Figuring out the units for the cosmological constant

• Mike2
In summary, the cosmological constant Lambda has the same units as the Ricci scalar curvature R, which is (1/length)^2. The metric tensor gab has no units, while the energy stress tensor Tab has units of energy density. The 8 Pi term, when not set to one, has units of m /J and when combined with the Einstein tensor, results in J/m^3, which is an energy density. The x-zero term in the metric is defined as c times proper time, which has the dimension of length. Proper time is measured in seconds, while x0 is measured in meters.
Mike2
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.

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.

1. What is the cosmological constant?

The cosmological constant is a term in Einstein's theory of general relativity that represents the energy density of the vacuum of space.

2. Why is figuring out the units for the cosmological constant important?

Understanding the units for the cosmological constant is crucial for accurately predicting the behavior of the universe and making meaningful comparisons between different theories.

3. How do scientists determine the units for the cosmological constant?

Scientists use mathematical equations and experimental data to derive the units for the cosmological constant. These units are typically expressed in terms of length, time, and mass.

4. Can the units for the cosmological constant change?

The units for the cosmological constant are a fundamental aspect of the theory of general relativity and are not expected to change. However, as our understanding of the universe evolves, the value of the cosmological constant may be refined.

5. What implications does figuring out the units for the cosmological constant have?

Understanding the units for the cosmological constant allows scientists to accurately predict the expansion rate of the universe and the effects of dark energy. It also helps to refine and test theories of gravity and the nature of the vacuum of space.

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