Luminosity

What is Luminosity?

We have an almost intuitive understanding of the relationship between light and temperature like if we look at the stove when it is turned on, then there would be no difference whether it is turned on or off especially when the temperature is not specifically high. As the temperature increases, we would see that a red glow of the flame blows about the stove and when the temperature continues to increase, the flame turns to a bright yellowish-white color, as you can see in the two images below.

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Now, if we look at other sources, like iron being heated turns blue, stars emit light, the electric bulbs emit energy in the form of small packets called photons.

Since the light has energy, therefore, the stove and the candle emit light energy.

So the light energy emitted by it per second is called its luminosity, denoted by symbol L.

The unit used for measuring the luminosity is Joules per second or J/s.

What Does Luminosity Mean?

Luminosity is described as an inherent property of objects that emit light such as a star, flame of a burning candle, iron rod on getting heated, electric bulbs. They all give off light energy in all directions every second.

So, this is the same as saying that the more the object emits light, the more it gives off the power in watts, the more will be its luminosity.

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What is the Definition of Luminosity?

We define luminosity as the total amount of energy given off by an object every second.

The graph below shows that the wavelength of the yellow light is at the peak.

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Let’s take a real-life example:

If an electric bulb gives off 100 watts of power, it means it gives off 100 Joules of energy every second.

We can see that the luminosity of the sun is 3.85 x 1026Watts which is a very high luminosity.

As you compare between the Sun and the bulb, the sun is four trillion times trillion more powerful light is being radiated by the Sun every second.

What is the Meaning of Luminosity?

Let’s talk about an astronomical body like a star (an imperfect blackbody); they are spherical objects, primarily made up of very hot elements .i.e., hydrogen and helium that emits light continuously.

We can’t measure the temperature of a star, all we can do is just observe the light with maximum wavelength and use the Wein’s displacement law which states that the temperature can be found at the point where the radiation curve peaks, i.e. 入peak = x 10\[^{m}\] nm (nano microns).

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As you can see, this graph shows the spectrum of a star and through this, we can determine the wavelength where the emission peaks are given by, 入peak.

By Wein’s formula:

               入peak = (0.29 cm K)/T

Here, T = 5500 K

Putting the value in the above equation, we get:

                 入peak = 5.27 x 10\[^{-5}\] m

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Then, compare to the spectra of computer models of stellar spectra of different temperatures (as you can see in Fig.2), and develop an exact color-temperature relation.

Luminosity Theory

In this theory, we will discuss the relationship of luminosity to temperature and the surface area of an object.

Let’s talk about the factors here:

1.  Luminosity: The total amount of energy emitted per second in Watts.

2. Apparent brightness: It determines how bright a star appears to be; the power per meter squared as measured at a distance from the star.

Its unit is Watt/meter\[^{2}\].

Luminosity is denoted by L.

So, LSUN = 3.85 x 10\[^{26}\] J/s or watts.

In the CGS system, it is 3.8 x 10\[^{33}\] erg/s.

So, the surface area of a star can be calculated as A = 4πR² 

Where R is the radius of a star.

If we consider a star to be a completely black body, the radiation emitted per second can be calculated by using the Stefan- Boltzmann law.

Stefan-Boltzmann states that the total energy emitted by a body is directly proportional to the fourth power of temperature.

Here s= Stefan’s constant whose value = 5.7 x 10\[^{-8}\] Wm\[^{-2}\] K\[^{-4}\], 

               A = Surface area of a star, and

               T = Absolute temperature of a star.

Let’s consider radius of a star like the Sun = 6.96340 x 10\[^{8}\] 

Surface area of star = 4πR² = 4 x 3.14 x(6.96340 x 10\[^{8}\])\[^{2}\]

TSUN = 6000 K

On calculating, we get

A = 6.09 x 10\[^{18}\] m\[^{2}\] 

Now using the equation(1):

  E = (5.7 x 10\[^{-8}\]) * (6.09 x 10\[^{18}\]) * (6000)\[^{4}\]

Therefore, we get the energy radiated by the sun = 4.4998 x 10\[^{26}\] J.