Is dτ Invariant Under Transformations Beyond Lorentz in Special Relativity?

[friend]

[Original reply replaced by...]

I think this part of his argument is questionable, since it ignores uniform dilations of the form $(t',x') = (e^\alpha x, e^\alpha t)$, where $\alpha$ is independent of $x$ and $t$. Including such dilations would lead to 2-parameter transformations. He dispenses with dilations implicitly a bit later in the paragraph after eq(9) where he describes the case $K(a)=0$ as pathological, and in the related footnote 10 he mentions the "static group" or "Carroll" group. So he's implicitly assuming a common scale in both frames.

Another way of looking at this is that for the case $v=0$, the old and new frames are one and the same, apart from a possible scale difference in their respective units of time and length -- which are promptly assumed to be standardized uniformly when the origins coincide.

It's easy for me to view equation (1) as a Taylor series expansion of the transformation functions between the primed and unprimed coordinates, expanded about (x,t)=(0,0). Then he seems to be trying to derive the coefficients of that expansion using his hypothesis. He thinks there is no loss of generality if he sets the first coefficients of those expansions to 0 so that equation (4) holds. But I'm not even seeing that.

He starts out with a good idea and an interesting approach to the problem. But he's not communicating it very well as far as I'm concerned.

 

[strangerep]

It's easy for me to view equation (1) as a Taylor series expansion of the transformation functions between the primed and unprimed coordinates, expanded about (x,t)=(0,0).

Eq(1) is not a Taylor series. He's simply writing down the most general form for a coordinate transformation in 1+1D spacetime that depends on various parameters $a_i$ .
The new (primed) coordinates are simply functions of the old (unprimed) coordinates and the parameters $a_i$. Nothing deeper than that at this stage.

Then he seems to be trying to derive the coefficients of that expansion using his hypothesis. He thinks there is no loss of generality if he sets the first coefficients of those expansions to 0 so that equation (4) holds. [...]

He's simply restricting the transformation equations to the class of "inertial frames with common spacetime origins", which can be achieved by applying spacetime translations to either the old or new coordinates. That's how he gets eq(4) -- from the restriction that the old and new origins coincide (achieved by using up the translational freedom).

He starts out with a good idea and an interesting approach to the problem. But he's not communicating it very well as far as I'm concerned.

It's hard to write a paper suitable for Am. J. Phys. which is also pedagogical, but I've seen far more obscure papers than this one. As with most research literature, one must persevere and try to understand and reproduce the steps in detail.

 

[friend]

I think I found a paper that does what I'm asking. It seems to derive the Lorentz transformations from the properties of spacetime alone. I'd appreciate comments on this paper, 7 pages, relatively easy math. Thank you.

One more derivation of the Lorentz transformation

He writes,

"I will take as a starting point the statement of the principle of relativity in a very general form: there exists an infinite continuous class of reference frames in spacetime which are physically equivalent."

Instead of accepting on faith this principle of relativity in a very general form, I'd like to offer a reason that would give rise to this "infinite continuous class of reference frames". It might be that whatever gives rise to this requirement might also help in understanding the other symmetries of spacetime that allow us to fully specify the Lorentz nature of the transformations in the reference frames.

It seems the dirac delta function specifies a diffeomorphism between coordinate transformations that constitute the reference frames.

Consider the defining property of the Dirac delta function,
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1
If we change coordinates to y = y(x) so that x = x(y), then dx = \frac{{dx}}{{dy}}dy.
Then we can use the notation
\frac{{dx}}{{dy}} = \sqrt {g(y)} = \sqrt {\frac{{dx}}{{dy}} \cdot \frac{{dx}}{{dy}}}
in order to be consistent with higher dimensional versions. We also have
x - {x_0} = \int_{{x_0}}^x {dx'}
And by using the transformation, y = y(x), so that {y_0} = y({x_0}), this integral becomes,
x - {x_0} = \int_{{x_0}}^x {dx'} = \int_{{y_0}}^y {\sqrt {g(y')} dy'}
And the original integral can be transformed to,
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx} = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (\int_{{y_0}}^y {\sqrt {g(y')} \,dy'} )\,\,\sqrt {g(y)} \,\,dy} = 1
where {y^{ + \infty }} = y(x = + \infty ) and {y^{ - \infty }} = y(x = - \infty ).

Then using the composition rule for the Dirac delta, we have
\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (\int_{{y_0}}^y {\sqrt {g(y')} \,dy'} )\,\,\sqrt {g(y)} \,\,dy} = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\left( {\frac{{\delta (y - {y_0})}}{{|\sqrt {g({y_0})} \,|}}} \right)\,\,\sqrt {g(y)} \,\,dy = } \frac{1}{{|\sqrt {g({y_0})} \,|}}\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,\sqrt {g(y)} \,\,dy}
where {{y_0}} is where \int_{{y_0}}^y {\sqrt {g(y')} \,dy'} = 0.

And then using the sifting property of the Dira delta we have
\frac{1}{{|\sqrt {g({y_0})} \,|}}\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,\sqrt {g(y)} \,\,dy} = \frac{1}{{|\sqrt {g({y_0})} \,|}}\left( {\,\sqrt {g({y_0})} \,\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,dy} } \right) = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,dy} = 1
So that finally,
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (y - {y_0})\,dy} = 1
where I assume that using an integration interval of - \infty \le y \le + \infty changes nothing.

So it appears here that any invertible coordinate transformation (diffeomorphism) leaves the integration of the Dirac delta unchanged in form, meaning it is diffeomorphism invariant.

The properties of the Dirac delta function are necessary in any reference frame or any space. They are true on principle alone. So if the Dirac delta is required by some logic, then that logic would appear to also specify an "infinite continuous class of reference frames in spacetime".

 

[strangerep]

[...]
So that finally,
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (y - {y_0})\,dy} = 1
where I assume that using an integration interval of - \infty \le y \le + \infty changes nothing.

You've proved nothing that is not already contained in the definition of the Dirac delta itself.
I can just as well write down:
$$
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} ~=~ 1
~=~ \int_{ - \infty }^{ + \infty } {\delta (x - y_0)\,dx}
~=~ \int_{ - \infty }^{ + \infty } {\delta (y - y_0)\,dy}
$$
where, in the last step, I simply renamed the dummy integration variable.

So it appears here that any invertible coordinate transformation (diffeomorphism) leaves the integration of the Dirac delta unchanged in form, meaning it is diffeomorphism invariant.

The properties of the Dirac delta function are necessary in any reference frame or any space. They are true on principle alone. So if the Dirac delta is required by some logic, then that logic would appear to also specify an "infinite continuous class of reference frames in spacetime".

The "infinite continuous class of reference frames in spacetime" is simply a consequence of the fact that the group of coordinate transformations which preserve the equation of inertial motion, i.e.,
$$
\frac{d^2 {\mathbf x}}{dt^2} ~=~ 0
$$
turns out to be a Lie group.

This does not have to be "accepted on faith". An unaccelerated observer enclosed in an opaque box cannot tell whether he is stationary or moving relative to some external point.

 

[friend]

You've proved nothing that is not already contained in the definition of the Dirac delta itself.
I can just as well write down:
$$
\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} ~=~ 1
~=~ \int_{ - \infty }^{ + \infty } {\delta (x - y_0)\,dx}
~=~ \int_{ - \infty }^{ + \infty } {\delta (y - y_0)\,dy}
$$
where, in the last step, I simply renamed the dummy integration variable.

Of course you can always change the dummy variable in any equation and get the same answer. But then you're stuck having to interpret the new dummy variable as if it were the exact same thing as the old dummy variable.

I don't think you can change x₀ to y₀ without specifying a function y(x). So I don't think you can go from the integral on the left to the middle integral. But if you do, then it is no longer a dummy variable change; you're actually specifying a function from x to y.

What I think I've shown is when there is a relationship between x and y, namely, y=y(x), you get the same equation back again. Then we can interpret x and y as different coordinates. And x and y can not have any arbitrary relationship but must be a diffeomorphism. I think this means that the Dirac delta function is diffeomorphism invariant, right? So if the dirac delta is required for some reason (TBD), then that requirement also specifies the existence of some properties of the space on which it resides, namely, that it admitts a diffeomorphic coordinate patches, right?

The "infinite continuous class of reference frames in spacetime" is simply a consequence of the fact that the group of coordinate transformations which preserve the equation of inertial motion, i.e.,
$$
\frac{d^2 {\mathbf x}}{dt^2} ~=~ 0
$$
turns out to be a Lie group.

This does not have to be "accepted on faith". An unaccelerated observer enclosed in an opaque box cannot tell whether he is stationary or moving relative to some external point.

All this does is render a mathematical description of the observations. It does not explain it. What I'm trying to understand is WHY space would have the properties it does that makes necessary these kinds of observations.

Even the author of the paper is trying to derive the Lorentz transformations from properties of spacetime. He starts with the a priori existence of diffeomorphic coordinate transformations on patches of an underlying manifold (paraphrased). I'd like to go even further back and explain the need of for this property of spacetime to begin with.

 

[strangerep]

I don't think you can change x₀ to y₀ without specifying a function y(x). So I don't think you can go from the integral on the left to the middle integral.

Yes I can. Both steps are valid, provided both $x_0$ and $y_0$ are somewhere between $-\infty$ and $+\infty$.

Anyway...

Although I'm happy to help you understand the content of published papers, I'm not going to follow you down a rabbit hole into crackpot land.

 

[friend]

Yes I can. Both steps are valid, provided both $x_0$ and $y_0$ are somewhere between $-\infty$ and $+\infty$.

Anyway...

Although I'm happy to help you understand the content of published papers, I'm not going to follow you down a rabbit hole into crackpot land.

I certainly hope it's not a "rabit hole". I was trying to find the simplest construction that relates a metric to a field. Then maybe that could be used in the derivation of both the fields of QFT and curvature of GR. And it occurs to me that fields consist of individual values at each point in a space. And the minimum section of space is a infinitesimally small flat portion at the point of interest. A metric is inherently needed to do integration. So what integration process uses the least portion of space and picks out individual field values? That would be the Dirac delta function. Intuitively that seems to me like a good place to start when trying to relate space and fields in terms of their smallest constituents. One would think that the integration of the dirac delta being one is a true statement independent of any physical reason to use it. So if it does prove useful in deriving physics, then we will have succeeded in deriving physics from inherent truth.

As I recall, the integration of the dirac delta is suppose to be one no matter how small the integration interval, as long as the interval contains the zero of the dirac delta's argument. And if the interval of integration is not infinite but is small instead, then my objection to your comments holds.

But even with an infinite interval of integration, I wonder if there must be a diffeomorphism between the x and y coordinates. For it seems that the field in both coordinates needs to be sufficiently well behaved in order to do the integration. Does that mean we should be able to construct a smooth and invertible function between x and y, a diffeomorphism?