Time and Gravity

1. Apr 13, 2007

Dravish

From my understanding of GR, time and gravity were shown to be essentially the same thing, is that the generally accepted interpretation in contemporary physics?

2. Apr 13, 2007

robphy

Contemporary physics does not interpret
Time and gravity as essentially the same thing.

What GR did is to assert that time and gravity are not generally independent concepts... they are related [in a specific way according to GR]... but they are not the same thing.

3. Apr 13, 2007

Dravish

In that case I must be missing something fundamental to the theory, how are gravity and time differentiated?

4. Apr 13, 2007

anantchowdhary

Gravity...is because of mass-energy and also other factors.Time is however not there because of mass

5. Apr 13, 2007

Dravish

To quote from ‘Time, Gravity and Quantum Mechanics, by W. G. Unruh’;
“A more accurate way of summarizing the lessons of General Relativity is that gravity does not cause time to run differently in different places (e.g., faster far from the earth than near it). Gravity is the unequable flow of time from place to place. It is not
that there are two separate phenomena, namely gravity and time and that the one, gravity, affects the other. Rather the theory states that the phenomena we usually ascribe to gravity are actually caused by time’s flowing unequably from place to place.”

It seems that perhaps gravity and time should be equated (though I personally think that they are aspects of the same phenomena seen from 'different angles' so to speak), surely this will lead to an evaporation of many of the issues that arise when trying to unify QED and GR :surprised

Any thoughts on this line of thinking would be appreciated as I am trying to put together a paper using this perspective

6. Apr 13, 2007

anantchowdhary

I dont think time exists because of mass.But gravity,the curvature of space-time depends on the presence of mass.

So it can be said that

Only the flow of time is affected by gravity or by the presence of huge mass,energy etc.

7. Apr 13, 2007

pervect

Staff Emeritus
What is behind what Unruh is saying is that one can often derive gravity from the metric coefficient g_00, a metric coefficient which in one common viewpoint is described as "gravitational time dilation". (I'd have to review the derivation to recall exactly which assumptions are needed to derive the "force" of gravity in this way, for starters one needs a static field, though. The force turns out the be proportional to the gradient, i.e. the rate of change with respect to position, of g_00 when the necessary assumptions are made).

However, it is possible and perhaps more accurate to take the view that time always flows at the rate of 1 second per second - in which case, it becomes more complex to describe exactly what g_00 is describing. We can say that it describes the "metric", of course, but that just invites other questions.

The usual way of saying this is to say that it is space-time curvature that causes gravity, not time that causes gravity. This approach can also cause arguments, however, which is perhaps why Unruh avoided it. (The issue here is that curvature is sometimes used losely to describe varying metric coefficients, but is othertimes used more strictly to indicate a non-zero Riemann curvature tensor. According to the first defintion of curvature, it makes sense to call gravity curved space-time, but it does NOT make sense if one uses the second, stricter defintion of curvature.)

Basically, the issue is one of semantics, but I don't think that it would correct to represent "time=gravity" as a standard view. (It might, from your quote, be reasonable to represent this as Unruh's view, I suppose - i.e. the view of one particular physicist, rather than a general consensus.)

To a certain extent, the problems of how to popularize GR are like the problems of how to describe an elephant to a blind man, i.e.

so it isn't really too surprising that on the popular level, different descriptions of GR might appear to be different to the layperson when they are actually describing the same theory.

Last edited by a moderator: May 2, 2017
8. Apr 13, 2007

Robert J. Grave

Mass/energy and space/time together is the universe. They cannot exist seperately. The shape of space/time is related to the mass in it. All aspects of space/time are altered by this shape change. Three spacial and one time dimension. This effect is the gravitational effect. The more accute the bending or distoting of space/time the more pronounced the effects are. The black hole discriptions speaks of time stopping inside it.
This is the extream of space/time warping on to it's self. - Robert

9. Apr 13, 2007

MeJennifer

Unruh, in http://arxiv.org/PS_cache/gr-qc/pdf/9312/9312027v2.pdf" [Broken] juxtaposes time in GR and time in Quantum mechanics.

In GR there is no background, particles do not live in spacetime they make spacetime, in other words, time and space are emergent properties of the gravitational field. Time is not like some infinite and perfectly flat bowling lane on which particles roll. Each observer can have a unique notion of time and space. This notion completely depends on his relative position in the combined mass and energy distribution of all causally connected particles and thus constantly changes.

In quantum mechanics it is quite different, here time is a background, here time is like an infinite and perfectly flat bowling lane on which particles roll.

Last edited by a moderator: May 2, 2017
10. Apr 13, 2007

Robert J. Grave

If time is flat in QM, how do we explane time dilation in relativity, an observed occurance? -Robert.

11. Apr 13, 2007

pervect

Staff Emeritus
How would we go about testing these statements? It seems to me this is more a matter of personal philosophy than something that can be put to experimental test, though I'm willing to be convinced otherwise if you can demonstrate some way of testing your statements.

12. Apr 13, 2007

jimbobjames

Here are a couple of statements showing the link between time and gravity:

1) A clock at the top of a tall building ticks faster than a clock at the floor of the same building. This follows directly from the equivalence principle, by studying the behavior of clocks at the front and rear of an accelerated rocket.

2) The path which a freely falling particle takes is that for which the proper time of that particle is greatest.

Strap a watch to a tennis ball and throw it up into the air, remembering that clocks higher up tick faster than those lower down. The velocity of the ball causes the watch to tick more slowly (a special relativistic time dilation effect) but the higher the ball goes, the faster it ticks (since it is then higher in the gravitational field) - the net result of both effects (slowing due to speed and speeding up higher in the field) is that the actual path taken is the one which gives the most ticks on that watch. Any other path would result in less ticks. Its an optimization.

The beauty of the geodesic equation is that it includes the SR effect as well as the field effect into one succinct coordinate invariant equation.

"Straight line paths" through curved spacetime (geodesics) are those for which the proper time is greatest.

Of course it is still interesting to consider why it is that clocks at different points in the field tick at dfferent rates - does anyone out there actually understand why this is? Does a deeper study of GR or String Theory or QFT provide the answer? (Id prefer to know up front so I can optimize my time usage!).

(To say the answer is simply "spacetime curvature" would be circular, because different clock rates at different places is curvature of the time part of spacetime).

13. Apr 13, 2007

pervect

Staff Emeritus
"Why" questions are always somewhat problematical, but here is an analogy which may help you.

Suppose you have a naval ship that can travel on the Earth's oceans at some constant rate - say 20 knots, where a knot is a nautical mile per hour.

At the equator of the Earth, 1 nautical mile is 1 minute of arc of longitude, so the speed of the ship is 20 nautical miles per hour, if the ship goes east-west.

Nearer the poles, at higher latitudes, however, the same ship, going at an identical rate of 20 knots, travels more than 20 minutes of arc per hour.

This is analogous to the way that time dilation acts. The ultimate cause is the same in both cases, in one case it is the curvature of the Earth, in the other case it is the curvature of space-time.

One way of describing things would be to say that the metric on the Earth's surface changes, and that distances are shorter at higher lattitudes.

But this may obscure the fact that a nautical mile is a nautical mile - it's really the coordinates (degrees of longitude) that change size, not the actual distances.

In GR, it's a similar situation. A second really is a second no matter where you go, but when you set up a coordinate system (similar to the degrees of longitude), you find that coordinate time intervals are not the same as the physical seconds one measures with a local watch.

14. Apr 13, 2007

intel

If this is so - particles make spacetime. Then what makes the gravitational field? I believed large masses were their cause, but small particles make big masses - so it still leaves what makes the gravitational field?

Last edited by a moderator: May 2, 2017
15. Apr 13, 2007

jimbobjames

Thats a good analogy, Pervect. Cheers.

The real physical time is the one on the watch, regardless of the coordinate system you choose to set up. Right.

"A second really is a second no matter where you go":

But the real rate of ticking really is different at different heights in the field, since if I bring those watches back to the same place later, the watch readings differ (regardless of coordinate choice). The relative number of physical ticks on the watches at different heights, which you can measure and compare when you reunite the watches, is different (and is independent of coordinate choice). And so one real second for me might correspond to 2 real seconds for you, it depends on our relative paths through spacetime. Or am I wrong about this?

I'd love to see a picture of the 2-d projection of the 4-d spacetime in the spherically symmetric (but weak) field near a planet. The 2 dimensions would be the radial (or height) and (coordinate-) time dimensions.

Geodesics on that curved 2-d surface - assuming the curvature to be the time curvature only - would show what we perceive as objects falling from large r to small r. What shape is that surface - can it be embedded into a flat 3-d space?

Have you - has anyone here - ever come across a visualization like that? I'm sure it would help me understand the analogy the relationship between gravity and time even better.

16. Apr 13, 2007

pmb_phy

I think it was Einstein who said it best when he said in Nature (Feb. 17, 1921, p. 783)
Pete

17. Apr 13, 2007

pervect

Staff Emeritus
No, you are not wrong. But you can set up an analogous situation in the analogy.

Naval ship Alpha starts at 0 degrees longitude on the equator, and goes east for 20 nautical miles. It winds up at 1 minute of longitude because, at the equator, one NM = 1 minute of longitude. (I may have gotten my signs wrong, maybe it's -1 minute of longitude, but it doesn't really matter.)

Naval ship Beta starts out at the equator, and travels north to 45 degrees latitude. It then proceeds east for 20 nautical miles. It then goes south back to the equator. Ship Beta finds that it has traveled MORE than 1 minute of latitude.

So, if degrees of longitude were time, if we image ship Alpha continuing on its trip along the equator, ship Beta would see ship Alpha as being "older" when the two ships unite.

You could explain it as "time" (distance) being differentat different latitudes, but it could just as well be described as being an artifact of the curvature of the Earth, and the comparison process.

Different authors take different approaches - Unruh choses to stress the fact that "a second is a second" much as one might chose to stress that "a mile is a mile". But it's also possible to view the situation as distances changing with lattitude.

While I've seen embedding diagrams of the Schwarzschild geometry, for

they usually show r and phi, and suppress t and theta, so they're not what you asked for.

I think the following *may* work, but I haven't checked it closely. Imagine a sphere,around which you wrap an ace bandage of zero thickness that's infinitely long.

The height of the ace bandage is limited in this analogy - unfortunately you need more than three dimensions to wrap a plane around a sphere. (I think you can do it in 4. Make the bandage too high, and it will intersect itself. But if you have another dimension, you can arrange matters so this doesn't happen.)

Anyway, the finite height of the ace bandage will be a spatial dimension, and the infinitely long dimension of the ace bandage will be the time dimension.

The gaussian curvature of this is positive (there is only one curvature component in 2d) and uniform, so I think it's the right analogy.

Last edited: Apr 13, 2007
18. Apr 13, 2007

MeJennifer

Spacetime is the combined gravitational field of all particles.

19. Apr 15, 2007

intel

Can I take this and say that spacetime, space and time are the direct result of particles or any matter which came into existence at t=0, the big bang?

20. Apr 16, 2007

jimbobjames

@pervect - that's a good example.

Im trying to figure out the equation of a geodesic on a surface like that of the sphere (or more generally a 2-d surface given as a height function z=f(x,y))

Am I correct in saying that I need to:

1) figure out the metric for the surface in question
2) work out the non-zero christoffel symbols
3) write down the geodesic equation(s) and interpret

For the sphere (assume radius 1), I guess the metric is the diagonal matrix with 1 and sine^2 (theta) on the diagonal.

If we were to assume that the equator represents the coordinate time axis, and the zero degree line of longitude were to represent a spacial coordinate, h say (for height), then the metric is diagonal with -1 and sine^2 (t) - right?

I get that only the 0,0,0 and 0,1,1 components of the christoffel symbols are then non-zero - and equal to sin(t)*cos(t) - but the geodesic equation seems a bit strange.

Its like:

dt/(dTau^2) + (sin(t)*cos(t)) [(dt/dTau)^2 + (dh/dTau)^2 = 0

but I may have made a mistake - I've read a bit, but I have'nt practiced until now. The result should be the equation of a great circle - like the equator itself - but I cant see that from this equation.

Any tips?

21. Apr 16, 2007

Mentz114

$$ds^2 = dr^2 + r^2d\theta^2 +r^2sin^2\theta d\phi^2$$
is the spherical metric for polar coordinates r, theta, phi. If you are restricted to the surface then set dr = 0.

But you want the 4D space

$$ds^2 = dr^2 + r^2d\theta^2 +r^2sin^2\theta d\phi^2 - c^2dt^2$$

Last edited: Apr 16, 2007
22. Apr 17, 2007

jimbobjames

Thanks Mentz114.

Actually I was looking for the metric of the 2-d space (sorry if I didnt make that clear - not a 3-1 or a 2-1, just a 1-1 space - I know its not very physical but I wanted to see geodesics on 2-d surfaces).

The only relevant coordinates are theta and phi as far as I can tell. I assumed r=1. And then I put theta = t (for this thought experiment - ie the equator represents the coordinate time axis). And I put r*dphi = dphi (r=1) = dh

thats why I got:

ds^2 = - c^2 dt^2 + sin^2 t dh^2

(but do correct me if this is wrong - will check out latex later).

Last edited: Apr 17, 2007
23. Apr 17, 2007

Mentz114

$$ds^2 = -c^2 dt^2 + sin^2( t )dh^2$$
is what you've written. Which is very weird. I'll work out the 3D problem you stated earlier.

I'm not sure what you're doing here.

To get the geodesics on the surface of a sphere you need the metric I posted earlier, with dr=0, r constant.

Last edited: Apr 17, 2007
24. Apr 17, 2007

Mentz114

For this metric -
$$ds^2 = d\theta^2 +sin^2\theta d\phi^2 - c^2dt^2$$
The Christoffel symbols are -

$$\Gamma^\phi_{\theta\phi} = cot\theta$$

and
$$\Gamma^\theta_{\phi\phi} = -sin\theta cos\theta$$

Last edited: Apr 17, 2007
25. Apr 17, 2007

jimbobjames

Thanks again.

What Im trying to see are geodesics on 2-d curved surfaces where time is one of those dimensions (and not an extra third dimension). So I am looking at 1 space and 1 time coordinate which I am calling t and h respectively (not 2 space plus 1 time like you have done).

If you put theta = t and phi = h then you get what I have written from what you have written.

$$ds^2 = -c^2 dt^2 + sin^2( t )dh^2$$

(But I am not an expert so now that youve hopefully understood what I am trying to do, can you check this please?)

Coordinate Time (t) along the equator corresponding to theta (because the radius is 1) and spacial height (h) going from the equator to the north pole. 1 space and 1 time dimension.

Cheers.