Instead of thinking of gravity as mass warping spacetime, could it be thought of as mass warping only time, whereby time would advance at faster rates at locations where more mass is present?

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    $\begingroup$ Doesn't the Lorentz transformation 'mix' the space and time coordinates? $\endgroup$ – Alfred Centauri Nov 8 '17 at 2:39

No. This kind of proposal can "control" at most one independent component of the Riemann curvature tensor (through a rate parameter), and we routinely meet metrics in gravitational theory that have pure spatial curvature as well as time components to the curvature tensor.

Witness, as an example, the FLRW metric wherefor:

$$\begin{array}{lcl}R_{tt}& =& - 3 \frac{\ddot{a}}{a}\\ R_{rr} &=&\frac{c^{-2}(a(t)\ddot{a}(t) + 2\dot{a}^2(t)) + 2k}{1 - kr^2}\\ R_{\theta\theta} &=& r^2(c^{-2}(a(t)\ddot{a}(t) + 2\dot{a}^2(t)) + 2k)\\ R_{\phi\phi} &=&r^2(c^{-2}(a(t)\ddot{a}(t) + 2\dot{a}^2(t)) + 2k)\sin^2(\theta)\end{array}$$

where one has two independent parameters, the scale factor $a(t)$ (which could be construed as the "control knob" in the OP's proposal) and the spatial curvature. It's true that one can normalize the equations, but you're still left with three fundamentally distinct possibilities $k=\pm1$ and $k=0$. Furthermore, one can of course find much more complicated, inhomogeneous metrics as valid solutions to the Einstein field equations.

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    $\begingroup$ Hi Rod, this answers my questions. Thanks so much!!! $\endgroup$ – Betvin Géant Nov 8 '17 at 16:10

Before we answer this question, I think it is proper to discuss what is asked. As Rod explains, there are certainly solutions to the Einstein Field equations that cannot be explained by a single parameter. However, the question as framed seems to suppose there is some universal way of defining how quickly time passes. What is meant by time dilation is that the proper time along two different timelike curves connecting two events differ. This can happen in Minkowski spacetime, where the universal reference frame makes things easier for us, and it can happen in curved spacetimes. In this view, the quesion should perhaps be reframed as:

Can the effects of spacetime curvature be reconstructed from knowledge of time dilation along all different timelike paths? Possibly by introducing new dynamic equations which link time dilation to other observable effects.

If this was the case, one could say that spacetime curvature could be entirely explained only by considering time dilation. Now, proper time along a curve, $\gamma$, is found by integrating the metric along the curve $$\Delta\tau = \int_\gamma d\tau = \int_\gamma\sqrt{g_{ab}dx^adx^b},$$ where we supposed units such that $c = 1$ and timelike signature, $(+---)$. This would in principle allow us to reconstruct the primitive of the metric components along our timelike curves, and the curvature is given by a rather lengthy expression involving second derivatives of the metric.

However, promising as this may look, the fact that we can only consider timelike curves, means that we have no way of investigating the dependency of metric components along spacelike or null curves (through time dilation). In slightly more mathematical language: we can only know the derivatives of the metric components with respect to timelike coordinates, and of those only the "timelike components:" If we choose coordinates $(t,x^1,x^2,x^3)$ such that $x^i$ ($i \in \{1,2,3\}$) are constant along a curve then $$\Delta\tau = \int_\gamma\sqrt{g_{tt}}dt,$$ and we may in principle reconstruct the primitive of $\sqrt{g_{tt}}$ with respect to $t$ only. And this is certainly not enough to reconstruct the curvature tensor, in any spacetime, even if all possible timelike coordinates are investigated.

  • $\begingroup$ Hi Erik, thanks for rephrasing my question and providing more details on the validity and limits of only considering time dilation! $\endgroup$ – Betvin Géant Nov 9 '17 at 2:29
  • $\begingroup$ @Erik Jörgenfelt: You want to check the question in any possible frame, but this is not required. It would be sufficient if the answer could be "yes" in one single frame (or better: one kind of frame, i.e. the frame of the far-away observer). - The question of Betvin Géant is excellent without need for correction, see edit 2 of my answer. $\endgroup$ – Moonraker Nov 9 '17 at 13:27
  • $\begingroup$ @Moonraker The point is that even if we somehow attain knowledge of every possible frame, it is not sufficient. So it is rather the opposite of what you imagine. $\endgroup$ – Erik Jörgenfelt Nov 9 '17 at 15:00

It is difficult to answer your question for all possible configurations of gravity, but within Schwarzschild metric it is easy to show that gravity may not only be represented as curved spacetime but - exactly as you suppose - also as gravitational time dilation in flat space.

The Schwarzschild metric is the most basic description of the spacetime curvature by gravity, it is $$ \mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta ~\mathrm d\Phi^2\right)$$

In contrast, the corresponding Minkowski metric (with flat spacetime) is $$ \mathrm ds^2 = -~ c^2~\mathrm dt^2 + \mathrm dr^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta~\mathrm d\Phi^2\right)$$

where $\mathrm dt$ is uncurved time and $\mathrm dr$ is uncurved radial displacement.

Comparing both, you find that in the Schwarzschild metric, time $\mathrm dt$ is multiplied by the constant

$$ \sqrt{1 - \frac{2GM}{c^2 r}}$$ and space $\mathrm dr$ is divided by the same constant. It is exactly this factor which represents the spacetime curvature. The constant is gravitational time dilation. If we set the constant = $C$, we can write Schwarzschild metric shorter in the following way: $$ \mathrm ds^2 = -~c^2 (C~\mathrm dt)^2 + {\left(\frac {\mathrm dr}{C}\right)}^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta~\mathrm d\Phi^2\right)$$ Comparing this short form with the equation above of Minkowski metric, the Schwarzschild metric differs from uncurved Minkowski metric only by one coefficient $C$ which is identic with gravitational time dilation. That means that the curved spacetime of Schwarzschild metric may also be described in terms of gravitational time dilation - in absolute, flat space!

Curved spacetime and gravitational time dilation in flat space are two equivalent models. For the latter, gravitation is expressed by the tendency of particles to maximize their own gravitational time dilation.

Edit 1: The Schwarzschild metric even provides a simpler answer to your question: in the Schwarzschild equation, the displacement coordinates dt and dr (as well as dΘ and dΦ) are not curved! As you see, the metric ds is the result of a multiplication of dt by C and the division of dr by C, providing a warped metric. However, the terms on the right side dt and dr are not distorted and not warped, they are flat polar coordinates, and the polar coordinate system may be transcribed into the corresponding Cartesian coordinate system of flat space.

Edit 2: Your question may be of high importance. If what I showed for the Schwarzschild metric would be true in general (e.g. Kerr metric etc.) this would mean that gravity may be described exclusively in terms of time modulation in flat space. This could be crucial with regard to the fact that the time parameter in quantum mechanics is no operator, it is classical. In my personal opinion, here could exist a path permitting to conciliate gravity and quantum mechanics.

  • $\begingroup$ Hi Moonraker, thanks for presenting this thorough explanation!! I wasn't aware of the Schwarzschild metric, so it's good to know that it can be described in terms of gravitational time dilation in flat space. Cheers! $\endgroup$ – Betvin Géant Nov 8 '17 at 16:09
  • $\begingroup$ Not only is your $C$ not a constant, as it depends on $r$, an easily rectified mistake, but your identification of $C$ with time dilation is ill motivated. If it was a constant then it would be equivalent to a coordinate change, and even though it is not, in what way can you identify it with gravitational time dilation? The fact that it appears before $dt^2$ just means that the time coordinate is not normalized. $\endgroup$ – Erik Jörgenfelt Nov 8 '17 at 22:48
  • $\begingroup$ We may theoretically be able to find $C$ through investigating time dilation, but only by using our knowledge of the Schwarzschild solution can we reconstruct the entire line element. In this way it becomes circular: using our knowledge of the curvature to reconstruct it. $\endgroup$ – Erik Jörgenfelt Nov 8 '17 at 22:57
  • $\begingroup$ @ Erik Jörgenfelt: Thank you for the feedback, you are right, the word "constant" may hurt. Naturally, it is not observer-invariant but it corresponds to the gravitational time dilation (τ/t) from the point of view of a far-away observer. $\endgroup$ – Moonraker Nov 9 '17 at 13:24
  • $\begingroup$ See my edit of the article and my question (thank you for the hint you gave me). No circularity, the logic goes from the right to the left side: It is the metric ds (not the coordinates r and t) which is curved. On the right side you find the uncurved coordinates (!) t and r. On the left the curved metric, which corresponds to the Minkowski metric, however t and r are distorted by gravitational time dilation, and as a result you get the curved metric ds. $\endgroup$ – Moonraker Nov 9 '17 at 13:25

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