:evil:
Or maybe they are so different that comparing them breaks the Apple/Bandicoot Rule.
I'm happy about this. I always like the ether theory. Not because it made sense. I just liked it for some reason. :-D
I think of matter as "folded" space, and when people talk about matter and energy I view it as densely packed space. And energy in this view is just how this packed space is moving. It's not a new or original perspective, but I like it.
Stuff like wormholes becomes possible with this perspective. Curvature in space-time makes for a bit interesting view with packed space.
I just picture a turtle trying to "understand" New York City.
Then I have another beer and kick back. :lol:
The luminiferous aether was thought to be a substance that had to support light, the thing that light was a wave in. The spacetime thing is the background geometry for which all of physics occurs in, and of course, it was discovered earlier that there are (mathematically) more geometries than the flat Euclidean. Special relativity begets general relativity when you consider what happens when you allow a locally Minkowskian metric to be globally curved and make a few educated guesses about the nature of how the curvature of spacetime is related to the local distribution of matter, and the result is that you see something that looks a whole lot like gravity. I see only the vaguest of similarities.
Sorry, Sol. The luminiferous aether theory, as actually formulated, is still dead and buried. It hasn't been resurrected. We know that it doesn't exist, because it cannot live in a world where the Michelson-Morley experiment is a null result (which implies we are stationary to the aether), and yet we still see the seasonal aberration of starlight (which implies at the same time that we are moving relative to it).
TL;DR version: Leo is right. You are comparing apples to bandicoots.
Quote from: "Hakurei Reimu"TL;DR version: Leo is right. You are comparing apples to bandicoots.
:-k
I see the universe as a big fucking pinball machine and one day, I'm gonna pull that lever............
The aether actually returned via Lorentz and his equations, but it wasn't accepted because Lorentz postulated that the aether occupied some privileged position, and the aether was in principle undetectable. So, with there being no possible evidence for Lorentz's aether reference frame and direct contradictions with experimental evidence for Einstein's relativity, tis dumped.
Though you still get people like WLC and his cronies who need Lorentz and his aether frame to support the Kalam Cosmological argument. They are also committed to the view (which they explicitly endorse) that time dilatation and length contraction only "apparently" occur.
I figured you guys would find that face-palmingly stupid, especially Joesephazalo. :rolleyes:
So, the OP is mostly nonsense.
:evil:
:evil:
Can you refer us to the TV show you're citing?
:evil:
:evil:
Quote from: "Solitary"Quote from: "Gawdzilla Sama"Can you refer us to the TV show you're citing?
Is this directed toward me? I never cited any TV show. :-? Solitary
Odd, it certainly sounds like it.
:evil:
I keep hearing Morgan Freeman.
:evil:
Well no. Lorentz was by no means stupid (he was quite the physicist), he was just wrong.
Quote from: "Solitary"Did you even read what I wrote? The aluminiferous aether theory was about space, the new ether theory is about space-time.
Which means that they really bear no real resemblance to each other at all.
Quote from: "Solitary"You are the one comparing apples to bandicoots.
Stop projecting.
Quote from: "Solitary"Just because teachers don't want to fry students minds by saying light waves in space-time doesn't mean it doesn't.
Light waves are in space-time in the same sense that everything else is in space-time. Your statement has no content.
Quote from: "Solitary"It seems though that the electromagnetic field (light) consists of wave-like distortions of the new ether (space-time.) Solitary
Only for people like you. Undulating distortions of space-time would be
gravity waves. It is not even possible for space-time to be the wave medium of photons, because the equations say that, should they exist, the quantum force carriers of such waves would be spin 2, whereas photons are strictly spin 1, if the disparage of interaction strength wasn't enough to clinch it.
We normally think of space and time being distinct things, independent and fundamentally different in a way that we can't really describe. We think of space being three-dimensional (x, y, and z), and time being one-dimensional (t). We describe an object by its spatial coordinates at a point in time, and then it may be at another set of coordinates at another point in time.
We could say that at t=1 an object is at ?1,2,3?, and at t=2 it is at ?2,4,6?. It seems like a small step to just bring the time dimension inside our space vector and make a spacetime vector, ?t,x,y,z?, but when this is done, the time coordinate must be multiplied by c (the speed of light) and i (square root of -1, this comes up later) so that the elements of the vector have the same units (distance). This makes the spacetime vector ?ict,x,y,z?, and we denote this vector by R.
If we want to look at the "speed" at which something moves through spacetime, it will be the derivative of R with respect to "proper time", denoted by ? (Greek letter tau). For some background on this, the special relativity node has some good information, but the bottom line is if person 1 is standing still and watches person 2 move past at a constant speed on a train, they see person 2's ticking more slowly than their own. Person 2, who is on the train, will see their clock ticking normally, but will see person 1's clock ticking slowly.
How slowly the two clocks appear to tick depends on how quickly the train is moving; the faster it is moving, the more slowly the clocks will appear to tick. Proper time is time it takes for one tick of either person's clock from their own perspective, and if ? is person 2's proper time, then person 1 sees person 2's clock taking time t to tick, which is related to ? through ? (Greek letter gamma) as such: t = ? × ?. Equivalently, ? = t/?. Gamma depends on person 2's speed relative to person 1. This relationship between ? and t makes it far easier to find a meaningful derivative of R, since we can turn the derivative with respect to an external proper time into a derivative with respect to the observed time of the object.
So after that digression, we can find the "speed" of the spacetime vector with respect to proper time, which is denoted by U. We differentiate the spacetime vector with respect to proper time:
U = dR/d?
? = dR/d(t/?)
? = ? × dR/dt
? = ??ic, dx/dt , dy/dt , dz/dt?
? = ??ic, ?(dx/dt) , ?(dy/dt) , ?(dz/dt)?.
This is the spacetime "speed" in vector form, but the magnitude of this vector is a little different from the magnitude of a spatial 3-vector. The magnitude of a spatial vector ?x,y,z? is found by s2 = x2 + y2 + z2. This is a geometric invariant, which basically means that you can rotate or move this vector however you like, or change the coordinate system, and its magnitude (length in this case) will stay the same. Similarly, our spacetime 4-vector ?ct,x,y,z? should be invariant through a change in inertial frames of reference, so that all observers will record the same magnitude of the vector, even if they see the components differently. The magnitude of the 4-vector is given by:
s2 = (ict)2 + x2 + y2 + z2
? = -(ct)2 + x2 + y2 + z2
It is conventional to reverse the sign of the right hand sign, making it s2 = (ct)2 - x2 - y2 - z2. This is also a geometric invariant, and the same form of the magnitude will be applicable to the derivative of the vector.
Now, the definition of ? is (1 - (v/c)2)-1/2 = c(c2 - v2)-1/2, so ?2 = c2(c2 - v2)-1, v being the speed of the object being observed.
So, finally, we can get an expression for the magnitude of the derivative of the spacetime vector, denoted by s:
s2 = (?c)2 - (?(dx/dt))2 - (?(dy/dt))2 - (?(dz/dt))2
? = ?2(c2 - (dx/dt)2 - (dy/dt)2 - (dz/dt)2)
? = ?2(c2 - v2), since v2= (dx/dt)2 + (dy/dt)2 + (dz/dt)2.
? = (c2 - v2) × c2(c2 - v2)-1, by substituting in the expression for ?.
? = c2
Therefore s = c, regardless of where you are or how quickly you are moving†.
Solitary
Everything moves through spacetime at the speed of light (//http://everything2.com/title/Everything+moves+through+spacetime+at+the+speed+of+light)
by museman
Quote from: "Solitary"It is conventional to reverse the sign of the right hand sign, making it (*) s2 = (ct)2 - x2 - y2 - z2. This is also a geometric invariant, and the same form of the magnitude will be applicable to the derivative of the vector.
Now, the definition of ? is (1 - (v/c)2)-1/2 = c(c2 - v2)-1/2, so ?2 = c2(c2 - v2)-1, v being the speed of the object being observed.
So, finally, we can get an expression for the magnitude of the derivative of the spacetime vector, denoted by s:
(**) s2 = (?c)2 - (?(dx/dt))2 - (?(dy/dt))2 - (?(dz/dt))2
? = ?2(c2 - (dx/dt)2 - (dy/dt)2 - (dz/dt)2)
? = ?2(c2 - v2), since v2= (dx/dt)2 + (dy/dt)2 + (dz/dt)2.
? = (c2 - v2) × c2(c2 - v2)-1, by substituting in the expression for ?.
? = c2
Therefore (***) s = c, regardless of where you are or how quickly you are moving†.
Solitary
The bold part and the labels (*), (**), (***) is of my doing.Sorry, but that result
(***) is wrong.
If
(*) s[sup:1bdyvdq8]2[/sup:1bdyvdq8] = (ct)[sup:1bdyvdq8]2[/sup:1bdyvdq8] - x[sup:1bdyvdq8]2[/sup:1bdyvdq8] - y[sup:1bdyvdq8]2[/sup:1bdyvdq8] - z[sup:1bdyvdq8]2[/sup:1bdyvdq8].
Then taking the derivative wrt time on both sides:
Right-hand side = 2sds/dt
Left-hand side = 2ct - 2x(dx/dt) - 2y(dy/dt) - 2z(dz/dt).
Equating these two, and you can eliminate the 2 on both sides, you get,
s(ds/dt) = ct - x(dx/dt) - y(dy/dt) - z(dz/dt), not your equation which I labelled by
(**) From this, you don't get
(***) s[sup:1bdyvdq8]2[/sup:1bdyvdq8] = c[sup:1bdyvdq8]2[/sup:1bdyvdq8]