So far I haven't found a derivation that leaves me (conceptually) any further forward than when I started. Any links, fair use or home-brewed versions greatly appreciated. I will do my best to attempt all.
-C2.
Thank you AlexG
(1) x-ct=0
(2) x'-ct'=0
then..
(3) (x'-ct')=k(x-ct)
or
0=k0
By my analysis k=absolutely anything and by the third line I am already lost
(1) x-ct=0
(2) x'-ct'=0
then..
(3) (x'-ct')=k(x-ct)
or
0=k0
By my analysis k=absolutely anything and by the third line I am already lost
Postulate: For any two events in space-time have an interval s² = (cΔt)² - (Δx)² which all inertial observers agree on.
Postulate: No massive body can travel faster relative to another body than the speed of light.
Postulate: Any two observers passing on another automatically agree on their relative speed.
Therefore, if the two events are separate such that a massive particle could get from one to the other, then we speak of the proper time between events as tau, τ. Obviously, Δτ = √(s²/c²) = √((Δt)² - (Δx/c)²) since we could toss a massive clock inertially from event to event and measure the time in the coordinates where Δx = 0. So this space-time interval is "real" in the sense it can be measured as s² = (cΔτ)² for the clock which moves inertially between the points.
Therefore, we have the twin paradox. Imagine 3 events, A then B, then C, all separations are such that a massive particle could proceed A->C or A->B->C. Let's work in coordinates such that A and C share the same x-coordinate. Then the coordinates of A, B and C are (t_A,0), (t_B,x_B), (t_C, 0). Then the proper times are τ_AC = t_C - t_A = (t_C - tB) + (t_B - t_A). τ_AB = √((t_B - t_A)² - ((x_B)²/c²)), τ_BC = √((t_C - t_A)² - ((x_B)²/c²)). Now if x_B = 0, then τ_AC = τ_AB + τ_BC -- but if x_B is different that zero, but still small enough so that the clock doesn't have to break the speed of light, we have τ_AB < t_B - t_A and τ_BC < t_C - t_B so that τ_AC > τ_AB + τ_BC. So the clock that moves A->B->C ticks less ticks than the clock that moves inertially from A->C.
Therefore we have time dilation. If someone zips past us with v = Δx/Δt then we calculate the ratio of their proper time between any two points on their path with our time, and we get ratio = (√((Δt)² - (Δx/c)²))/Δt = √(1 - (v/c)²). Where our clocks measure 1 second, their clocks measure less than 1 second.
In a same way as we defined proper time in terms of ticks of a clock moving inertially along a path between events, we can talk about two events where no particle traveling at or below the speed of light could every be at both events. Then we could speak of the proper distance between the points, d = √(-s²) = √((Δx)² - (cΔt)²). For our twin who moves at some velocity from A to B, we know their clocks tick less ticks than the clock moving from A to C. Since both observers agree on the relative velocity, it follows that to the moving clock the actual distance from A to B is not v Δt = Δx, but instead v Δt √(1 - (v/c)²) = Δx √(1 - (v/c)²). This is length contraction in the direction of movement.
I don't like this, but it's one way to the same math.
Postulate: No massive body can travel faster relative to another body than the speed of light.
Postulate: Any two observers passing on another automatically agree on their relative speed.
Therefore, if the two events are separate such that a massive particle could get from one to the other, then we speak of the proper time between events as tau, τ. Obviously, Δτ = √(s²/c²) = √((Δt)² - (Δx/c)²) since we could toss a massive clock inertially from event to event and measure the time in the coordinates where Δx = 0. So this space-time interval is "real" in the sense it can be measured as s² = (cΔτ)² for the clock which moves inertially between the points.
Therefore, we have the twin paradox. Imagine 3 events, A then B, then C, all separations are such that a massive particle could proceed A->C or A->B->C. Let's work in coordinates such that A and C share the same x-coordinate. Then the coordinates of A, B and C are (t_A,0), (t_B,x_B), (t_C, 0). Then the proper times are τ_AC = t_C - t_A = (t_C - tB) + (t_B - t_A). τ_AB = √((t_B - t_A)² - ((x_B)²/c²)), τ_BC = √((t_C - t_A)² - ((x_B)²/c²)). Now if x_B = 0, then τ_AC = τ_AB + τ_BC -- but if x_B is different that zero, but still small enough so that the clock doesn't have to break the speed of light, we have τ_AB < t_B - t_A and τ_BC < t_C - t_B so that τ_AC > τ_AB + τ_BC. So the clock that moves A->B->C ticks less ticks than the clock that moves inertially from A->C.
Therefore we have time dilation. If someone zips past us with v = Δx/Δt then we calculate the ratio of their proper time between any two points on their path with our time, and we get ratio = (√((Δt)² - (Δx/c)²))/Δt = √(1 - (v/c)²). Where our clocks measure 1 second, their clocks measure less than 1 second.
In a same way as we defined proper time in terms of ticks of a clock moving inertially along a path between events, we can talk about two events where no particle traveling at or below the speed of light could every be at both events. Then we could speak of the proper distance between the points, d = √(-s²) = √((Δx)² - (cΔt)²). For our twin who moves at some velocity from A to B, we know their clocks tick less ticks than the clock moving from A to C. Since both observers agree on the relative velocity, it follows that to the moving clock the actual distance from A to B is not v Δt = Δx, but instead v Δt √(1 - (v/c)²) = Δx √(1 - (v/c)²). This is length contraction in the direction of movement.
I don't like this, but it's one way to the same math.
Thank you rpenner. Still looking at it - I've been working and I can't think and work at the same time.
Incidentally - the OP wasn't intended to discourage comments from others interested in Lorentz Transform derivations.
-C2.
Incidentally - the OP wasn't intended to discourage comments from others interested in Lorentz Transform derivations.
-C2.
FWIW there's a flock of them here :-
http://www.physicsforums.com/archive/index.php/t-123103.html
I'll try them when I'm done with rpenner's.
Anyone else .. please report if you find a good one..
-C2.
http://www.physicsforums.com/archive/index.php/t-123103.html
I'll try them when I'm done with rpenner's.
Anyone else .. please report if you find a good one..
-C2.
Postulate (see rpenner's post (above)):-
s² = (cΔt)² - (Δx)²
a) I think this the same (?) as :-
s² = (cΔt)² + (iΔx)²
If one were used to using complex notation as a representation of phase one might infer that time and distance are 'at right angles'.
b) Alternatively one might write:-
(Δx)² + s² = (cΔt)²
..and I'm left wondering where the funny s² component is coming from.
----------------------
I favour time and distance being 'at right angles' - comments/thoughts (even criticism!) most welcome.
-C2.
s² = (cΔt)² - (Δx)²
a) I think this the same (?) as :-
s² = (cΔt)² + (iΔx)²
If one were used to using complex notation as a representation of phase one might infer that time and distance are 'at right angles'.
b) Alternatively one might write:-
(Δx)² + s² = (cΔt)²
..and I'm left wondering where the funny s² component is coming from.
----------------------
I favour time and distance being 'at right angles' - comments/thoughts (even criticism!) most welcome.
-C2.
Distance and time being at right angles works fine as long as there is only one guy with a clock. But when that one guy looks at another guy with a clock who is in relative motion, it can be established that that other guy's time and space aren't at right angles as measured by the first guy. In practice this gets severe only at speeds achieved by elementary particles (like muon showers from cosmic rays), but then again "right angle" is defined by picky mathematicians to be an exact right angle.