# Physics of tsunami: the relationship between wavelength, sea depth and the height of the water

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?

• Here is something on Tao's blog. terrytao.wordpress.com/2011/03/13/…
– MBN
Mar 16 '11 at 4:04
• @MBN, I would appreciate if you could post this as an answer; I think this comment is good enough to be an answer. Mar 16 '11 at 4:05
• @Graviton It's a comment, of course. Usually people post a link as a comment if they don't have a lot of time. If they want to go in more depth, they post an answer with their link and then summarize its contents. Mar 16 '11 at 4:25
• @Mark: Should I delete the answer?
– MBN
Mar 16 '11 at 4:29
• @MBN It's your answer; I don't know. I don't think this is some official thing - just my observations. You could start a meta thread if you want to clarify the difference between comments and answers. Mar 16 '11 at 4:45

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.