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Why is the Time component in the Space Time Interval negative? 
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#19
Jul1511, 01:20 AM

P: 85

[QUOTE=hmsmatthew;3402349]



#20
Jul1511, 03:29 PM

PF Gold
P: 706

Another way of looking at it is that when you are talking [itex](x,y)[/itex] coordinate distance, you can imagine concentric circles around either particle and ask, what concentric circle the other particle is on. The answer involves a [itex]r=\sqrt{x^2 + y^2}[/itex]
With [itex](x,t)[/itex] event distances, you can imagine "concentric" hyperbolas around one of the events and ask which hyperbola the other event is on. The answer involves a [itex]\tau = \sqrt{t^2x^2}[/itex] or [itex]s=\sqrt{x^2t^2}[/itex] (depending on which quadrant you're in.) 


#21
Jul1711, 06:30 PM

P: 292

There are a lot of good but complicated answers to this here, but I think in simple terms, that the signature is related to the observation that you can only travel in one direction in the time dimension of spacetime.



#22
Jul1711, 06:33 PM

Mentor
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#23
Jul1811, 09:58 AM

PF Gold
P: 706

The statement that you can only travel in one direction of time is somewhat ambiguous, since a particle can travel between any two timelike separated events, and whether two observers agree that it's traveling in the same "direction" through time is not a particularly well defined statement, mathematically. Sure, they are both aging, but from the oberver's point of view, the observer is aging faster, and from the particle's point of view, the observer is aging slower. I could just as easily say there are any number of directions of time through spacetime. But all those directions are within the lightcone. 


#24
Jul1911, 07:53 AM

PF Gold
P: 706

On further reflection, it occurred to me that if there were only one direction for time in spacetime, the metric should be:
[tex]d \tau^2 = dt^2 + 0 dx^2 + 0 dy^2 + 0 dz^2[/tex] rather than [tex]d \tau^2 = dt^2  \frac{1}{c} dx^2  \frac{1}{c} dy^2  \frac{1}{c} dz^2[/tex] The fact that c is such a large number (300 million meters per second) means that the two equations look identical in our experience, so it seems like there is only one direction in time. For visualization purposes, imagine what this diagram would look like if c were set to 300 million meters per second (by setting the vertical scale to seconds, and the horizontal scale to meters), instead of c=10 The hyperbolas on the top and bottom would flatten into horizontal lines, and the ones on either side would disappear altogether. The lines would mark off t≈τ=1, t≈τ=0, t≈τ=1, t≈τ=2, etc. Eight observer dependent directions: I may be belaboring the point, but, with this in mind, it should be understood that there are basically eight observerdependent directons. Up (z), down, left, right (x), forward(y), and backward are what we are familiar with, and these are all obviously observer dependent, because if we are facing different directions, you know that my Δx' and Δy' are different from your Δx and Δy. What's not at all obvious in our daily experience is that future and past are also observer dependent directions. If you and I are traveling at different velocities in the x direction then my Δx' and Δt' are different from your Δx and Δt. Our intuition that comes from years of living in a world where speeds of even .01c are inconceivably high, is that there is only one direction in time, and that this direction is NOT observer dependent. 


#25
Jul1911, 08:39 AM

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#26
Jul1911, 02:57 PM

PF Gold
P: 706

Here is how I would go from equations (3) and (4) to equation (5), though it might not be the same as how Einstein did it. Sorry this is a bit long, but I want to make the relationship to hyperbolic geometry perfectly clear. [itex]\begin{align*} dx' &=\gamma(dxvdt) \\ &= \gamma(dx\frac{v}{c} c dt) \\ &= \gamma dx  \beta \gamma c dt\\ &= \begin{pmatrix} \gamma & \beta \gamma \end{pmatrix} \begin{pmatrix} dx \\ c dt \end{pmatrix} \end{align*}[/itex] and [itex]\begin{matrix} dt'= \left (\gamma dt\frac{v dx}{c^2} \right )\\ \begin{align*} c dt' &=\gamma \left (c dt  \frac{v}{c} \cdot dx \right ) \\ &= \gamma c dt  \beta \gamma dx \\ &=\begin{pmatrix} \beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} dx \\ c dt \end{pmatrix} \end{align*} \end{matrix}[/itex] hence [tex]\begin{pmatrix} dx'\\ c dt' \end{pmatrix} =\begin{pmatrix} \gamma & \beta \gamma\\ \beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} dx\\ c dt \end{pmatrix}[/tex] Since β has a range of (1,1) we can define a new variable called the "rapidity" φ such that tanh(φ) = β ( A visualization of the relationship between φ, cosh(φ) and sinh(φ) can be seen here: http://en.wikipedia.org/wiki/File:Hy..._functions.svg ) Then [tex]\begin{align*} \gamma &= \frac{1}{\sqrt{1\beta^2}} \\ &= \frac{1}{\sqrt{1\tanh^2(\varphi)}} \\ &= \frac{1}{\sqrt{1 \frac{ \sinh^2(\varphi) }{ \cosh^2(\varphi)} }} \\ &=\frac{\cosh(\varphi) }{\sqrt{\cosh^2(\varphi) \sinh^2(\varphi)}}\\ &=\cosh(\varphi) \end{align*}[/tex] and similarly [tex]\begin{align*} \beta \gamma &= \frac{\beta}{\sqrt{1\beta^2}} \\ &= \frac{ \tanh(\varphi) }{\sqrt{1\tanh^2(\varphi) }} \\ &= \frac{\tanh(\varphi)}{\sqrt{1 \frac{ \sinh^2(\varphi) }{ \cosh^2(\varphi)} }} \\ &=\frac{\cosh(\varphi) \tanh(\varphi)}{\sqrt{\cosh^2(\varphi) \sinh^2(\varphi)}}\\ &=\sinh(\varphi) \end{align*}[/tex] Now, [tex]\begin{align*} (c dt')^2 &= (\sinh(\varphi )dx +\cosh(\varphi)c dt)^2\\ &=\sinh^2 (\varphi)(dx)^2 2 \cosh(\varphi)c dx dt +\cosh^2(\varphi)(c dt)^2 \end{align*}[/tex] and [tex]\begin{align*} (dx')^2 &=\left( \cosh(\varphi )dx \sinh(\varphi)c dt\right)^2\\ &=\cosh^2 (\varphi)(dx)^2 2 \cosh(\varphi)c dx dt +\sinh^2(\varphi)(c dt)^2 \end{align*}[/tex] Subtracting the two gives [tex]\begin{align*} (c dt')^2  (dx)^2 &=\left[\cosh^2(\varphi)  \sinh^2(\varphi) \right] (c dt)^2 +\left[\sinh^2(\varphi)  \cosh^2(\varphi) \right] (dx)^2 \\ &= (c dt)^2 (dx)^2 \end{align*}[/tex] So this quantity will always be the same. 


#27
Jul1911, 04:49 PM

P: 432

In the most simplest of terms, you and I see each other in motion at inertial v. I move vt (per you) while the photon moves ct (per you). You recognize that I hold myself as stationary, ie velocity =0. Therefore, when you consider how I hold said interval, you must subtract out the material motion you record of me from the light's motion ... hence, cv, and therefore ctvt. So the polarity of ct and vt must be opposite. Unfortunately though, it's not that simple. We're dealing with vectors here, and so the magnitude of the subtraction must abide by Pythagorean's theorem. Hence ... s^{2} = (ct)^{2}(vt)^{2}Or ... s^{2} = (ct)^{2}+(vt)^{2}where (vt)^{2} = x^{2} + y^{2} + z^{2} GrayGhost 


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