StringJunky Posted April 20, 2017 Posted April 20, 2017 (edited) I am prompted by something mordred said: space is volume. I have no problem with this but I was wondering, when we see the co-ordinates xyzt, does this mean that time is a part of space or just that it is treated in conjunction with space but time is actually a property of the fields within space, bearing in mind the prior assumption that space is just volume? Hope this makes sense. I didn't know where to put this as it's a general question rather than necessarily to do with Relativity specifically; quantum physics may have a position on it. Edited April 20, 2017 by StringJunky 1
geordief Posted April 20, 2017 Posted April 20, 2017 "What do you understand by "space is volume"? 3D space ? As I picture it 3D space is an idealization (and a mathematical model). So that phrase "space is volume" would have to be rewritten to "a volume in spacetime is the region enclosed by 2 events in the model"
studiot Posted April 20, 2017 Posted April 20, 2017 (edited) It's a good question that deserves a good straight answer. +1 Time is not a property of space. Yes it can be considered a 'property' of fields, but that suggests there is no time where there are no fields which presents a philosophical difficulty. The notion that Time is a necessary property of fields was first appreciated by Faraday and extended by Maxwell in their treatment of the interaction between material bodies and their rejection of 'action at a distance', inherent in Newton's and Coulomb's theories. Edited April 20, 2017 by studiot 3
timo Posted April 20, 2017 Posted April 20, 2017 I'm going to propose the opposite straight answer Time is a coordinate dimension akin to the known spatial dimension. Hence, you speak of the coordinates x, y, z and t or the coordinate tuple (x,y,z,t). Fields are functions that are defined on the 4-dimensional space described by these coordinates, written i.e. functions of type f(x,y,z,t) (or [math]\psi(x,y,z,t)[/math] or something similar if you prefer fancy greek letters). 1
studiot Posted April 20, 2017 Posted April 20, 2017 I'm going to propose the opposite straight answer Time is a coordinate dimension akin to the known spatial dimension. Hence, you speak of the coordinates x, y, z and t or the coordinate tuple (x,y,z,t). Fields are functions that are defined on the 4-dimensional space described by these coordinates, written i.e. functions of type f(x,y,z,t) (or [math]\psi(x,y,z,t)[/math] or something similar if you prefer fancy greek letters). But that 4 dimensional 'space' is a mathematical construct and does not correspond to the 3D space or volume mentioned by the OP. So yes you are right if you construct a phase space to include a time axis then time is necessary by tautology.
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 But that 4 dimensional 'space' is a mathematical construct and does not correspond to the 3D space or volume mentioned by the OP. So yes you are right if you construct a phase space to include a time axis then time is necessary by tautology. Yes, I'm talking about simple 3D space, as we know it, with time. Basically, if we remove all the fields, is time still there?
studiot Posted April 20, 2017 Posted April 20, 2017 Yes, I'm talking about simple 3D space, as we know it, with time. Basically, if we remove all the fields, is time still there? I would say yes, by considering this thought experiment. Say you have a very very powerful magnet whose field extends a substantial distance (ie throughout a substantial volume). Now say you get you super blow torch out and heat this magnet up past the Curie point so the field disappears. How will you distinguish the state of the system with and without the field, if time was not still there?
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 I would say yes, by considering this thought experiment. Say you have a very very powerful magnet whose field extends a substantial distance (ie throughout a substantial volume). Now say you get you super blow torch out and heat this magnet up past the Curie point so the field disappears. How will you distinguish the state of the system with and without the field, if time was not still there? I see what you mean but isn't that because there is a time-dependent observer present to measure the difference? The observer is a collection of fields themselves, aren't they? I know there's not an an easy answer to this and my basic knowledge is limited but I can try.
kjelleman Posted April 20, 2017 Posted April 20, 2017 I guess we focus on physical time here, not aging or evolutionary. In physics time is defined through Newton's force law. It is then easy to understand both space and time by utilizing relational mechanics which was first postulated by Leibniz. Here space is just distance between objects and time corresponds to the motion of the objects. Then it is clear that there is no absolute time nor absolute space, all must be related to something, a problem we encounter when trying to define meter and second. Having this view of space and time, relativity e.g. becomes a piece of cake, both SR and GR. 1
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 I guess we focus on physical time here, not aging or evolutionary. In physics time is defined through Newton's force law. It is then easy to understand both space and time by utilizing relational mechanics which was first postulated by Leibniz. Here space is just distance between objects and time corresponds to the motion of the objects. Then it is clear that there is no absolute time nor absolute space, all must be related to something, a problem we encounter when trying to define meter and second. Having this view of space and time, relativity e.g. becomes a piece of cake, both SR and GR. Yes. I've read somewhere that without objects there is no space.
Strange Posted April 20, 2017 Posted April 20, 2017 Yes, I'm talking about simple 3D space, as we know it, with time. Basically, if we remove all the fields, is time still there? I would say yes, because GR (where the idea of space-time comes from) doesn't include fields as essential components. They, or rather their energy, can be added in the equations. Yes. I've read somewhere that without objects there is no space. Again, in GR, you can have space-time completely empty of energy or matter. This is a useful class of models for solving problems in GR. 1
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 I would say yes, because GR (where the idea of space-time comes from) doesn't include fields as essential components. They, or rather their energy, can be added in the equations. Again, in GR, you can have space-time completely empty of energy or matter. This is a useful class of models for solving problems in GR. That's two votes for yes, although it would be easier if the opposite were true.
koti Posted April 20, 2017 Posted April 20, 2017 I'm wondering...when dealing with GR we're dealing with a concept of spacetime which is inseparable. Does it make sense to ask if one is the property of the other?
Mordred Posted April 20, 2017 Posted April 20, 2017 It's a good question that deserves a good straight answer. +1 Time is not a property of space. Yes it can be considered a 'property' of fields, but that suggests there is no time where there are no fields which presents a philosophical difficulty. The notion that Time is a necessary property of fields was first appreciated by Faraday and extended by Maxwell in their treatment of the interaction between material bodies and their rejection of 'action at a distance', inherent in Newton's and Coulomb's theories. Good answer +1 though every answer in this thread thus far has been good. ok lets play model construct. Start with a universe with no fields or matter. You just have a volume. There is no anistropy regions so you will not have time dilation. To model this universe you would use 3d Galilean relativity. So although time does exist, there is no need to include an independent variable for time. You have an effective 3D universe. Now add fields and other particles you now develop anistropy regions and spacetime curvature comes into play. (In regards to Studiots reply, particles are field excitations). So Now you require the independent variable for time. Short answer is time is always present but how you model time depends on the fields involved and content distribution as to whether you require 3d or 4d to accurately model that universe. 1
MigL Posted April 20, 2017 Posted April 20, 2017 Can you remove ALL the fields ? I remember something AJB once said... 'In GR geometry is the field' And geometry is space-time.
Mordred Posted April 20, 2017 Posted April 20, 2017 (edited) Can you remove ALL the fields ? I remember something AJB once said... 'In GR geometry is the field' And geometry is space-time. yes he is correct a field is any collection of objects/coordinates or events. So under GR you would still have a field (coordinate/event field). Good catch I should have stated remove all matter and force/matter fields. Edited April 20, 2017 by Mordred
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 (edited) yes he is correct a field is any collection of objects/coordinates or events. So under GR you would still have a field (coordinate/event field). Good catch I should have stated remove all matter and force/matter fields. When I used the word fields, I was using it to represent everything besides space, including matter , since it is also made up of fields. i could have been more explicit about that. As you say: matter is excitations in a field. In conclusion then: we have to say time is always present even in a toy universe with nothing in it but it behaves the same everywhere since their is nothing to change the rate that it ticks in any given spatial co-ordinate? Edited April 20, 2017 by StringJunky 1
Mordred Posted April 20, 2017 Posted April 20, 2017 (edited) correct this would also hold true in any homogeneous and isotropic distribution. Both observer and emitter IF frames would be under the same conditions. Edited April 20, 2017 by Mordred 1
StringJunky Posted April 20, 2017 Author Posted April 20, 2017 correct this would also hold true in any homogeneous and isotropic distribution. Both observer and emitter IF frames would be under the same conditions. Thank you.
koti Posted April 20, 2017 Posted April 20, 2017 (edited) But our "toy" universe could never be really empty as there are things happening within spacetime on the quantum level right ? Edited April 20, 2017 by koti
Mordred Posted April 20, 2017 Posted April 20, 2017 According to QM's Heisenburg uncertainty principle the lowest possible energy state would be the zero point energy at [latex]1/2\hbar w [/latex]. So yes you will always have fluctuations. However I should note those fluctuations have less than a quanta of energy so will not have observable "action" unless you exceed a quanta under boundary confinement. At this point you now have a real particle not a virtual 1
koti Posted April 20, 2017 Posted April 20, 2017 According to QM's Heisenburg uncertainty principle the lowest possible energy state would be the zero point energy at [latex]1/2\hbar w [/latex]. So yes you will always have fluctuations. However I should note those fluctuations have less than a quanta of energy so will not have observable "action" unless you exceed a quanta under boundary confinement. At this point you now have a real particle not a virtual Right. Thanks for the explanation Mordred.
Strange Posted April 21, 2017 Posted April 21, 2017 But our "toy" universe could never be really empty as there are things happening within spacetime on the quantum level right ? But if the "toy universe" is purely described by GR then there won't be any thing happening on the quantum level. 2
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