# I have an error in integrating to calculate the age of the Universe

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• Buzz Bloom

#### Buzz Bloom

Gold Member
TL;DR Summary
I am seeking help in finding my error. I am integrating the Friedmann equation (with just H_0, Ω_m, and Ω_Λ) expecting a result of about 13.8 x 10^9 years. My result and how I calculated it is in the main body of this post.
NOTE: I am attempting to convey the equations in this post into LaTerX format in Post #19.

My result is way off. It is about 7.44 x 10^9 years. The values I use are:
1/H_0 = 14.4 X 10^9 years,​
M = Ω_m = 0.3103, and​
L = Ω_Λ = 1 - Ω_m = 0.6897.​

dt = (1/H_0) (1/(M/a^3 + L)^(1/2)) (da/a)
I make the following substitutions.
x = a^3 and a = x^(1/3),​
da = (1/3) dx / x^(2/3)​
Q = (1/2) M/L and M/L = 2Q​
This leads to the following equation.
dt = (1/3) (1/H_0) (1/L^1/2)) dx / (x^2 + 2Q x)^(1/2)
Integrating produces
t = (1/3) (1/H_0) (1/L^(1/2) [log ( x+Q + (x^2 + 2Qx)^(1/2) )].​
Note: I did differentiate the above integration result to confirm it leads to the previous differential equation.
Calculating the value of t uses the integration limits between x=a=0 and x=a=1. The result is the following.
t = (1/3) (1/H_0) (1/L^(1/2) log (W(1) / W(0)), where​
W(1) = 1+Q + (1+2Q)^(1/2), and​
W(0) = (Q+2Q)(1/2).​
Now comes the numerical calculations.
(1/3) (1/L^(1/2) (1/H_0) = (1/3) (1/0.6897)^1/2 14.4 x 10^9 years​
~= 5,780,000,000 years.​
W(1) ~= ~2.43​
W(0) ~= 0.896​
W(1)/W(0) = ~2.71​
log(W(1)/W(0)) =~1.00​
t = (1/3) (1/L^(1/2) (1/H_0) log (W(1) / W(0))​
~=5,780,000,000 years X ~1.00 ~=5,770,000,000 years​

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0.997656155687048

Of for goodness sake. You can start by making it easier on people who might want to help you by proper rounding. Make it easy for them, not hard.

I would much appreciate your specific suggestion about "rounding". I do not have any instinct about how much is enough.

Regards,
Buzz

Hi @PeterDonis:

It has been a very long time (years) since I last used LaTeX. My memory about it has vanished. Would you please tell me how I can find the instructions about how to use LaTeX?

Regards,
Buzz

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You should know that those numbers have way too many digits of precision. If you don't, you need to take a step back and brush up on basic mathematics before tackling cosmology. Whether the correct number of digits is three or four or something else is less relevant that 156752675.6786876877686876876167861876 is clearly not right.

Peter is also right - a second reason that your post is unreadable is you didn't bother to make it clear by using LaTexX.

If you make it hard on the people who you want to help you, you will get less help.

I do not have any instinct about how much is enough.
Age of the universe? Order of magnitude, second digit is excessive/hubris.

Regards,
Buzz

Hi @Bystander:

Your comment "Order of magnitude, second digit is excessive/hubris," does not communicate to me at all. I just do not get what you are trying to tell me.

Regards,
Buzz

Would you please tell me how I can find the instructions about how to use LaTeX?
There's a "LaTeX Guide" link at the bottom left of the edit window when you're making a post.

Buzz Bloom
Age of the universe
Is 13,700,000,001 years old. I know that because last year I read it was 13.7 billion years old.

Steve4Physics, malawi_glenn, timmdeeg and 3 others
Hi @PeterDonis:

I am editing Post #1 to try to convert equations to LaTeX, but I am so far not able to succeed on my first equation. Can you help me?

Regards,
Buzz

I am editing Post #1
You know better than that! Write a new messgae - don't change messages that have already received replies.

Buzz Bloom and berkeman
I am editing Post #1 to try to convert equations to LaTeX, but I am so far not able to succeed on my first equation.
It looks like you already got a harder equation right. You just need to be patient.

Also, @Vanadium 50 is right that you should not be editing the OP. Instead add new posts with equations in LaTeX as you do them.

I make the following substitutions
That looks more complicated than necessary. Pulling the ##a## in the denominator (from ##a H_0##) inside the square root in your integrand gives for the integral (we can also pull ##H_0## out in front of the integral since it's a constant):

$$t =\frac{1}{H_0} \int_0^1 \frac{da}{\sqrt{\frac{M}{a} + L a^2}}$$

Evaluating this integral should give you a much closer answer.

Hi @PeterDonis:

I very much appreciate your comments and your advice. I will start a new post with a gradual intent to attempt to use LaTeX to complete the sequence that explains the integration.. When I edited Post 1, I alternated editing and then posting so I could see if the LaTeX I used for an equation was OK. I ran into a problem in that when I posted the edit, it did not show the LaTeX form. I have decided that in order to see the result in Latex form, I will need to log out of the Physics Forums and then re-enter.

Regards,
Buzz

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THIS POST IS NOW COMPLETED.
Note that I have finally corrected my errors and have calculated an expected result.

$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
I make the following substitutions.
##{x = a^3,}## ##{a = x^{1/3}},## ##{da = dx / 3x^{2/3}},## ##and## ##{M/L = 2Q}.##
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$
$$t = { \frac {1} {3H_0 \sqrt L}} {\ln ( {x+Q + \sqrt {x^2 + 2Qx} } ) }$$
NOTE: Post #28 shows the differentiation of the above resulting in the dt equation above.

Evaluating the numerical integration range is for x=a=0 to x=a=1.
$$t = {\frac {1} {3H_0 \sqrt L}} {\ln ( {\frac {1+Q + \sqrt {1 + 2Q} } {Q} ) }}$$
The values I use are:
##{1/H_0} =14.4 giga-years,##

##M = {Ω_m} = 0.3103 ##,

##L = {Ω_Λ} = {1 - Ω_m} = 0.6897, ##

## \sqrt {L} = 0.8305,## and

##Q = {M/2L} = 0.2250.##

##{\frac {1+Q + \sqrt {1 + 2Q} } {Q} } = \frac {{1.2250} + \sqrt {1.45}} {0.2250}## = 10.80

##{\ln ( { \frac {1+Q + \sqrt {1 + 2Q} } {Q} } )} = {2.38} ##

##t = {\frac {2.38} {3 H_0 0.8305}} = {\frac {2.38} {2.4915}} ({1/H_0}) = 0.9552 ({1/H_0})##

##t = {13.76}## giga-years

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Buzz Bloom
I have decided that in order to see the result in Latex form, I will need to log out of the Physics Forums and then re-enter.
No, just refresh the page with the little spinny arrow thingy:

PF uses an optimization called "lazy loading" for LaTeX (I think), which means that the LaTeX is not rendered until you refresh your browser at least once.

(It's a feature, not a bug...)

malawi_glenn and Buzz Bloom
Hi @pbuk:

I am very familiar with the reference you posted. My first LaTeX equation in Post #19 is based on the equations in your reference, except I omit ##Ω_r## and ##Ω_k##.

Regards,
Buzz

I am very familiar with the reference you posted.
In that case you will have seen the derivations for approximations to ## a(t) ## neglecting all but one of ## \Omega_M, \Omega_R \text{ and } \Omega_\Lambda ## (as well as ## \Omega_k ##). There is no derivation neglecting only ## \Omega_M ## (as well as ## \Omega_k ##) because there is no analytic form for this integral.

In that case you will have seen the derivations for approximations to ## a(t) ## neglecting all but one of ## \Omega_M, \Omega_R \text{ and } \Omega_\Lambda ## (as well as ## \Omega_k ##). There is no derivation neglecting only ## \Omega_M ## (as well as ## \Omega_k ##) because there is no analytic form for this integral.
The case that @Buzz Bloom considers here, ##\Omega_R = 0## and ##\Omega_M + \Omega_\Lambda = 1##, does have invertible closed form analytic solutions, i.e., there are closed form expressions for both ##a\left(t\right)## and ##t\left(a\right)## in this case.

pbuk and PeterDonis
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$
$$t = { \frac {1} {3H_0 \sqrt L}} {\ln ( {x+Q + \sqrt {x + 2Q} } ) }$$

I have not checked anything else, but differentiating
$$t = \frac {1} {3H_0 \sqrt L} \ln \left( x+Q + \sqrt {x + 2Q} \right)$$
does not seem to give
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }.$$

According to Dwight's book, formulae 380.001, the result of the integration is:

##t=\displaystyle\frac{1}{3H_0\sqrt{L}}\ln[2(x+Q+\sqrt{2Qx+x^2})]##

Hi @George Jones:

Thank you for responding to my problem. I should have made clear that I have not completed the presentation I am working on in Post #19.. The equation with ##t = ## has errors due to my lack of experience using LaTeX I have not fixed yet. The equation should be
$$t = { \frac {1} {3H_0 \sqrt L}} {\ln ( {x+Q + \sqrt {x^2 + 2Qx} } ) }$$

Regards,
Buzz

When I differentiate the
$$t = { \frac {1} {3H_0 \sqrt L}} {\ln ( {x+Q + \sqrt {x^2 + 2Qx} } ) }$$
equation, I get
$$dt =({ \frac {1} {3H_0 \sqrt L}}) ({\frac {1} {x+Q + \sqrt {x^2 + 2Qx} }}) ({1 + {\frac {2x+2Q} {2{\sqrt {x^2 + 2Qx}}}}})$$
This simplifies to
$$dt =({\frac {1} {3H_0 \sqrt L}}) ({\frac {1} {x+Q + \sqrt {x^2 + 2Qx}}}) ({\frac {x+Q + \sqrt {x^2 + 2Qx}} { \sqrt {x^2 + 2Qx}} } )$$
and
$$dt =({\frac {1} {3H_0 \sqrt L}}) ( {\frac {1} { \sqrt {x^2 + 2Qx}} } )$$
The above is the integrand I originally integrated.

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##t=\displaystyle\frac{1}{3H_0\sqrt{L}}\ln\frac{1+Q+\sqrt{1+2Q}}{Q}##
##L=0.6897## ##\sqrt{L}=0.8305##
##t=13.75## Gyr

Hi @mbond:

Thank you for your post. I would much appreciate it you would post the reference of the integral with the x variable which leads to your "t = ..." equation. The equation I reached is
$$t = { \frac {1} {3H_0 \sqrt L}} {\ln ( {x+Q + \sqrt {x^2 + 2Qx} } ) }$$.
If this is wrong, can you tell me where my error is?
Keep in mind that the numerical integration is from x=0 to x=1.
Regards,
Buzz

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If this is wrong, can you tell me where my error is?
I already showed you a simpler way of proceeding, that results in a different integral than the one you are doing. See post #17. The integral I gave there gives the correct answer; if you are doing a different integral, that would explain why you are getting the wrong answer.

Hi @PeterDonis:

I do not understand why the integrand I used is not equivalent to the integrand you posted.
Mine is
$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
Yours is based on
$$dt = \frac {da} {H_0 \sqrt{M/a + La^2} }$$.
The difference is multiplying the numerator and denominator of your integrand with ##\sqrt {a}##.

$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
I make the following substitutions.
##{x = a^3,}## ##{a = x^{1/3}},## ##{da = dx / 3x^{2/3}},##

(I am not going to use ##Q##.) The substitution that I prefer for this is
$$\frac{L}{M} a^3 = \sinh^2 u .$$
I make this substitution in order to make use of the identity ##\cosh^2 u - \sinh^2 u = 1##. Differentiate:
\begin{align} 3\frac{L}{M} a^2 da &= 2\sinh u \frac{d}{du} \sinh u du \\ &= 2\sinh u \cosh u du \\ da &= \frac{2M}{3L} a^{-2} \sinh u \cosh u du \end{align}
Substituting these into the integral gives
\begin{align} \frac {{\sqrt {a}} da} {H_0 \sqrt {M + L a^3}} &= \frac {{\sqrt {a}} da} {H_0 \sqrt {M} \sqrt{1 + \frac{L}{M} a^3}} \\ &= \frac {{\sqrt {a}} \frac{2M}{3L} a^{-2} \sinh u \cosh u du} {H_0 \sqrt {M} \sqrt{1 + \sinh^2 u}} . \end{align}
Now (if you want to proceed with this method), combine the ##a## s, and then replace them by something in terms of ##u## using my original substitution. Also, use the (hyperbolic) trig identity that I gave.

This should result in an expression for ##t\left(a\right)## that involves an inverse ##\sinh##, and that, if desired, can reexpressed using ##\ln## s.

Mine is
$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
Not here it isn't:

$$dt =({\frac {1} {3H_0 \sqrt L}}) ( {\frac {1} { \sqrt {x^2 + 2Qx}} } )$$
So which one are you integrating?

Hi @PeterDonis:

I do not understand why the integrand I uses is not equivalent to the integrand you posted.
Mine is
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
Yours is based on
$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.
The process of integrating involves a sequence of substitutions which are intended to create an integrand which is finally directly integrated. The issue is whether the substitutions I made are OK, or whether they include an error. The final integrand is then integrated, and I have shown in Post #28 that differentiating the integration result leads to the integrand which was integrated. This demonstrates that the integration step was correct. The final part of the presentation in Post #19 is to use the desired variable values and arithmetically calculate the numerical value of t explained in Post #30. Note that the arithmetic conclusion is based on
$$\int_0^1 dt = {t(1) - t(0)}.$$

Regards,
Buzz