QUOTE=Austin0;3940936]Thank you for your explication but as I fully understood the meaning of the interval and the metric I am afraid it completely missed the point.
That point being the meaning of the
word invariant. In this context it simply means constant,unchanging, across all
inertial coordinate systems
This necessarily implies the existence of other frames.
Would you disagree?
It does not apply to other coordinate systems within a single frame. I.e. changing from orthogonal to polar coordinates for eg.
It does not apply to local measurements as they apply within the frame.
It takes local coordinate measurements and outputs a value that is meaningful and constant in all other frames.
As that output value is related to the input values by the gamma factor it would appear it was a de facto transformation, semantic quibbles notwithstanding.
Yes[/QUOTE]
stevendaryl said:
Yes, I would disagree. The length of the hypotenuse of a triangle is, by Pythagorus, equal to √(A2 + B2), where A is the length of one leg, and B is the length of the other leg. Do you think that that definition implies the existence of other frames? The invariant interval in SR, τ = √((ct)2 - x2) is a geometric relationship, just like the length of the hypotenuse of a right triangle. It doesn't have anything to do with "frames". The metric doesn't have anything to do with frames. Curved surfaces have associated metrics, and that doesn't have anything to do with frames.
Well you completely disregarded the meaning of the word invariant.
τ = √((ct)
2 - x
2) this expression itself implicitly assumes and requires an orthogonal coordinate structure.
You have two points floating in space in front of you.
On this alone how do you define or express a metric? You can not even assign a distance of any kind without a ruler. I.e. a one dimensional coordinate system.
The thing that is fundamental about the metric is the signiture and that falls out of the gamma function. Out of the intrinsic structure of the world.It has nothing to do with historical precedent.
Without this fundamental aspect of reality there would be no Minkowski metric or gamma function and we would be living in a Galilean/Newtonian world.
gamma + Cartesian 4 D coordinates ===> Minkowski metric
gamma + Galilean transforms ====> Lorentz transforms
How do you possibly imagine anyone arriving at the metric without the knowledge of the gamma aspect of reality?
stevendaryl said:
Yes, it certainly does. In cartesian coordinates, the metric tensor is defined by
(in 2-D spacetime):
gtt = 1
gxx = -1/c2
gyy = -1/c2
(all other components are zero)
In polar coordinates, we have:
gtt = 1
grr = -1/c2
gθθ = -r2/c2
So the invariant interval is
ds2 = dt2 - dr2/c2 - r2dθ2/c2
You have misunderstood.
Of course within a frame every possible coordinate system will have a relevant version of the Minkoiwski metric. We are in the middle of a discussion regarding the -version for cylindrical coordinates right now. So it should be obvious that this is not what I meant.
Within a frame transformation of events between different coordinate systems does not invlove the Minkowski metric.
It only comes into play when regarding things moving relative to the frame. I.e. different coordinate frames.
Within a Minkowski 2 D chart :
Regarding the rest frame the geometry is purely Euclidean with the Euclidean metric in a Cartesian chart. Yes? the normal Pythagorean relationships apply, yes?
It is only regarding the moving frame that the Minkowski metric with it's specific form of the Pythagorean operation applies ,,,,Yes?
Applying that metric returns the proper time value for a segment of the moving particles world line.
That time is related to the time of the rest frame described by the vertical interval of the triangle by the gamma factor Yes?
So it accomplished a geometric gamma transformation or do you disagree?