 An overview of active galaxies

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# 7.8 Example 2 and questions

Example 2

Show that the synchrotron spectrum produced by a power-law distribution of electron energies N(E) dE = N0E−s dE, is described by You may use the approximation that all the power radiated by the electron has frequency = max, where max is given by Equation 15.

Solution

The flux emitted at a particular frequency, , is determined by the number of electrons emitting at that frequency, N( ), and the power, P, emitted by each of these. Expressing this mathematically: The power P is given by Equation 16, and the number of electrons is determined by Equation 17, but we must transform this distribution from the number of electrons with a particular energy, E, into the number of electrons emitting at a particular value of . This is accomplished by noting that the electrons and their properties are unchanged irrespective of how we decide to label them, so that i.e. where we have used Equation 18 in the final step.

Collecting these expressions for P and N in terms of E and substituting into Equation 20 we obtain where we have substituted for N(E), and collected powers of E to arrive at the final result. The only remaining work is to express the result in terms of rather than E, and this can be accomplished using Equation 15, which tells us that which can be recast as Substituting for E, therefore, we obtain as required.

The observed power-law spectrum described by Equation 19 is often more concisely written as where the new constant ( = (s – 1)/2)) is known as the power-law index. In Activity 4 we explore the production of power-law synchrotron spectra.

There is a generally useful trick to finding and measuring power-law dependencies. Taking logs of Equation 19 for F we obtain which can be rewritten using the properties of logs as Hence it is clear that the overall synchrotron spectrum shown in Figure 16, which shows log10(F ) versus log10( ), appears as a straight line with gradient -(s – 1)/2 or more concisely with gradient − , where = (s – 1)/2. The trick of using logarithmic axes makes it easy to recognise when the variables being plotted on the graph are related by a power law. If the variables are related by a power law, the power-law index can be determined by measuring the gradient of the straight line on the ‘log–log’ graph. This is a widely used technique, and astronomers often plot observed quantities on a log–log graph to check quickly whether there is a power-law relationship between them.

## Question 4

Figure 17 shows the spectrum of synchrotron emission from a particular astronomical source. What is the particle exponent describing the distribution of electron energies in that source? Figure 17 The spectrum of radio emission observed at a particular location in the sky

Choosing two convenient points on the straight line, for example (x = 9.0, y = 1.75) and (x = 10.0, y = 1.25), we can calculate the slope of the line. The slope is Putting in the values we get Comparing this with the information in Section 7.5, we see that in this case, the power-law index, , is = 0.5. But we must note that the question asked instead for the particle exponent, s, so we need to determine s using the relationship = (s−1)/2.

Rearranging, we have s = 2 + 1, i.e. s = 2(0.5) + 1,

Our final answer is that the graph shows synchrotron emission from a distribution of electrons with a particle exponent s = 2.

## Question 5

What is the power-law frequency dependence of the Rayleigh–Jeans tail?

That is, if we write B ,Rayleigh–Jeans  p, what is the value of p?

There are three ways you might have approached this question. A method which the material in this course might have suggested is to use Figure 9, noting that it has logarithmic axes. The Rayleigh–Jeans tail is the part of the Planck function which appears as a positive gradient straight line in this graph. A power-law relationship will always appear as a straight line in a log–log plot. Since the line has a positive gradient, p must be positive: both facts imply that Bν, Rayleigh–Jeans increases as increases. We can use this to check that our answer makes sense.

Clearly, the Rayleigh–Jeans tail is easiest to measure for the uppermost curve plotted on Figure 9 because it is (a) longer, and (b) less cluttered by other lines which might confuse the issue. To measure the gradient of the straight line, we need to:

• choose two points which are widely separated, and read off the (log10 , log10B ) values for each;

• subtract them to work out the values of Δlog10 and Δlog10B for each; and

• (iii) divide to evaluate which is the gradient of the line in the graph.

Applying this method, the points marked on Figure 18 illustrate one sensible choice. Reading off the (log10 , log10B ) values, we have the lower point at (log10 = 5, log10B = −19.5) and the upper point at (log10 = 16, log10B = 2.5).

Therefore Δlog10 = 16 − 5 = 11 and Δlog10B = 2.5 − (−19.5) = 22

Hence The question asked us ‘if we write B , Rayleigh–Jeans  p, what is the value of p?’. Taking logarithms of both sides of the expression we have where we used the properties of logarithms in the second step. Hence if we make the substitutions y = log10B and x = log10 , and recall that a straight-line graph represents the general equation we see that a graph of log10B versus log10 will be a straight line of gradient p.

Thus, using the result evaluated in (iii) above, we see that p = 2.

Therefore the Rayleigh–Jeans tail of the Planck function is a power law, B , Rayleigh–Jeans  2.

Another more straightforward way to answer the question would have been to use the expression for the Rayleigh–Jeans tail given in Equation 11 to deduce p = 2.

## Question 6

(a) With a magnetic field strength of 10−9 tesla, what values of electron energies are required in order to produce synchrotron radiation at radio frequencies of a few GHz?

(b) What values of electron energies are required for radio frequency production if the magnetic field is 10−10 tesla?

(a) Using a magnetic field strength of 10−9T, to produce synchrotron radiation with radio frequencies, i.e. a few × 109 Hz, requires input electron energies of a few × 109 eV, i.e. a few GeV. This means that very energetic electrons are needed to produce even radio synchrotron emission which lies at the low-energy end of the electromagnetic spectrum.

(b) If the magnetic field strength is reduced to 10−10T, then even more energetic electrons, with energies of at least 10 GeV (1010 eV), are required to produce radio synchrotron emission.

So far we have seen that the synchrotron emission from a power-law distribution of relativistic electrons is a power-law spectrum with increasing flux at decreasing frequencies. Like all good things, this must come to an end: absorption of photons by the synchrotron-emitting electrons can also occur. This process is known as synchrotron self-absorption (SSA), and is particularly important for photons with low frequency, . For these low-frequency photons, SSA can mean that the probability of a photon escaping from the synchrotron emitting region is small, i.e. at low frequencies the synchrotron emitting region becomes optically thick. This modifies the spectrum of the escaping radiation at low frequencies from to , as Figure 19 shows. This downturn in the spectrum at low frequencies is known as a low-frequency cutoff. The frequency at which self-absorption becomes important depends on the magnetic field strength, the density of electrons and the particle exponent. The details of these dependencies are beyond the scope of this course, but an obvious consequence of them is that the low-frequency cutoff can occur at a range of different frequencies, depending on the particular parameters of the emitting region. If an astronomical source has a spatially varying magnetic field, density, or electron energy distribution, then the overall emitted spectrum could be a sum of individual spectra like that shown in Figure 19, but shifted in frequency relative to each other. This is how so-called flat-spectrum radio sources are thought to arise. You will read about these in Activity 5. Figure 19 The full synchrotron spectrum from a power-law distribution of electron energies. At high photon energies the emission is optically thin, and the power-law dependence on the particle exponent holds

At low photon energies the emission is optically thick, and self-absorption causes a F   5/2spectral shape.

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