How much Sun is expected to curve spacetime?

Question

Is there any proper way of calculating that how much a space time will curve for any object.

Answer 1

I think there are two questions here:

1. how do we calculate the curvature caused by the Sun?

2. how can we observe the consequences of this curvature i.e. what does the curvature we calculate in part (1) physically mean?

Calculating the curvature is pretty straightforward if we're allowed to make a few approximations. If we start with a spherically symmetric non-rotating mass then we can solve Einstein's equation to get the spacetime geometry around the object. This was done by Karl Schwarzschild only a few months after Einstein published his theory of general relativity, and the solution is the Schwarzschild metric:

$$ ds^2 = -\left(1 - \frac{2GM}{c^2r}\right)c^2 dt^2 + \frac{dr^2}{1 - \frac{2GM}{c^2r}} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 \tag{1} $$

Now the Sun isn't spherical, and it is rotating, but it isn't very aspherical and it isn't rotating very fast so to a good approximation our equation (1) will describe the spacetime curvature around it. In fact equation (1) is the equation used by Einstein when he achieved one of GR's first triumphs by calculating the perihelion shift of Mercury.

To find the observable consequences of this curvature we need to take another step. When spacetime is flat freely moving objects obey Newton's first law i.e. they move in a straight line at constant speed. When spacetime is curved freely moving objects obey a different equation called the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} \tag{2} $$

This is kind of a version of Newton's second law, which for a freely moving body is:

$$ \frac{d^2x^a}{dt^2} = 0 $$

Comparing this to equation (2) the big difference is that in equation (2) we have terms involving $\Gamma^\mu_{\alpha\beta}$ on the right hand side instead of zero. These terms $\Gamma^\mu_{\alpha\beta}$ are called the Christoffel symbols and they are the most immediately obvious consequences of the curvature because they control how the object moves in the gravitational field. For example it's the non-zero values of the Christoffel symbols that makes things fall when you drop them. A detailed calculation showing this is given in How does "curved space" explain gravitational attraction? though I suspect that's far more detail than you're looking for.

But all this is a bit abstract, and I suspect you're after a more intuitive feel for how does spacetime curvature cause straight lines not to be straight? Well the most obvious definition of a straight line is the path of a light ray, because we all learn in school that light travels in straight lines. And the obvious demonstration of this is the bending of light rays by the Sun.

If we shine a light ray so that it just grazes the Sun's surface then that light doesn't travel in a straight line. We can calculate the deflection using the geodesic equation and the values of the Cristoffel symbols, and we find the light ray is bent by about $1.75$ arcseconds. But this is a tiny, tiny amount. $1.75$ arcseconds is about the angle subtended by a baseball at a distance of $9$ kilometers - you couldn't even see a baseball $9$ km away!

To summarise:

1. calculating the curvature involves some scary equations but it's surprisingly straightforward to do. General relativity isn't as hard as you might think.

2. in everyday terms the curvature of spacetime (as measured by the bending of light rays) is almost immeasurably small.