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Is spacetime a function of itself, objects within it, or both? I am struggling to understand just what is spacetime without objects in it (or theoretical reference points) and thus no frame of reference except itself. It seems that spacetime without objects (on its own) is more of an abstraction and would be hard if not impossible to quantify without someTHING in it by which it could be measured. I feel like I must be missing something.

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Voting to close as not a real question. This is much too vague and general to have a meaningful scientific answer. –  Ben Crowell Apr 22 '13 at 20:48
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Perhaps I could rephrase? What I mean to ask is: is space time the backdrop within which objects move so that it is not possible to declare space time itself a relative function of the object in question? –  Argo Apr 22 '13 at 20:57
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Two points, Zach. First rephasing like---whle encouraged---that should be edited into the body of the question, not put in a comment. Secondly you will then have a physics question, but it will be one contradicted by nearly 100 years of experimental evidence. All the weak field results of general relativity (gravitational red shift, gravitational time dilation, frame dragging, gravitational lensing) have been tested and found to agree with the theory. If you want to edit the question I will re-open, but there isn't really much point. –  dmckee Apr 23 '13 at 16:08
    
dmcmke, I wouldn't mind rephrasing if that is okay. I think you may feel I have a presupposition that I'm trying to get affirmed that is contrary to GR. believe me this is not the case. I'm merely trying to clear up some of my own confusion on the matter. I apologize if my question was vague. Could I repost? –  Argo Apr 23 '13 at 21:15
    
OK, good luck. BTW, if you want someone to be notified of a comment on a post they did not write use an at-tag like, @Zach. –  dmckee Apr 24 '13 at 4:31
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1 Answer

It is General Relativity that changed our concept of space and time. Even special relativity assumes that time and space exist as coordinates even though it changes the metric with respect to the Galilean (Newtonian) relativity.

As physics started to be rigorously mathematically formulated in the eighteenth century , space and time were considered as Euclidean coordinates, conceptually preexisting whether there were any masses or energies in the universe.

With General Relativity, the mathematical formulation:

eq1

The space description mathematically is on the left of the equation,

eq2

it is the curvature scalar. It says how much differentially spacetime curves.

On the right is the energy-momentum tensor. If there is no energy or momentum in the universe then things become more complicated.The solution to the second equation with the right hand side=0 depends on the model of general relativity. If there exists a cosmological constant or it is zero.

eq3

Taking the trace of both sides reveals that the constant of proportionality k for Einstein manifolds is related to the scalar curvature R by

eq4

eq5

Therefore, vacuum solutions of Einstein's equation are (Lorentzian) Einstein manifolds with k proportional to the cosmological constant.

In the case of the cosmological constant being zero one gets zero curvature, and recovers the pseudo-euclidean space of special relativity, which of course is of no descriptive of nature use, since there is no energy or momentum, i.e. observers, to observe this. spacetime.

Is spacetime a function of itself, objects within it, or both?

Spacetime as a term depends on its definition. I have described what the definition is for physics. One can mathematically define coordinates in infinitely many ways. As far as observations go up to now the spacetime nature has chosen is General Relativity's as described above which stops existing if the energy momentum tensor is zero since, for physics, no massless/energyless observers can exist, and physics is about observations.

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Anna, very very helpful. Thank you for taking your time to spell this out for me. I think my problem is that I was attempting to view spacetime apart from the mathematics; I now realize that, as you say, without a value existing within the coordinate system, the only "value" left is zero. And, as you point out, you cannot observe zero. Physics without tha math is purely philosophical, and it is impossible to observe a thing which is not, either literally or mathematically OBSERVABLE. Thanks again. –  Argo Apr 24 '13 at 14:03
    
I just realized I probably could have grasped this question myself if i had thought a bit more logically. Spacetime cannot be a function of spacetime (redundant); it cannot exist as a product of something else which is not a function of spacetime (contradictory). Thus it must be a mathematical abstraction. –  Argo Apr 24 '13 at 20:28
    
Well, it is OK to ask someone else, it is the method of transferring knowledge accumulated over centuries. I should have added in the concluding paragraph that: that is how the SR framework blends with the GR, as the very weak gravitational sources become more important with size and location. It is the weakness of the gravitational constant that allows the SR framework to be very good for studying particle physics, as an example; with only SR, as if they are in a vacuum as far as GR is concerned. –  anna v Apr 25 '13 at 4:04
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