How are tidal effects explained by general relativity?

[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.

 

[russ_watters]

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.

 

[A.T.]

I could not understand completely the tidal forces explained by the curvature of spacetime.

"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:

This is my own non-animated way of looking at it:

• • Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  • B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
    
    B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  • Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[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...

 

[A.T.]

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

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.