49
$\begingroup$

What is the connection between special and general relativity? As I understand general relativity does not need the assumption on speed of light constant. It is about the relation between mass and spacetime and gravity. Can general relativity be valid without special relativity?

$\endgroup$
  • 11
    $\begingroup$ I am so tempted to say "trivial" and "Levi-Civita"... $\endgroup$ – Gonenc Mogol Nov 13 '17 at 18:47
  • 2
  • 2
    $\begingroup$ On the first lecture when I took GR, as part of a quick run-down of the basics for the course, the lecturer wrote up "GR = SR + (gravity "=" accelration)" on the blackboard. After having taken the course, that pretty much sums it up. Special relativity, while excellent at handling accelration, has no gravity in it. The equivalence principle(s) says that locally, gravitation behaves like an accelrated system. If you put that on top of "special relativity is true locally", and work through the math, general relativity falls out. $\endgroup$ – Arthur Nov 14 '17 at 8:10
  • 1
    $\begingroup$ If someone is able to answer your question, arguably they won't need the Wikipedia links to understand what you are talking about... $\endgroup$ – DaG Nov 14 '17 at 21:34
  • 2
    $\begingroup$ Special relativity is the "special" case of "general" relativity. No pun intended. $\endgroup$ – Dr. Ikjyot Singh Kohli Nov 15 '17 at 1:28
90
$\begingroup$

Suppose we start by considering Galilean transformations, that is transformations between observers moving at different speeds where the speeds are well below the speed of light. Different observers will disagree about the speeds of objects, but there are some things they will agree on. Specifically, they will agree on the sizes of objects.

Suppose I have a metal rod that in my coordinate system has one end at the point $(0,0,0)$ and the other end at the point $(dx,dy,dz)$. The length of this rod can be calculated using Pythagoras' theorem:

$$ ds^2 = dx^2 + dy^2 + dz^2 \tag{1} $$

Now you may be moving relative to me, so we won't agree about the position and velocity of the rod, but we'll both agree on the length because, well, it's a chunk of metal - it doesn't change in size just because you are moving relative to me. So the length of the rod, $ds$, is an invariant i.e. it is something that all observers will agree on.

OK, let's move onto Special Relativity. What Special Relativity does is treat space and time together so the distance between two points has to take the time difference between the points into account as well. So our equation (1) is modified to include time and it becomes:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{2} $$

Note that our new equation for the length $ds$ now includes time, but the time has a minus sign. We also multiply the time by a constant with the dimensions of a velocity to convert the time into a length. Just as before the quantity $ds$ is an invariant i.e. all observers agree on it no matter how they are moving relative to each other. In fact we give this spacetime length a special name - we call it the proper length (or sometimes the proper time).

By now you're probably wondering what on Earth I'm rambling about, but it turns out we can derive all the weird stuff in Special Relativity simply from the requirement that $ds$ be an invariant. If you're interested I go through this in How do I derive the Lorentz contraction from the invariant interval?.

In fact the equation for $ds$ is so important in Special Relativity that it has its own name. It's called the Minkowski metric. And we can use this Minkowski metric to show that the speed of light must be the same for all observers. I do this in my answer to Special Relativity Second Postulate.

So where we've got to is that the fact the speed of light is constant in SR is equivalent to the statement that the Minkowski metric determines an invariant quantity. What General Relativity does is to generalise the Minkowski metric, equation (2). Suppose we rewrite equation (2) as:

$$ ds^2 = \sum_{\mu=0}^3 \sum_{\nu=0}^3 \,g_{\mu\nu}dx^\mu dx^\nu $$

where we are using the notation $dt=dx^0$, $dx=dx^1$, $dy=dx^2$ and $dz=dx^3$, and $g$ is the matrix:

$$g=\left(\begin{matrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right)$$

This matrix $g$ is called the metric tensor. Specifically the matrix I've written above is the metric tensor for flat spacetime i.e. Minkowski spacetime.

In General Relativity this matrix can have different values for its entries, and indeed those elements can be functions of position rather than constants. For example the spacetime around a static uncharged black hole has a metric tensor called the Schwarzschild metric:

$$g=\left(\begin{matrix} -c^2(1-\frac{r_s}{r}) & 0 & 0 & 0 \\ 0 & \frac{1}{1-\frac{r_s}{r}} & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2\sin^2\theta \end{matrix}\right)$$

(I mention this mostly for decoration - understanding how to work with the Schwarzschild metric needs you to do a course on GR)

In GR the metric $g$ is related to the distribution of matter and energy, and it is obtained by solving the Einstein equations (which is not a task for the faint hearted :-). The Minkowski metric is the solution we get when there is no matter or energy present${}^1$.

The point I'm getting at is that there is a simple sequence that takes use from everyday Newtonian mechanics to General Relativity. The first equation I wrote down, equation (1) i.e. Pythagoras' theorem, is also a metric - it's the metric for flat 3D space. Extending it to spacetime, equation (2), moves us on to Special Relativity, and extending equation (2) to a more general form for the metric tensor moves us on to general relativity. So Special Relativity is a subset of General Relativity, and Newtonian mechanics is a subset of Special Relativity.

To end let's return to that question of the speed of light. The speed of light is constant in SR so is it constant in GR? And the answer is, well, sort of. I go through this in some detail in GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light but you may find this a bit hard going. So I'll simply say that in GR the speed of light is always locally constant. That is, if I measure the speed of light at my location I will always get the result $c$. And if you measure the speed of light at your location you'll also get the result $c$. But, if I measure the speed of light at your location, and vice versa, we will in general not get the result $c$.


${}^1$ actually there are lots of solutions when no matter or energy is present. These are the vacuum solutions. The Minkowski metric is the solution with the lowest ADM energy.

$\endgroup$
  • 4
    $\begingroup$ Hi, your answer is very good as always, and I have upvoted it. I just have a question/remark: you say that "The Minkowski metric is the solution we get when there is no matter or energy present." Shouldn't be more correct to say that it is A solution? Other than all the crazy solutions that people find when they allow the topology of spacetime to go totally nuts, we also have the spherical and hyperbolic metrics, that are vacuum solutions. They're just different from Minkowski metric, as curvature is different. Or is there some reason to exclude those solutions? $\endgroup$ – Salvatore Baldino Nov 13 '17 at 21:02
  • 3
    $\begingroup$ "For example the spacetime around a black hole has a metric tensor called the Schwarzschild metric:" That's rather misleading, as the tensor you give is not in Euclidean coordinates. (It's also for a non-rotating, electrically neutral black hole, but that's a bit more nit-picky.) $\endgroup$ – Acccumulation Nov 14 '17 at 0:15
  • 1
    $\begingroup$ @SalvatoreBaldino: thanks, I've added a footnote to clarify. I originally just ignored all but the Minkowski vacuum solution because the answer is targeted at non-GR heads and I didn't want to make it any more complicated than it had to be. $\endgroup$ – John Rennie Nov 14 '17 at 6:07
  • 1
    $\begingroup$ @Acccumulation: the answer is targeted at people who have no knowledge of GR so I've done my best to keep it simple. Maybe I've oversimplified in some places, but I'm not sure it's easy to be more rigorous and still keep the answer comprehensible. $\endgroup$ – John Rennie Nov 14 '17 at 6:09
  • 2
    $\begingroup$ This is the most concise, clear summary of Physics 2 and Physics 3 that I've ever seen. I'm saving this answer for when my children will have to start those courses. Thank you! $\endgroup$ – dotancohen Nov 15 '17 at 11:04
14
$\begingroup$

Special relativity is the "special case" of general relativity where spacetime is flat. The speed of light is essential to both.

$\endgroup$
  • $\begingroup$ This should be a comment. $\endgroup$ – Jannik Pitt Nov 15 '17 at 19:28
  • $\begingroup$ @JannikPitt This is a partial answer, and answers or partial answers should not be comments. If you think this answer is too short to be a good answer, please comment about how this answer could be improved. $\endgroup$ – JiK Nov 16 '17 at 13:39
4
$\begingroup$

The best connection between the two theories regards how they deal with different observers or frames of reference. Special Relativity (SR) postulates that all inertial observers are equivalent whereas General Relativity (GR) assumes that a wider class of observers are equivalent. More precisely, all non rotating frames are equivalent. Thus, GR is more general than SR (aka Restricted Relativity) and therefore cannot be valid without SR. In another words, as theories, GR implies SR but the converse is not true.

$\endgroup$
  • 1
    $\begingroup$ I would tentatively disagree with the statement that all observers are equivalent in GR. In special relativity, inertial observers are equivalent insofar as there's no experimental way to distinguish one frame from another. In GR, there is certainly an observable difference between a rotating frame on the surface of the earth and an observer freely falling in deep space. $\endgroup$ – gj255 Nov 13 '17 at 18:22
  • $\begingroup$ @gj255 Yep, that's true. Thanks for pointing this out. $\endgroup$ – Diracology Nov 13 '17 at 18:27
  • $\begingroup$ Graviry cannot be explained as spacetime curveture without the C assumption ? $\endgroup$ – Sharon Salmon Nov 13 '17 at 18:46
2
$\begingroup$

I like to think of special relativity as first order or "local" general relativity.

One of the fundamental things underlying all relativity is the equivalence principle. But it kind of "disappears" in the mechanics, procedures and theory of general relativity. It is actually encoded in the "choice of building materials" for GR. That is, we think of spacetime as a manifold as opposed to other mathematical objects we might postulate it to be (such as a Variety - this is a somewhat silly example, because it's "almost" a manifold, but it is simple and meant to show that it is not a done deal that we must choose a manifold - there is real, in-principle-measurable physics affected by the choice). A manifold is a mathematical object that is everywhere locally Euclidean, or, in the case of GR, Minkowski. If we "zoom in" to the manifold at high enough magnification, we can make spacetime as near as we like to flat, Minkowski spacetime. More formally what this means is that we can always define a tangent space to every point. Here's the key to this answer:

As long as we don't go too far from this spacetime point and keep within a small neighborhood (it might have to be very small in highly curved space, but this is a theoretical possibility and our magnification can be any finite value), all relativistic calculations can be done with special relativity with the tangent space approximating spacetime in the neighborhood.

Inertial frames at the point in question are those momentarily comoving with objects and frames undergoing geodesic, torque-free motion in the more general, curved General Relativistic manifold and all of these are equivalent modulo a Lorentz transformation, just as in special relativity. Curvature is a second order notion, not definable in terms of the tangent space to one point alone. Einstein's original conception of the equivalence principle was that, to first order, there is no difference between experimental results carried out within a lab accelerated relative to these inertial frames defined by the tangent space. Whether they be accelerated by a rocket, or accelerated because the laboratory has bumped into and thus stuck to the surface of a planet, one cannot tell unless one looks outside the laboratory.

$\endgroup$
  • $\begingroup$ I really like this answer, but I think it could be improved by adding an explanation of how manifolds generalize flat (euclidean) spaces. Even for readers familiar with differential geometry, I think it would serve to clarify your analogy (that GR is to SR as [euclidean] manifolds are to euclidean spaces). $\endgroup$ – Jacob Maibach Nov 18 '17 at 17:14

protected by Qmechanic Nov 13 '17 at 20:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.