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My question might be based on a false premise, so here's why I asked. If you look up the meaning of the moment magnitude scale for measuring earthquake size, the articles usually say that each increase in magnitude by 1 point represents a tenfold increase in the amplitude of the wave, but a 31.6 increase in energy. Which suggests the energy is going up as the three halves power. With other waves the energy goes up with the square of the amplitude, so what is going on here?

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The law looks like a great fundamental law that should have a neat derivation and the power $3/2$ looks like an accurate number. However, you will be disappointed, it is just a phenomenological estimated formula developed by Beno Gutenberg and Charles Richter: $$ \log_{10}(E/{\rm erg}) = 11.8 + 1.5 M_S $$ Note that $M_S$ is calculated from a bandwidth between 18 and 22 seconds. See

http://earthquake.usgs.gov/learn/topics/measure.php

The shaking amplitude shown by the seismometer - and its base-ten logarithm is how the Richter scale is defined - is something between an "energy reading" and an "amplitude reading". For the latter, you would square the amplitude to get the energy; for the former, the exponent would be one. So in reality, the relationship is something in between and the exponent is $3/2$ but it is just an approximate estimate. It's a messy problem and one would have to make an analysis of the energy that can be released in any allowed frequency.

While we can measure how much seismometers move, we can't really reliably measure the energy released by an earthquake.

Note that Gutenberg's name was ultimately omitted from the name of the Richter scale - it's because Gutenberg didn't like interviews. The term "magnitude" was chosen by Richter because of his self-evident interest in astronomy which uses the same term for a similar logarithmic scale.

To see the actual paper by these two men that introduced the scale in the 1930s, read

http://www.sciencemag.org/content/83/2147/183

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Hmmm. Modern accelerometers could solve this problem. –  Carl Brannen Mar 14 '11 at 21:57
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We have confusion because the Richter scale is very popular, but only measures the energy of the initial second or two of the earthquake. Great quakes like this one take a few minutes to fully rupture and release strain energy over a large area, so the correlation between the sharpness of the initial break and the total energy released wouldn't be expected to be too great. The more modern measure is called seismic moment M, and attempts to account for all the energy. Many journalist hear magnitude, and automatically append "Richter scale" even when it is inappropriate. If you look up the moment magnitude scale on Wikipedia, you see that the 3/2 power is explicitly built into the formula, i.e. the log10 of the energy is estimated as M0 and then multiplied by 2/3 to come up with the reported moment. I presume this was done to make the energy/magnitude of the two scales as similar as possible (in the range where they both have reasonable validity).

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I hope you don't mind that I added a link ;) –  David Z Mar 14 '11 at 17:37
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My guess is the energy does not go up with the three halves power in the first place. I think it only seems so at 600 km distance from the epicenter: $M_L = M_W \Rightarrow log_{10}(A_{max}) = \frac{2}{3}log_{10}M_0-10.7 \Rightarrow M_0 = \sqrt{A_{max}}^3e^{10.7\frac{3}{2}}$ At different distances there is a correction term in the formula for the Richter scale.: $M_L = log_{10}(A_{max}) - A_0(\delta)$.

$A_0(\delta)$ being some empirical function, the answer would be: what the left part of the $M_L$ term "makes wrong" in terms of wave energy, the right one compensates for.

The momentum scale was designed to yield values similar to those from the Richter scale:

The scale was developed in the 1970s to succeed the 1930s-era Richter magnitude scale (ML). Even though the formulae are different, the new scale retains the familiar continuum of magnitude values defined by the older one.

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Okay, but why is the relationship between energy and amplitude one where E is proportional to A to the 3/2? –  Donald Mar 14 '11 at 12:54
    
I also see that the wording of my question was poor--I should have asked why the energy seems to vary with the amplitude to the three halves power. Hopefully the slightly incoherent phrasing won't keep people away. –  Donald Mar 14 '11 at 12:56
    
I think it's not the energy that varies with 3/2 but its "evaluation" in terms of the scale. Like with temperature: you state it in terms of a scale (e.g. Fahrenheit, Celsius or Kelvin etc.,) No matter which scale you choose it's still the same temperature. As a mechanical measuring device, the Wood-Anderson torsion seismometer has already some evaluation pattern built into it, which the momentum scale tries to approximate. –  artistoex Mar 14 '11 at 13:41
    
So, the 3/2 ultimately should stem from the torsion coefficient of the torsion wire being used in the seismometer, the dampening field, the calibration and maybe the rules from reading the device. –  artistoex Mar 14 '11 at 14:15
    
That sounds plausible--in hindsight I should have phrased the question differently, as it seemed unlikely that the energy really does vary with the 3/2 power. –  Donald Mar 14 '11 at 15:20
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