Calculating the critical density

[robertjford80]

Homework Statement

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

km(s)/Mpc/kg*m^3

 

[cepheid]

Homework Statement

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

km(s)/Mpc/kg*m^3

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!

 

[robertjford80]

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.

 

[kloptok]

The 3.086 in the denominator appears squared. So we get 0.08769/3.086²=0.0092

 

[cepheid]

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.

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