Query regarding bent space-time in General Relativity

Question

I am just a beginner in this topic and I accept I haven't gone through whole of the content of GR. However after going through some of the basic ideas, I encountered a problem.
If I did my homework well, there is a statement, "mass tells the space-time how to curve and curvatures in the space-time tell mass how to move".
Also, in every video they represent the 4-D space-time as a 2-D flat surface. They put mass over there and the flat surface bends.
But, it's the gravity that bends that 2-D flat surface when mass is placed over there (analogous to mass in 4-D space-time). This creates confusion as gravity is needed prior to the placement of mass in the space-time in-order to bend it. But, they say gravity doesn't exist without mass.
To make the problem more clear let's place "the 2-D flat representation of space-time" in outer space (where gravity is absent, ISS will work). Then, keep a massive object (say an iron ball) over the flat surface. Here, the flat surface does not produce any curvatures (in absence of gravity).
Is this approach of visualizing space-time "wrong"? What is the right approach?

Answer 1

I'm afraid the right approach to visualizing GR is mathematics.

Absolutely nothing else does the job properly and it's not, honestly, easy.

The simplest mathematics you can learn to help you (skipping all the details of how you get there) is the Schwarzschild metric.

Concentrate on learning what the different parts mean, without worrying too much about the complexities of tensors and the rest of stuff that make the mathematics of GR such "fun" to learn. This equation will give you a feel for the way stuff works and in particular the way time (and proper time) fits into the scheme of things.

Answer 2

It's not wrong because it's just a metaphor.

("This creates confusion as gravity is needed prior to the placement": This is admittedly an acute observation; but still, it's just a visual metaphor.)

Answer 3

There is no easy way to visualize it.

There are numerous ways to provided mathematical meaning to curvature (because $R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$ is unfortunately not very intuitive, and neither is $R^\sigma_{\kappa\mu\nu}=2\partial_{[\mu}\Gamma^\sigma_{\nu]\kappa}+2\Gamma^\sigma_{[\mu|\lambda|}\Gamma^\lambda_{\nu]\kappa}$), perhaps the most intuitive one is to consider the behaviour of parallel lines.

In euclidean space, parallel straight lines remain parallel, and non-parallel straight lines drift apart or together at a constant rate.

In curved spacetime, a curve $\gamma(\tau)$ is called a geodesic, if for its velocity vector $T=d\gamma/d\tau$ it is true that $$ \frac{d^\nabla T}{d\tau}=0, $$ eg. the curve has no acceleration. The derivative $d^\nabla/d\tau$ is something that in most respects behave like a usual derivative, but its actual form is different (it is a covariant derivative).

Given a one-parameter family of geodesics, $\gamma_\lambda(\tau)$ (for every $\lambda$, the curve is a geodesic), we can define a deviation vector as $Q=d\gamma/d\lambda$. We can take the relative velocity of the geodesic congruence to be $d^\nabla Q/d\tau$, eg. how the deviation vector changes as we move along the curves.

The relative acceleration of the congruence is then $(d^\nabla)^2Q/d\tau^2$, the second derivative.

After a calculation, one can obtain that $$ \frac{d^\nabla}{d\tau}\frac{d^\nabla Q}{d\tau}=R(T,X)T, $$ where $R(\cdot,\cdot)$ is an object that takes two vector fields into a linear operator (and this "taking" is also bilinear), which then acts on $T$.

The point is, of $R$ is zero, then two geodesics (analogues of straight lines) will always drift apart or together at a constant rate, but if $R$ is nonzero, then there will be such geodesics that drift apart or together at a varying rate, they'll accelerate.

This $R$ is called the curvature operator or curvature tensor.

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The point is, curvature is essentially a measure of how euclidean geometry fails to hold. Spacetime is curved, if geodesics deviate in an accelerating manner. Spacetime is curved if keeping a rod parallel to itself while moving it along a loop will fail to return the rod to its original orientation. Spacetime is curved if triangles fail to have their inner angles added to $\pi$.

It is not possible to visualize it, and while some visualizations may help, they will probably hinder as much as they help. In the end, what you have is that you can calculate $R$, and if it is zero, your spacetime is flat. If it isn't then you have nonzero gravity.