geophysicstsunami
If I understand correctly, when an earthquake occurs, energy will be transferred to the water, resulting in water waves. As the waves reach seashore, because the sea depth is getting shallower and wavelength is getting shorter, the height of the wave gets push up, resulting in tsunami. In other words in deep sea, water won't get pushed up as high as the water in shallow seashore.
Is my understanding correct? Is there a quantitative way to express the physics behind all this?
The positive or negative elevation of the leading wave is related to what happened to the ocean floor during the earthquake. If the seabed was raised, a crest should lead the tsunami; if it was lowered then a trough leads.
In the 2004 Sumatra tsunami, a piece of seabed of about 1000 km x 100 km went down a few meters. Another strip, located to the west of the first and with roughly the same size, went up. Because of this, the wave moving east (towards Thailand) was led by a trough. The one moving west, which reached Sri Lanka, India and Africa, was led by a crest.
This is seen in the simulation of that tsunami shown below (by Kenji Satake, of the Active Fault Research Center in Tsukuba, Japan). In the red areas the water surface is higher than normal, and in the blue ones it is lower.
A similar animation, with some data added, is at http://es.ucsc.edu/~ward/indo.mov
Best Answer
A tsunami is basically a shallow-water wave, even in deep seas. This means its velocity is $v=\sqrt{gH}$, where $H$ is the water depth and $g$ is the gravitational acceleration.
The energy of the tsunami scales as the square of its amplitude $A$, and thus the energy flux $S$ goes as $S\sim A^2 \sqrt{H}$. Conservation of energy then implies that the wave amplitude depends on the sea depth as $$ A \sim H^{-1/4}$$ a result known as Green's law.
For example, Green's law predicts that a tsunami with amplitude $A=1$m at $H=5000$m will run up to $A=4$m if the depth becomes $H=20$m.