Favorite Visualization of General Relativity?

[Orodruin]

If there is no spacetime curvature in B1 and B2, wouldn't that mean there is no gravity?

Gravity and acceleration are locally indistinguishable. It would be more appropriate to say that there are not tidal forces.

 

[A.T.]

If there is no spacetime curvature in B1 and B2, wouldn't that mean there is no gravity?

No, it means that there is no tidal effect. B1 and B2 represent an uniform gravitational field. C represents an non-uniform gravitational field.

 

[PAllen]

Note the waves in the following, must be gravitational, due to Einstein :

https://www.google.com/search?safe=.....1...33i299.3ec-iYgiDeU#imgrc=0ZG6HU3kReBiiM:

 

[Orodruin]

Note the waves in the following, must be gravitational, due to Einstein:

https://www.google.com/search?safe=.....1...33i299.3ec-iYgiDeU#imgrc=0ZG6HU3kReBiiM:

For once I get to point out that those are Einstein's gravity waves, not gravitational waves, instead of the other way around ...

 

[jambaugh]

Much better than the trampoline! One frustration I have with visualizations like this, though, is that they leave out time. Which seems to leave out the equivalence principle.

Yes but you cannot properly embed the indefinite geometry of space-time into an intuitive picture which exists in our definite geometry of curved surfaces and such. There is a trick I worked up in Special relativity where you switch proper time and coordinate time to get an Euclidean metric structure. Problem is that points on the picture can represent events occurring at distinct times. It is useful for demonstrating the resolution of the Twin's paradox.

So you start with a standard coordinate graph but you label the axes $x$ and $\tau$ for coordinate spatial position and proper time. Draw a curve representing an accelerating observers world line, say, starting at the origin but keeping it monotone-non-decreasing in the tau direction. (No fair letting your observer's proper time run backward.)

Now the coordinate time can be calculated as the arclength since: (in c=1 units) $dt^2 = dx^2 + d\tau^2$.

But you must be careful with assumptions in this system. Two observers are at the same space-time event and thus can causally communicate immediately with each other if they are at the same x coordinate and at the same distance along their arc-length.

What's nice about this picture is that there's no "well sorta" the quantitative effects are exactly represented and not qualitatively analogized. The problem is that the graph points are not single space-time event points so you can't then go and invoke GR by curving the surface.

 

[A.T.]

There is a trick I worked up in Special relativity where you switch proper time and coordinate time to get an Euclidean metric structure. .

This is also the approach Epstein uses in his book mentioned above. The space-proper-time diagrams for both twins (traveler has constant acceleration) would look like this:

Here is a comparison of Minkowski and Epstein diagrams for the 3 inertial frames of the twins (traveler has constant speed with instantaneous turn around):
http://www.adamtoons.de/physics/twins.html