4
$\begingroup$

I just watched this video where it is explained how time dilation causes gravity. It is said in this video that time dilation (caused by masses such as the earth) is the only cause for gravity.

But then, what about the curving of space? Does this not contribute to gravity (by objects following the straightest line as depicted in this image)?

Is gravity only caused by gravitational time dilation, if yes, why does the curving of space not have any effect and if no, how much of gravity is caused by time dilation compared to the curving of space?


Note: I am aware that one cannot actually split the curving of spacetime into two things (the curving of time, and space seperately). I could also ask my question as:

How much of the attractive "force" (well, it is not actually a force) felt by an object in the gravitational field of another object like a planet or star is caused by clocks closer to the object runnning slower (compared to clocks further away) compared to the object moving toward the heavier object due to curved geodesics?

$\endgroup$
7
  • $\begingroup$ Does this answer your question? How much of gravity is caused by time dilation? $\endgroup$ – Jon Custer Jan 21 at 15:30
  • 2
    $\begingroup$ @JonCuster the other one has been flagged as a dupe of this one :) I'm sure the repost was an accident. $\endgroup$ – Nihar Karve Jan 21 at 15:30
  • 1
    $\begingroup$ @JonCuster Sorry about the duplicate question. It was indeed not intentional; I will delete the other question $\endgroup$ – Jonas Jan 21 at 16:07
  • 1
    $\begingroup$ In this paper it is explained that gravity may not only be expressed in the form of spacetime curvature but also alternatively in the form of gravitational time dilation in flat, uncurved space. But this is not mainstream physics. $\endgroup$ – Moonraker Jan 21 at 18:28
  • 1
    $\begingroup$ You may like my answer to a related question: physics.stackexchange.com/a/587025/123208 $\endgroup$ – PM 2Ring Jan 22 at 2:22
2
$\begingroup$

No, it isn't only time curvature that can cause gravitational effects, but the temporal curvature terms (i.e. ones that include time and space, as opposed to just spatial directions) are the only ones we typically notice in everyday life. The Earth curves space and time together in comparable amounts (see this for the exact values) and in theory all of these curvature terms do have effects on trajectories of objects/light, but some we notice and some we don't.

Most everyday objects on Earth (humans, falling apples, etc.) have a much larger temporal component of velocity than spatial component of velocity $\dfrac{d(ct)}{d\tau}>>\dfrac{dx}{d\tau}$. So when finding a geodesic trajectory, the spatial only curvature terms do not significantly contribute.

Basically, you only notice curvature if you move significantly in a curved direction, even if all directions are curved equally. For example, if you walked all the way around the Earth along the equator you could measure that your distance walked is many millions of metersticks different from $2\pi R=2\pi$(distance walked from North Pole to Equator) and therefore you would notice curvature along latitudes. But if you only walked 10 meters North from the equator, it would be difficult to measure whether your arc length deviated from that of a flat space or not, so you would not notice any curvature along longitudes (even though the Earth's surface has equal curvature in all directions). Such is the case in General Relativity for objects that move fast through time, and slow through space. e.g. $\dfrac{d(ct)}{d\tau}>>\dfrac{dx}{d\tau}$

$\endgroup$
3
$\begingroup$

The video is just nonsense. Early on, he starts talking about "curvature in time" and saying "time can curve." He just has no idea what he's talking about. Curvature is curvature of spacetime, not time or space separately.

Gravitational time dilation isn't generically even something that can be defined in general relativity. It can only be defined for certain kinds of spacetime that are called static. What is true is that in such a spacetime, the time dilation is closely related to the gravitational potential. And even in this special case, it's not a question of "causing" gravity.

Do yourself a favor and stop trying to learn about science from videos. Read a book.

$\endgroup$
3
  • 7
    $\begingroup$ There's no need for hyperbolic claims like "he has no idea what he's talking about" or insults to the OP about not learning from videos. Also, you can have curved space separate from time. The surface of a sphere is a curved spatial manifold with no time curvature elements necessarily introduced. $\endgroup$ – David Santo Pietro Jan 21 at 23:41
  • 2
    $\begingroup$ @DavidSanto: "Read a book" is excellent advice here. Helpful advice is pretty much the opposite of an insult. $\endgroup$ – WillO Jan 22 at 2:10
  • 2
    $\begingroup$ OP will get to decide if it was insulting or not. But telling someone that came to a stack exchange to "read a book" is an insult in my book. As if their question isn't even worthy of discussion. $\endgroup$ – David Santo Pietro Jan 22 at 3:05
3
$\begingroup$

We perceive gravity as a fictitious force as we are pushed off our geodesic by the surface of the Earth, according to the geodesic equation (the version that uses time):

$$ \ddot x^{\mu} = -\Gamma^{\mu}_{\alpha\beta} \dot x^{\alpha}\dot x^{\beta} + -\Gamma^0_{\alpha\beta} \dot x^{\alpha}\dot x^{\beta}\dot x^{\mu}$$

The point of this equation, here, is to show that it's not just time curvature that matters. Alpha and beta run over all the indices.

However, when you are on the Earth's surface sitting at rest:

$$ \dot x^{\mu} \approx (c, 0, 0, 0) $$

and $\alpha$ (or $\beta$) $= 1,2,3$ just don't get an opportunity to enter the calculation.

Also, when you plug in Christoffel symbols for the Schwarzschild metric at $r\gg r_s$, you get the nice result that:

$$ \ddot x^r = -\frac{GM}{(x^r)^2} $$

which is Newton's law of gravitation.

If you get closer to the Schwarzchild radius, move at relativistic velocities, or give the gravitating body lots of angular momentum, then you need more terms than $\Gamma^r_{tt}$ to quantify gravitation.

Back in the weak field of Earth, you can imagine time curvature by considering that path dependence of time as follows:

Two researchers (A & B) in the basement at NIST have atomic clocks, and agree to meet in the top floor conference room in 1 year. A takes the elevator up and waits in the conference room, B sits in the basement for a year, and only then takes the elevator up top. He sees A who then says, "You're late"....their two right angle paths do not meet at the same point in spacetime.

$\endgroup$
3
$\begingroup$

How much of the attractive "force" (well, it is not actually a force) felt by an object in the gravitational field of another object like a planet or star is caused by clocks closer to the object runnning slower (compared to clocks further away) compared to the object moving toward the heavier object due to curved geodesics?

The geodesics are paths in the 4 dimensional spacetime, not in 3-D space only. They include time effects.

The video tries to explain the accelerated movement associated to gravity to differences in the clock rate at different point of an object. It is not the case.

Suppose an object thrown upward. Its velocity will decrease gradually until stops, and then changes direction, and gradually increases until it has the same initial speed, (but opposite direction).

Compare that movement with a plane flying from Paris to Seattle (both cities with similar latitudes about $48^\circ$. Looking at a globe, it is clear that the shortest path (that is part of a great circle, called geodesic) requires keeping a north component of the velocity, that gradually becomes smaller until reach a maximum latitude (about $64^\circ$). Then a small south component appears, that are getting greater until reach Seattle.

If we compare with the previous example, taking latitude as height and longitude as time, the situation is similar to the object thrown upward.

For the plane, the direction of the velocity changes during all the trip due to the curvilinear coordinates. So the movement could be called accelerated. But we know that it is as straight as possible.

For the object, the path is also as straight as possible (a geodesic), but it is shown to us as accelerated, due to the curvilinear coordinates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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