# Backround independence and the definition of distance

1. Jun 17, 2011

### jfy4

Hi everyone,

General relativity gives us the definition of the distance between two "events"
$$d=\int \sqrt{g_{\alpha\beta}dx^\alpha dx^\beta}.$$
I don't think this will complicate things but lets say its between two balls. Now like I said, with GR we have the definition of the distance between these two balls, which is possible through the gravitational field $g_{\alpha\beta}$. However, as far as I know the gravitational field does not prescribe the distance between the gravitational field and the balls. This troubles me since, the gravitational field is clearly a quantity that depends on what's close by, so to speak. However, there should not be a background distance to "tell" the gravitational field what it is "close" to.

In summary, while the gravitational field provides a metric for the distance between balls, it does not provide a metric, for the metric and the balls. How does the gravitational field "know" where the balls are?

Thank you,

2. Jun 17, 2011

### Passionflower

Do not mistake the spacetime distance between two events with the spatial distance between two objects.

Imagine the distance between two boats on a lake without any wind, one places a very light string over the flat surface of the water between the ships and and reads the length of the string to determine the distance. Now imagine the water having waves, you cannot really take this string over the waves between the boats because the waves are constantly moving and changing and the boats are as well.

3. Jun 17, 2011

### jfy4

My mistake, but to make my point in your language, how does the water know how close it is to the boats?

4. Jun 17, 2011

### Passionflower

It does not know how close the boats are, all that is happens is to 'know' what is immediately surrounding for gravity to work, we could integrate all those surroundings in stationary spacetimes to get some sense of distance between the boats but in non stationary spacetimes we can not.

In the case of two balls the spacetime is not stationary, in some solutions with two balls we even have gravitational waves interact at infinity and then 'coming back' in the solution.

5. Jun 17, 2011

### PAllen

I'll just add a little to what Passionflower said (which I agree with).

First, even mathematically, you integral is over some specified path. To describe a length along this particular spacetime path, the path must be spacelike. While in SR, there is unique geodesic which can unambiguously be called the distance, in GR, unless the points are close together, there will be multiple geodesics. Thus 'distance' rather than proper length of a path, is ill defined in GR. Further, a physically meaningful distance would have to be described in relation to some measuring method and apparatus, which would impose some definition of simultaneity, and that would specify the path you should integrate.

Ok, so that is just about distance between events in spacetime whose geometry is determined by the mass/energy around. If you are talking about 'balls', they are going to have timelike paths through spacetime. If small and not subject to any forces, these will be timelike geodesics. If they are not small, the Einstein field equations would need to be solved for the whole system including them (producing gravitational waves, and generally requiring numeric or perturbative methods of solution). It is the field equations that describe how each influences the geometry and responds (with finite propagation time) to the changing geometry produced by motion of other bodies. The whole point of field theory is that nothing know how far away something is. Each object produces influences with finite propagation speed and responds to local influences propagating from other objects, all in a complex, non-linear way (in general).

6. Jun 17, 2011

### pervect

Staff Emeritus
I'm not quite sure what your question is. Viewed as a geometric entity, the metric gives you the distances and the proper times between any two nearby points. You haven't talked at all about the tricky issue of splitting space time into a spatial part and a time part, the mere existence of a metric as a geometric entity won't necessarily perform that task for you. I'm not sure if it's useful to go into more on this point other than to point out it's something that needs to be accomplished, and that the metric of GR is a 4-dimensional entity, not a 3-dimensional one.

If you add to the metric Einstein's field equations (which you haven't mentioned at all yet), the metric also tells you the distribution of matter that is present - given enough mathematical sophistication and ability.

You'd need to take the metric, bang it around with some high-powered math, and get it to spit out the various tensors that describe its curvature. At the end, you'll come up with the Einstein tensor, which will be proportional to the matter density. The matter density won't be a single number, either, for reasons that are probably two complex to get into. But you will have, at the end, starting with the metric, using Einstein's equations, an expression for the matter density, even if it's more complex than what you're used to.

After you've done this, you can say "aha - this metric describes two balls, and their gravitational field". But it doesn't tell you really how to figure out what the metric you want is to get the two balls, it does tell you if you write down a metric if it describes two balls.

There's some more advanced stuff you can do - if you take the ADM route, you can specify some 3-d hypersurface at "some instant of time", with some three dimensional metric (which appears to be more in lines of your thinking, anyway, the GR metric includes space and time and you seem to be thinking only space). But you need something else, something that describes how the spatial surface is "curved" in its time embedding. (Technically it's K_ab, a rank two tensor, according to Wald). Given this data, you can in theory predict the global time evolution of the manifold, it's 3d metric, and the K_ab tensor field.

By "can do it" I mean that mathematicaly there is in theory a unique solution. Finding it numerically so that it makes physical sense and the errors stay within reasonable bounds is a job for a team of specialists with supercomputers.

Last edited: Jun 17, 2011
7. Jun 17, 2011

### jfy4

Thanks for the responses, I'm familiar with the curvature tensors etc... but that does not answer my question, probably since you couldn't tell what the question was. I'll try again, sorry for not being more clear. I'm going to do this completely by analogy with the hopes that this doesn't fail miserably.

If we consider and upside down canoe representing the curvature of spacetime, we assign a metric to the canoe. Now if we place two marbles on the canoe, yes they will start to move and the whole situation is full of math, but we can also put two tacks in and put a string between the tacks which gives the distance between these tacks. I hope we are ok so far.

My question is about that there is another distance here that we aren't considering, and that is the distance between the marbles and the canoe. I mean, we set them on there, but I mean in real-life, how do bodies know "where" they are relative to the gravitational field? Is there a mechanism responsible for a bodies relative location to the gravitational field?

8. Jun 17, 2011

### atyy

When we use the gravitational field as a spacetime metric, it is typically not background independent.

The reason is that we use test particles or ideal clocks to measure the proper time between two events. We assume that the test particles or ideal clocks do not contribute to spacetime curvature, and simply measure it. The background independence is due to spacetime curvature caused by all other mass-energy in the universe excluding our test particles and ideal clocks.

So in the canoe example, the most fundamental thing is we do not know what the "shape" of the canoe is until we put the balls in. However, we also do not have a meaning for "shape" until we can measure with string and tacks.

So more properly, the analogy is that we do not know what the "X-field" of the canoe is until we put the balls in. After the balls are put in, and we know the X-field, we interpret the X-field as "shape" by assuming that our string and tacks only measure the X-field without changing it. This latter assumption is wrong, since strings and tacks have mass-energy like balls and should also change the X-field. However, usually the change due to the strings and tacks is small enough to be ignored for practical purposes.

Last edited: Jun 17, 2011
9. Jun 17, 2011

### jfy4

Thanks, that's very enlightening. I'll think about that for a bit.

10. Jun 17, 2011

### Passionflower

We are OK in that the distance you mention is the spacetime distance between two events NOT locations.

Points in spacetime are events not locations!

The worldlines of the (free falling) bodies follow the local curvature.

11. Jun 17, 2011

### DrGreg

I don't understand what you are getting at here. Isn't it obvious that the distance between the marbles and the canoe is zero?
The "gravitational field" is everywhere, so the distance between any body and the "gravitational field" is zero.

12. Jun 17, 2011

### jfy4

Yes, this seems very much like the case, I'm not attempting to say I can't see that that isn't true, I'm wondering if there is a reason for this. Given that there is no background metric to discern this "distance" between the gravitational field and a body, how does a body know where it is relative to the gravitational field.

atyy response seems to suggest an answer, I'm mulling it over.

Thanks for the response.