Understanding Stars and Black Body Radiation

Hello and welcome to today's lesson where we're going to look at stars as black bodies which falls into the astrophysics option of the AQA A-level physics specification. Now in today's lesson we're going to try to deduce the emissions of stars by considering stars as black bodies. So we're going to try and define Stefan's law and Vines displacement law. We're going to try and draw the general shape of black body curves and use Vines displacement law to estimate the temperature of stars and then finally use Stefan's law. to compare the power output and temperatures of stars, which falls into the following section of the AQA A-Level Physics Specification in the Astrophysics option, Classification by Temperature and Blackbody Radiation. We are used to the terms red hot and white hot when applied to sources of heat. We're also aware that hot objects emit radiation at infrared wavelengths. This behaviour of hot objects and the electromagnetic radiation they produced was investigated at the end of the 19th century and contributed to the development of quantum theory by Max Planck. Now this is important when observing stars as all the information we get about stars comes from the electromagnetic radiation they emit and we receive. So for example an analysis of the intensity of the light at different wavelengths allows astronomers to measure the black body temperature of a star. Using this and an estimation of the power output of a star allows its diameter to be determined. So to fully understand the radiation emitted by a star, we assume all stars are black bodies. Now this is a reasonable approximation to consider stars and some astronomical sources as these black body radiators. Now we can use black body radiation curves to make estimations about an object's temperature and other properties. So a black body is an object with a pure black surface that emits radiation strongly and in a well-defined way. It absorbs all light instant on it. Now they are black bodies because they do not reflect light. So a black body is a body that absorbs all electromagnetic radiation of all wavelengths and can emit all wavelengths of electromagnetic radiation. So an analysis of the intensity of the electromagnetic radiation emitted from a black body at different wavelengths produces the following result, which we call a black body radiation curve. Now there are three lines on this curve for three different temperatures. Now because they emit all wavelengths of electromagnetic radiation, black bodies have a continuous spectrum of electromagnetic radiation. leading to this continuous curve produced in the blackbody radiation curves. So all blackbodies have the same shaped curve, and the graph that shows the hottest object, P, has a peak at the shortest wavelength. So a blackbody radiation is radiation produced by heated objects, particularly from a blackbody. So a blackbody is an object that absorbs all radiation that falls on it. This means it will also radiate at all frequencies. that heat and from the heat energy produces and so physicists try to find the mathematics to explain how this worked so in this graph this links the intensity of the black body radiator output with the wavelength of the radiation produced i.e the color now like we mentioned before black body radiators are perfect emitters of radiation so they emit all radiation they contain they absorb all radiation they encounter they do not reflect they do not refract any radiation Now an example of a blackbody radiator is any star like the Sun. So a blackbody radiator does not have to be black, it just has to be very good at absorbing and emitting radiation. So because all stars emit lots of radiation, we can consider them blackbody radiators. Now the following graph shows how close the Sun's output is to a blackbody radiator, with grey being the idealised blackbody radiator. radiation curve from the mathematics and the orange being the actual observed output from the sun. Now it is noted that the peak wavelength of radiation is linked to the temperature of the black body radiator. Now the higher the temperature of the object the higher the peak wavelength of emission so the more shifted towards the blue end of the spectrum the object colour is. So the higher the temperature of the object the higher the peak wavelength of emission. the more shifted towards the shorter wavelengths of the spectrum the peak is. Now, so when the temperature of a blackbody increases, the intensity of radiation increases, but it increases more so at the smaller wavelengths and shifts the peak wavelength to the lower wavelength. Now it's very important to note that if the temperature of the object is increased the intensity of radiation it emits is greater at every wavelength but it's not an equal increase. The shorter the wavelength the greater the increase in intensity at that wavelength as shown on this particular graph indicated. So when the temperature of the object increases you've got to first note that the intensity of all wavelengths increases but the increase is greater for shorter wavelengths. So like mentioned before, increasing the temperature of a blackbody radiator shifts the peak wavelength of emission to smaller wavelengths, which is very important for our observations because the peak wavelength determines the colour of that object. So therefore the hotter the object, the bluer or the shorter the wavelength the object is, which is only true for the colour due to the emission of the radiation. So therefore the cooler the object, the redder or the longer the wavelength the object is. Now hotter stars will therefore produce more of their light at the blue or violet end of the spectrum and therefore will appear white or blue-white to observers. Now cooler stars will look red as they produce more of their light at longer wavelengths. So when analysing the light from stars in this way, it's assumed that the star is acting as a black body so therefore no light is absorbed. scattered by material between the star and the observer. And also it's important to note that a hotter star may not appear as bright as a cooler one as it may mostly emit radiation that's not in the visible part of the electromagnetic spectrum. Now if a cooler star emits more radiation in the visible spectrum than a hotter one it will appear brighter to optical telescopes which is why it's important to observe stars in both optical non-optical telescopes. So this is this explained why different objects glow at different temperatures when they are heated. So when something looks white hot it's bluer in color so as a higher temperature and shorter peak wavelength while something is redder in color it has a lower temperature but a longer peak wavelength. So this is why you can have blue hypergiant stars and red giant stars. So this leads to something called The stellar classification system which classifies stars based on their colour and therefore effective temperature which we'll look at in upcoming lessons and allows us to classify different stars according to their colour which we'll cover in the upcoming lessons. So the Sun is a G-type star which is a relatively cool star which is shown by its longish peak wavelength and appears predominantly orange. So the Sun has a peak emission of 5,800 Kelvin which gives us this trademark colour of yellow orange. So as we mentioned before a hotter blackbody will emit more radiation than a cooler one and its peak wavelength will be will be a shorter wavelength than the cooler star. So it also it has a higher total power output if we assume that the hot star and the cooler star have the same surface area. Now we can formalize this equation mathematically by saying that the peak wavelength or lambda max times by the absolute temperature in Kelvin is equal to 2.9 times 10 to the minus 3 meters Kelvin. Now please note the unit there is the amalgamation of the wavelength unit and the temperature unit. It is not in fact millikelvin. So this idea of 2.9 times 10 to the minus 3 meters Kelvin is called a constant value, which we call the vine constant. So this shows that as temperature increases, the peak wavelength of emission decreases. So lambda max indicates the peak wavelength of emission for a star, which determines the colour of the star. Whilst T is the surface temperature of that star, which is calculated in Kelvin. It's a common examination mistake to calculate the value in Celsius. You always convert it to Kelvin. Now this equation assumes that the star is acting as a black body. And we call this equation... the vine displacement law and it links the black body temperature, the surface temperature, with the peak wavelength of emission. Now remember the surface temperature of a star will also determine its colour so this equation details the colour of a star also and this equation is given to you in your examination book. So let's look at an example question. An astronomer has observed radiation from a star and recorded the peak wavelength to be in the visible light region at 520 nanometres. Assuming that the star behaves as a black body, estimate the absolute temperature of the star. Well, step one, you want to place your wavelength into meters, so you convert from nanometers into meters. You then use Vines'displacement law, where lambda max times by t is equal to 2.9 times 10 to the minus 3. Rearrange to get t to be the subject, pop your values in, and you get an answer of 5,600. 600 Kelvin. Now we can also link the surface temperature of a blackbody with the power of emission. Now remember, the power output of a star is the total energy, the radiation it emits every second. So this is a very important idea. The power output is proportional to the fourth power of the star's surface temperature, and is directly proportional to the surface area of the star. This means we can formalise this with a mathematical equation. where power is equal to the Stefan constant times by the surface area of the star in metres squared times by the surface temperature of the star in Kelvin to the power 4, where the Stefan's constant is 5.67 times 10 to the minus 8 watts per metre squared per Kelvin to the power of 4, which we call Stefan's law. Now this equation and the Stefan constant is given to you in your examination equation book. So this equation shows that the power output of a star is directly proportional to t to the power 4 and the surface area of the star. So this means a cooler star needs to be a lot bigger to have the same power output. Now remember when calculating the surface area of a star we assume the star to be a sphere. So therefore in this equation the surface area is equal to 4 pi squared, 4 pi r squared even. Now Another term for power is luminosity, so therefore Stefan's law will also determine the luminosity of a star. Now, it's a common in examination questions for you to determine the relative surface area of the star when compared to other stars with this equation. So here's an example question. A star has a surface area of 4.1 times 10 to the 13 meters squared and produces a blackbody spectrum with a peak wavelength of 115 nanometers. What is the power output of the star? Well, part one, you find the temperature of the star, so it's going to be lambda max times by T equals 2.9 times 10 to the minus 3. Rearrange that to find T, which will be in Kelvin. So you pop that in, you get 25217.39. Remember, we aren't going to go to our significant figures yet because we're halfway through the question. Then the second part, you will use Stefan's law, which is P equals Stefan's constant times by A T to the power of 4. So you pop your values through and you get power is equal or luminosity to 9.40 times 10 to the 23 watts. Now we can use the previous laws to compare two different stars, such as star A and star B. Now the laws tells us that if two stars have the same blackbody temperature, so therefore they are the same spectral class, the star with the brighter absolute magnitude must have the larger diameter. Now it's also quite common for Stefan's law to be used qualitatively. when comparing the size of stars. So for example, Antares and Proxima Centauri are both M-class stars, so although they have approximately the same black body temperature, Antares is one of the brightest stars with an absolute magnitude of minus 5.3, and Proxima Centauri is one of the dimmest with an absolute magnitude of 15. So therefore, you can then say the diameter of Antares is much larger than that of Proxima Centauri. Now we can also quantify. the ratio of the diameters of two different stars. Because if we consider stars with black body temperatures T1 and T2, if the ratios of their power outputs is known, P1 over P2, therefore Stefan's law can be used to calculate the ratio of the diameters, D1 and D2. So you can say D1 over D2 is equal to T2 squared over T1 squared times by the square root of P1 over P2. and then you can use that to find how much bigger certain stars are compared to others. So here are the key facts for this particular topic. The luminosity of a star depends on its size and its temperature. Stars may be regarded as black bodies with a characteristic black body curve depending on their surface temperature, where Stefan's law states that P is equal to sigma A T to the 4 for a black body radiator, where P is the total power radiated by an object of surface area A. T is the absolute temperature of the object and sigma is a constant of proportionality called the Stefan-Boltzmann constant. And Vine's displacement law relates the peak wavelength, lambda max, of a star's black body spectrum to its temperature. Where we say in that lambda max times by T is equal to a constant which is 2.90 times 10 to the minus 3 meters Kelvin. So if we've been successful and we've learnt in today's lesson, we should be able to state Stefan's law in the Vine displacement law. Understand the general shape of black. body curves and the use of Vines displacement law to estimate blackbody temperatures of sources have an idea of the equation that lambda max times by T is equal to a constant. We assume that the star is a blackbody and we can use Stefan's law to compare the power output temperature and size of stars with the equation P equals Sigma a t to the 4. So we should be able to define Stefan's law and Vines displacement laws, draw the general shape of blackbody curves and use Vines displacement law to estimate the temperature of stars. and finally use Stefan's law to compare the power output and temperature of stars. I hope you've enjoyed today's lesson looking at stars as black bodies, and I hope you have a lovely day.

We're going to try and draw the general shape of black body curves and use Vines displacement law to estimate the temperature of stars and then finally use Stefan's law. to compare the power output and temperatures of stars, which falls into the following section of the AQA A-Level Physics Specification in the Astrophysics option, Classification by Temperature and Blackbody Radiation. We are used to the terms red hot and white hot when applied to sources of heat.

We're also aware that hot objects emit radiation at infrared wavelengths. This behaviour of hot objects and the electromagnetic radiation they produced was investigated at the end of the 19th century and contributed to the development of quantum theory by Max Planck. Now this is important when observing stars as all the information we get about stars comes from the electromagnetic radiation they emit and we receive. So for example an analysis of the intensity of the light at different wavelengths allows astronomers to measure the black body temperature of a star. Using this and an estimation of the power output of a star allows its diameter to be determined.

So to fully understand the radiation emitted by a star, we assume all stars are black bodies. Now this is a reasonable approximation to consider stars and some astronomical sources as these black body radiators. Now we can use black body radiation curves to make estimations about an object's temperature and other properties.

So a black body is an object with a pure black surface that emits radiation strongly and in a well-defined way. It absorbs all light instant on it. Now they are black bodies because they do not reflect light.

So a black body is a body that absorbs all electromagnetic radiation of all wavelengths and can emit all wavelengths of electromagnetic radiation. So an analysis of the intensity of the electromagnetic radiation emitted from a black body at different wavelengths produces the following result, which we call a black body radiation curve. Now there are three lines on this curve for three different temperatures. Now because they emit all wavelengths of electromagnetic radiation, black bodies have a continuous spectrum of electromagnetic radiation.

leading to this continuous curve produced in the blackbody radiation curves. So all blackbodies have the same shaped curve, and the graph that shows the hottest object, P, has a peak at the shortest wavelength. So a blackbody radiation is radiation produced by heated objects, particularly from a blackbody.

So a blackbody is an object that absorbs all radiation that falls on it. This means it will also radiate at all frequencies. that heat and from the heat energy produces and so physicists try to find the mathematics to explain how this worked so in this graph this links the intensity of the black body radiator output with the wavelength of the radiation produced i.e the color now like we mentioned before black body radiators are perfect emitters of radiation so they emit all radiation they contain they absorb all radiation they encounter they do not reflect they do not refract any radiation Now an example of a blackbody radiator is any star like the Sun.

So a blackbody radiator does not have to be black, it just has to be very good at absorbing and emitting radiation. So because all stars emit lots of radiation, we can consider them blackbody radiators. Now the following graph shows how close the Sun's output is to a blackbody radiator, with grey being the idealised blackbody radiator. radiation curve from the mathematics and the orange being the actual observed output from the sun. Now it is noted that the peak wavelength of radiation is linked to the temperature of the black body radiator.

Now the higher the temperature of the object the higher the peak wavelength of emission so the more shifted towards the blue end of the spectrum the object colour is. So the higher the temperature of the object the higher the peak wavelength of emission. the more shifted towards the shorter wavelengths of the spectrum the peak is.

Now, so when the temperature of a blackbody increases, the intensity of radiation increases, but it increases more so at the smaller wavelengths and shifts the peak wavelength to the lower wavelength. Now it's very important to note that if the temperature of the object is increased the intensity of radiation it emits is greater at every wavelength but it's not an equal increase. The shorter the wavelength the greater the increase in intensity at that wavelength as shown on this particular graph indicated.

So when the temperature of the object increases you've got to first note that the intensity of all wavelengths increases but the increase is greater for shorter wavelengths. So like mentioned before, increasing the temperature of a blackbody radiator shifts the peak wavelength of emission to smaller wavelengths, which is very important for our observations because the peak wavelength determines the colour of that object. So therefore the hotter the object, the bluer or the shorter the wavelength the object is, which is only true for the colour due to the emission of the radiation. So therefore the cooler the object, the redder or the longer the wavelength the object is.

Now hotter stars will therefore produce more of their light at the blue or violet end of the spectrum and therefore will appear white or blue-white to observers. Now cooler stars will look red as they produce more of their light at longer wavelengths. So when analysing the light from stars in this way, it's assumed that the star is acting as a black body so therefore no light is absorbed. scattered by material between the star and the observer. And also it's important to note that a hotter star may not appear as bright as a cooler one as it may mostly emit radiation that's not in the visible part of the electromagnetic spectrum.

Now if a cooler star emits more radiation in the visible spectrum than a hotter one it will appear brighter to optical telescopes which is why it's important to observe stars in both optical non-optical telescopes. So this is this explained why different objects glow at different temperatures when they are heated. So when something looks white hot it's bluer in color so as a higher temperature and shorter peak wavelength while something is redder in color it has a lower temperature but a longer peak wavelength.

So this is why you can have blue hypergiant stars and red giant stars. So this leads to something called The stellar classification system which classifies stars based on their colour and therefore effective temperature which we'll look at in upcoming lessons and allows us to classify different stars according to their colour which we'll cover in the upcoming lessons. So the Sun is a G-type star which is a relatively cool star which is shown by its longish peak wavelength and appears predominantly orange.

So the Sun has a peak emission of 5,800 Kelvin which gives us this trademark colour of yellow orange. So as we mentioned before a hotter blackbody will emit more radiation than a cooler one and its peak wavelength will be will be a shorter wavelength than the cooler star. So it also it has a higher total power output if we assume that the hot star and the cooler star have the same surface area.

Now we can formalize this equation mathematically by saying that the peak wavelength or lambda max times by the absolute temperature in Kelvin is equal to 2.9 times 10 to the minus 3 meters Kelvin. Now please note the unit there is the amalgamation of the wavelength unit and the temperature unit. It is not in fact millikelvin.

So this idea of 2.9 times 10 to the minus 3 meters Kelvin is called a constant value, which we call the vine constant. So this shows that as temperature increases, the peak wavelength of emission decreases. So lambda max indicates the peak wavelength of emission for a star, which determines the colour of the star. Whilst T is the surface temperature of that star, which is calculated in Kelvin.

It's a common examination mistake to calculate the value in Celsius. You always convert it to Kelvin. Now this equation assumes that the star is acting as a black body.

And we call this equation... the vine displacement law and it links the black body temperature, the surface temperature, with the peak wavelength of emission. Now remember the surface temperature of a star will also determine its colour so this equation details the colour of a star also and this equation is given to you in your examination book. So let's look at an example question.

An astronomer has observed radiation from a star and recorded the peak wavelength to be in the visible light region at 520 nanometres. Assuming that the star behaves as a black body, estimate the absolute temperature of the star. Well, step one, you want to place your wavelength into meters, so you convert from nanometers into meters. You then use Vines'displacement law, where lambda max times by t is equal to 2.9 times 10 to the minus 3. Rearrange to get t to be the subject, pop your values in, and you get an answer of 5,600. 600 Kelvin.

Now we can also link the surface temperature of a blackbody with the power of emission. Now remember, the power output of a star is the total energy, the radiation it emits every second. So this is a very important idea.

The power output is proportional to the fourth power of the star's surface temperature, and is directly proportional to the surface area of the star. This means we can formalise this with a mathematical equation. where power is equal to the Stefan constant times by the surface area of the star in metres squared times by the surface temperature of the star in Kelvin to the power 4, where the Stefan's constant is 5.67 times 10 to the minus 8 watts per metre squared per Kelvin to the power of 4, which we call Stefan's law.

Now this equation and the Stefan constant is given to you in your examination equation book. So this equation shows that the power output of a star is directly proportional to t to the power 4 and the surface area of the star. So this means a cooler star needs to be a lot bigger to have the same power output. Now remember when calculating the surface area of a star we assume the star to be a sphere.

So therefore in this equation the surface area is equal to 4 pi squared, 4 pi r squared even. Now Another term for power is luminosity, so therefore Stefan's law will also determine the luminosity of a star. Now, it's a common in examination questions for you to determine the relative surface area of the star when compared to other stars with this equation.

So here's an example question. A star has a surface area of 4.1 times 10 to the 13 meters squared and produces a blackbody spectrum with a peak wavelength of 115 nanometers. What is the power output of the star? Well, part one, you find the temperature of the star, so it's going to be lambda max times by T equals 2.9 times 10 to the minus 3. Rearrange that to find T, which will be in Kelvin. So you pop that in, you get 25217.39.

Remember, we aren't going to go to our significant figures yet because we're halfway through the question. Then the second part, you will use Stefan's law, which is P equals Stefan's constant times by A T to the power of 4. So you pop your values through and you get power is equal or luminosity to 9.40 times 10 to the 23 watts. Now we can use the previous laws to compare two different stars, such as star A and star B.

Now the laws tells us that if two stars have the same blackbody temperature, so therefore they are the same spectral class, the star with the brighter absolute magnitude must have the larger diameter. Now it's also quite common for Stefan's law to be used qualitatively. when comparing the size of stars.

So for example, Antares and Proxima Centauri are both M-class stars, so although they have approximately the same black body temperature, Antares is one of the brightest stars with an absolute magnitude of minus 5.3, and Proxima Centauri is one of the dimmest with an absolute magnitude of 15. So therefore, you can then say the diameter of Antares is much larger than that of Proxima Centauri. Now we can also quantify. the ratio of the diameters of two different stars.

Because if we consider stars with black body temperatures T1 and T2, if the ratios of their power outputs is known, P1 over P2, therefore Stefan's law can be used to calculate the ratio of the diameters, D1 and D2. So you can say D1 over D2 is equal to T2 squared over T1 squared times by the square root of P1 over P2. and then you can use that to find how much bigger certain stars are compared to others. So here are the key facts for this particular topic.

The luminosity of a star depends on its size and its temperature. Stars may be regarded as black bodies with a characteristic black body curve depending on their surface temperature, where Stefan's law states that P is equal to sigma A T to the 4 for a black body radiator, where P is the total power radiated by an object of surface area A. T is the absolute temperature of the object and sigma is a constant of proportionality called the Stefan-Boltzmann constant.

And Vine's displacement law relates the peak wavelength, lambda max, of a star's black body spectrum to its temperature. Where we say in that lambda max times by T is equal to a constant which is 2.90 times 10 to the minus 3 meters Kelvin. So if we've been successful and we've learnt in today's lesson, we should be able to state Stefan's law in the Vine displacement law. Understand the general shape of black. body curves and the use of Vines displacement law to estimate blackbody temperatures of sources have an idea of the equation that lambda max times by T is equal to a constant.

We assume that the star is a blackbody and we can use Stefan's law to compare the power output temperature and size of stars with the equation P equals Sigma a t to the 4. So we should be able to define Stefan's law and Vines displacement laws, draw the general shape of blackbody curves and use Vines displacement law to estimate the temperature of stars. and finally use Stefan's law to compare the power output and temperature of stars. I hope you've enjoyed today's lesson looking at stars as black bodies, and I hope you have a lovely day.

Transcript for:Understanding Stars and Black Body Radiation

Transcript for:
Understanding Stars and Black Body Radiation