# Generalisation of Time Dilation

1. Apr 11, 2015

### Candidus

Hi. If calculations of the Lorentz factor are by thought experiments using the constancy of the speed of light in inertial frames (I also know of using simultaneous equations for a light signal in two different frames with a conjectured Lorentz factor to calculate it but this also involves light!) then how is it possible to generalise the result to all objects in inertial frames not involving light, unless it is the reflection of light that is of concern, but I've never seen it derived this way? Forgive my naivety.

2. Apr 11, 2015

### Staff: Mentor

Hi Candidus, and welcome to PF!

The Lorentz transformations keep the speed of light invariant, but in order to even measure the speed of light, you need things that aren't light, like clocks and rulers. So if the transformations only applied to light, but not to other objects, we wouldn't measure the speed of light to be the same in all inertial frames. But we do, so the transformations must apply to everything, not just light.

3. Apr 11, 2015

### Ibix

Einstein proposed that the speed of light was invariant for all observers (he didn't pull that out of the air - he put a lot of puzzle pieces together before anyone else and realised he could explain a lot of things if this was true). The Lorentz transforms explain what must be true for observers (and their clocks and rulers) if that is the case.

Rephrasing that slightly: Light comes into the derivations because its speed is the thing that doesn't change. But, as PeterDonis says, you measure its speed with a clock and a ruler. So, despite all the focus on the light, the things we are actually thought-experimenting with are the clocks and the rulers measuring its speed.

4. Apr 11, 2015

### robphy

You can also use a clock and light signals... using a radar method. In the Galilean view, clocks and rulers are primitives. From a special relativity view, clocks and light signals are likely more primitive.

5. Apr 11, 2015

### bcrowell

Staff Emeritus
Relativity doesn't have anything to do with light. The c in relativity isn't really interpreted as the speed of light; it's better thought of as a conversion factor between our units of space and time. The fact that Einstein's 1905 axiomatization of special relativity talks about light is just a historical accident. Physicists today do not think of light as playing any special or fundamental role in SR or in physics in general. It's unfortunate that people continue to teach relativity using Einstein's original formulation of the postulates, because it leads to all kinds of confusion, of which this thread is a good example. Better approaches are now available, as presented, for example, in this paper http://arxiv.org/abs/physics/0302045 .

6. Apr 11, 2015

### robphy

In my opinion, and in the way that I teach relativity, there are two distinct roles for "the speed of light".
(1) As bcrowell says, it is a conversion factor between units for space and time.... [edit: ...any speed could have been used to do that.]
(2) It happens to be equal to the maximum signal speed for special relativity, which is more fundamental for relativity. [The key point is that the maximum signal speed isn't infinity, as it is for Galilean relativity.]
The distinction is often not made in many presentations of relativity.

Last edited: Apr 11, 2015
7. Apr 12, 2015

### Ibix

Are you sure about that? I would take that to mean that you want to replace the (cdt)2 term in the interval with a (Vdt)2 term, which wouldn't be invariant.

8. Apr 12, 2015

### Ibix

Pal's approach is very elegant and straightforward. I see that he shows that there are only two possibilities, which are that K=0 or K>0. The K=0 case obviously reduces to the Galilean transforms without further ado, but the K>0 case still requires some determination of the value of K, surely. I appreciate that that value is arbitrary in some sense (it's 1 in natural units - in fact, that's the definition of the natural unit of speed, I think), but you'd have to go to Maxwell's equations, or clocks-and-rulers style measurement to show that light travels at that invariant speed.

Or are you just saying that it's irrelevant that there exists something that travels at the speed limit? That if we lived in a universe where everything had mass there would be nothing that travelled on null paths, but SR would work the same?

9. Apr 12, 2015

### robphy

The point of the conversion constant is so that you can
so use "x and Ct" or "x/C and t" [or even "Ax and Bt" so that Ax and Bt have the same units].

When I said use "any speed", i mean a constant-value with units of speed, like the "speed of light", or "speed-of-sound" or even "10m/s".
If you choose "speed of light", then your equations look simple...
if you choose any other, then your equations carry a dimensionless constant [which doesn't transform] like "speed of light"/"speed of sound",
which will be an annoyance but not incorrect.

So, as you wrote, "Vdt" would be bad because it transforms... but "(10m/s) t" is fine but annoying.

I raise this issue because when I teach relativity, using Galilean relativity, as bridge from Euclidean geometry to Minkowski geometry,
I express my distances in units of "x/c" and and times in "t", where c is the speed "3e8 m/s".... even though c has no real importance in Galilean relativity.
In other words, instead of using "meters", I use "light-seconds"....even in Galilean relativity.
(Note: I can't use the invariant speed of infinity in Galilean relativity.)
Then, in my method, I can more easily make the analogies between the three geometries.

Last edited: Apr 12, 2015
10. Apr 12, 2015

### Staff: Mentor

"Fine" in what sense? You still have to have an extra factor in the equations if you use any conversion factor other than $c$. For example, if you write the Lorentz interval using units of (10 m/s) t, you still need to use $ct$ instead of $c$, just with $c$ having a different numerical value to adjust for shifting the units of $t$.

The way I would describe all this is that there is an invariant speed, $c$, and that speed is what we use as the conversion factor between time and distance units. Then you just show that any object with zero invariant mass must travel at the invariant speed, and observe that light is such an object.

11. Apr 12, 2015

### robphy

"Fine" in the sense that the physics and mathematics is still correct.
However, the equations you write down are understandably ugly because of a nuisance numerical factor that gets propagated all over the place.

Elaborating on a point I mentioned earlier,
when dealing the Galilean spacetime (or trying to carefully define a limiting case of Minkowski),
it is good to decouple "the invariant maximum signal speed" (associated with the eigenvector of a boost)
from "a conversion constant" which allows you to form spacetime-vectors from space- and time-components.
The Galilean case suggests that the two are logically decoupled since we can't use an infinite speed as a conversion constant.

Last edited: Apr 12, 2015
12. Apr 12, 2015

### Staff: Mentor

But you also can't make spacetime 4-vectors in the Galilean case; spacetime in that case is not a 4-dimensional geometry, because there are no invariant intervals between events at different times. So in the Galilean case there is no "conversion constant" picked out by the physics; any choice of units of time and distance is arbitrary.

13. Apr 12, 2015

### bcrowell

Staff Emeritus
You don't need Maxwell's equations to prove that light moves at the universal speed c, nor do you need to appeal to experimental determinations of the speed of light. Experiments show that light is massless. Relativity tells us that massless particles have to move at the universal speed c.

Proof: Suppose that a massless particle had |v| < c in the frame of some observer. Then some other observer could be at rest relative to the particle. In such a frame, the particle’s momentum p is zero by symmetry, since there is no preferred direction for it. Then E 2 = p 2 + m 2 is zero as well, so the particle’s entire energy-momentum vector is zero. But a vector that vanishes in one frame also vanishes in every other frame. That means we’re talking about a particle that can’t undergo scattering, emission, or absorption, and is therefore undetectable by any experiment. This is physically unacceptable because we don’t consider phenomena (e.g., invisible fairies) to be of physical interest if they are undetectable even in principle.

It's relevant and of interest that light travels at c. However, we don't need to assume it in order to develop the foundational aspects of relativity. We can develop the foundational aspects of SR and then use them to prove that light travels at c.

14. Apr 12, 2015

### robphy

You actually can...
There is a not-very-well-known spacetime-geometry underlying the position-vs-time graph of PHY 101....

It is called the Galilean geometry (a special case of a Newtonian or Newton-Cartan spacetime)
https://archive.org/details/ASimpleNon-euclideanGeometryAndItsPhysicalBasis
It is one of nine Cayley-Klein Geometries of the plane [which includes Euclidean, Spherical, Hyperbolic, Minkowski, Galilean, DeSitter and Anti-Sitter and their (degenerate) Newtonian-limits (Newton-Cartan)]. (see page 218 in Yaglom's book).
In the constant-curvature case, these are studied in projective geometry
(Galilean is called a https://www.google.com/search?q="doubly-parabolic+geometry" ).

In relativity, these are studied by authors like
Malament http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf
Ehlers http://pubman.mpdl.mpg.de/pubman/item/escidoc:153004:1/component/escidoc:153003/328699.pdf
Trautman http://www.fuw.edu.pl/~amt/CompofNewt.pdf