Calculating uncertainty when using the distance modulus equation

[badhex]

Hello all,

We've covering errors and uncertainties as part of the Astronomy course I'm doing. I'm comfortable calculating total error on, for example, colour index, where it's simply subtracting one value from another, meaning I would use the formula:

But I'm completely lost with how to calculate error when distance modulus is involved.

We have been told that the uncertainty on log10(d) is given by:

 

So, I think, I need to plug in my distance (d) and (delta d), multiply that by 0.4343. Then I multiply that again by 5 because distance modulus is 5logd-5

Then I take that value and plug it into the first formula, for example as (delta x), where (delta y) would be the uncertainty on my apparent magnitude? 

So with an example:

(delta y) = 0.165

d = 3930

(delta d) = 50

 

0.4343 * (50 / 3930) = 0.005525445293

0.005525445293 * 5 = 0.02762722646

SQRT((0.02762722646)^2)+(0.165^2)) 

To make matters worse, excel is giving me two different outcomes depending on whether I run the whole calculation as one or broken up!

When I run it all as one calculation (I've checked the brackets, they are all in the right place) I get 0.1672969326 which seems to be likely to be the correct answer as the error on  distance is very small, so should not modify the error on magnitude so much.

When I break it up, I get 0.02817435337 which I'm sure cannot be correct, it's smaller than the original magnitude error.

 

Am I as dense as a neutron star? Help!

[George Jones]

What are you given the error in? Apparent magnitude? Absolute magnitude? Distance modulus? Something else?

What are you trying to calculate the error in? Apparent magnitude? Absolute magnitude? Distance modulus? Something else?

 

9 hours ago, badhex said:

When I break it up, I get 0.02817435337 which I'm sure cannot be correct, it's smaller than the original magnitude error.

In this case, did you forget to take the final square root?

[badhex]

16 minutes ago, George Jones said:

What are you given the error in? Apparent magnitude? Absolute magnitude? Distance modulus? Something else?

What are you trying to calculate the error in? Apparent magnitude? Absolute magnitude? Distance modulus? Something else? 

Apologies, I know the error in apparent magnitude and distance in parsecs, and I need to calculate the error in absolute magnitude. Along the way, I think I need to calculate the error in distance modulus first which I think should be:

5 * (0.4343 * (50/3930)

Once that's done, I use the first equation and plug in the error in distance modulus as (delta x), and the error in apparent magnitude as (delta y). Is this correct? 

21 minutes ago, George Jones said:

In this case, did you forget to take the final square root?

Hmm, it seems like that might be the case, I'll check my calculations but I'm sure it was in there, or maybe I misplaced a bracket on the last step. 

 

[George Jones]

I have had a chance to look at this in more detail. Looks good to me; you have clearly laid out your steps.. I worked the example independently, and I get the same result. I got a symbolic expression for Delta M, and then plugged in all the numbers at the end.

Not sure if this is of any interest, but 0.4343 is 1/(ln10), and is involved in the conversion from log base 10 to nature logs. The error in a ln is easy to approximate using a derivative.

[badhex]

Thanks! Very helpful to have someone smarter than me confirm it! 

I'm struggling a bit with the maths on this course in places, I used to be very good at maths but it has all dribbled out of my ear at some point.