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Calculating the critical density

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data


    I can't get how they calculate that.

    I'm not sure if there including H0 which I think is 70 or not but I've tried with and without and both don't work.

    I get (3 * 70^2)/(8pi*6.67 * 10^-11) which is way off.

    I also can't get the units that they arrive at

    I get

    km/s/Mpc * s^2/m^3kg which is

  2. jcsd
  3. Jun 1, 2012 #2


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    The definition of small h that I'm familiar with is that H0 = h(100 km/s/Mpc), which is useful, because now you have dimensionless constant that you can set to different values depending on how different you think H0 is in value from 100 km/s/Mpc. I think that this was especially relevant back in the day when there was a lot of uncertainty in the value of H0, and introducing this parameter allowed you to express theoretical results in terms of a dimensionless constant that could then be set to the desired value whenever this was well-determined. So in other words, you had your answer "up to" a multiplicative constant. Now we know with a fair amount of certainty that h = 0.71 (I think, or maybe 0.72, not sure what the latest and greatest value is). In other words, H0 = 71 or 72 km/s/Mpc.

    All that having been said, I notice that the small h in THIS problem has a subscript: h70, which suggests to me that its definition is given by H0 = h70(70 km/s/Mpc). I.e. the assumed base value is 70 rather than 100. In fact, I have very good reason to believe that this is the case. So let's proceed with the calculation under that assumption. THen we have:$$\rho_c = \frac{3h_{70}^2}{8\pi} \frac{(70~\textrm{km} \cdot \textrm{s}^{-1} \cdot \textrm{Mpc}^{-1})^2}{G}$$ $$= \frac{3h_{70}^2}{8\pi} \frac{(7.0\times 10^4)^2~\textrm{m}^2 \cdot \textrm{s}^{-2} \cdot \textrm{Mpc}^{-2}}{6.67 \times 10^{-11}~\textrm{N}\cdot\textrm{m}^2\cdot\textrm{kg}^{-2}}$$Let's convert newtons into SI base units:$$= \frac{3h_{70}^2}{8\pi} \frac{4.9\times10^9~\textrm{m}^2 \cdot \textrm{s}^{-2} \cdot \textrm{Mpc}^{-2}}{6.67 \times 10^{-11}~\textrm{kg}\cdot\textrm{m}\cdot\textrm{s}^{-2}\cdot\textrm{m}^2\cdot\textrm{kg}^{-2}}$$Now let's get rid of the mess of numbers by multiplying them all together. Let's also simply the units by doing as many cancellations as we can:$$=(0.08769\times 10^{20}) h_{70}^2 \frac{ \textrm{Mpc}^{-2}}{\textrm{m}\cdot\textrm{kg}^{-1}}$$The final thing that is going to be a bit of a pain in the butt is to convert square megaparsecs to square metres. I will leave it as an exercise for you to show that 1 Mpc = 3.086e22 m. Therefore, we have:$$=(0.08769\times 10^{20}) h_{70}^2 \frac{ \textrm{kg}}{\textrm{m}\cdot (3.086\times 10^{22}~\textrm{m})^{2}}$$Now, notice that the power of 10 in the numerator is 10^20, and in the denominator we have (10^22)^2 = 10^44. When we divide them, we get 10^-24. I've also multiplied all the numbers together again to obtain:

    $$= (0.00921 \times 10^{-24}) h_{70}^2 \frac{ \textrm{kg}}{\textrm{m}\cdot \textrm{m}^{2}}$$Multiply by 100 and subtract 2 from the exponent (these two operations cancel each other out):$$=(0.921 \times 10^{-26}) h_{70}^2 \frac{ \textrm{kg}}{\textrm{m}^{3}}$$Close enough if you ask me!
    Last edited: Jun 1, 2012
  4. Jun 2, 2012 #3
    Wow!! Excellent help! You really rock! I'm just having trouble with one step:

    $$=(0.08769\times 10^{20}) h_{70}^2 \frac{ \textrm{kg}}{\textrm{m}\cdot (3.086\times 10^{22}~\textrm{m})^{2}}$$

    $$= (0.00921 \times 10^{-24}) h_{70}^2 \frac{ \textrm{kg}}{\textrm{m}\cdot \textrm{m}^{2}}$$

    right off the bat 8.7/3.1 = 2.8 not .92 (ignoring the orders of magnitude) So it must be the case that you're doing something extra there.
  5. Jun 2, 2012 #4
    The 3.086 in the denominator appears squared. So we get 0.08769/3.0862=0.0092
  6. Jun 2, 2012 #5


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    Glad to be of help. As far as the above, yeah, what kloptok said.
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