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Your students likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small path along the latitude (squared) + a small path along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

Your students likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small arc along the latitude (squared) + a small arc along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

You student likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small path along the latitude (squared) + a small path along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

Your students likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small path along the latitude (squared) + a small path along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

You student likely know about pythagor theorem. Because of the triangle inequality, we know it is the shortess distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface: ds² = dx² + dy²

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element ds² will be a small path along the latitude (squared) + a small path along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere: ds² = (Rdθ)² + (Rsin(dθ) * dΦ)²

You student likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small path along the latitude (squared) + a small path along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$