
#1
Nov2112, 10:01 AM

P: 962

Can the metric of special relativity be derived from requiring the infinitesimal line segment, dτ, to be invariant in space and time? If we parameterize a line segment by the variable τ marked off along the line (that exists in space and time dimensions) is the length in τ of that line segment only invariant with respect to the Lorentz transformations? Or are there other coordinate Xformations for which dτ would also be invariant? Thank you.




#2
Nov2112, 04:18 PM

PF Gold
P: 4,081

This thread http://www.physicsforums.com/showthread.php?t=651640 will be of interest.




#3
Nov2812, 11:42 AM

P: 962

I'm given to understand that dτ^{2}=dt^{2}dx^{2} = dt'^{2}dx'^{2} when (t',x') are the Lorentz transformation of (t,x). Perhaps it's instructive to consider in what circumstances dτ should want to be considered invariant wrt to coordinate changes. Maybe those requirements are the driving force behind the necessity of the Lorentz transformations. For example, the most obvious use of dτ is in the calculation of the line integral, [tex]\int_{{\tau _0}}^\tau {d\tau '} = \tau  {\tau _0}[/tex] which is the length of a line measured in terms of segments marked off along the length of the line. Then, of course, we can always place this line in an arbitrarily oriented coordinate system and express τ in term of those coordinates. So the question are: 1) when do we want to use the coordinates (t,x), and 2) when would we want ττ_{0} to be invariant wrt to those coordinates? As far as 1) goes, usually, we specify a curve in space by parameterizing the space coordinates with an arbitrary variable, call it "t". But since the x and t coordinates are arbitrarily assigned, the length of the curve can depend on the (t,x) coordinates. But if you specify that the length of the curve is invariant, then this requires the Lorentz transformations between coordinate systems. But for 2) what should require the length of the curve to be invariant? Perhaps if we have a more fundamental requirement like [tex]\int_{{\tau _0}}^\tau {f(\tau  {\tau _0})d\tau } = a[/tex] this will require the length of ττ_{0} to be invariant wrt to coordinate changes in (t,x). For example, maybe [itex]{f(\tau  {\tau _0})}[/itex] might be a probability distribution along a path so that its integral along the path must be 1 in any coordinate system. Did I get this all right? I would appreciate comments. Thank you. 



#4
Nov2812, 12:24 PM

PF Gold
P: 4,081

Does SR = invariance of dτ
Before I attempt to answer your questions, I should point out that the LT is the only known transformation that retains the causal structure of SR. Timelike, null and spacelike intervals retain this property under LT. Also, crucially, identifying the proper interval as the time recorded on a clock travelling on the curve in question eliminates all clock paradoxes (timebomb paradoxes) because the clock times are invariant. I know you're most probably aware of this.
[tex] \int_{{\tau _0}}^\tau {f(\tau  {\tau _0})d\tau } = a [/tex] Right now I can't see any meaning in this. I'll have to think about it. 



#5
Nov2812, 12:44 PM

Physics
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PF Gold
P: 5,507





#6
Nov2812, 03:48 PM

P: 962

I appreciate your comments. I can see that you are knowledgeable in SR. However, I don't think you understand what I'm trying to do here. I'm trying to explain where the SR metric, with its speed of light, comes from to begin with. So I'm trying to avoid starting with the observed speed of light as in the usual development of SR. And I'm trying to find principles that would give rise to the metric of SR, with its implied speed of light.
So my thinking is that since the Dirac delta function is used in the development of QM, it might also be used to develop SR. And here's how I figure. Since we have [tex]\int_{  \infty }^{ + \infty } {\delta (\tau '  {\tau _0})d\tau '} = 1[/tex] we can embed this line integral in a background spacetime (t,x), where [itex]\tau [/itex] then becomes a function of the coordinates (t,x). And then it seems that the requirement that this integral remain constant wrt changes in coordinate system requires the coordinate transformations to be Lorentzian to keep [itex]{\tau  {\tau _0}}[/itex] invariant. Does this sound right to you? 



#7
Nov2812, 04:08 PM

PF Gold
P: 4,081

The speed of light is in the metric itself because for a null curve 0 = dt^{2}dx^{2} implies (dx/dt)^{2}=1. 



#8
Nov2812, 04:39 PM

Physics
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PF Gold
P: 5,507

If the latter is really your question, the best explanations I've seen that don't depend on any assumptions about the speed of light, start from the assumptions of translation and rotation invariance. That is enough to narrow down the possibilities to either the Lorentz transformation or the Galilei transformation, which is just the limit of the Lorentz transformation as c > infinity. More precisely, translation and rotation invariance are enough to show that there must be some "invariant speed" c that appears in coordinate transformations, but which may be infinite (Galilei transformation) or finite (Lorentz transformation). The only way to choose between these two options is by experiments like the MichelsonMorley experiment, which show that the invariant speed is finite. 



#9
Nov2912, 10:35 AM

P: 962

[tex]\int_{  \infty }^{ + \infty } {\delta (\tau  {\tau _0})d\tau } = \int_{  \infty }^{ + \infty } {\delta ((\tau '  \tau {'_0})/a)\frac{{d\tau }}{a}} = \int_{  \infty }^{ + \infty } {a \cdot \delta (\tau '  \tau {'_0})\frac{{d\tau '}}{a}} = \int_{  \infty }^{ + \infty } {\delta (\tau '  \tau {'_0})d\tau '} = 1[/tex] So it seems this integral is insensitive to changes in the [itex]\tau [/itex] coordinate. Does this mean that ANY changes in the backgound (t,x) could change the [itex]\tau [/itex] coordinate in arbitrary ways that still results in the integral being 1? In other words, the integral no longer depends on the [itex]\tau  {\tau _0}[/itex] not changing. Wouldn't that mean that this integral has no power to restrict the background transformations to the Lorentz transformations? Or does this scaling property allow only changes in [itex]\tau [/itex] by a constant factor? Can I still get the Lorentz Xformation from this? Perhaps it would be better to integrate a different function other than the Dirac delta. Then I'd not be dealing with the scaling property and the integral could restrict the background transformation of (t,x) to the Lorentz type. 



#10
Nov2912, 11:36 AM

Physics
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PF Gold
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#11
Nov2912, 12:25 PM

P: 962

Why the integral? As I said in post 3, the most obvious use of dτ is in the calculation of the line integral, ∫dτ′=τ−τ0 And when would τ−τ0 be invariant. Well you could have τ−τ0=constant. But then you ask why. So you notice that this is just another way of saying f(τ−τ0)= constant. And I suppose that this could always be put in the form ∫g(τ'−τ0)dτ′=constant And one thing to consider is the integral of the Dirac delta equal to one, always, in any coordinate system, right? 



#12
Nov2912, 12:56 PM

Physics
Sci Advisor
PF Gold
P: 5,507

Part of the problem here may be the issue of what counts as an "explanation". If I say that physics has to be translation, rotation, and boost invariant, does that count as an explanation of why only Galilei or Lorentz transformations are allowed? If I need experiments to tell me that the invariant speed is finite instead of infinite, does that count as an explanation of why Lorentz transformations, not Galilei transformations, are the right ones to use? To me these are questions about words, not about physics. Of course you could continue to ask questions, like "why does the invariant speed have to be finite rather than infinite?". AFAIK nobody has a good answer to that question, other than "that's what we find experimentally". Does that mean we don't have an explanation? Again, that seems to me to be a question about words, not about physics. 



#13
Nov2912, 01:22 PM

P: 2,890

The invariance of a finite c was Einstein second postulate of SR, there is no logical necessity for a postulate, it just happens to be backed up by experiments. Also invariance of dτ is not equal to SR since it is shared by GR for instance. EDIT: I see Peter has already answered, sorry about the overlapping. 



#14
Nov2912, 02:42 PM

P: 962

Perhaps the only reason that there must be a finite c is because if it were infinite, then everything would happen all at once and there would be no distinction between cause and effect. How's that sound? It's a little more basic than measuring c. 



#15
Nov2912, 03:13 PM

Physics
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PF Gold
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#16
Nov2912, 04:11 PM

P: 2,890





#17
Nov2912, 04:46 PM

P: 962

If c were infinite, you could cause and emission over there by forcing an absorbsion of a photon over here. My arguement is that it is not logical for nature to ever be ambiguous about cause and effect. And one way to do this (if not the only way) is to ensure that c is not infinite. Some people even try to derive the Lorentzian metric from the requirement of causality alone. Although I think they are presupposing causality and not arguing for its necessity to begin with.
EDIT: To continue with these thoughts... Suppose a particle could travel with infinite speed, it would be impossible to say where the particle was at that instant. It could be anywhere from the edge of the universe to your present location. It would turn a point particle at one instant into an 13.75ly line an infinitesimal instant later, which would be a totally different thing than a particle. Further, there would be no frame of reference in which the particle is at rest. Would this violate the differential structure of the universe? Doesn't causality require a different position for each value of time? 



#18
Dec112, 12:53 PM

P: 962

I guess my question is whether diffeomorphism invariance requires the choice of Lorentz transformations above Galileans transformations. Is diffeomorphism invariance between coordinate transformations a more basic necessity to physics than measuring the speed of light? I mean, if the laws of physics necessarily require diffeomorphisms between frames of reference, does that necessarily exclude the infinite speeds used in the Galilean transformations? Can there be a diffeomorphism between two frames of reference when one of them has and infinite speed? And if not, does that make diffeomorphism invariance more basic than the speed of light? Thanks.



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