Timeline for Can I fix a point in Minkowski space to give it a vector space structure?
Current License: CC BY-SA 3.0
21 events
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| Jan 27, 2015 at 21:55 | vote | accept | Stan Shunpike | ||
| Jan 12, 2015 at 20:36 | comment | added | Qmechanic♦ | Related post by OP: math.stackexchange.com/q/1083207/11127 | |
| Jan 12, 2015 at 20:21 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Jan 4, 2015 at 0:10 | comment | added | David Hammen | Sure you can. You're looking at it wrong. You pick this point as an origin, I pick that point as an origin, moving at some velocity v with respect to yours. We both assign coordinates to some third point in spacetime. What is the transformation from your representation of that point to mine (or from mine to yours)? | |
| Jan 3, 2015 at 22:07 | comment | added | Stan Shunpike | @DavidHammen So just to clarify...based on what has been said in this thread, we can "fix a coordinate system on our affine space" (ie choose Earth, lab rocket frame etc) but it only behaves as a vector relative to the Lorentz group. For more general transformations (eg spacetime translations) we cannot fix a point because this would be like having an absolute frame of reference. | |
| Jan 3, 2015 at 21:06 | comment | added | David Hammen | Exactly. Your statement in the question "the basic principle of Special Relativity is that no point in X is a preferred reference frame" seems to imply that you think there are no reference frames in SR. That's just wrong. A better way to look at the "no preferred frame" (first meaning) in SR is that all frames of reference are equally valid. You can use any frame you want. | |
| Jan 3, 2015 at 20:59 | comment | added | Stan Shunpike | @DavidHammen Oh, maybe I see your point. I remember reading in Wheeler and Taylor's book Spacetime Physics. They would always go between different frames. They had the Earth frame or the rocket frame or the lab frame. And obviously one could in this sense choose a preferred frame since all were defined in terms of the metric. But there was no absolute frame. I think that's the distinction you are making if I am understanding correctly. | |
| Jan 3, 2015 at 20:55 | comment | added | David Hammen | @StanShunpike - Absolutely. You used the phrase "preferred reference frame" to mean some frame that somehow stands out as unique (e.g., the aether frame that the Michelson–Morley experiment didn't find would have stood out as unique). There's another very different meaning of preferred frame, which is the frame in which one prefers to work. For example, the lab frame. Just because there is no preferred frame (first meaning) does not mean that there are no preferred frames (second meaning). That second meaning means one can use algebra instead of geometry. | |
| Jan 3, 2015 at 20:38 | comment | added | Stan Shunpike | @DavidHammen I am confused. Is your comment in agreement with the conditions stated in Phoenix87's answer? | |
| Jan 3, 2015 at 20:03 | comment | added | David Hammen | @CuriousOne - Re Is this the same idea by which one can define an "absolute" space in classical mechanics? - Not at all. There is no physical mechanism that can pick c! Sure there is. Point at it! Point at any spot you want! We do this all the time: Arbitrarily pick some point as an origin, and voila! you can assign coordinates to any point in an affine space. How else are you going to assign coordinates to a point? | |
| Dec 28, 2014 at 15:21 | comment | added | Stan Shunpike | Excellent. This was my impression and it seems to be the consensus on Physics SE as well. | |
| Dec 28, 2014 at 15:14 | comment | added | CuriousOne | Hmmm... it looks like to me they are characterizing the structure that has no physical meaning, at least not in the standard interpretation. | |
| Dec 28, 2014 at 14:03 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Dec 28, 2014 at 10:03 | answer | added | Phoenix87 | timeline score: 6 | |
| Dec 28, 2014 at 9:35 | comment | added | G. Paily | There should not be an physical meaning to picking a point. But you could think of it as an origin associated with a (very general!) class of observers. Of course, there is nothing to single out any particular inertial observer. | |
| Dec 28, 2014 at 9:07 | comment | added | Stan Shunpike | (Continued) In case $k$ is the field of the real numbers $\Bbb{R}$, $(X,V,\Bbb{R})$ is called real affine space. | |
| Dec 28, 2014 at 9:05 | comment | added | Stan Shunpike | Sure, let me see if I can provide some. On page 6-7, they say " the axiom system for $n$-dimensional affine space over a division ring $k$ consists of a nonempty set $X$, an $n$-dimensional left vector space $V$ over the division ring $k$, and an "action" of the additive group of $V$ on $X$. The elements of $X$ are called points and are denoted $x,y,z,...$; the elements of $V$ are called vectors and are denoted by $A,B,C,...$. Scalars are ways written on the left of vectors. The affine space defined by $X,V,k$ and the action of the additive group of $V$ on $X$ will be denoted by $(X,V,k)$. | |
| Dec 28, 2014 at 8:53 | comment | added | CuriousOne | Does the text give any more context? How are the authors exploiting the vector space property of this construction? | |
| Dec 28, 2014 at 8:46 | comment | added | Stan Shunpike | I had similar thoughts, but I really have no idea so that's why I posted the question. | |
| Dec 28, 2014 at 8:21 | comment | added | CuriousOne | Is this the same idea by which one can define an "absolute" space in classical mechanics? If it is, then it probably runs into the same problem as in classical mechanics: there is no physical mechanism that can pick c! All such "fix points" are equally unphysical, even though nothing stops us from using them in all of our calculations. In the end all they are expressing by their equal meaninglessness is the very degeneracy that makes the theory "relative" to begin with. Having said all of this, there is a high likelihood that I am completely misunderstanding the idea. | |
| Dec 28, 2014 at 8:02 | history | asked | Stan Shunpike | CC BY-SA 3.0 |