if we move this term to the left side, the left side is multiplied by J /m. Since the Einstein tensor has units of 1 /m^2, the product of both thus becomes J/m^3, thus an energy density, which has thus the same units as the energy stress tensor. So everything is consistent. Note also that pressure and energy density have the same units. So, even when terms look different in the energy stress tensor, they have the same units.
Doesn't g00 have to be in units of c2 so that multiplying by dt2 will give the same units as the rest of the metric which is in units of meters2, right?
 
Mike2 said:
Doesn't g00 have to be in units of c2 so that multiplying by dt2 will give the same units as the rest of the metric which is in units of meters2, right?
The metric is the matrix which multiplies the dx-dx terms, which results in flat spacetime in the diagonal matrix (-1,1,1,1) and these numbers have no units (while the dx terms themselves have units of meter). The x-zero term is defined as c times t-conventional, so that also x-zero has the dimension of a length.
 
notknowing said:
The metric is the matrix which multiplies the dx-dx terms, which results in flat spacetime in the diagonal matrix (-1,1,1,1) and these numbers have no units (while the dx terms themselves have units of meter). The x-zero term is defined as c times t-conventional, so that also x-zero has the dimension of a length.
So what is that called such that g00dx0dx0 has the units of length? Is dx0 in this context called the proper time so that it differs in units of time in dt from "proper time" in c2dt*dt which is units of lenth?
 
Mike2 said:
So what is that called such that g00dx0dx0 has the units of length? Is dx0 in this context called the proper time so that it differs in units of time in dt from "proper time" in c2dt*dt which is units of lenth?
Sorry for the late reply (have been sick).

Proper time^2 is DEFINED as -ds^2/c^2. So if we set dx1,dx2,dx3 equal to zero, the proper time^2 becomes dx0^2/c^2, or c^2 dt^2/c^2 or dt^2
So proper time is clearly measured in seconds, x0 in meter, the metric without units.
 

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