Is it possible to measure gravity other than cause and effect. Gravity is a principle which can only be measured by cause and effect. Ie since an object falls we assume there must be something pushing it down and since this principle holds true for a variety of scenarios, we hold it as true. Is that all we have to go on with gravity? To put it into perspective electricity can be measured by other methods than turning on a light switch and observing the filament glowing. Infrared waves can be measured by heat. For gravity other than seeing an object fall, what other proofs do we have? (Info sourced by RMMS).

  • 1
    $\begingroup$ possible duplicate of What was the Law of Gravity better explained by? $\endgroup$ May 30, 2014 at 8:28
  • 2
    $\begingroup$ The question I've linked may not seem an obvious duplicate, but my answer to it also answers your question. Both Newton and Einstein's theories of gravity are mathematical models. Their proof or disproof is a matter of experiment. So far no experimental disproofs of general relativity have been found. $\endgroup$ May 30, 2014 at 8:30
  • $\begingroup$ @JohnRennie thanks for that answer cleared up some things. I re-edited the question to make it clearer. Sorry for the trouble $\endgroup$
    – TreeKing
    May 30, 2014 at 9:38
  • $\begingroup$ I've answered what I think is your modified question. Let me know if I've misunderstood what you're asking. $\endgroup$ May 30, 2014 at 10:02

1 Answer 1


Gravity is due to the curvature of spacetime and is described by the metric tensor. In principle we can directly measure the metric tensor (though in practice the curvature is usually much too small to be measured in this way).

Suppose you take a vector and you parallel transport it round a small square:

Parallel transport

Obviously when you finish going round the square you'll be back at your starting point and the vector will still be pointing in the same direction.

But this is only true when the spacetime you're moving through is flat. If you try this on a curved spacetime you'll find that after completing the square you won't be back at your starting point and the vector may have rotated away from it's original direction.

Parallel transport on sphere

In this diagram the red vector is the initial vector at $P$, and we find that we get back to our starting point after only three sides of the square (I call it a square because the angle at each corner is 90° - it's really a triangle of course) and when we get back to $P$ the angle of the vector has changed.

The change in the vector is described by the Riemann tensor, and in General Relativity the Riemann tensor is what determines the gravitational field. So we can measure gravity by measuring the geometry.

In real life there are problems with this. As I've mentioned already the curvature is generally far too small to measure in this way, but a bigger problem is that in GR time is curved as well as space and you can't take a vector round a loop in time.

  • $\begingroup$ That went way over my head. But I'll take your word for it. I should have clarified that I am a novice in this. But would this be an example of measuring it's effect and not gravity itself. IE if there are three steps; person drop object, gravitational pull makes object fall, can hits ground. In order to measure this arent we only taking the first step and last step and assuming through an equation the middle step. We dont know first hand that there is a force pulling it down. Basically I just want to know if I understand this, are we measuring gravity through it's effects only? $\endgroup$
    – TreeKing
    May 30, 2014 at 10:18
  • $\begingroup$ @armoose: Gravity is spacetime curvature and we can measure spacetime curvature directly. If you choose other methods to measure gravity, like dropping things, then yes they will be indirect. However since we understand the theory (General Relativity) that predicts how the dropped object will fall even the indirect method is a good way to measure gravity. $\endgroup$ May 30, 2014 at 10:28
  • $\begingroup$ o ok thnx, sorry for the misunderstanding $\endgroup$
    – TreeKing
    May 30, 2014 at 10:44

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