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what about a theory where no time exists, only discrete space, steady states and occasion chains. what would speak against this?


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closed as not a real question by Sklivvz, Georg, Deepak Vaid, Shog9 Apr 5 '11 at 16:06

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Results from a no-time theory should be translated into a timed-reference to be tested in timed-laboratories (only known labs till now), then the complete theory would include that time translation front end, then it would include time. – HDE Apr 5 '11 at 14:40
why is this not a real question? in thermodynamics you also throw away the concept of time as you look to all states of a phase space at once. I think its funny how different people react here. thanks and repekt to those who waste a short thought about this crazy idea. – headkit Apr 6 '11 at 8:38

I am not sure this is a completely nonsensical question, though it is phrased in an odd way. Julian Barbour is pretty much into the notion that time does not exist. This is based largely on the Wheeler DeWitt equation ${\cal H}\Psi[g]~=~0$, which is a quantum version of the Hamiltonian constraint in ADM relativity.

I could well enough imagine presenting how time exists, but space does not. We could presume there is some one dimensional space, a line or curve, and there is a fibration on that space by a three dimensional space. This internal space is a symmetry of the dynamics of this one dimensional parameterized space we label as time. This then connects to relativity when we consider the metric line element $$ ds^2~=~-c^2dt^2~+~g_{ij}dx^idx^j, $$ where mixed time-space metric components are not included. We have here two notions of time. The first is the proper time $\tau~=~s/c$, which is the invariant of relativity. The other time is a coordinate time $t$, which is not an invariant.

The obvious question to ask is whether $ds$ is real. We can multiply it by $mc^2$ and define an action according to the extremal principle of the proper interval $$ S~=~mc\int ds, $$ which appears real in some sense. It has units of action, or angular momentum, which is a measurable quantity. Yet there is something a bit troublesome about all of this. How does the observer on this world line actually measure this interval? A clock is employed which must have some system of oscillations, such as a spring. Yet this is measuring the invariant interval according to something carried on that world line that deviates from the world line. Hence some sort of nongeodesic motion is being used to define or measure an interval along a geodesic path. Of course I am thinking primarily of a mechanical clock, but an atomic one still appears to hold for an EM field must be applied to knock electrons in the Ce atoms.

This Lagrangian is measured according to something which is not invariant. So we might then consider that action as $dS~=~pdq~-~Hdt$. Now we have some Hamiltonian, which might include a part for the dynamics of the clock. Hamiltonians must be specified on some Cauchy surface of data with a coordinate time direction. Yet this has gotten us into some funny issue, for to define an invariant interval it appears that we need a coordinate defined clock.

So far we have some identification of $Hdt$, or the square of this, with the $c^2dt^2$ in the interval above. We then have that the bare action term $pdq$ is identified with $mc\sqrt{g_{ij}dx^idx^j}$. So we have a bare action given by our fibration, but we also have some constraint, where $H$ acts as a Lagrange multiplier. So we then have our one dimensional curve defined in a spacetime, where the space is the space of fibration and the Lagrange multiplier determines the symmetry of that fibration which is the Lorentz group.

Now to make things curious, we could imagine this picture as dual in some ways to the picture where time does not exist, but space does. The duality might then have a noncommutative coordinate geometric content in quantum gravity.

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True. This might not be a completely nonsensical question. But I'm quite sure that the OP's desire was not to elicit a reasoned, scientific response. Read the question again and tell me if you detect any signs that the OP cares for a discussion at this level. So while, yes, nice answer, -1 because you're simply encouraging crazy questions. – user346 Apr 5 '11 at 16:07
I Always think that the best questions born out crazy, and they get saner later. +1 – lurscher Apr 5 '11 at 16:22
The basic question is alright. The suggestions about chains and other alternatives do not make much sense. I see the question is closed. – Lawrence B. Crowell Apr 5 '11 at 18:07
thanx for your thoughts! I am happy someone thinks this is a real question. :-) – headkit Apr 6 '11 at 9:29
what we measure is not time (as it seems to be an imaginary construct) but states of energy in an uncertain-discrete space. or measurment of time depends on changing states of positions and energy levels as you said. but what keeps the game in motion when there is no time? could't it be some kind of cause-and-effect-chain, pushing every unstable state in the phase space into its next "position". maybe there is some kind of time, but it is discrete, some kind of steps of phase space states which overlap in some uncertain range. – headkit Apr 6 '11 at 9:29

a physical theory is meant to describe observations. It may be an important pastime of an idle, crazy pseudo philosopher to ponder over imaginary speculations, but the job of physicists are much more prosaic and constrained. You have asked what would be against your "hypothetical theory". Nothing much, except that it is a good for nothing.

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@sb1 why do you waste your energy providing answers to such nonsensical question? Also -1 is not mine. But, again, what are you saying that you couldn't have said in a comment. I say this as a friend. – user346 Apr 5 '11 at 14:24
@Deepak Hmm maybe you are right. I just start typing in the white area. I don't care if I get a -1 for this. – user1355 Apr 5 '11 at 14:36
@sb1, good answer, but as Deepak said, such questions should be ignored. The downvote is from th Qm, I bet. And: some of the most advanced theories in cosmology sound equally "wild" to a layman. So he may come to the idea to "create" other conjectural theories "without" something. – Georg Apr 5 '11 at 14:40
@Georg some of the most advanced theories in cosmology sound equally "wild" to a layman ... absolutely. And this is the wonder that draws you into physics in the first place. The hard part is realizing how much mud you have to trod through to eventually get to a place of understanding. – user346 Apr 5 '11 at 14:42
@ Deepak : Per aspera ad astra! :=) or :=( ? – Georg Apr 5 '11 at 15:00

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