week 8 solutions

Week 8, Part A, Question 6 - DetailsAnonymous · 2 weeks ago

Question 6 If a type-Ia supernova with luminosity 1010L⊙ were to be detected in RD1, what

would its apparent brightness be as observed from Earth? Express your answer in terms of the

Solar constant b⊙ and round to two significant figures.

Vote this post up 2 Vote this post down

Loh Siu Yin · 2 weeks ago

I'm not getting this. Given b=L4πD2L and DL=D0(1+z)let Lsn be the luminosity of the supernova and Lsun and bsun be the Solar luminosity and

brightness respectively

then

bsnbsun=LsnLsun(DsunDLsn)2What am I getting wrong here?


Comments

Daniel Grabianowski · 2 weeks ago

I'm doing the same thing, and also getting it wrong. I'm pretty sure that I'm doing Q5 right, but I

am getting it wrong as well. Dr. Plesser corrected a mistake in Q4 earlier, so perhaps there is

also a mistake in the autograder for this question. We'll see tomorrow, I guess. Either that, or

these are some pretty complicated questions.


Mark Polak · 2 weeks ago

Like Dan, I'm getting both 5 and 6 wrong. They seem reasonably straightforward to me

(especially Q5). Has anyone got these right yet?


A comment was deleted.


Thank you for your hint Prasad, x.x e -xx .

It was a rounding error. I had to round down to get answer accepted by the autograde. I had

previously rounded up to get x.y e -xx.


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S V Prasad (Student) · 2 weeks ago

The formula is correct. But I am not sure why are you dividing the Luminosity value with Sun's

Luminosity. It should be something like this b / b_sun =

((L_given*L_Sun)/(4*pi*(D_0(1+z))^2))/b_sun. You should get x.xe-xx


Comments

Oscar Orta (Student) · 2 weeks ago

S V: I am also stuck here, but in your formula at the end of it, you have again /b_sun.....you have

it already in the right side .....just take it as an observation......




Oscar the requirement is to describe the answer in solar constant and that is what is explained.

The right hand side of the formula is correct. I already got the correct answer to this problem.


Ioanna · 2 weeks ago

Oscar what S V Prasad says is correct. Basically we find the brightness of this object (given its

luminosity and distance). But then we want this brightness as multiple of b_sun, rather than in

W/m^2. So we divide by b_sun. (doing it in one step, means dividing both sides of the equation.

You can not just divide one side by a number).

We want b= x b_sun ie b/b_sun =x

we find the brightness the normal way, so say b = y W/m^2

Then b/b_sun = (y W/m^2)/ b_sun


Anonymous · 2 weeks ago

The formula by Prasad (4*pi*(D_0(1+z))^2), whether (D_0(1+z)) is the answer from Q3? I applied

the Distance from Q3 but is not accepted by the autograde. Please share your thoughts.



The formula for Brightness is b = L / (4*pi*D_L^2) ---> This is what we studied in our class. We

know D_L = D_0(1+z) From Q3 we already got the value for D_0 the coordinate distance. We

know the value for z =5.34 and L in terms of Sun's Luminosity. After you get the values divide the

brightness by solar constant (that is what is requested) you are all set. Keep in mind you need to

convert light years to meters. Hope this helps.


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Satish Pisharody (Student) · 2 weeks ago

Prasad, D_0 would be the answer to Q3 or Q4?


Comments


For Q3 you are asked to find Co-ordinate distance so it is D_0. For Q4 it is the other distance you

need to find out.




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A post was deleted.

Ingrid Paulussen · 2 weeks ago

First I have to convert D_0 tot other units, right?


Comments

Mateusz Wielgosz · 2 weeks ago

So D_0 should be in AU?



I think D_0 should be in meters.


https://www.coursera.org/user/i/3e4afcbdaee7a7eb2b71e8b82f00f9fb

https://www.coursera.org/user/i/717a598ddc8d7b7b34c84726939863dd



But isn't L_sun given for Earth at 1AU? therefore in numerator you have just 10^10, since L_sun

will be cleared out.



I agree, but D_0 as we calulated is in lightyears,



L_sun is in W (Watts). We just call 3.83E+26 W = 1 L_sun. Luminosity does not depend on

distance, it is an intrinsic property of an object: how much energy it is radiating per second ie J/s

= W.

Now brightness does depend on distance. For a given luminosity, an object closer is brighter,

and on object further away is dimmer.



OK let's do it step by step. Imagine we converted D_0 to AU then Q6 is:

[ 10^10 L_sun / 4 pi (D_L)^2 ] / [ L_sun / 4 pi (D_sun)^2 ] = [ 10^10 / (D_L)^2 ] / [ 1 / (D_sun)^2 ]

Since we agreed on AU, then D_sun is just 1:

Q6 = [ 10^10 / (D_L)^2 ] = [ 10^10 / (D_0)^2 * (z+1)^2]

Right?



Yes, that also works. But this gives the answer you have to truncate, as Gary mentioned. If you

do it like S V Prasad said, where you just divide by the solar constant given in TOC, you do not

need to truncate.


Σταύρος A. Αθανασιάδης · 2 weeks ago

@ Mateusz Wielgosz




https://www.coursera.org/user/i/9c9c5f51640b76f934d53b4dfc60c63b

Right. Plain and straightforward ratios. The result should be in b⨀ units, so need to mess up

with the exact value of that, namely the value of b⨀Vote this post up 0 Vote this post down

Frank Astier · 1 week ago

@Mateusz - best and simplest method.


João Romário Fernandes Filho · 3 days ago

That's what I've been trying to do since the beginning, but my result is not accepted! I get

x,y4849E-AB...



Oh, I can't believe!!! All the time I was using the wrong distance!!! I was considering the past one

insted of the "current"... Now I got it!



Oh, I can't believe!!! All the time I was using the wrong distance!!! I was considering the past one

instead of the "current"... Now I got it!


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stephen gould · 2 weeks ago

you just need to make sure the distance you are using in the denominator and numerator are in

the same units. The suns distance is 1 au but it can be converted to km, m ly or anything else.

The distance to the object was expressed in ly in earlier questions and can be converted

similarly. Of course the more conversions you do the better chance you will get some rounding or

input errors. And despite knowing this and agreeing with the above equations which do simplify I

can't get the right answer. Reader beware. This is sort of a pain and a waste. B/bsun should =

L/Lsun or 10 **10 times the ration of distances squared or so I think. The trick is the denominator

distance D_L


Comments

Sotiris Batsakis (Student) · 2 weeks ago

https://www.coursera.org/user/i/aef7c473edcc02d6f63a3929a2789a58



Converting distance of question 3 from ly to AU. Also comments from Loh Siu Yin and Prasad

above about rounding are helpful


Gary Lloyd-Rees · 2 weeks ago

Stephen, I used the same approach as you - don't forget to "redshift the distance" and you also

need to truncate the answer....


+ Add New Comment


In confirmation of others - to get the "correct" answer by the autograder you need to "truncate" to

2 sds instead of rounding. Wow the autograder is really given us a workout.....

Edit to add - this applies if you are using the luminosity to brightness ratios approach from Week

4.


Comments




Thanks Gary, I got this Qu right at last.. i wish I could up vote your post more than once..



and finally the truncating did the trick!


Juan F. González Hernández · 2 weeks ago

Truncating???



Instead of rounding as you should you just cut last digits.



https://www.coursera.org/user/i/8d3052195f65f3eaa23e0826d6d54a00




I did not truncate, an my answer was accepted (I never round anything, other what I enter into

the autograder). My 3rd figure is 4, so just rounding worked for me. If the calculations are

different and you have a 5, then you would have to truncate.



Ok, I see if one uses solar constant, no need to truncate. If one uses ratio of distances (basically

recalculating b_sun, but cancelling some stuff) then one needs to truncate.



Ionna, if you use the straightforward ratios equation of L/Lo to b/bo route (with no interim

rounding and using TOC conversions to AUs) then you get x.x5614...... which truncates to the

"correct" answer of x.x



Yes I tried that too, and indeed one needs to truncate in that case. I had done it same as S V

Prasad, that was easier for me: find b. Divide by solar constant 1.37 × 10^3 W m^−2 (this is

rounded to 2 sig figs, so will slightly different result the L/Lo to b/bo route). The solar constant

way gives x.y428ez, so I had just rounded normally.


+ Add New Comment


I calculated in this way:

$$b/(b_S))=(L/L(sun))(1/(1+z)^2)(1A.U/D(Q4))^2$$

I got x.yze-wt and the bot says I am wrong...What do you mean by "truncate"? Please, answer if

it is a bot issue too...

Vote this post up -1 Vote this post down

Comments




If you find something like (this is an example) 2.29e18...., truncation to two s.d. means using

2.2e18 instead of rounding to 2.3e18.



OK; truncating worked for me!!! Only Q5 remains...


+ Add New Comment


By the way, what is the order of magnitude you get? Just curious...I am desperated to finish this

one since I know I am right with my numbers and I want to finisth the QB too, but not till I end this

one...

Crazy bot! Hahahaha... I know you... Can you hear me dear nasty bot? Hahaha... Just joking, ...


+ Add New Comment

Stoica Dorian - Bogdan · 2 weeks ago

and we obtain the ratio . My approach was to insert both

distances in A.U., and before that, calculate using Slide 5, clip 2. (redshift is given, the

angular distance should be known :) )


Comments

Wheeler Huneycutt · 2 weeks ago

Stoica, you are the Man! Keep Calm and Love Physics!!

I can not believe PoliSci majors at Dook can do this stuff!


Sohan P Jain · 1 week ago

Stocia, Just to confirm again, in your ratio, 10^10*(D_SUN^2/D_L^2), D_Sun is in A.U. and D_L

is in L.Y. Is that right? (I'll convert to a common unit, prreferably to A.U.)


+ Add New Comment



https://www.coursera.org/user/i/0b87f06637764ec94693ebbb2c89e5fb

https://www.coursera.org/user/i/fa8d2a3f19322f52f90237cd7fb58ee0

Stoica Dorian - Bogdan · 2 weeks ago

@ Juan : Wish i could give you the exact order of magnitude. However, is a negative number,

bigger than -30, smaller than -19.


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Stuart Aitken · 2 weeks ago

Wow. This too me so long. I tried everything - every distance, every variety, etc, etc. Turns out I

was dividing my final answer by the solar luminosity instead of the solar constant. Hah!!!


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Javier (Student) · 1 week ago

The autograder and me do not get along. I have tried the formula: Bsuper/Bsun= (Lsuper/Lsun)

(1/(1+z)^2)(1 / Q4)^2, of course converting Q4(already OK) to AU,..I have truncated the Q4

formula, but also the autograder says no way,... I'm lost,


+ Add New Comment

Wheeler Huneycutt · 1 week ago

Javier, I think you need to move the 1+z inside the DA squared so the 1+z is to the fourth power,

1+z is squared if you using the D0 . Q4 was DA as the answer, no? HTH


Comments


Thanks so much Wheeler, I admit I have just done the mathematics and now I need to

understand the physics,.. but definitively the auto grader and me are again friends. Thanks for

your help!!


+ Add New Comment


Javier, look at Clip 2 page 5 on the pdf . It gives DLD0DA relations as each multiplied by (1+z)

and then squared.

I leave it to you to explain to us a simple way to remember which is which. It is measurements of

apparent luminosity, angular size and coordinate distance.

Angle looks bigger than coordinate cause the light left early?


https://www.coursera.org/user/i/9952d00d5a8914c0deab952aacfb48eb



Luminosity drops one more z cause photons get stretched.


Comments


Wheeler, let me see if I've got it, Luminosity requires as shown by Clip 2 page 5 a redshift

(1+z)^2, and then there is the relation between distances squared that add additional (1+z)^2,..

then that's why (1+z)^4. is that a correct deduction?.. for me it seems clear now, but I prefer

double check


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Anna Czarina V. Cabe · 1 week ago

And I wondered why I could not ever get this right... I mistook a(t) for z. How brilliant. My two

hours were for naught. Lesson learned.Always label your answers. Thanks everyone!


+ Add New Comment

Judit · 1 week ago

I'm utterly confused by all these distances and indices. I have no idea which one is which. Could

anyone help me pls with sorting them out? I have distances in Q3, Q4, and a radius and angular

radius in Q5. Which one do I need in the brightness formula? Thanks!


Comments

Hy D Tran · 1 week ago

Judit, you want to look at clip 2, around slide 5. D0 is the distance today (now). The observed

brightness (apparent brightness) is b from the equation on the upper right of the slide.


Anna Czarina V. Cabe · 1 week ago

Judit, Hy is right. Do not forget to multiply your answer from number 3 with the red shift

(1+z)=this is D_L. Mateusz' solution from above should help. Here it goes:

10^10*(D_sun^2/D_L^2).


Judit · 1 week ago

https://www.coursera.org/user/i/422b55230770c6d3148897fe6e774f63

https://www.coursera.org/user/i/22dd0443dddb23bbb159bb07b2a55914


Thanks guys! I got the formula I just didn't understand what all the distances we discussed meant

and which one should be plugged into the formula. For some reason this took me a lot of time to

figure out.


+ Add New Comment


This is really complicated. I am going to try to explain it.

D0 is where the thing is today. Distance from here to there today.

DA=D01+z D_A is the distance we measure when we measure angular size of "bound"

objects.

DL=D0∗(1+z) D_L is the distance we measure when we measure Luminosity or brightness.

Therefore: DL=DA∗(1+z)2Now here's how to figure it out.

Grab a relatively big galaxy in your hand and hold it about a foot away (thats about 30cm OK?).

Hold it up where you see it real good about a foot away. Pretend that this is a while back in time,

back when space was only half as big radially as it is today. Also pretend you are a big guy and

can do this. Now notice the stars in this galaxy, some are near the edges of the galaxy. They are

emitting photos, that is to say, the stars are shining. And some of the photons are aimed right at

your eye. Pay close attention to the angular diameter of this galaxy, it should be about the angle

between your eye and your thumb and forefinger. Now push the galaxy away to two feet (that's

about 60cm). You just expanded space by double. Now imagine that that is where the galaxy is

right now, but the light when it was only one foot away is the light reaching you now.

D_0 is two feet. D_A is one foot. z is one. Space expanded by 1+1 or 2, double. (I want to say

everything moved twice as far apart, but I know many will quibble with that.)

Any instrument you have to measure the angular diameter is going to be measuring the photons

emitted at one foot away. And it will measure the angular diameter today when the galaxy was

one foot away. Now today the galaxy is two foot away. Now when you expanded space by

double, think of it as your universe was one foot in radius of a sphere. And you made the sphere

two foot radius. Well, when you expanded space, should you not have spread your thumb and

forefinger apart by double, too? Well, yeah! you should. Try it with both hands and you'll see they

just naturally separate by double, hold them at 90 degrees, that helps. Well why did the galaxy

not spread apart? Because it was gravitationally bound together. It may look like a bunch of

separate stars, but those stars are held together by gravity. And gravity is much stronger than

any old expanding space. Well, space is not really expanding, what is really going on is the Big

Bang blew everything to smithereens. Only the galaxies were able to hold everything together,

thank goodness! If you can get a tennis ball, you can try this experiment using that as a sample

galaxy, you can see how it holds together as space expands. Also, go get a handful of flour or

dust or salt, and try pitching it about two feet up in the air. Neither Dust nor flour nor salt are

gravitationally bound. This would be like measuring the angle between Arcturus and Betelgeuse,


they are not bound so they separate when you expand space. Well, that's not true, but you know

what I mean. Things at great distance and great angular diameter spread apart. Well, it seems

like a whole lot of bloviating just to explain the D_A.

What about the D_L? It's even more complicated cause there's two things going on.

First, we've got the same thing as measuring the angular size. Start out at one foot. You are

holding a star or a supernova. We are going to measure the Luminosity which is a fine thing to

measure, it is light itself. Well, actually, we are going to measure the brightness which spreads

out as the light travels with the square of the distance. This we did way back when in week 2

maybe. It is sort of a like a sphere too, spreading out by area , distance squared. This part is

easily taken care of like we did in week 2 with the formula b = L / 4 pi are squared, where are is

D_L , the distance where the photons were emitted.

Which would be D_A, EXCEPT, we have to modify two things.

One, this star or supernova is firing photons at you. and they are redshifted, AND TWO they are

delayed in frequency too! So you have to multiply D_A by (1+z) twice! Once for the number of

photons observed, cause the source is moving away at (1+z) times speed, the bam, bam, bam

frequency of photon after photon is decreased, that is seems slower as it moves away. Also, the

very vibrational frequency, the color of the photon itself is lowered, due to the source moving

away, this is the hum of the light that is too high pitched to hear, we can only see it as color or

feel it as warmth, well, sort of anyway:)

So 1+1=2 etc.

Here's some "z" factors:

http://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects

It mentions M104, the Sombrero as being .004 the second galaxy redshift measured, next to

Andromeda which has too much peculiar motion.

Here's what SIMBAD says about M104 : z=.003642

http://simbad.u-strasbg.fr/simbad/sim-id?Ident=M104


Comments


Judit · 1 week ago

(edit) Thanks for the detailed explanation, Wheeler! So let me see if I got this right. I can either

use D_now or D_A, correct? Could I just get D_L (the distance I need to use in the brightness

formula) by multiplying the distance I calculated in Q3 (which is where the galaxy is now) by

(1+z)?

D_L = D_now*(1+z) where D_now is Q3, right?

Should give the same result as D_A*(1+z)^2, where D_A would be Q4.


Judit · 1 week ago

Apparently right, because my calculation was accepted by the grader. Wooohoo!!! For some

reason this one was very hard for me to comprehend. Thanks again for the help!

(P.S: I remember you from the hangout! :) )


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John Owen · 1 week ago

What in heaven's name am I doing wrong ?

I get D_L around e15 AU and an answer in the e-21 region (I've ? 3 right)

I'm going to have a beer

I blame it all on the person who, in Week 7, mentioned the pataphysical sciences, and I cannot

get 'Joan was quizzical, studied pataphysical sciences at home' out of my head


Comments

Siamak Kazemi · 1 week ago

Same here John. I'm stuck where you are.



Hey John, I found out what was wrong. It's the Excel. Make sure you're dividing 10,000,000,000

by the denominator. You'll get an answer in the order of magnitude of e-22. That'll do it. Good

luck.


Morin olivier · 1 week ago

Oooops John, i'm sorry, 'cause i think i'm the one, precisely. "Joan was quizzical/ studied

pataphysical / Science in the home / Late nights all alone with a test tube / Oh, oh, oh, oh..."


+ Add New Comment

https://www.coursera.org/user/i/12ee2583bca8607350f0a3a6b35b1c7e

https://www.coursera.org/user/i/3adadd8c9f792d226ba773ffb8ef4892



Thanks, Siamak, you really got me to LOOK.

Aaaaaaargh... I was using the value for a(t) instead of Z

I must have had at least 20 tries,,,,,,,,,,,,,,,,,,,,,,

On to 7


Comments


You're welcome John.


+ Add New Comment


Morin

You are forgiven you

I can't stop singing 'Clerk-Maxwell's Silver Hammer'

Regards

John


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Anonymous · 2 days ago

Hi,

I recall the formula given in the lecture (and we can redemonstrate/prove it by going back to the

definition of Luminosity)

LL0=bb0∗D2, where D is in AU (Astronomical Unit)

So bb0=LL0∗1D2LL0 is given in the question and LL0=1010So this is straightforward. But my question is: what is D ? This is the distance between the

Supernova and the Earth, obviously. Is it the answer of the question 3 or the answer of the

question 4 ? I guess this is the answer of the question 4 (including the redshift) but my answer is

not accepted by the autograder. All is OK with units (distance everything in AU), I guess.

I found something like this: x.yye-19. And you ? Where am I wrong, please ?





Comments

Sohan P Jain · 2 days ago

Hi, though I am not sure I can give you a practical advice - try both - that is why up to 100 tries

are allowed.


Anna Czarina V. Cabe · 2 days ago

D is computed by multiplying D_0 to (1+z)... do not forget to square the whole expression.

Hope this helps.


+ Add New Comment

MALINA ELEONORA COCIOCEANU · 8 hours ago

Please help with this one, I belive my conversion is wrong.So, we have b/b_sun=L/L_sun*1/D^2

D=1/(D_0*(1+z))^2.


+ Add New Comment

MALINA ELEONORA COCIOCEANU · 7 hours ago

Never mind, rounding error :)


Comments

Dan Murphy · 7 hours ago

Glad you got it



https://www.coursera.org/user/i/94257ed5f52ddfbe7c3e4642b6a476da


https://www.coursera.org/user/i/5347d7d6d3559463b24e990ccf9e0a94

Week 8, Part A, Question 7 - DetailsAnonymous · 2 weeks ago

Question 7 The luminosity of a type-Ia supernova declines by a factor of six from its maximum

within about 40 days. How long will it take for the brightness of the supernova from Question 6 to

decline by a factor of six from its maximal value? Express your answer in days and round to two

significant figures.



Would this just be = number of days / (1+z)^2 ? So, 40/(1+z)^2

The logic being, brightness inversely scales to the square of the distance & the Supernova is

receding by a factor (1+z) ?

Any thoughts?


Comments

Jeroen van der Graaf (Student) · 2 weeks ago

We are talking about an event that is very far away from us and happens on a very short

timescale (in cosmological terms), so the dimming effect due to the distance increase in that time

would be negligible.

This question is about something that takes 40 days in a frame of reference comoving with the

supernova. In our frame of reference the supernova is receding from us with relativistic velocity.

What does special relativity have to say about time in such cases?



I haven't sent my results to autograde yet, but there is no reason to square the (1+z) factor.

We're not talking about brightness, just about time. So only elongation of period due to

cosmological redshift should apply. That's first thing.

Second thing if you divide 40 by 1+z you'll get shorter time, and that doesn't seem right. So I'd

say it'll make more sense to multiply time by (1+z). However what bothers me is that with z =

5,34, we'll get time waaay longer than actual. But maybe that's just because that's how data for

questions was chosen.




Totally agree, not luminosity so the (1+z) factor must not be squared. Also time is dilated so

answer must be a greater value than 40.



+1 Mateusz Wielgosz

You are more than right IMHO




Thanks, Jeroen, Mateusz, Σταύρος A. Αθανασιάδης for your inputs. But this is so counter-

intuitive... the Luminosity of an object, receding from us at relativistic velocities, falls to a sixth of

its maximum luminosity in 40 days, but the apparent brightness takes much longer to reduce to

1/6th? I'm assuming this is because each succeeding photon has to travel a much longer

distance to reach us & the information that the luminosity is falling also takes longer to get here -

the information gets stretched out. Can't see this happening to the tail lights of a car moving

away from me, but I guess cars don't travel fast enough. May be I should change my intuition, as

Ronen advised in one of the relativity video clips:)



Satish, I don't think the question says "falls to a sixth"; rather, it says "falls by a sixth" and

therefore "falls to 5/6" of the maximum value. Another issue: if the luminosity falls by a sixth of

the maximum in 40 days then it will become zero in just 240 days - too short on astronomical

scale! Does it make sense? (Note: the maximum luminosity must be unique.) I am missing

soimething here.


Satish Pisharody (Student) · 1 week ago

Sohan, the question does say "falls by a factor of 6", which to my mind means becomes 1/6th...

however, this is not pertinent in answering the question. We are asked to find out the "time

dilation" associated with the fall in luminosity as seen from Earth, which is the fall in "apparent

brightness", b.

Remember, Type I supernova is essentially a carbon star that goes out in one massive

explosion, converting itself into Iron or whatever. Which may be the reason for the short


luminosity (just days, a very short period in cosmic terms), not much of a star is left after the

almighty blast, is my guess. Would like Justin or Dr.Plesser to comment on this.


+ Add New Comment


To any one out there,

I approached this question as if it is asking me when do i see this dimming effect? To answer this

it seems to require quite a few steps considering co motions and the like. Comments please.


+ Add New Comment

William Byars Samson · 1 week ago

The way I've approached it is to say that if an event takes 40 days and it has a redshift z, I am

inclined to multiply that time interval by 1+z to give a longer time interval as special relativity

would suggest. Unfortunately the auto-grader doesn't agree with my answer, so I'm stumped.


Comments

Eric Dunbar · 1 week ago

Hmm. My logic is this.

We get b=L/(4πD2L) from an earlier problem. Since 4π is a constant and, at the time scale

we're talking about, DL is effectively static we can simplify the equation to b=gL where g is a

constant and g=(4πD2L)−1. Thus we know that b∝L so the brightness should diminish at

the same rate as luminosity.

We are told that luminosity decreases to a 1/6th over 40 days.

That said, we need to remember that our here-and-now experience of how long this reduction in

luminosity will take will be coloured (pun intended) by how long it took the light of the supernova

to reach us. We learnt that the red-shift is a crude correction.

Thus, William, I think you're completely on the right track. Perhaps you're not using the right

adjustment with z. The correction ought to be quite simple.



One additional thought: your reasoning seems 100% correct to me. Are you perhaps missing

brackets? Or using an incorrect z-value (5.34). At one point I had typed in the incorrect z-value,

and, using logic that I knew had to be correct I couldn't solve it. One quick check of my values

and I realised my mistake :).

https://www.coursera.org/user/i/519f7c2c66b78a598013ccd98d1b40fc


William Byars Samson · 1 week ago

You are absolutely right! For some reason I used z = 5.84 in this calculation so my answer wasn't

close enough. Old age - failing eyesight - tiny fonts - AARRGGHH!!!

Your sensible advice is much appreciated, Eric. Isn't it strange how we tend to look for something

subtle when there's nothing subtle about this error.



That's the same mistaken z-value that I used.


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Anonymous · 1 week ago

William Byars Samson ,

I have put this whole redshift concept as answering the question: "An event that occurs THEN is

what event that occurs NOW? You set up an equation accordingly, remembering that t=t(0)=1 at

NOW. It seems that you are on the right track. Hope this helps


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Radek Krahl · 1 week ago

Using formula for time dilation:

t′=t/(1−v2/c2)−−−−−−−−−√for obtaining v causing time dilation calculated from cosmological redshift (1+z) (t == 40) and

then Hubble formula:

v=H0Di've obtained distance to SN in question equal to ~4Gpc (~1 Gpc smaller than distance calculated

in Q3) .

I know that this may be stupid question - but where's the difference coming from?


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https://www.coursera.org/user/i/519f7c2c66b78a598013ccd98d1b40fc

What's the logic behind multiplication with (1+z)? Why can't we use time dilation formula

(t=t01−v2/c2√)? (Velocity could be found from λλ0=1+v/c1−v/c−−−−−√=1+z→vc=(1+z)2−1(1+z)2+1).


Comments

Troy Williams · 1 week ago

I believe the cosmological redshift takes into account relativistic effects. If you take a look at

this:http://astronomy.swin.edu.au/cosmos/c/cosmological+redshift

I would be interested to find out if both methods agree.



What I found that the solution is independent of the information "factor of six" -- it could be six or

six hundred or six million! I don't know why it is so. I want to hear from my coursera class-fellows.



Sohan,

You probably found that it is a function of time. As so an entity in the cosmos experiences an

event (like luminosity dimming) and as time passes as such, 40 days. It will happen regardless of

anything we do. The only consequence that we can realize is a change in apparent brightness.

That is what we see, however, the only question that we answer is "when in time to we see it"?



Sohan,

typo "when in time do we see it"



Yes, that is what I found. The solution depends on "40 days" but not on "factor of six". I don't

know exactly why it is so.



Sohan,

Without spending the time I, myself, need to fully understand this I am thinking about the redshift

parameter z (how far back we are looking in time) is related to the scaling factor a(t) and a is

related to t. The larger the z value, the further back in time we are detecting. We see that the

radiation density was greater at z > 3300 (as said in the lectures) and the time past since the BB

could approx 2,500< t <55000 years, as a rough guess. Again all these parameters are a relation

to time on a cosmological scale. These distances of objects (galaxies, clusters, etc.) from us are

on such a massive scale (Mpc and lyrs, etc.) that we quantify them in "time" to get there (lyrs).

Equally, for us to observe or detect those cosmological events, it is a matter of time. I do not

pretend to know all but you can ask yourself this question "what should it depend on if not on

time"? I believe you will find the answer that will satisfy you. Chow


David Gerstl · 1 week ago

Sohan,

The reason that "40 days" matters and not "factor of 6" is that we're interested only in the time

dilation. It could be any event taking 40 days. Assume someone was standing on the surface of a

planet at the same position/relative velocity with a flag that they raised for 40 days (in their local

time). We'd get the same answer for "how many days is the flag raised (when observed from

Earth with a telescope). Just happens that the supernova change is an event that is actually

observable from earth.


+ Add New Comment


I thought this was straightforward, but I've spent hours on this and tried rounding up and down,

and am not getting a correct answer. The most immediate solution that came to mind is

40*(1+z)=40*6.34=253.6=2.5e3 I've tried numerous variation on this theme, and can't think of

another way to get this. I feel rather silly.


Comments


Never mind. It's hilarious. Just posting it here was enough to figure out the error of my ways. I

should have complained a lot earlier!


Oscar Orta (Student) · 1 week ago

if you found that your answer is 253.6 and you try to tell the grader as 2.5e3 you are making a

mistake.....2.5e3 = 2500 , not 250....check your arithmetic only.....just a simple mistake. Hope

this helps.


+ Add New Comment


So I get this one ... sort of.

Here's my problem:

(1+z) = \lambda_observed / \lambda_emitted

Now, most of this is due to relativity, but some should be doppler. Shouldn't we be removing that

factor, or is it still small relative to the time dilation?


+ Add New Comment

Ignac Fetser · 1 week ago

Could someone please answer Anonymous' question? "What's the logic behind multiplication

with (1+z)? Why can't we use time dilation formula (t=t0/(1−v^2/c^2)^.5)? (Velocity could be

found from..." The difference between the two methods is huge - the result with the time dilation

formula is about 42 days while the t(1+z) is ~253 - why is that?


Comments

Ignac Fetser · 1 week ago

ok i think i get it now - we would use the time dilation formula if the space were not expanding.

but it does so that 40 days long light packet gets stretched out on its way here by the factor the

universe has expanded between the time from its emission to observation.

https://www.coursera.org/user/i/0c235189506e864e54da793930dcaf5e

https://www.coursera.org/user/i/0c235189506e864e54da793930dcaf5e

Week 8, Part A, Question 8 - Details

Patty Allison · 2 weeks ago

Question 8 What was the temperature of the Cosmic Microwave Background when the

supernova exploded in RD1? Express your answer in K and round to three significant figures.



Using formula at slide 2 page 5. T_obs is given at slide 6 page 4


Comments


This is kind of you, Σωτήρη! :-) Clip 6 is very enlightening..



Formula at slide 2 page 5?

T_obs at slide 6 page 4?

What videos are referred to please?



2 and 6



C'mon do a bit of work Anon..... "Slide" = Video....


https://www.coursera.org/user/i/bd2af5a1f1a2a1264002b76ab7603234



Dave Fattori · 1 week ago

What has the formula in slide 2 clip 5 to do with the question? It doesn't have T in it.


Enith Paternina · 3 days ago

T=2.726(1+z)


+ Add New Comment


OK I just did this one, and I can say: for current CMB temperature use 2,725K, not 3K, because

you won't get the right result.


Comments


you probably mean T=2.726 K



For the non-continental Europeans, FYI, Mateusz's comma is a decimal point.....



Hi Σταύρος, still waiting for a resolution on Q5 - are you ready to tackle Part B?



Hi Gary, not yet. I can leave it for another day. I ain't got no time fighting with the typoed

autograder...


William Gunn · 2 weeks ago

Apart from messing up my units after Q1 & Q2 (and the wierd thing on Q6) the only remaining

part A question I'm struggling with is Q5




Bill



You shouldn't. Its straightforward once you use there a redzhifted distance..



Yeah - I don't understand it - mind you - I've had a lot of distractions with "real work" last couple

of days - probably done something stupid (again) !!!

Bill



Maybe this is a bit off topic but here it is: small angle formula. Find that angle α in arcsec.

Nominator: radius of galaxy (in whatever). Denominator: our distance (in whatever) from the

galaxy when the light was emitted (reminds of a previous Qu, right?). Whatevers cancel out. Oh,

don't you forget to multiply all that with our good ol' 206265 cuz you want this α in arcsec, not

radians. This post will self-destruct in 5''. Good luck Bill.



LOL, thanks, but been there, done that! Part of my problem was thinking "this is one of the

simplest Qs" - I can't believe it though - I have put all the questions for all the weeks into a VB

program - and - I copy each question text as a comment - then I have to (marginally) edit the

pasted question lines (just so they fit properly) Errrrrrr........... well yes.... typo Bill, typo. If I had re-

read the question text rather than relying on my pasted/edited copy..... oh well.... Ronen's not the

only one who is a "spent force on this course"... ROFL. Oh well, gonna be a busy day at work

today - but I must try to get a few more of part B done, would be nice to tie this all up before the

weekend! Cheers, Bill



Week 8, Part B, Question 1 - Details

No tags yet. + Add Tag


The ionization energy of a Hydrogen atom is Eion=2.17896e−18J.Find the temperature T at which Eion=kB∗T.

Express your answer in K and round to two significant figures.

This should not cause us any problem..



There is also this :

https://class.coursera.org/introastro-2012-001/forum/thread?thread_id=2214

Since noone has posted there, seems noone has a problem with it so far...


Comments


I hadn't noticed that Ioanna. I am deleting this one....



Week 8, Part B, Question 2, Details

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No tags yet. + Add TagAnonymous · 2 weeks ago

Question 2 In fact, Hydrogen recombination is considered to have occurred at z=1100. What was

the temperature of the Cosmic Microwave Background at this time? Express your answer in K

and round to two significant figures.



Is the answer to this = Ans to Q1/(1+z) ? Need some conceptual help on this.


Comments

Daniel Scandiuzzi · 1 day ago

Actually you have to multiply (1+z) and Q1. I looks like this: Ans to Q1*(1+z)


+ Add New Comment


thought it might be T*(1+Z) but of course this doesn't work as most of my efforts on this

homework


+ Add New Comment


Same approach as HWA Q8 should work


Comments

Dave Fattori · 1 week ago

It doesn't work for me.


Patty Allison · 1 week ago

T_c=T_o x ( 1+z ) using the T from Q8


Gonzalo Ignacio Andía Olivares · 1 week ago

Use T_o the actual temperature of CMB


+ Add New Comment


use the T from HWA Q8


Comments

Pascual Pérez Cuenca · 1 week ago

I thought this was the easiest question of all, but at the end it is the only one I'm now totaly lost. I

thought that the only thing to do was multiply the answer of q1 for 1101. The aotoograder say no

all time. Can someone give me an idea of what I'm doing bad.?



Latifis Konstantinos · 1 week ago

Patty correctly says: use the T from HWA Q8. It's only a multiplication, but don't use the answer

of q1. You can find this T in Clip 6, slide 4...


Pascual Pérez Cuenca · 1 week ago

Thanks Latifis I'll do after watching the clip 6 once more.




https://www.coursera.org/user/i/627f8937ca1e8840ae1c028818826df1

https://www.coursera.org/user/i/627f8937ca1e8840ae1c028818826df1

Konstantinos Siettos · 1 week ago

Latifis you mean the type from 8A not the T


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sio (Student) · 1 week ago

Can't get this one. I got all the 8 other questions but this last one. I followed the Patty's post and

used the T found in HWA Q8 and multiplied it by 1101 but still it is not good. What is the order of

magnitude do you have ? Is that something x10^4?


Comments


T_o = 2.7255 K. z= 1100 so following the formula, you get the answer. Rounding to 2 significate

digits and I didn't even convert to x.ye3.


sio (Student) · 1 week ago

Arf, ok, got it now. How makes things more complex than they are... Thanks for the help Patty.


+ Add New Comment

stephen weir · 4 days ago

okay, totally lost, Patty, when you say HWA Q8 are you referring to WK8, part A, Q8 which dealt

with a temperature of a supernovae in a galaxy far, far away. What does that have to do with Part

B, Q2?


+ Add New Comment

Patty Allison · 3 days ago

when solving problem 8 of part a of this week, we have to use the CMB temperature which is

2.7255 K. At the time I wrote the first hint, I was trying to avoid coming right out and saying the

temp. They both have the Cosmic Microwave Background.

https://www.coursera.org/user/i/0271af33ba350013b721af372cade715




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Question 4 The wavelength of acoustic waves in the Cosmic Microwave background was given

in clip 8 as 201 kpc at z=1100. Estimate the size of such a perturbation today, in Mpc. Round to

two significant figures.



Simple. Space expanded by some factor and the wavelength too.


Comments


Requiring this size of perturbation in Mpc almost gives away what are we supposed to do with

that 1+z in our hands ...


+ Add New Comment


Anybody have any help to offer on this one? Just need a starting point..


Comments


Although it may not seem so at first look, it just about comparing sizes (length in this case) at

different times. We are told how big something was in the past (at z=1100). (This something is a

wavelength, that was 201 kpc long.) How big (long) is it today? (The universe has expanded, and



so has this length). So you just calculate the scale factor, and then convert the length then to

length now.



Thanks! I was having a brain cramp! I will try it again in the morning.



Got this one, thanks Ioanna!


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Eric (Student) · 1 week ago

This one should be straightforward, but can't get the grader to accept my answer.

We have z = 1100, L = 0.201 Mpc (same as 201 kpc)

Shouldn't the answer in Mpc be L times the square of the inverse of the scaling factor ?

I have x.y..E+5, where x+y = 6


Comments

Radek Krahl · 1 week ago

Eric - you're scaling linear (one dimensional) value - length. So you must use first power of

scaling factor (only one dimension to scale).


+ Add New Comment


Oops, you're right, my bad !

Thanks !


+ Add New Comment


Dear classemates... Am I tired or silly?



Scale factor shouldn't be 1/(1100+1)?

and the wave that i have to multiply by this scale factor shouldn't be 0.201Mpc?


Comments

Hy D Tran · 1 week ago

Olivier,

You have most of the concepts correct. One more thing to consider: The scale today is 1 (by

definition). When the wavelength was 201 kpc, the scale was (1/(1100+1)). Therefore, what is

the wavelength today?



Arfffff, thank you Hy D Tran...i've entered my answered as answer of Q5

I'm finally tired to be silly and silly to be tired like that.



This does not make sense to me.... 201KPC = (1/(1100+1), ?=1, so it is just 201/(1/(1100+1))

right???

https://www.coursera.org/user/i/22dd0443dddb23bbb159bb07b2a55914

Week 8, Part B, Question 3, Clarification


For photons should we use the radiation density in the lecture slides 4.765e-5 or does this

involve both photons and neutrinos?


ronen plesser INSTRUCTOR · 2 weeks ago

Use that number as the energy density in photons. Neutrinos are an interesting issue. We are not

sure about their masses, so do not know how relativistic they are - in particular, some species

may be massive enough to be nonrelativistic at the neutrino background temperature of about

2K.


Comments


I used that number but things did not work well. I did the following: 1.Multiplied \Omega of

radiation by the critical density given in the correction page s 9.47e-27 kg/m^3 2.Multiplied by

(1+z)^4 to get the density at the time of interest 3. Multiplied by c^2 to have the density as J/m^3.

4. Computed the temperature at the specified z and then used KT to find the energy per photon.

5. Divided the energy density by the energy per photon I checked the computations several times

but the grader did not accept my trails.



I did that differently and it worked. Clip 4, slide 5: there is a formula for energy density of

radiation, that depends only T (and we have T. Well, we calculate it for z=1400, since we need

it.) and g for photons is g=2 (mentioned in the video). My answer with this method, was

accepted.



Yes you are right. What about the third number? It should be straightforward but the grader

rejects. I got for the number of ionizing photons x.yze7, where x=(y+z)/2



I also have a problem with the 3rd number. I get x.yze7 (so same power) but not x=(y+z)/2 (but

close). I have two different answers, depending on what I use for T today (2.7K or 2.726K. This

affects T at z=1400). Both are wrong though. My 2nd number remains correct using either T

(only the last sig fig changes by 1, and both possibilities are accepted). So it can not be an issue

with T.

I made another thread (called 'details' since it is more detailed) and also included the question

(so it is easier to see what we are talking about).

https://class.coursera.org/introastro-2012-001/forum/thread?thread_id=2234



Thank you Ioanna for your immense contributions.


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Question 5 Imagine that the entire baryonic mass of the universe were converted to radiation.

Find the temperature of the resulting radiation field. This is an alternative way of thinking about

Olbers's paradox. Express your answer in K and round to two significant figures.



I used the omegas of baryonic matter and radiation in the lecture. The ratio is the ratio of

temperatures to the fourth. Knowing the current temperature of radiation (2.726 K), we can get

the temperature if baryonic mass were converted to radiation. Did not work! What is wrong with

this?


Comments


Hmm.. It says "Find the temperature of the resulting radiation field". I interpreted that as find T of

radiation field created by baryonic matter converted to radiation. So the only radiation is that from

the baryonic matter becoming radiation. I did not take into account the existing CMB and its

temperature, to try and average the temperatures. My answer was also wrong by the way.


+ Add New Comment


I have no clue how to even start on this - any suggestions?!


+ Add New Comment

Nicholas · 2 weeks ago

Having trouble with this one - here's my method.

Baryonic density today = 4.17e−28kg/m3 Convert to energy using E=mc2 ... getting

the radiation field (kg/m3∗m2/s2=(kg∗m2/s2)/m3=J/m3)Use radiation density formula from Clip4,Slide5 and solve for

T. (rho=g∗sigma∗T4/(4c) implies

T4=(4∗c∗rho/(g∗sigma))Units seem to work out here: (J/m3∗m/s∗K4∗m2/W)=(K4∗J/s/W)=(K4) I use g=2

to solve, and take the fourth root, getting a K between 0 and 100.

No luck getting a correct answer yet - should I be using a different approach or different formula

for energy conversion or energy density, or a different value of today's baryonic density, or a

different value of g?


Comments


Well, if even you can not get a correct answer here, maybe the autograder is indeed wrong.. :)

I also get something between 1 and 100 K, which is not accepted.




Easiest way - to multiply the one-fourth power of Ωbar/Ωrad to the CMB temperature. It will

give the same answer as your method, if you correct the formula for energy

density: ρ=4σcT4 (see course corrections page). But the grader do not accept this answer too.



I thought I understood but not quite from slides omega rad 4.765e-5 and omega baryonic

was .256 (Omega b/omega r) to exp(.25) times K today : K result between 0 and 100 but did not

work.

last one to go - help



Nicholas: I do not know if you already saw this: but check your T^4 formula.....I saw a correction

in the course corrections section.....and the grader works.!. Maybe you already know this, but it is

send with the best intentions.



Nicholas

Did you get this to work. I used your approach with the correct and incorrect energy density

formuales and even then added the existing Temperature from radiation to the resulting

amounts. None worked. I think the formulaes make sense in terms of units since they all result in

reasonable temperature results. Not sure if I am making a math error, there is some scaling

factor off , auto grader off but it says it is fixed. Also wanted to make sure someone had this right

to eliminate the auto problem.



After converting to energy with E=mc^2, and then using the corrected formula to get T^4, (and

then T, of course), the result is accepted by the grader. Hope this helps.



Ahhhh - thanks - I've been distracted today - I'll check the corrections.... ty

Bill


+ Add New Comment

A post was deleted.

Shloka Ananthanarayanan · 2 weeks ago

The autograder seems to be fixed now!


+ Add New Comment


About this question and the cosmic destiny. Baryon burning to give "new life" and "heat" the

universe with new radiation can not last forever. :( The amount of heat released is not well

enough to make the Universe live forever! :(


+ Add New Comment

Wong Wo Chai · 2 weeks ago


https://www.coursera.org/user/i/66d375d7c0288f65c94117d92940d6ef

E=mc^2=sigma*T^4/(2c), even I've used the corrected data for density, answer not accepted.

Have I missed something?


Comments


Look at the course corrections for the energy density of radiation (the formula is corrected there)


Wong Wo Chai · 1 week ago

I got it, tks loanna


+ Add New Comment

Richard Torson · 1 week ago

Look to course corrections......correct the density using E=mc^2....m=baryonic density(make sure

to use c in meters...equate this to the corrected formula for pressure=(g*2*sigma*T^4)/c...use the

value for a given in the course corrections to get a value for (g*2*sigma)/c....take the fourth root

of this number to determine the answer....no need for the T_cmb.....we are just looking for the

temperature resulting from the radiation.


+ Add New Comment


Even though I went through all the hints above, I seem to be missing something still.

I know how to extract the temperature from the energy density eρ=4σT4/c in [J/m3], obviously.

However, I have seen an equation above relating E=mc^2 [J] to eρ, which are not

dimensionally consistent. Also, what is the mass used in E ? From these pieces of

information, I cannot calculate eρ.

I could easily get eρ from the given baryonic density in [kg/m^3] and the

number of J/kg of baryonic matter, but I can't find this number.

I am obviously missing something very simple. I would appreciate any additional hint, thanks.


Comments

Ioanna · 1 week ago

E=mc^2 can also convert mass density to energy density, ie

https://www.coursera.org/user/i/66d375d7c0288f65c94117d92940d6ef

https://www.coursera.org/user/i/c273d11394788b1826dd05d89df9e2da

energy density (J/m^3) is =ρc^2

where ρ is density of baryons (which is given earlier in qu3 (or you can calculate it by

Ω_m x ρ_c = ρ_m,b

The value for ρ_c (critical density) has a course correction, so do not use the value in the slides

but the corrected one.


Σταύρος A. Αθανασιάδης · 1 week ago

This is a brilliant explanation!


Richard Torson · 1 week ago

Nicolas did a great job with the dimensional analysis above....check it out, and as Ioanna stated

above...she is the belle of the discussion forum!!!



Of course, this is exactly what I missed : J/kg = (m/s)^2, so I should have guessed the c^2

factor !

Thank you Ioanna !

If there is a medal for the best helper, you should get it !


+ Add New Comment


If Eρ=g2σT4c then T=(Eρc2gσ)1/4 (eq.1),where Eρ I call energy density in Jm3, Τ is in K, c and σ from the TOC.

BUT E=mc2→Eρ=EV=mVc2=ρc2→Eρ=ρc2 (eq.2), where ρ is mass density

in kgm3.

(eq.1)Λ(eq.2)→T=(ρc32gσ)1/4 (eq.3) →⎡⎣⎢⎢⎢kgm3m3s3Jsm2K4⎤⎦⎥⎥⎥1/4=[sm2K4kgm3Jm3s3]1/4==[kgm2K4s2J]1/4=[JK4J]1/4=[K4]1/4=KRadiation means photons so for a gas of photons it is g=2, so:

(eq.3)→T=(ρc34σ)1/4and I find a T in [12K, 18K]. Am I on the right track? I'd rather ask all of you than the autograder.


https://www.coursera.org/user/i/c273d11394788b1826dd05d89df9e2da



Comments


The method looks correct. I was about to say your previous result (e-2) did not look right. Should

be e1. What you have now looks ok.



Ioanna, thanks for your feedback!

I tried all this elaborate reasoning which probably looks correct and I previously entered c

in km/s in Excel instead of m/swhich is the proper unit. A minor detail can keep you struggling.




+ Add New Comment

Sean Storey · 1 week ago

Thought this was going to be straight forward but cannot get the autograder to accept my

answer.

I'm calculating the energy density from the baryonic density as 8.511e-10

Does this look right ?

The rest should be straight-forward e.g. mulitple by 1/a( Radiation constant from course

corrections) then take the 4th root


+ Add New Comment


That doesn't look right Sean (it is 1 order of magnitude off). I did your calculation backwards, and

I think you have used the critical density (ρ_c), rather than the baryonic density. The density of

baryons is given in qu3 a), or you can calculate it more accurately by Ω_b x ρ_c (and look at

course corrections for the correct value of ρ_c).


Comments

Sean Storey · 1 week ago


Thanks - that's it.


+ Add New Comment

Chiok Ching Chaw (Student) · 1 week ago

Hi. Got a related question. Radiation constant is given by a = 4*sigma/c. However substituting

sigma = 5.670e-8 and c = 1.49597870691e11, I am not able to get a = 7.565767e-16 as stated in

the Course Correction notes, am I missing something?


Comments

ronen plesser INSTRUCTOR · 1 week ago

Where did that c come from? We have c = 2.998e8 m/s



your measure of c is wrong.....it should be c= 2.997924 e 8 m/s...then you will get it right...I hope

this helps.


Chiok Ching Chaw (Student) · 1 week ago

Opps! My bad! I got mixed up with the value for AU! Thanks!


+ Add New Comment


I've tried this so many times, I can't see where I've gone wrong. I've used density = 4.17e-28 ,

c=2.99792458e8, sigma= 5.6703e-8 and I keep coming up with a figure of the order e-2. I've also

tried using the calculated version of density which is very close to but just a bit smaller than the

given density and get pretty much the same answer. Can anyone see my mistake?


Comments




+ Add New Comment

serge · 13 hours ago

What expression for the radiation density (equation of state for phonons) shall we use now?

Whether the erroneous one in the slides, or, the corrected one in the course corrections which

differs by a factor 8? (that is whether the autograder has been adjusted to corrected equation or

it remains with this mistake)?


Comments


ronen plesser INSTRUCTOR · 13 hours ago

Use the corrected expression.


+ Add New Comment

Anonymous · 40 minutes ago

I think frustration is setting i and I am missing something. . Converting density into E_p: Baryonic

Density * (2.9979e8)^2 E_p = (2*g*sigma*T^4)/c; Solving for T^4 = (E_p*c)/(2*g*sigma) since g =

2 then T^4 = (E_p*c)/(4*sigma); since (4*sigma) /c = 7.565767e-16 then the inverse =

1.32174305e15 which is equal to c/(4*sigma); T= (Baryonic Density * (2.9979e8)^2 *

1.32174305e15)^.25 It seems my power is correct e^1 but my coefficent is incorrect. What am I

missing? Thank you

https://www.coursera.org/user/i/85fc292ba5f5b73d08d8cdc3b2e216d9


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Question 6 In this set of linked problems we will take a look at exponential expansion - which we

think describes both the distant future of our universe and an important phase in its very early

past. In exponential expansion with a(t)=eH0(t−t0) the coordinate distance - the distance now -

between the position of a photon at time t1 and its position at time t2 is

D=cH0−1(e−H0(t2)−e−H0t1). Consider a photon observed at t0. Find the distance at which it was

emitted and write this in terms of the observed redshift z. Your answer should be an expression

containing z, H0, and c.



A necessary clarification in the statement of the problem: I can calculate the proper distance

integrating the scale factor inverted in the following way:

d=c∫t2t1dt′a(t′)Inserting the scale factor a(t′)=exp(H0(t′−t0)), we get

d=c∫t2t1dt′exp(H0(t′−t0))=c∫t2t1exp(H0(t0−t′))dt′=cH0(exp(H0(t0−t2)−exp(H0(t0−t1))Set t0=0 and we have the condition Ronen gave us in the problem. Calculus and integrals are

cool (and rocks)like bow-ties!!!! LOL

Let me know if I am right or not...It is relevant in order to get the right expression relation

c, H0 and the redshift z. Moreover, it is also relevant in the calculation of the "particle horizon".


Comments


I did the same, but integrated from t to t0. I obtained cH∘(eH0(t0−t)−1). The exponent is

readily expressible in terms of z. What is the formula for the particle horizon?



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You don't need calculus for that. It's straight plain n simple algebra.


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I know, but I was having fun with advanced topics in Cosmology I am studying now, hahaha...


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Alexander Edward Drew · 2 weeks ago

Is it correct to start by setting a(tem)=1/(1+z)=exp(H0(tem−t0)) and then solve

for exp(−H0tem) ? Using this approach and substituting a constant for H0t0 I reach an

incorrect answer.


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A post was deleted.


I assumed H0t0→0 and it worked miracles but am I justified assuming this? I am in a

conceptual loss here, although the grader likes it and says it's OK. Anyone with a good insight

about my assumption?


Comments


If z is high then t0 is large. i.e. we are in an old universe and e−H0t0 is small.





With the new definition Ronen gives below, t0=0. So H0t0=0Also a(t0)=eH0(t0−t0)=eH00=1 (ie scale factor today is 1)




And generally the scale factor at time t is a(t)=eH0tVote this post up 0 Vote this post down

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Am stuck up on this problem & the succeeding linked problems. Appreciate help with the steps!


Comments


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A few clarifying comments on this one: First, in going from the first expression to the second I

have tacitly taken t0=0. In an exponential universe this is a natural way to measure time

because there is no obvious big bang - singularity is at t→−∞. I could have been more clear on

this.

Second, in order for the signs to make sense as I wrote them you have to

have t2<t1 (otherwise D is negative).

To pick nits, I also have a completely unnecessary set of parentheses around t2.

The particle horizon at time t1 (about which I did not ask) would include the most distant events

from which a photon could have arrived by t1. To find this, fix t1 and adjust the emission

time t2 to maximize D. The result is interesting.

The event horizon at time t2 is the maximal distance from which a photon emitted at t2 will ever

reach us. To find this you fix t2 and adjust t1 (when in the future we will receive the photon) to

maximize D.


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Thanks, Dr.Plesser. Back to the drawing board on this one to see if I can get back on track. The

concept still eludes me, may be will become clear as I go through this set!


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Thanks for the clarifications Ronen, but I am still confused with this. So we are saying t goes

from t→−∞ to t0=0. So then t is always -ve, right?. (Well if t0 is today, we have -ve t in the

past, and +ve t in the future)

So the a(t) formula gives us scale factor a(t)<1, for -ve t (so times earlier than t0). And a(t)>1 for

+ve t, ie times later than t0 (ie in the future).

But if t is -ve, and we already have a '-' in the exponents in the 2nd formula, the exponents are

then +ve? (for times earlier than t0). But for -ve t (times earlier than t0) the scale factor is <1. And

the +ve exponent will give us scale factor >1.

And is t2 is earlier than t1? You mention t2<t1 but it depends if these are +ve or -ve to find

what is earlier.

I have already answered the question 'correctly', but I am still not sure I understand the answer.. I

basically guessed it, rather than find out what it is the proper way..


Comments


Ok, further to my post above, I 'think', with -ve t's, we still need to keep the '-' in the exponents in

the 2nd formula.

We have that the scale factor is a(t)=eH0t. Then e−H0t=1a(t), and everything seems to

work ok. I think...


Alexander Edward Drew · 2 weeks ago

I think the fact that t can go back to negative infinity is only needed in this problem in so far as to

let you know that we are taking t0=0 and that tem is some negative time prior to that that

could be any magnitude. You can then replace tem in the distance equation in terms

of z and t0=0 and solve for D in terms of z, c, and H0.


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These are the outskirts of calculus and we take, thanks to Mr. Ronen and this excellent question,

a small idea of how useful this section of mathematics can be.

But I know some basic n rusty algebra and it helped me through this course. I would produce a

train equation series, as my friend Gary from Canada would call it, but I guess Prof. Plesser's

comments above are more than detailed.


Comments


Thanks, Σταύρος A. Αθανασιάδης for all your contributions to this forum. You & Ioanna have

made the forum a great place for after-class learning... I am yet to completely grasp the

intricacies of week 8, and find myself struggling a bit. Hopefully the autograder will approve of my

efforts when I get to it. Now for Q8 & Q9 & I should be done:)




I am yet, if I ever, to completely grasp W8 myself, too...


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Carlos Sarno · 2 weeks ago

I have an answer which was accepted but I can't see why. The eqn seems to be appropriate for

distance now (i.e as given in the question), not distance when emitted. I.e. I omitted 1/(1+z)

from my original (unaccepted) answer and then it was accepted. This is the only question in the

whole course where I've been puzzled!!!


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Brian Ritchie (Student) · 1 week ago

Well I'm very gradually becoming aware of what this question entails but I'm still far from solving

it. This is what i ended up with, which is wrong. c/H_0/(1+z) - c/H_0 I must admit I'm having

difficulty getting my head around what's going on here lol


Comments


Start off by t0=0, as Prof. Plesser states above. Then it's −H0t0=0, thus e−H0t0=e0=1.

Then, treat t2 as tand t1 as t0.




(delete & repost)

Aaargh. I can't make heads nor tails yet of this question.

a(t)=eH0(t−t0)Since t0=0 that means that 11+z=eH0(t) and 1+z=e−H0(t)D=cH−10(e−H0(t2)−e−H0t1)Next, if we are to treat t2 as t and t1 as t0 then I should be able to simplify the above

expression for D


https://www.coursera.org/user/i/6ad868a53d4ca73641a90a58940a1951


Well, slap self on head again. I forgot that you have to explicitly tell the autograder where the

multiplication sign appears. Remember those pesky *s. This question is doable.

PS My son is now up. Good, solid three hour nap and now hopefully we'll have him down by 8

(three hours is pretty long for him nowadays so he might stretch things out to 20.30 ;-).


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Σταύρος - thanks for that - I'll give it a go after another cup of coffee!


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Hmmm - that gives me the same equation. Looks like there is something wrong with the way I'm

introducing z? (exp)-H_0t = 1/1+z ???


Comments

Carlos Sarno · 1 week ago

exp(-H0*t) = 1+z not 1/(1+z)


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Looks that that coffee was not so strong. It is exp(H_0*t) = 1/(1+z) or the same in

Latex eH0∗t=11+z. Be careful with the math formality. Do not forget those * and where to

put brackets.

I ended up on the conclusion that H_0*t_0 should be 0, namely t_0 should be 0, otherwise I

couldn't have a formula with the symbols/quantities required. So, I said, let t_0 be zero, then

worked some algebra, then I found an expression which looked to simple to be true and correct. I

asked the grader and it replied it was OK.


Comments



Σταυρο, I think your fraction should be inverted? We have:



a(t)=11+zAnd a(t)=eH0t Then e−H0t=1a(t)So e−H0t=1+zVote this post up 7 Vote this post down


Right on, Ioanna, I just corrected it.


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ie 1/1+z is 11+z whereas 1/(1+z) is 11+z, namely, much different expressions.


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Hey Stavros,

My expression in the parentheses is 1/(1+z)-1, yet the answer is wrong.

Any Ideas?

Thanks


Comments


Yep.

a(t)a(t0)=11+z→a(t)1=a(t)=11+z→a(t)=eH0∗(t−t0)=eH0∗t∗e−H0∗t0=eH0∗t∗1=11+z→eH0∗t=11+z→e−H0∗t=1+zVote this post up 5 Vote this post down

Carlos Sarno · 1 week ago

Also obtained simply from Ronen's a(t) eqn directly by setting t0=0 and inverting.


Colin Derek Dando · 1 week ago

Carlos, someone else from Colchester, UK





https://www.coursera.org/user/i/9643c445dfa60a3e249b4eb81b274415



Inside my brackets I get (1+z)-(1+z)... how do i differentiate between t1 and t2???


Carlos Sarno · 1 day ago

Hi Colin. Only just seen your post. Moved on to other courses once I finished this one. I'm

actually in Colne Engaine near Halstead.


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Yes, I understand that (I missed out the brackets in my post above). When I review it it looks OK

but the grader is not accepting my answer.


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Paul Robinson · 1 week ago

I spent hours on this yesterday, looking back at my notes I came very very close to correct.

After a double espresso and reviewing comments from Σταύρος A. Αθανασιάδης I stumbled

upon the right answer - and couldn't believe how simple it was until the autograder gave me a

wink back ;-)

Thank you again Σταύρος A. Αθανασιάδης


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Ah, yes - thanks Stavros - one extra coffee did it :) - the minus sign oh dear. lol


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With all that coffee that we all drank during this course, I hope our stomachs will get better after it

finishes. If this course ever assumes a sponsor, it would probably be a coffee corporation.


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https://www.coursera.org/user/i/5e711b929a7aaf167fbfd62fd1537306


Stoica Dorian - Bogdan · 1 week ago

I still don't get it...why don't you use Clip 2 Slide 4????


Comments


That's probably because we like drinking coffee, I guess...:-))


Nancy Nelson · 1 week ago

And getting only 6 hrs of sleep over a span of 2 days does not help either...but that is probably

why I had to drink so much coffee...


Eugene · 5 days ago

I also like drinking coffee, but it doesn't help now!


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Kristina Mois · 1 week ago

Enough coffee, I just poured me some wine to celebrate my completion of the final problem set!

Thank you for your help, Ionna, Σταύρος and many others who have been contributing here. You

saved me a lot of headaches on this problem set. You banged your heads against the wall with

autograder issues and conceptual issues, so that others like me could waltz right through after

you.


Comments


I tried to bang my head less with the autograder cuz my time available for this kind of adventure

is limited :-) Nonetheless, this kind of struggle with the autograder shows that there are students

here on this course who are quite energetic regarding the homework that are given and do not

take it in a passive attitude. I think this is great and all of us owe them a lot. I am sure this attitude

helped Prof. Plesser refine and better the material given for the rest of the students.


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https://www.coursera.org/user/i/f6c4e7518a7e5ae47b6b94d7bd7968b8



Yes, same here - thanks to you all! I finished the final set yesterday. :) You were right Stavros, it

just wouldn't have been right to get so close and not complete the course - would have bugged

me no end. Now I need to re-watch the videos at leisure - lots of interesting stuff which I never

really absorbed. lol


Comments

Sue Terwilliger · 1 week ago

I was going to settle for 90%, but you guys talked me into plugging away! I finished all but 2

questions yesterday, and I have 2 weeks to get them!



Sue, go for 100%. Plenty of friendly assistance around, and believe me you will feel so great

when you get there :-)


Sue Terwilliger · 1 week ago

Got it, and it didn't take the whole 2 weeks! Now I have to concentrate on Calculus!


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I'm going to take the liberty of combining Ronen's post with the question so I can see everything

in the original, glorious Latex (I can't wrap my head around t1, t2 and t0).

Ronen's clarification:

A few clarifying comments on this one: First, in going from the first expression to the second I

have tacitly taken t0=0. In an exponential universe this is a natural way to measure time

because there is no obvious big bang - singularity is at t→−∞. I could have been more clear on

this.

Second, in order for the signs to make sense as I wrote them you have to

have t2<t1 (otherwise D is negative).

To pick nits, I also have a completely unnecessary set of parentheses around t2.

The particle horizon at time t1 (about which I did not ask) would include the most distant events

from which a photon could have arrived by t1. To find this, fix t1 and adjust the emission

time t2 to maximize D. The result is interesting.

https://www.coursera.org/user/i/3d8838537082a9019bd507fcc73464cf


https://www.coursera.org/user/i/3d8838537082a9019bd507fcc73464cf

The event horizon at time t2 is the maximal distance from which a photon emitted at t2 will ever

reach us. To find this you fix t2 and adjust t1 (when in the future we will receive the photon) to

maximize D.

The question:

In this set of linked problems we will take a look at exponential expansion - which we think

describes both the distant future of our universe and an important phase in its very early past.

In exponential expansion with a(t)=eH0(t−t0) the coordinate distance - the distance now -

between the position of a photon at time t1 and its position at

time t2 D=cH−10(e−H0(t2)−e−H0t1).Consider a photon observed at t0. Find the distance at which it was emitted and write this in

terms of the observed redshift z. Your answer should be an expression containing z, H0, and c.


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Someone, please clarify the expresion for D: Does H_0(t_2) stand for H_0*t_2 or H_0 as a

function of t_2? Similar question for H_0t_1: Does it stand for H_0*t_1? (The two similar

subexpressions are written differently.)


Comments

Philip Tay (Student) · 1 week ago

Sohan, that pair of brackets around t_2 was unnecessary - see Ronan's thread above. I thought

that was naughty of him but ok he admitted it.


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All's well that ends well but that was confusing with so many ts.


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Tjerk Gauderis · 1 week ago

Ronen's extra question about the particle Horizon has really a mind blowing result... "hard to

wrap my mind around it"


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I wonder if someone can help me with this. I've come back to it time and time again and I'm still

not getting it... conceptually or mathematically

If I treat t2 as t and t1 as t0 I end up with e−H0t1=1 and e−H0t2=e−1/(1+z)I've read the post above and I think it's just that I don't understand what I'm aiming at here. Week

8 has been really tough!


Comments


Must be something about displaying silly mistakes in pubic that make them glaringly obvious!!

Got it.


ronen plesser INSTRUCTOR · 3 days ago

Tell me about it...:)


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Question 7 Assuming the value we measure for H0 find the distance to a source with z=5.34 (you

can compare your answer to the results you found for a dust-filled universe in Part A). Express

your answer in light-years and round to two significant figures.


Dimitris Nikolaidis · 1 week ago

Should the result be the same as the question 3 in part A ?


Comments


It is true that both results are on the same order of magnitude but not exactly

identical..Remember, that was a dust-dominated flat universe in Part A.


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If you are having unit problems with this question, I can recommend going to the Q9 thread and

picking up and using Stoica's H0conversion.


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Vlad Alexandru Niculescu · 1 week ago

huge lapsus here. Where is the formula for the distance in this case? i can't seem to locate it

anywhere..


https://www.coursera.org/user/i/e1676bdd87e13dc313a78b55676b977c


Comments


It's just the answer to the previous QuB6. You found D=D(c,z,H0), remember? You are

given z, you know c and H0, find Distance D expressed in lightyears. Careful on the unit

conversions.


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This is the only question that I have not managed to find the correct answer. My answer on QB6

accepted by the A.G and the unit convertions are correct (I suppose). I had follow Stoica's

convertions at QB9. Any ideas, what I can doing wrong? Thanks for your advice


Comments


Is your c/H0 of order e10 (in light years)?



Dimitri,

If you cannot manage this one but you dealt with 8B3 with success, then something may be

wrong with your units or your powers or both. Check again.



Dimitri, Ioanna,

Same thing here: all my other answers are accepted by the AG. Since H0 had to be used before,

I suppose the conversion is fine. My c/Ho however is not in the order of e10 (in Ly) as Ioanna

mentions? Since'"just" plugging in the data to my accepted answer of Q6 doesn't work. Do I

oversimplify this thinking that just the equation of q6 is needed here..? I wonder what am I not

grasping here? Any thoughts would be highly appreciated



No Ioanna, my results (all) are of order e11 ly







I reached a result of the order of x.ye10 with x + y = 5 but the result by the AG was wrong again


+ Add New Comment


Since you have qu6 correct, it means you are using the right formula. And it only involves, z, c,

H0. So check your calculations:

Giving rounded numbers as a guide,

H0 = 2.4e-18 s^-1

1/H0= 4.3 e17 s

1/H0= 1.38 e10 yrs

(remember this is the estimated age of our universe, and you know it is ~ 13.8 billion years old.

So your 1/H0 must be around 1.38 e10 yrs, else something is wrong).

Once you have it in years, you can use the approach that speed of light c, is 1 ly per year, so

c/H0 should be around 13.8 billion light years. And you then just increase increase it by z. (you

would need to use more accurate numbers than these that I mentioned as guide, else the the

answer may be out of range of autograder).


Comments


Ioanna, Σταύρο thanks a lot. Finally I managed this question, It was so simple, repeated wrong

convertions that I can't see. Again thank you very much for your assistance.


Konstantinos Siettos · 1 week ago

How do we know that H_0=2.4e-18 s^-1?



Konstantinos, the question states "Assuming the value we measure for H0" - this is a "real world"

measurement established by measuring redshift in distant galaxies. The Wikipedia article on



https://www.coursera.org/user/i/0271af33ba350013b721af372cade715

Hubble's Law has this to say "As of 20th Dec 2012 the Hubble constant, as measured by NASA's

Wilkinson Microwave Anisotropy Probe (WMAP) and reported in arxiv

(http://arxiv.org/pdf/1212.5225.pdf), is 69.32 ± 0.80 (km/s)/Mpc".

To get from (km/s)/Mpc to s^-1, we take 69.32 and divide by the number of km in a Mpc,

3.09e19, which gives us 2.24e-18 s^-1 (slightly different from Ioanna's value, I think the Hubble

Constant is given in the video lectures as something slightly different from the 69.32 found at

Wikipedia).


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Ioanna,

Thought I found my error. (as we were warned in advance by Σταύρος). Your last comment

confirmed it.. And so does the AG. Thanks both of you for dragging me over the finish..

Best regards


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Question 8 For a photon emitted at time t find the distance today of the most distant point it will

ever reach. This is the event horizon at time t. Your answer should be an expression in terms of

H0, c, and t.



Shouldn't the answer also depend on t0?


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In Clip 5, Slide 4, the horizon is given as proportional to a. If this formula is correct, then there

may be dependence of t_0. Nevertheless, the grader refused my solution with t_0 and mentioned

that it is a symbol not part of the problem.


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Nicholas · 2 weeks ago

I believe at the time (t_0) at which the photon will reach it's most distant point is consistent with

t_0 approaching infinity.


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According to Wiki, the event horizon is defined as: the event horizon is the largest comoving

distance from which light emitted now can ever reach the observer in the future. So it involves

integrating from t0 to infinity. This should give cH0, which of course is not accepted by the grader.


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A post was deleted.


In the exponential universe, H is a constant so H0 would be the same no matter when you

measured it. As a result t0 will not appear in the answer to this one. The answer contains a

factor of the form exp(x) but also a prefactor.


Comments


Thanks Ronen. It now works. I was using t (not t0), but I had a problem with my exponential term

which I managed to fix (by guessing, mainly. If it not one then it is the other, sort of thinking).



Prof. Ronen, I do not want to just make trial and error. What is the event horizon? Isn't

it c∫∞tdta?



I am sort of shocked but I just created an expression based on what we were given as simply as

possible and it worked and then multiplied out an awful math thing for 9 and it worked as well.

Now back to the question 5 but if Ioanna can't get it how can us mortals get it to work

Oh and Ioanna thnaks - I don't usually get help from others but you were terrific and clearly have

a good thought process


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If H0t0 appoaches infinity, then e−H0t0 approaches 0. So, t0 is not included in the requested

expression. Thank you guys for your insight.


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A post was deleted.


Σταυρο, I think in the thread for qu6, it was said t0=0 (so it does not approach infinity)


Ok, I think we are still looking at the formula given in qu6:

D=cH−10(e−H0t2−e−H0t1)t2 is the time of emission, and t1 is the later time. The formula gives us the distance D, that light

emitted at t2, will reach at time t1 (the distance from where it was emitted).

So for this question we have that it is emitted at t (so t2=t). At some later time t1, it will be at

distance D. What is the maximum distance it will reach? Well the more time passes the further

away it goes, so max D is when t1→+∞ So we use the above formula with t2=t (time of

emission) and t1=+∞ (so t0 is not involved here at all)

(That is not how I did it originally, I had no clue what to do and I somehow managed to solve qu9

first but constructing my own formula, which I then solved for D, and used it to answer qu8.

Going completely backwards in other words ...)


Comments



Well... thank you Ioanna! I only did this:

t1→+∞→−t1→−∞→−H0t1→−∞→e−H0t1→e−∞=0→e−H0t1=0→−e−H0t1=0→e−H0t−e−H0t1=e−H0t→c(e−H0t−e−H0t1)=ce−H0t→→cH0(e−H0t−e−H0t1)=cH0e−H0t,∀t<t1→+∞Both parts of the last equation have length dimensions.

So, give me that time t of the photon emission and I will find you the distance today of the most

distant point it will ever reach, namely the event horizon at time.. Sounds poetic, doesn't it?



Thank you, Ioanna -- good and clear explanation!



Σταύρος A. Αθανασιάδης, I get the same answer you found above, but the grader doesn't think

it's poetic enough! May be, it's the way I've entered the answer in the grader that's the issue,

especially the exponent part...or is it? Let me bang my head against the wall some more, it's

bloody but unbowed....




I almost give away the answer, so be careful on what you re entering in your submission. Careful

with * and exp(x) stuff..


Judit · 2 days ago

Exp...? :S I tried e^( ) and it didn't work, I was reprimanded for including e at all :*( How do I enter

"e to something" correctly?



Justin Johnsen STAFF · 2 days ago

Sorry, wrong e. For the exponential function, use exp(x), where x is whatever you are putting as

the power. So, exp(3) = e^3


Judit · 2 days ago

Worked like a charm. Thanks!


Christoffel Johannes Lombard · 16 hours ago

So how do I enter infinity?


Christoffel Johannes Lombard · 10 hours ago

Please ignore my previous question. I have got the equation right, thanks for your help. I could

not have done it without your posts.


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Hello, in what form will the grader accept H_sub_0 to be written? Thanks! Julie


Comments


Paul Robinson · 4 days ago

H underscore 0


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Bernard Colloff (Student) · 11 hours ago

Hello, Still having difficulty with this question. I would appreciate another review. Thanks for your

help.


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serge · 7 hours ago

We have the working formula given in the Question 6 (the question itself, not the answer is

meant) - the formula for distance D that the light emitted at time t1 reaches at the time t2. Note

that this is NOT just c(t2−t1). Why it is not so? Because the universe is expanding, and all

possible means for measuring distance do so. You can check however, that for small times

(small values H0t) this expression acquires more common form (use the calculus approximate

formula exp(x)=1+x+… valid for x≪1 - surprisingly, it will turn to c(t2−t1). So

everything beyond is a cosmological correction depending upon the cosmological model

accepted. The part A of this week related to the "dust-dominated" Universe with other

dependence of distances traveled by light (but again which would reduce to common

formula D=ct at small times). Now we have the model of exponentially expanding Universe

which is supposedly the case of its later evolution from now on. This strange formula for distance

is related to the scale dependence on time, a(t) which is also related to the cosmological model

and can be found by solving Friedman equation by neglecting different terms in it. For dust

dominated Universe we had a(t)=(afactor)⋅t2/3, for the time t0 "now" it is 1. For

exponentially expanding Universe it is given a(t)=exp(H0t); notice that at t=0 ("now") it is

again 1; at times "before" with t<0 it is a number between 0 and 1; at times "after" with t>0 it

is a number above 1. And third variant is given for the radiation era (W8, clip4, slide 4),

where a(t)∼(√t). But we are interested in exponential era now. The formula for the distance

given above and in Q6 is just the integral c∫t2t1dta(t). If a were constant, this is just the

statement, that the light for a small interval time t in some remote past travels as we perceive it

from now not the distance ct, but this distance scaled by factor a: ct/a. For changing a(t) we

need a calculus with this integral to arrive to the formula for D given here as ready-made.

Anyhow, we have this formula for distance passed by light between times t1 and t2 - ALREADY

corrected for cosmological expansion and as viewed from today. That is this is the distance

which we and now measure for this light. Now let us think, what is the maximal distance the light

ever travels if it is emitted at some time moment t. This is the time horizon. From outside this

distance no event has chance to reach us and to be detected. So just take the formula for

distance measured in given cosmology and consider what will be the maximal distance of light


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path for an event (flash of light, or, say, star explosion) occurred at time t. We can think of a

partial case. Consider a light flash "now" at some distance from us. Will it be possible to see it at

some time later? If it is close enough, the light from it will reach us after some time. If it is beyond

that critical distance (time horizon) - the light will never reach us because the Universe at that

point will be expanding faster than the light travels. This critical distance can be given from the

formula if we set one time to "now", and other time - to very big value. And the general case is

considered similarly, but instead of "now" we use a given time t.


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Question 9 At what time t will this become of the order of the distance to the Andromeda galaxy

(2.5 Mly)? Express your answer in years (counting from today) and round to two significant

figures.


Jens-Peter Imohr · 2 weeks ago

for this one I could need some math-help. How to solve an equation

like: exp(−A∗t)=B for t ?


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Shlomo Eshet · 2 weeks ago

LN(exp(-A*t)=LN(B) <=> -A*t=LN(B) <=> t=LN(B)/(-A)


Comments

Jens-Peter Imohr · 2 weeks ago

Thanks a lot. I think now I can solve the last Question



I have problem to understand LN? Please help explain.


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ln(x) is the natural logarithm of x and is the inverse function of ex or exp(x) and vise versa.

So, ln(exp(x))=ln(ex)=x and eln(x)=exp(ln(x))=x.

So, if a is any known number and x is the unknown:

ex=a→ln(ex)=ln(a)→x=ln(a) and you find x whatever that maybe, e.g. x = t, etc...


Joanna Loesch (Student) · 1 week ago

Thank you so much Σταύρος & Ioanna :)

With your invaluable help I finally managed to wrestle down set B. I swear I feel 15cm taller now

and I have the urge to go around, grab random people by lapels and shoult "I did it!" :D


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I think I'm messing up the units here a bit... I've converted the Distance (2.5 mill ly) from ly to

parsec. H_0 is in kms/sec per megaparsec, which I've converted into kms/sec per parsec as well.

c is in kms/sec.

I land up with an expression like what Jens-Peter mentions exp(−A∗t)=B which then I'm trying to

solve for t along the lines Shlomo has indicated.... I land up with a number (x.ye4, x+y = 10) I'm

unable to figure the units of, and, of course, the grader doesn't like it either.

What am I missing? I'm not too comfortable with Ln, so may be I'm making a mistake there. Or

else in unit conversions. Help needed. Thanks!


Comments

Lauren Suzanne Gonzalez · 1 week ago

I had that same thing (except my exponent was -4) with x+y=10, and it was dimensionless...due

to the fact that H_0 can be converted to units related to 1/time, and it is multiplied by time.



Satish, remember when you have exp(something), the something is unitless (so just a number).

In our case it would be H0t, so (1/s)*s=1, no units, as it should be.


https://www.coursera.org/user/i/5ad28f30c6d8842a7a8fc7dee70f946b

When you take ln of something, the something is also unitless. Because lnx=y means 'to what

exponent (x) do we need to raise e to get y'? so x is the exponent so unitless, so is its ln, and so

is y.

And ln[exp(something)]=something

When dealing with ln and exp it is useful to remember we are dealing with ratios (and ratios are

unitless, as long as they are in the same units so km with km, or pc with pc, but not say pc with

km, or sec with years).

we are basically saying that a ratio of distances (so unitless, as long as we have converted both

numerator and denominator to the same unit of length) is equal to exp(something), also unitless,

and where the something is a ratio of times.



Thanks, Ioanna. You've been a great one to have around:)

I sorted this one out a few hours back. It was, as I thought, a problem with unit conversions -

sometimes when in a hurry, one tends to do things mechanically.... now that I'm done with HW8

A&B and the course, want to go over all the math to see if I can derive a more deeper meaning to

some of the HW problems (some, I admit I completed with only a shaky understanding of the

underlying physics, at best). I suppose, the purpose of the HW (and indeed, of my taking up this

course) is for us to gain a holistic understanding of the subject. It's in this area that these forums,

with people like you in it, have been immensely helpful.

Math might be great tool to understand & explore the physical sciences, but there are times I

feel, too much math boggles the mind, makes the concepts far too abstract. Week 8 was one

such, where, frankly, the conceptual understanding is trailing the HW completion by some

distance! Hence, a little bit of 'mulling over' is in order, want to spend some time untangling the

equations in my mind & understanding the underlying principles:)

't was great to have you guys in the "classroom", will sure bump into some of you in the other

Coursera courses:)



Satish, agreeing you with you I would stress that I kinda liked more the "find the expression"

questions rather than "you have km, now find it in ly, then find it in mm and then find it in Mpc..

Oh! Not in Mpc but in kpc!" questions. Sometimes I found myself struggling with unit conversions

(God Save Excel) and in the meantime I lost all the underlying physics concepts. Found a

number in kpc and I almost completely lost its meaning. I know I exaggerate a little and I am not

saying that "gimme a number" questions were a bad thing but...


...on the other hand, "find the expression"questions focused me more on the symbols and the

quantities they represented and what was their relation. What's that ρ? What's that R there?

Hmm.. radius of a sphere! Well, may I know it's mass? May I know its volume? What t stands for

in relation to t prime? I could present numerous similar examples. Those questions really

stressed on the concepts and kept me really focused. And, most of all: no rounding errors, no

marginal differences due to different procedures of calculation. Either you are ok and the grader

says you're ok or you are not. One and only one expression could be true.



I read the feedback on keeping track of units correctly. In case of HW8B/9, t has an expression

that involves four subexpresssions ln(H_0), ln(2.5Mly), Ln(c) and H_0, all with different units. I

had no problem resolving units without Ln function (for example, in case of HW8A/2 and

HW8A/3). My first choice is to convert each of them to years as the answer is required in years.

(Even ln(H_0) and H_0, logically, have different units. H_0 has units km/sec/MPC and ln(H_0)

has units ln(km/sec/MPC). Isn't it? I tried different ways but I had no luck so far. Sometimes, I get

negative value of t. I want to understand the correct logic instead of depending on autograder. (I

have become so much dependent on autograder, I don't know if I will ever be sure about the

accuracy of a solution without the autograder.) I am 100% sure my expression for t is correct.


+ Add New Comment


I got a number ~e6, which seems reasonable, given the distance of 2.5MLY, but my answer was

not accepted. Can anyone confirm a similar answer? There are so many conversions in this

problem, and so many places to make mistakes.


Comments


Oops, pc to mpc and a mis-typed number changed my order of magnitude from 6 to 8. Still

wrong. Ideas?


+ Add New Comment


Lauren, your order of magnitude is still too small.




My approach was to convert all the inputs into km for distance and seconds for time. Then at the

end convert my seconds into years.


Comments


Ah, that did it! Thanks so much!!



Glad to give a little back, after getting so much help along the way from fellow students :-)


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Dimensional analysis:

c∗H−10∗e(−H0∗t)=DSince H0 has units of s−1, H−10 has units of skms∗s∗e(s−1∗s)=Dkm=DVote this post up 0 Vote this post down

Comments


Thanks, Paul. Had tied myself up in knots with all the unit conversions. Your dimensional

analysis brought some sanity into it:)



But H_0 has units km/s/MPC. Isn't it?



H_0 is sometimes expressed in km/s per Mpc as an easy to use number. This is distance/time *

1/distance = 1/time







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I expressed H_0 in seconds, and then in

years. . Ugly numbers, i

know, but because we have the distance in Mly, we can set c=1 lightyear / year, and the result

will be in years.


Comments


Excellent - very useful here and in Q7



Stoica: That idea was great....thank you for it.Helped a lot to solve the equation correctly.



I have problem to understand "ln" and to solve exp (-A*t) = B for t? Please help.



Dudes: Type this into search box: "71 km/s = ? Mpc / yr"



Stocia,

Thank you. It works!


MALINA ELEONORA COCIOCEANU · 3 days ago

Thank you, Bogdan.Very useful your post!





SANTIAGO ARÉVALO GONZÁLEZ · 1 day ago

No me da! La respuesta me da en orden de E12 o E13 pero no la acepta el autograder


Judit · 1 day ago

Thanks Stoica, your insight regarding c helped a lot.


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A post was deleted.

Yandi Jaya · 1 week ago

Holy stars and garters, I'm done. I'M DONE!


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1. Anon, is equivalent with . So .

2. The general rule is: if then .

3. Example 1: if then .

4. Example 2: if then .

5. This should be useful: http://en.wikipedia.org/wiki/Natural_logarithm


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Dudes!! and Dudettes!!!

Veni Vedi Fini !!!!


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Nirmala Sonathi (Student) · 1 week ago

https://www.coursera.org/user/i/74de8cfb4ad29befd50ef534b1014718



oops! This one is taking me on a joy-ride.I am getting xye9, where x+y=7. 2.5Mly=2.5* 10^6*

9.460536207e12 km. I use c=2.99792458e5 km/s, value of H_0=2.30095e-18/s. I get my answer

in sec which i convert to years. Where am i doing wrong? This last one seems easy but

consuming all my time.Can someone help.


Comments


Stoica, thanks for your advices on the links, they are truly helpful.


derek young · 1 week ago

I too am using the same values ... c=2.9979e5 km/s H_0=2.30095e-18 /s D=2.365e19 km

equation from qu8 becomes (log((D*H_0)/c))/H_0 and answer is x.ye10 to 2sig figs ... (x+y=7) ...

which is wrong ...

Driving me nuts ... Where am I going wrong?


Samarth Agarwal · 1 week ago

i think it should be ln and not log. (here 'e' is 2.71 not our 10^x thing!!)

hope u get the right answer then :)


derek young · 1 week ago

aaaaaaaaargh. /facepalm.


Nirmala Sonathi (Student) · 1 week ago

Thanks Samarth.Actually i was also doing log. Chalo now i can move on to my next course.


Samarth Agarwal · 1 week ago

:)


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https://www.coursera.org/user/i/252334c21c4896d48312ecb9b05a642d


Derek, I have the same units as you but my quotient is inverted because of the -ve power

(exponent) from Q8.

Hence (ln(c/D*H_0))/H_0. What do you think?


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Pierre FERNANDEZ · 1 week ago

Bravo to Ronen and his staff, and to all those who communicated on the forums. Thanks to you, I

learnt, understood, and responded with success to all the HW (except two or three questions

where I mixed the units! Shame on me!).

The chapter 8 concerning the cosmology is excellent. I ignored everything on the scale factor

and other elements presented by Ronen. But the subject is difficult, and I did not really assimilate

all the concepts developed in this section.

For example, I am very surprised by my result in this question 8B9, (near 100 times the age of

the universe) and I really doubted before subjecting it to the autograder ! The future belongs to

photons !


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Richard F Riccelli · 3 days ago

I am sad to see this adventure end! Sincere thanks to Ronen and Justin for all their effort in

making this such an engaging and rewarding experience. Also thanks to Ioanna, Σταύρος (aka

Greek Guy), Anonymous (you know who you are) and all the various forum contributors who

have helped me and others find the way when needed. And let's not forget the auto-grader... :-) I

hope to see many of you again at Professor Ronen's next opus on Coursera. (hint!) Rick R.


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Anna Czarina V. Cabe · 1 day ago

Yay! Finaly! Thank you very much Stoica! And this ends IntroAstro Homeworks!


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Judit · 1 day ago

I have the following prob. I get a positive nr for ln(B). Shouldn't that be negative in order to get a

positive t when divided by -H0?



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Comments


Stoica's post above should save you from a lot of conversion factors, and it makes c=1, so the

resulting expression is not too messy.



Judit, yes, ln(B) should be negative.


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Wheeler Huneycutt · 1 day ago

Judit, You are correct , use -H0 but make sure H0 is in 1/year terms both for taking the ln on top

as well as bottom term. and put c in ly/year too.


Comments

Judit · 1 day ago

Thanks! Indeed, I messed up something around H0, I got some weird exponents there. Also, it

helps if one doesn't feed 25 Mly into Excel as 2,5*10^6.... Unit conversions are the root of all evil.

And commas. And periods. And late night calculations.

I was finally able to sort out my numbers and get the correct answer. Phew.


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Jim Olson · 1 day ago

Finished finally! So long and thanks for all the hints...


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GERARDO IVAN VELAZQUEZ ALVARADO (Student) · 23 hours ago

So, this is it, thanks to Professor Ronen Thanks to Ioanna, Σταύρος, Wheeler, and all of the crue,

see you again when it happens, so be it!




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week 8 solutions

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