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 Author Topic: Blackbody radiation curves for different temperatures  (Read 67179 times) 0 Members and 3 Guests are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: February 25, 2007, 02:39:04 pm »

Planck's law describes the spectral radiance of electromagnetic radiation at all wavelengths from a black body at temperature T. As a function of frequency ν, Planck's law is written as
$I(\nu,T) = \frac{2 h\nu^3 }{c^2} \frac{1}{e^{h\nu/kT}-1}$ or

It can be converted to an expression for I'(λ,T) in wavelength units by substituting ν by c / λ and evaluating

$I'(\lambda,T) = I(\nu,T)\left|\frac{d\nu}{d\lambda}\right|$ or

$\frac{2 h\nu^3 }{c^2}=\frac{2 h(c\lambda)^3 }{c^2}=\frac{2hc}{\lambda^3}$
and from $\nu= c\lambda$, we have
$d\nu=-\frac{c}{\lambda^2}d\lambda$
so
$I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$ or
The above equation is energy per unit wavelength per unit solid angle.

This applets will show six black cureves of blackbody radiation curve betwen Tmin and Tmax.
Another curve in red is also shown (it's temperature can be adjusted with left slider bar)
Maximum wavelength shown can be adjusted with right slider bar.
You can use it for study the intensity for blackbody radiation.
If you want to study different temperature range, You can change Tmin and Tmax, to change the temperature range,too.
The wavelength unit in the simulation is Å (angstrom).
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is the spectral energy density function with units of energy per unit wavelength per unit volume.
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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #1 on: March 15, 2009, 08:04:49 pm »

another quality simulation from
a good applet too

just wondering if it is a good idea to allow temperatures from 300K minimum.

i got time to remix for you
 « Last Edit: March 15, 2009, 08:15:08 pm by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #2 on: March 16, 2009, 08:23:07 am » posted from:Taipei,T\'ai-pei,Taiwan

If you want to study different temperature range, You can change Tmin and Tmax, to change the temperature range,too.
Tmin an Tmax are text field, so you can enter your own range. But the intensity will change a lot.
So you might want to change to log scale to view all the range.
The yscale can be changed between 0-100 with slider at the right side.
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lookang
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 « Embed this message Reply #3 on: March 16, 2009, 08:59:12 pm »

question: where can i find the equations to verify the calculations used? like wikipedia or other physics sources.

h = PLANCK = 6.6252 E-34
h4 = 4.*h*1.e47   = 2.65008.e14    // what is this?
c =the speed of light = 299 792 458 m / s
k =Boltzman k, 1.38E-23 J/K
cst = h*c/(k*1.e-10)  = 1.43.e8  // // what is this constant?
fT = "h4/((Math.exp(cst/(r*T))-1)*(r*r*r))"

what i am trying to do.
change the lamda to nanometre nm instead of 1 Ångström   ≡   1×10-10   metre http://en.wikipedia.org/wiki/Metre

i am guessing i need to change h4 = 4.*h*1.e46  for the nanometer calculation, i am guessing your calculation are for Ångström metre.

am i correct?

some references i look at.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c4

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmod/uvcatas.gif

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmod/bb7b.gif

Thanks!
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Fu-Kwun Hwang
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 « Embed this message Reply #4 on: March 16, 2009, 10:25:28 pm » posted from:Taipei,T\'ai-pei,Taiwan

cst= hc/k; add 1.e-10 because I use Ångström instead of m as unit for length.
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lookang
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 « Embed this message Reply #5 on: March 16, 2009, 11:16:04 pm »

thanks!
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lookang
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 « Embed this message Reply #6 on: March 17, 2009, 08:56:12 am »

another question

is it possible to make the x axis number always 100, 200, 300, 400, nm
instead of currently auto calculation now 0.1 , 0.2 , 0.3 , 0.4 X10^3 nm

what is desired

what is current

i tried changing the X format to 0000 but doesn't seem to work the way i hoped.
 Picture 2.png (6.54 KB, 1476x92 - viewed 12661 times.)  Picture 3.png (5.48 KB, 1469x90 - viewed 14661 times.)  Picture 4.png (53 KB, 753x615 - viewed 12640 times.) « Last Edit: March 17, 2009, 09:01:49 am by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #7 on: March 17, 2009, 09:00:25 am » posted from:Taipei,T\'ai-pei,Taiwan

I guest you were using auto-scale in the drawingPanel.
Set up your xmin,xmax properly. You should be able to get what you want.
Try it by yourself first. You will get to know it better.
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lookang
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 « Embed this message Reply #8 on: March 17, 2009, 09:21:43 am »

but is it possible to display 4000 instead of 4.0x10^3 ?

look at my attached pictures

it is already not autoscale X,

rmin and rmax are already setup
rmin = 50
rmax = 3000

ok i try again
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Fu-Kwun Hwang
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 « Embed this message Reply #9 on: March 17, 2009, 10:02:28 am » posted from:Taipei,T'ai-pei,Taiwan

There are X Format/Y Format properties, but I can not find any document at paco's web site.
http://www.um.es/fem/EjsWiki/index.php/Main/ElementsPlottingPanel

I have never used that before.
If I really need it, I will use drawingPanel  instead and I can draw those grid lines/labels with build in GUI elements.
May be it can be done by setting proper value for X Format, but I do not know. Sorry!

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lookang
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 « Embed this message Reply #10 on: March 17, 2009, 11:42:39 am »

Not to worry.
Maybe it can't be done yet.
I will explore abit more and move on to other parts of XML .

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lookang
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 « Embed this message Reply #11 on: March 18, 2009, 01:20:09 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

yes it is confirm it currently cannot be done http://www.um.es/fem/EjsWiki/index.php/FeedbackEn/00026
quote:
Resolution
The axes do not allow formatting the ticks. The Format X property refers to the way the coordinates of the point appear (in a yellow box at the lower left corner of the panel) when you click on a point in the panel.
unquote:

BTW

I am puzzled by the equation used.

I look at the equation and i look at the equation in the XML fT = "h4/((Math.exp(cst/(r*T))-1)*(r*r*r))"

shouldn't fT = "2*h*c*c/((Math.exp(cst/(r*T))-1)*(r*r*r*r*r))" ?

strangely i tried to implement the new fT ="2*h*c*c/((Math.exp(cst/(r*T))-1)*(r*r*r*r*r))" but the graph is very low and flat and near zero in value.
 « Last Edit: March 18, 2009, 01:23:40 pm by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #12 on: March 18, 2009, 05:53:53 pm » posted from:Taipei,T'ai-pei,Taiwan

The above equation is a normalized function, i.e., integration will give you 1.

So you can multiple it by any number.
If you change the y-axis to log-scale, you will know how big/small it is.
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lookang
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 « Embed this message Reply #13 on: March 18, 2009, 07:23:29 pm » posted from:Singapore,,Singapore

The above equation is a normalized function, i.e., integration will give you 1.
So you can multiple it by any number.
ok i understand normalized function http://mathworld.wolfram.com/NormalDistribution.html
ok i also understand the area under this normalized function is 1

but i am  puzzled by the missing r*r.

equation on http://en.wikipedia.org/wiki/Planck%27s_law

equation in your XML : fT = "h4/((Math.exp(cst/(r*T))-1)*(r*r*r))"

notice your XML : fT has only r*r*r but the equation on wikipedia has r*r*r*r*r

unless when normalized the r*r is absorbed? then it make sense to me now! I don't really know why now but at least i can accept the logic. (it that correct ?)......hahaha.

If you change the y-axis to log-scale, you will know how big/small it is.
log scale i understand. i try to experiment more

Maybe i lack the prior knowledge, sorry!
i don't remember encountering this equation for black body radiation, only vaguely remember
Stefan–Boltzmann law Main article: Stefan–Boltzmann law

This law states that amount of thermal radiations emitted per second per unit area of the surface of a black body is directly proportional to the fourth power of its absolute temperature. The total energy radiated per unit area per unit time j^{\star} (in watts per square meter) by a black body is related to its temperature T (in kelvins) and the Stefan–Boltzmann constant σ as follows:

where sigma=5.67 x 10-8Wm-2K-4
http://en.wikipedia.org/wiki/Black_body
 « Last Edit: March 18, 2009, 07:40:29 pm by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #14 on: March 18, 2009, 09:44:08 pm » posted from:Taipei,T\'ai-pei,Taiwan

Sorry! It should be proportion to λ-5.
The previous version was not correct. It is fixed now.
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lookang
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 « Embed this message Reply #15 on: April 12, 2009, 07:37:28 pm »

hi prof, based on your new equations

what is the unit of the intensity?

I am trying to verify both your equations with the

phet has the units of MW/m^2/µ.m

can explain what is your applet's?

Thanks!

attached is my remix copy of your original first version which i modify by looking at the 2nd version
 bodybodyqn01.PNG (26.16 KB, 816x498 - viewed 987 times.)  blackbody-spectrum.swf (36.4 KB, 640x480 - viewed 11935 times.) *** There are 1 more attached files. You need to login to acces it! Logged
Fu-Kwun Hwang
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 « Embed this message Reply #16 on: April 12, 2009, 10:17:02 pm » posted from:Taipei,T\'ai-pei,Taiwan

I modified the code again. The unit for the intensity is change to W/(m2-Å), where Å is angstrom 10-10m.
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lookang
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 « Embed this message Reply #17 on: April 13, 2009, 10:28:16 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

$I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$ or
The above equation is energy per unit wavelength per unit solid angle.
is spectral energy density function with units of energy per unit wavelength per unit volume.

Question1:
In school, i taught intensity = power / area.
so actually what the y-axis is displaying is intensity/ wavelength ?
is spectral energy density function with units of energy per unit wavelength per unit area volume.

Suggestion:
Should the applet called it energy density function with units of energy per unit wavelength per unit area

Question2:
eightpihc2 = 8*pi*h*c*c*1.e40
but i don't understand why has only 1 c, if you code is 2 c's

Is the image wrong?

Question3:
what is the url of the wikimedia.org? for the picture http://upload.wikimedia.org/math/3/6/c/36c7c70624f4ed44af2f004c87bf22c4.png
maybe i can go read a bit to understand more

Where did you find this equation
i look through
http://en.wikipedia.org/wiki/Black_body
but no
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lookang
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 « Embed this message Reply #18 on: April 13, 2009, 09:34:40 pm » posted from:Singapore,,Singapore

ok i found out Question3
http://en.wikipedia.org/wiki/Planck%27s_law
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Fu-Kwun Hwang
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 « Embed this message Reply #19 on: April 13, 2009, 10:38:36 pm » posted from:Taipei,T\'ai-pei,Taiwan

What is shown in the simulation is .
It is the emitted power per unit area of emitting surface, per unit solid angle, and per unit wavelength (in unit of Å :angstrom). If integral over all solid angle, it need to be multipled by 4*π.
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lookang
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 « Embed this message Reply #20 on: April 14, 2009, 10:15:55 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

oic....thanks for the clarification.

by the way, i being trying to figure out how to add text to the AnalyticCurve Family to show

dynamically on the plottingpanel.
TS[0] = 1000
TS[1] =
TS[2]=
.
.
TS[9]= 6000

Any tips which text do i used? A 2D text ? or a set of 2D text?

I am having difficulty nailing dynamically the position of the text to be at the peak of the function fT

i suspect conceptually, i need to differentiate fT w.r.t. math_failure (math_unknown_error): \lambda \ , equate to zero for the turning point for the highest position PosX, sub back to get the PosY. But i dunno how to do in programming.

the end result is something like this

I notice you are doing of things with other members, keep up the good work:)
 blackbody03.PNG (33.14 KB, 698x311 - viewed 11017 times.) « Last Edit: April 14, 2009, 10:30:01 am by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #21 on: April 14, 2009, 10:43:45 am » posted from:Taipei,T'ai-pei,Taiwan

If TS[n] are strings to be displayed at TX[n], TY[n] (where TX[n],TY[n] are coordinate for the string-- you need to calculate those in a loop.)

Assign:
# of Elements: n
Pos X: Tx
Pos Y: Ty
Text : TS
Offset: WEST
Pixel Size: true

You can calculate the differentiation directly by yourself or find out equation for the highest point from wiki page: Tmax=constant/lambda.
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lookang
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 « Embed this message Reply #22 on: April 14, 2009, 04:54:41 pm » posted from:Singapore,,Singapore

amazing ! i will study you codes with the new additionals of text.

It is really cool now! will report back my findings
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lookang
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 « Embed this message Reply #23 on: April 14, 2009, 10:08:50 pm » posted from:Singapore,,Singapore

amazing codes you have.

My attempt at making a remix for nanometre 10-9 m instead of angstrom 10-10 m.
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1037.0

basic code needed to make text appear are:

// code to appear Tmax top position
// taken from http://en.wikipedia.org/wiki/Planck%27s_law
// This function peaks for hc = 4.97λkT, a factor of 1.76 shorter in wavelength (higher in frequency) than the frequency peak. It is the more commonly used peak in Wien's displacement law.
// xm is λ where peak occurs
1.  xmred=TXred=h*c/(4.97*k*T)*1.e9; // cos lookang doing nm instead of Am

2. TMSGred="T="+((int)((T*10+0.5)/10.))+"K, peak at "+(int)((xmred*10+0.5)/10.) +" nm";

3.  TYred=twohc2/((Math.exp(cst/(xmred*T))-1)*(xmred*xmred*xmred*xmred*xmred));

4. Text properties are:
PosX is  TXred
PosY is TYred
Pixel Size is True
Text is %TMSGred%

Amazing learning from this codes, lucky you made the text, i still can't create these codes myself.....sigh

Thanks!
 backbodyradiation.PNG (31.41 KB, 702x308 - viewed 786 times.) « Last Edit: April 14, 2009, 10:26:48 pm by lookang » Logged
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There is a better way to do it; find it. ..."Thomas Edison(1847-1931, American inventor, 1093 patients)"