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Figuring out the units for the cosmological constant

  1. Oct 27, 2006 #1
    How does one figure out the units of measure for the cosmological constant? In the Einstein Field Equation:

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

    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?

  2. jcsd
  3. Oct 27, 2006 #2


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    Dimensionally, [tex]\Lambda[/tex] has the same units as the Ricci scalar curvature [tex]R[/tex], which has units of (1/length)^2, each (1/length) factor introduced by a derivative operator in the Riemann tensor.
  4. Oct 27, 2006 #3
    So there are no units of measure for the metric gab?
  5. Oct 29, 2006 #4
    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.
  6. Oct 29, 2006 #5
    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?
  7. Oct 30, 2006 #6
    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.
  8. Oct 30, 2006 #7
    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?
  9. Nov 1, 2006 #8
    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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