Does SR = invariance of dτ

friend

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.

Mentz114

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

friend

Thank you. I'm trying to keep up with that thread. There's some interesting stuff there. I tried asking some questions there. But they seem to be focused on their own goals in that thread. So I'd like to post my questions here to see if I can generate interest and get comments here without cluttering up someone else's thread.



I'm given to understand that

dτ²=dt²-dx² = 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 τ-τ₀ 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 τ-τ₀ 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.

Mentz114

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. Time-like, 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 (time-bomb paradoxes) because the clock times are invariant. I know you're most probably aware of this.

It seems to me that the answer to 2) is 'always - if SR is to be consistent'. I don't understand the first question. We have to use some coordinates or other in this methodology.

[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.

PeterDonis

Yes.

As Mentz114 said, the answer to this is "always". The reason is that [itex]d\tau[/itex] is a direct observable; it corresponds to the elapsed time on a clock following the given worldline for an infinitesimal segment. Direct observables must have the same value in all coordinate systems, so their values have to be invariant under coordinate changes.

Exactly; that's the main point of that other thread Mentz114 referenced.

It's also the time elapsed on a clock following the given worldline from [itex]\tau_0[/itex] to [itex]\tau[/itex]. So as I said above, it's a direct observable.

Yes, and the value of [itex]\tau[/itex] must be the same regardless of which coordinates you express it in.

As opposed to what other coordinates?

Always.

Actually, [itex]\tau[/itex] itself can be used to parameterize the worldline. The coordinates (t, x) are then functions of the parameter [itex]\tau[/itex].

No, they're not, because this...

...is false. The length of the curve is an observable. See above.

Yes.

Because it's a direct observable. See above.

friend

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?

Mentz114

My first take on this is that there's nothing wrong with your argument in the second paragraph, but it doesn't tell us anything we didn't know. Incidentally, is the prime on the differential a typo or significant ?

The speed of light is in the metric itself because for a null curve 0 = dt²-dx² implies (dx/dt)²=1.

PeterDonis

This still doesn't completely clear up what you're trying to do, at least not for me. When you say "the SR metric", it seems to me that you really mean "the Lorentz transformation", since that's what preserves the spacetime interval. So it seems to me that the question you're really trying to ask is "where does the Lorentz transformation come from?" and find an answer that doesn't depend on any assumptions about the speed of light.

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 Michelson-Morley experiment, which show that the invariant speed is finite.

Not really. The delta function [itex]\delta ( \tau - \tau_0 )[/itex] is zero unless [itex]\tau = \tau_0[/itex]. I don't see what that's supposed to mean physically; it looks like you're saying that no proper time elapses at all except at the single event [itex]\tau_0[/itex], where *all* of the proper time on the entire worldline of the object elapses. That doesn't seem reasonable to me.

friend

Ideally, I don't want to start with any assumptions about what kind of transformations apply. I'd like the transformation to be derived from some more fundamental requirement not based on any measurement of anything physical. I have reason to believe that the Dirac delta function is a mathematical representation of causality. At least in linear systems a system is considered causal depending on how it responds to an impulse = dirac delta. So if there is some logic for starting with the integral of the Dirac delta, then perhaps the constant speed of light and the Lorentz transformations can be derived from that logic.



I'm not entirely confident that I want to start with the Dirac delta function. For if we change the coordinate on the curve to [itex]\tau ' = a\tau [/itex] , then [itex]d\tau ' = ad\tau [/itex] , and [itex]\tau = \tau '/a[/itex] , and [itex]d\tau = d\tau '/a[/itex] . Then we have [itex]\tau - {\tau _0} = \tau '/a - \tau {'_0}/a = (\tau ' - \tau {'_0})/a[/itex] and [itex]\delta (\tau - {\tau _0}) = \delta ((\tau ' - \tau {'_0})/a)[/itex] . And this gives us by the scaling property of the Dirac delta,
[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.

PeterDonis

The assumptions of translation and rotation invariance aren't assumptions about what kinds of transformations apply. They're very general assumptions about the physics, that happen to have as a consequence the restriction on possible transformations that I gave.

I'm not sure why translation and rotation invariance wouldn't count as "fundamental requirements".

This seems like a strange thing to want. In order to test whether *any* fundamental assumption actually holds, you're going to have to make physical measurements and compare them with what you would expect to see if the assumption were true.

That's not the same as the delta function itself being a representation of causality. The representation of causality is in the *response* function, not the function describing the initial impulse.

Why are you starting with integrals in the first place? If you're looking for fundamental requirements, it seems to me that you would want to start with something local, i.e., to only look at physical quantities and their derivatives (or infinitesimals) at a single point. The integral representation of physics over a finite length or in a finite region is built up from the local representation; it's not fundamental in itself.

friend

My question is: what logical necessity could there possibly be for the invariance of [itex]d\tau [/itex] that could give rise to the Lorentz metric with its constant speed of light. So in effect I'm trying to explain the LT and c. I don't think an explanation of such can start by measurements of c, that would be a circuilar argument. Yes, of course, whatever theory one derives must be confirmed by experiment. But I don't think experiment is an explanation for the experiment. That would be finding some math that fits the data without an explanation for the data or the math, IMO. If you are not comfortable with this approach to the problem, you don't have to contribute.

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?

PeterDonis

Because [itex]d\tau[/itex] represents a direct physical observable, the time elapsed on a clock over a very small portion of its worldline. It seems reasonable to want your mathematical model of physics to represent direct physical observables by quantities that are invariant under coordinate transformations. Whether that counts as "logical necessity" seems to me to be beside the point; every physical theory has to be based on some assumptions that can't be proved, but just have to be taken as given.

That's where translation and rotation invariance comes in; and I also should have included invariance under boosts, i.e., under changes of inertial frame, because of the principle of relativity: physics should look the same to two observers who are inertial and are moving at a constant velocity relative to each other.

You don't have to start with measurements of c if you start with translation, rotation, and boost invariance; those requirements are enough to narrow down the possible transformations to Lorentz or Galilei transformations, as I said before. But you do have to know about measurements of c if you want to determine whether Lorentz or Galilei transformations are the right ones. I don't know of any other way to make that determination.

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.

TrickyDicky

The LT (together with the Galilean transformation when not including the second postulate of SR) as commented by Peter Donis is based first of all on the reasonable assumption of isotropy and homogeneity, geometrically this is a feature of constant sectional curvature spaces, this assumption introduces the choice of a metric and since the motion that is dealt with in this case is inertial motion the logical constant curvature metric to choose (also remember this is 1905 for Einstein's SR and 1907 for Minkowski coming up with the natural space for SR, Einstein hadn't introduced the idea of curvature in physics yet) was the flat one, this leads to the linearity of both Galilean and Lorentz transformations.

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.

friend

OK, so what I'm beginning to understand is that the invariance of [itex]d\tau [/itex] only implies the flat metric locally. However, does this mean that the invariance of [itex]\int_{{\tau _0}}^\tau {d\tau ' = } \tau - {\tau _0}[/itex] would specify a more global flat space, with a Lorentzian or Galilean transformation? If c is not infinite, then the background metric is Lorentzian and the same along the trajectory, is this right?

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.

PeterDonis

Yes.

Yes. It also assumes that the curve parametrized by [itex]\tau[/itex] is a straight line in that global flat spacetime (I use that word in preference to "space" since time is one of the dimensions in this "space".)

In a flat spacetime, yes.

I'm not sure that works, because Newtonian physics has an "infinite c", and everything doesn't happen all at once in Newtonian physics.

TrickyDicky

Right, what actually happens in Newtonian physics rather than "happenning all at once" wich would imply a block-like interpretation is that there is no relativity of simultaneity. Time and space are not related in the same way they are in SR.

friend

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?