# How are tidal effects explained by general relativity?

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1. Nov 27, 2014

### Leonardo Muzzi

I can fairly understand the concept of gravity as a curvature in space time in general relativity, but so far I could not understand completely the tidal forces explained by the curvature of spacetime.

When the moon is on one side of the earth, the oceans on this side come closer to the moon, AND the oceans on the opposite side goes far from it, causing high tides on both sides.

Now, I can understand the effect on the opposite side explained by newtonian gravitation: the earth itself suffer a force stronger than the water on the opposite side. So far so good. But I cannot explain the same effect thinking in a spacetime curvature. I can see why a more stretched spacetime near the moon would cause high tides on the side directly facing it, but I cannot visualize how the curvature would cause the same effect on the other side.

2. Nov 27, 2014

### Staff: Mentor

Well, Relativity is reducible to Newtonian gravity, so they work the same. I find it helpful to visualize the earth and two bulges as three objects connected with string. The tidal force is the tension on the string.

3. Nov 27, 2014

### A.T.

"Tidal forces" refers to an effect , where nearby free falling objects converge or diverge. This is visualized in Fig. C below, for two objects falling radially:

4. Nov 27, 2014

### Leonardo Muzzi

That's actually a very interesting draw to point out our perception of why particles move the way they do in a curved spacetime.

Still, I cannot understand how this has do to with the high tide on earth's opposite side relative to the moon.

Giving some thought, I'm imagining the answer is that the earth itself is a little more stretched in the direction of the moon than the water on the opposite side... maybe...

5. Nov 27, 2014

### A.T.

If the two particles in Fig. C were connected with a string, it would be stretched. And so is the Earth being stretched along the line radial line.

Last edited: Nov 27, 2014