# Gravity/Orbit Question

1. Dec 16, 2006

### ||spoon||

In GR Einstein said that the sun (and other planets) bend space-time yeh.. and we are on the verge of that "indentation" and that is why we (the earth) orbits the sun?. If thats right arent we being accelerated towards the sun because wouldnt we move down the slope?

Also if space is bent by the sun or planets does that mean that time is also?

thanks,
-spoon

2. Dec 16, 2006

### LURCH

We aren't exactly "on the verge", we are within the area that is being deformed by the Sun. Mars is further away than we are, but it is still inside the so-called "gravity well. Even as far out as Pluto and well into the Oort cloud, the Sun's gravitational influence dominates the motion of objects.

Truth is, there is no real "verge", or edge of the area that is curved by gravity, there's just a decreasing amount of influence gravity has on things (with increase in distanc, of course) untill the effect of gravity is so slight that it can be ignored.

So, in answer to your first question, I will say both yes and no. Yes, we are being accelerated toward the Sun, but no, we're not getting any closer to it. As we travel around the Sun our momentum ("centrifugal force", one might say) is accelerating us away from it. The Sun's gravity is accelerating us toward it. The two cancell out, and we go nowhere fast (about 40 miles per second, I believe).

In answer to your second question, an unqualified "Yes, gravity bends space and time", or "gravity bends spacetime".

3. Dec 16, 2006

### ||spoon||

is it possible to draw this in 3D? because there would be a "well" form every direction would there not?

4. Dec 16, 2006

### ||spoon||

i mean not just drawing what the planets would be "sitting" on but also what is around them?

5. Dec 16, 2006

### pervect

Staff Emeritus
A full drawing would have to be in 4-d, not 3-d, because it is space-time that is curved (you asked about this earlier).

It is possible to draw smaller dimensional "parts" of the entire picture, these are known as "embedding diagrams".

In fact, it is what can be roughly described as the "time" part of the curvature that is the most important. The effects of this curvature exhibit themselves as "gravitational time dilation".

Newton-Cartan theory, an intermediate step on the way to full General Realtivity, can in fact be used to derive Newtonian gravity only from the existence of gravitational time dilation (and the principle of least action). Some knowledge of Lagrangian mechanics (the principle of least action) would be needed to follow the derivation, though.

http://www.eftaylor.com/leastaction.html has a lot of resources discussing the princple of least action, I couldn't find anything elementary that talked about Newton-Cartan theory, though, i.e. something that does the derivation I talked about above.

Last edited: Dec 16, 2006
6. Dec 16, 2006

### MeJennifer

Looks like I need to become a bit more rigorous and exact in my understanding of GR because I was under the impression that gravitational time dilation is expressed by both the curvature of the spatial and temporal dimensions.
I certainly do not want to learn the wrong thing.

Could you perhaps explain how this is the case that only the curvature of the temporal dimension contributes to gravitational time dilation?

7. Dec 16, 2006

### Thrice

The schw. metric is not time dependent so you can split up a distortion in spacetime into its space and time components. The space distortion ensures that the radial coordinate no longer measures distance & the time distortion ensures that the proper time between two events is no longer t2-t1.

8. Dec 16, 2006

### ||spoon||

thanks for the input :-)

9. Dec 16, 2006

### DaveC426913

I realize you're trying to simplify things, but there's no need to throw myths such as centrifugal force in.

It is our momentum - our inertia - that keeps Earth wanting to go in a straight line. This straight line path would carry Earth away from the sun.

10. Dec 16, 2006

### MeJennifer

So you would say that for instance when we observe a photon coming from a star it is redshifted because of curvature of the time dimension alone?

11. Dec 17, 2006

### pervect

Staff Emeritus
Let's take the simplest possible example.

Suppose we have a 2 dimensional metric, one space coordinate, one time coordinate, and that we have a time dilation factor g00 that depends on position, but not on time. This gives us a metric of the form

ds^2 = g00(x) dt^2 - dx^2

I suppose I should add that I've assumed that c=1. The rindler metric for the "uniform gravitational field" of an accelreated observer can be put into this format, for example.

Then the function x(t) which extremizes (maximizes, usually) proper time will be the function which extremizes

$$\int ds = \int \sqrt{g00 - \left( \frac{dx}{dt} \right) ^2} dt$$

This function must satisfy the Euler-Lagrange equations:

http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html
http://en.wikipedia.org/wiki/Euler-Lagrange_equation

with
$$f(t ,x ,xdot) = \sqrt{g00 - xdot^2}$$

and we must note that g00 is a function of x, and xdot is defined as dx/dt, the coordinate velocity.

Now we will make tons of simplifications - feel free to work out the full details if you can and if you really want to. These simplifications make the problem essentially Newtonian, so we can compare it with the Newtonian answer, and easy to work.

We will assume that g00 is very nearly unity. This is the normal state of affairs, gravitational time dilation on the Earth's surface is only a part per billion, for instance.

We will also assume that xdot is very small. Since we have assumed that c=1, this means that we assume that we are dealing with velocites much smaller than the speed of light, which is also usually the case.

We can then say by using a Taylor series approximation that

$$f(t,x,xdot) \approx \sqrt{g00} - \frac{1}{2 \sqrt{g00}} xdot^2$$

The Euler-Lagrange equations tell us that the path that extermizes proper time is then approximately.

$$\frac{d}{dt} ( \frac{\partial f}{\partial xdot} ) = \frac{\partial f}{\partial x}$$

With our simplifying assumptions, this leads us to

$$- \frac{d}{dt} xdot = \frac{1}{2} \frac{\partial g00(x)}{\partial x}$$

Since (d/dt) xdot is the acceleration, we see that the acceleration is just
$$-\frac{1}{2} \frac{\partial g00(x)}{\partial x}$$

We see that the rate of change of time dilation with respect to position acts just like a "force" as far as the equations of motion go. If time dilation is not a function of position, bodies following geodesics (i.e. bodies extermizing proper time, which means that they satisfy the Euler-Lagrange equations) do not accelerate. When time dilation is present and varying with position, we can interpret the equations of motion to be the same as if we had an actual force present. We call this force "gravity".

Last edited: Dec 17, 2006
12. Dec 17, 2006

### MeJennifer

So let me confirm Pervect that the redshift of a photon coming from a star can be explained completely by the curvature of the time dimension alone.

Out of curiosity Pervect, how many more situations are the in GR where we conveniently can isolate the Lorentz metric in separate time and spatial components?

Last edited: Dec 17, 2006
13. Dec 17, 2006

### Thrice

Gravitational time dilation, yeah. I'm trying to avoid answering it tautologically. If the metric isn't time dependent, you can define distances because the integral of some spatial element dl doesn't depend on the chosen world line. Then you turn up the mass & find the radial distance dR > the coordinate differential dr. Works about the same for time. I'll post more if I can track down my class notes.

14. Dec 17, 2006

### Thrice

Heh. Thank you for that. It was a pleasure to read.

15. Dec 17, 2006

### Thrice

Metric

$ds^{2} = (1-\frac{r_G}{r}) c^2dt^2 - (1-\frac{r_G}{r})^{-1}dr^2- r^2 (d\theta^2 + sin^2\theta d\varphi^2)$

spatial line element (dt =0)

$dl^{2} = (1-\frac{r_G}{r})^{-1}dr^2 + r^2 (d\theta^2 + sin^2\theta d\varphi^2)$

radial distance ($d \theta = d \varphi =0$)

$dR^{2} = (1-\frac{r_G}{r})^{-1}dr^2$
$\Rightarrow dR > dr$

proper time

$d\tau = \sqrt{g_{00}} dt= \sqrt{1-\frac{r_G}{r}} dt<dt$

(i guess this would be a good place to plug my https://www.physicsforums.com/showthread.php?t=147741" :p)

Last edited by a moderator: Apr 22, 2017
16. Dec 17, 2006

### MeJennifer

Well it seems that both Pervect and Thrice think that gravitational time dilation is due to the curvature of the time dimension only.

Sorry but I am not convinced in the least.

To me gravitational time dilation is caused by the combined curvature of time and space.

17. Dec 17, 2006

### pervect

Staff Emeritus
Actually, that's not quite exactly what I was trying to get across. I was trying to think about how to explain my point better, but the thread, and your thoughts ran on without me ....

Here is a quote that might help, talking about the Newton-Cartan theory that I was mentioning before.

As far as GR goes, what I am trying to say is that usually the curvature of space is not terribly important. Important excpetions are very fast moving objects, like light. (Remember some of the threads where we talked about the fact that light is deflected twice as much in GR as it is in quasi-Newtonian theories).

If you only want to explain Newtonian gravitation for slow-moving objects (and not add in the refinements of genral relativity), you don't need space-curvature at all. The Newtonian component of GR, what we usually think of as "gravitational force" is all contained in the time component. Only some usually (but not always) small relativistic "corrections" are contained in the space part of the curvature.

Onto the quote (from MTW) about Newton-Cartan theory:

and to note that you have curvature (in the exacting sense of a non-zero curvature tensor) when you parallel transport a vector around a closed loop, and it changes.

So when I'm talking about the time component of curvature being important, I'm saying that part of the loop that you parallel transport the vector around must be a transport in time (a time-like line on a space-time diagram), not a transport in space in order for there to be any effect on a vector in Newton-Cartan theory.

In GR (as opposed to Newton-Cartan theory, which the above quote is about) there is some curvature of the spatial slices as well. But as I said before, it's basically a second order effect - it's not needed to explain the Newtonian component (gMm/r^2) of gravity.

Last edited by a moderator: May 2, 2017
18. Dec 18, 2006

### Thrice

Well you can't fault me for not trying. Work out the math see what you get.

19. Dec 18, 2006

### pervect

Staff Emeritus
I think, on reflection, that my post 11 may not have been as clear as it could have been. Let me skip lightly over the details of the math, and go over the outline of what I was trying to accomplish once again. (The intent is that one might then re-read the post with a better understanding of what its goal is).

The overal goal is this. We want to use the princple of least action to determine how particles move in a curved space-time. We will model the curved space-time in a standard manner, using a space-time metric. This may not be totally clear unless one is already familiar with metrics, I suppose.

So I'll talk briefly about metrics, but I'm assuming that this is a review, not the first time the reader has heard the term, because I don't want to write an entire book. The reader who isn't familair with metrics can either give up now (he's probably given up long ago), or do some more reading :-).

Metrics can be used to model curved surfaces (such as the surface of the Earth) in applications other than relativity. The sole purpose of a metric is that it allows to allow us to calculate the distance between two nearby points (in relativity, we calculate not the distance but the Lorentz interval) when the surface (or, in relativity, space-time) is curved.

Next off, we know (or have heard) that when particles follow geodesics in space-time, that their path extremizes their proper time. So, given the path of the particle, position as a function of time, or x(t), how do we calculate the proper time? In special relativity, we would calculate the proper time between two nearby points as being equal to the Lorentz interval between those points. In general relativity, we need to modify the SR calculation by using the metric to calculate the Lorentz interval. It is still true, however, that the proper time is equal to the timelike Lorentz interval.

OK, so by applying the metric, we now have a function which computes the proper time interval given the trajectory x(t), which we've done by calculating the Lorentz interval ds between two nearaby points, and integrating. This is basically the process of using an intergal to find the length of a curve (but in this case, the "length" of the curve is the Lorentz interval, a time interval rather than a space interval).

Given this intergal, how do we find the trajectory which extermizes proper time? This is a problem in the calculus of variations. Rather than get into the details, I provide some web references to the standard, already worked-out answer.

The answer to this problem is that if the particle obeys the Euler-Lagrange equations, it extremizes the associated function (some more exacting authors will call this a functional).

Don't tell anyone (well, OK, go ahead), but I've forgotten the details of how the Euler-Lagrange equations were originally derived anyway - I just use them as a tool in my mathematical "toolbox" without thinking about exactly where they came from.

At this point, we've worked out our function (or functional) which gives us the proper time, and we know that the equations of motion are given by applying the Euler-Lagrange equations to our function.

The remaining step is the actual solution to these equations, during which we make many approximations to simplify the process.

The end result is the equations of motion for a particle in a curved space-time. By inspecting this solution, we see that it is equivalent to Newton's laws with an added "force" term. This added "force" term can be interpreted as being phyically real as the "force" of gravity. It can also be equally viewed in the manner in which we derived it, as an abstract term resulting from solving for the equations of motion of a particle in a curved space-time by using variational principles.

Last edited: Dec 18, 2006
20. Dec 18, 2006

### MeJennifer

I assume you goal was to answer my question as formulated in the prior posting:
Right, and the Lorentz interval includes both time and space.

Right, so to me that means that if someone would ask me if gravitational time dilation is caused by the curvature of time only, the answer would be simple and resounding "no".

Given the non linear nature of GR I would never state it in these terms.

By the way explaining GR in terms of Newtonian theory approximations and low energy examples only is destined to give people the wrong understanding IMHO.

Last edited: Dec 18, 2006