# Coordinate systems

Discussion in 'Physics & Math' started by arfa brane, Mar 18, 2017.

1. ### arfa branecall me arfValued Senior Member

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I claimed recently that the lengths of timelines, or timelike trajectories I guess, are frame dependent.
But, no, says rpenner, a frame is an imaginary construction.

But what about spacetime diagrams, like these two equivalent diagrams that "explain" the twin paradox.

I claim they support what I said: the lengths of timelines depend on which frame you're in.

3. ### karenmanskerHSIRIBanned

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I tend to agree with arfa brane . . . . . . however, it may depend upon one's relative perspective

5. ### rpennerFully WiredRegistered Senior Member

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Well, let's label the events of the diagrams with A for the shared start, B for the shared finish and C for the turnaround point for Mr. Traveller.

$\begin{array}{c|rr|rr} \textrm{Event} & x & t & x' & t' \\ \hline \\ ? & \frac{5}{4} x' + \frac{3}{4} c t' & \frac{3}{4} x' / c + \frac{5}{4} t' & \frac{5}{4} x - \frac{3}{4} c t & -\frac{3}{4} x / c + \frac{5}{4} t \\ A & 0.0 \, c \cdot \textrm{years}& 0.0 \, \textrm{years} & 0.0 \, c \cdot \textrm{years}& 0.0 \, \textrm{years} \\ C & +3.0 \, c \cdot \textrm{years} & 5.0 \textrm{years} & 0.0 \, c \cdot \textrm{years}& 4.0 \, \textrm{years} \\ B & 0.0 \, c \cdot \textrm{years}& 10.0 \, \textrm{years} & - 7.5 \, c \cdot \textrm{years} & 12.5 \, \textrm{years} \end{array}$

Then we can compute coordinate differences and the Invariant interval.
$\begin{array}{c|rrr|rrr} \textrm{Event Pair} & \Delta x & \Delta t & c^2 (\Delta t)^2 - (\Delta x)^2 & \Delta x' & \Delta t' & c^2 (\Delta t')^2 - (\Delta x')^2 \\ \hline \\ AB & 0.0 \, c \cdot \textrm{years} & 10.0 \, \textrm{years} & 100.00 \, c^2 \cdot \, \textrm{years}^2 & - 7.5 \, c \cdot \textrm{years} & 12.5 \, \textrm{years} & 100.00 \, c^2 \cdot \, \textrm{years}^2 \\ AC & +3.0 \, c \cdot \textrm{years} & 5.0 \textrm{years} & 16.00 \, c^2 \cdot \, \textrm{years}^2 & 0.0 \, c \cdot \textrm{years}& 4.0 \, \textrm{years} & 16.00 \, c^2 \cdot \, \textrm{years}^2 \\ CB & -3.0 \, c \cdot \textrm{years} & 5.0 \textrm{years} & 16.00 \, c^2 \cdot \, \textrm{years}^2 & - 7.5 \, c \cdot \textrm{years} & 8.5 \, \textrm{years} & 16.00 \, c^2 \cdot \, \textrm{years}^2 \end{array}$

Finally, we can compute elapsed proper time for each trajectory:
$\begin{array}{c|r|r} \textrm{Trajectory} & \Delta \tau = \frac{1}{c} \sum \sqrt{c^2 (\Delta t)^2 - (\Delta x)^2 } & \Delta \tau' = \frac{1}{c} \sum \sqrt{c^2 (\Delta t')^2 - (\Delta x' )^2 } \\ \hline \\ AB & 10 \, \textrm{years} & 10 \, \textrm{years} \\ AC & 4 \, \textrm{years} & 4 \, \textrm{years} \\ CB & 4 \, \textrm{years} & 4 \, \textrm{years} \\ AC + CB & 8 \, \textrm{years} & 8 \, \textrm{years} \end{array}$

So while x and t and even Δx and Δt are dependent on a choice of coordinates, c²(Δt)²−(Δx)² = c²(Δτ)² is not.

Both the interval between any two events of space-time and the total elapsed proper time for any time-like trajectory, no matter how complicated are independent of the choice of inertial coordinate system. They are frame-invariant quantities.

We can prove it is invariant for any inertial trajectory, given a Lorentz transform which means we can write the coordinate differences of one frame in terms of another,
$\Delta x' = \frac{ \Delta x - v \Delta t }{\sqrt{1 - v^2/c^2}}, \; \Delta t' = \frac{ -(v/c^2) \Delta x + \Delta t }{\sqrt{1 - v^2/c^2}}$
So
$c^2 (\Delta t')^2 - (\Delta x')^2 = c^2 \frac{ (-(v/c^2) \Delta x + \Delta t)^2 }{1 - v^2/c^2} - \frac{ (\Delta x - v \Delta t)^2 }{1 - v^2/c^2} \\ \quad \quad = \frac{ \left( \frac{v^2}{c^2} (\Delta x)^2 - 2 v (\Delta x)(\Delta t) + c^2 (\Delta t)^2 \right) - \left( (\Delta x)^2 - 2 v (\Delta x)(\Delta t) + v^2 (\Delta t)^2 \right) }{1 - v^2/c^2} \\ \quad \quad = \frac{ \frac{v^2}{c^2} (\Delta x)^2 + c^2 (\Delta t)^2 - (\Delta x)^2 - v^2 (\Delta t)^2 }{1 - v^2/c^2} \\ \quad \quad = \frac{ v^2 (\Delta x)^2 + c^2 c^2 (\Delta t)^2 - c^2 (\Delta x)^2 - c^2 v^2 (\Delta t)^2 }{c^2 - v^2} \\ \quad \quad = \frac{ ( v^2 - c^2 ) (\Delta x)^2 + c^2 ( c^2 - v^2) (\Delta t)^2 }{c^2 - v^2} \\ \quad \quad = c^2 (\Delta t)^2 - (\Delta x)^2$

Since the Lorentz transform doesn't change y or z when they are orthogonal to the direction of the change in velocity, the full four-dimensional invariant interval is also invariant.

Last edited: Mar 18, 2017
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7. ### rpennerFully WiredRegistered Senior Member

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Linear algebra.

Here I chose a particular matrix N, its inverse L and an undefined diagonal matrix M.
I want to consider the product LMN.
$L = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2c} & - \frac{1}{2c} \end{pmatrix} \\ M = \begin{pmatrix} b & 0 \\ 0 & a \end{pmatrix} \\ N = \begin{pmatrix} 1 & c \\ 1 & -c \end{pmatrix} \\ L N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ L M N = \begin{pmatrix} \frac{a + b}{2} & c \frac{a - b}{2} \\ c^{-1} \frac{a - b}{2} & \frac{a + b}{2} \end{pmatrix}$
It has an inverse provided neither a or b is zero. And its determinant is the same as that of M which means it is 1 only if a and b have a special relationship. Finally, because of the choice of N, it respects particular directions in space-time as eigenvectors.
$(L M N)^{-1} = N^{-1} M^{-1} L^{-1} = L M^{-1} N = L \begin{pmatrix} b^{-1} & 0 \\ 0 & a^{-1} \end{pmatrix} N = \begin{pmatrix} \frac{a^{-1} + b^{-1}}{2} & c \frac{a^{-1} - b^{-1}}{2} \\ c^{-1} \frac{a^{-1} - b^{-1}}{2} & \frac{a^{-1} + b^{-1}}{2} \end{pmatrix} \\ \textrm{det} \, L = \frac{1}{2 c} , \quad \textrm{det} \, M = a b, \quad \textrm{det} \, N = 2c \\ \textrm{det} (LMN) = 1 \quad \Rightarrow \quad b = a^{-1} \\ L M N \begin{pmatrix} +c \Delta t \\ \Delta t \end{pmatrix} = a \begin{pmatrix} +c \Delta t \\ \Delta t \end{pmatrix} \\ L M N \begin{pmatrix} -c \Delta t \\ \Delta t \end{pmatrix} = b \begin{pmatrix} -c \Delta t \\ \Delta t \end{pmatrix}$

So LMN is a stretch of "a" for light-like trajectories in the +x direction and a stretch of "b" for light-like trajectories in the -x direction (not surprising since it was constructed to be an example of eigenvalues and eigenvectors), and when $b = a^{-1}$ it has determinant of 1.

From here on, let's use only this special form.
$M = \begin{pmatrix} d^{-1} & 0 \\ 0 & d \end{pmatrix} \\ L M N = \begin{pmatrix} \frac{d + d^{-1}}{2} & c \frac{d - d^{-1}}{2} \\ c^{-1} \frac{d - d^{-1}}{2} & \frac{d + d^{-1}}{2} \end{pmatrix} \\ d = e^{\rho} \quad \Rightarrow \quad LMN = \begin{pmatrix} \cosh \rho & c \sinh \rho \\ c^{-1} \sinh \rho & \cosh \rho \end{pmatrix} \\ d = \sqrt{ \frac{c + v}{c - v} } \quad \Rightarrow \quad LMN = \begin{pmatrix} \frac{c}{\sqrt{c^2 - v^2}} & \frac{c v}{\sqrt{c^2 - v^2}} \\ c^{-2} \frac{c v}{\sqrt{c^2 - v^2}} & \frac{c}{\sqrt{c^2 - v^2}} \end{pmatrix} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \begin{pmatrix} 1 & v \\ c^{-2} v & 1 \end{pmatrix} \\ \begin{pmatrix} \Delta x' \\ \Delta t' \end{pmatrix} = LMN \begin{pmatrix} \Delta x \\ \Delta t \end{pmatrix} = \begin{pmatrix} \frac{d^2 + 1}{2d} \Delta x + \frac{d^2 - 1}{2d} c \Delta t \\ \frac{d^2 - 1}{2d} c^{-1} \Delta x + \frac{d^2 + 1}{2d} \Delta t \end{pmatrix} \\ N \begin{pmatrix} \Delta x \\ \Delta t \end{pmatrix} = \begin{pmatrix} \Delta x - c \Delta t \\ \Delta x + c \Delta t \end{pmatrix} \\ \begin{pmatrix} \Delta x' - c \Delta t' \\ \Delta x' + c \Delta t' \end{pmatrix} = N \begin{pmatrix} \Delta x' \\ \Delta t' \end{pmatrix} = N L M N \begin{pmatrix} \Delta x \\ \Delta t \end{pmatrix} = (NL) M \left( N \begin{pmatrix} \Delta x \\ \Delta t \end{pmatrix} \right) = M \begin{pmatrix} \Delta x - c \Delta t \\ \Delta x + c \Delta t \end{pmatrix} = \begin{pmatrix} d^{-1} ( \Delta x - c \Delta t ) \\ d ( \Delta x + c \Delta t ) \end{pmatrix} \\ c \Delta t' + \Delta x' = d ( c \Delta t + \Delta x) \\ c \Delta t' - \Delta x' = d^{-1} ( c \Delta t - \Delta x) \\ c^2 ( \Delta t' )^2 - ( \Delta x' )^2 = ( c \Delta t' + \Delta x' ) (c \Delta t' - \Delta x' ) = d ( c \Delta t + \Delta x) d^{-1} ( c \Delta t - \Delta x) = c^2 ( \Delta t)^2 - ( \Delta x )^2$
So that's three ways to write the Lorentz transform where
$d = e^{\rho} = \sqrt{\frac{c + v}{c - v}} \\ v = c \tanh \rho = c \frac{d - d^{-1}}{d + d^{-1}} = c \frac{d^2 - 1}{d^2 + 1} \\ \rho = \tanh^{-1} \frac{v}{c} = \ln \sqrt{\frac{c + v}{c - v}} = \ln d$

And we have a demonstration that it preserves the invariant interval.
Therefore no choice of inertial coordinates can change the invariant interval — it is independent of choice of frame.

Last edited: Mar 18, 2017
8. ### arfa branecall me arfValued Senior Member

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In the context of the diagrams, a would be in the outward leg (of lightlike trajectories), and b in the inward leg? This is where the two twins communicate with light signals IOW?

Could we say that tensor algebra is linear algebra with indices, or is that too naive a question?
More to get one's head around in 4-dimensions, events are related and the three relations pertain to causality (or 'causal structure'); the relations between pairs of events are timelike, spacelike, or null (e.g. event x is null-related to event y if light is 'communicated' between x and y.

Naive question #2: why are there only three causal relations between pairs of events? Does it have anything to do with the number of dimensions?

Last edited: Mar 19, 2017
9. ### arfa branecall me arfValued Senior Member

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I'll try to answer the second question: in a Minkowski diagram with 1 + 1 dimensions, the speed (or velocity) of light separates the set of timelike related pairs of events from the set of spacelike related pairs (so the related points are on timelike or spacelike trajectories resp.), there are only three sets in 2 + 1, or even in n + 1 dimensions. It seems that having a time dimension is important (because MS being flat makes it a velocity space and velocity is a time derivative), but it's important to have one or more space dimensions, a weaker condition perhaps.

10. ### rpennerFully WiredRegistered Senior Member

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In post #3, the outward leg of the traveller is identified as AC and the inward leg of the traveller is identified as CB.
In post #4, only the math of the Lorentz transform, independent of application to the diagram, is discussed to explain why it guarantees independence from choice of coordinates of the invariant interval.
Since the transform from coordinates where AB is considered at rest to coordinates where AC is considered at rest is given by the particular Lorentz transform where $v = -\frac{3}{5} c, \, d = a = b^{-1} = \frac{1}{2}, \; \rho = - \ln 2$ this is the only way I see to relate "a" and "b" of post #4 to the diagram.
The light-like signals between Earth and the traveller are not discussed in either post #3 or #4. Would you care to label the coordinates of one and discuss it in terms of post #3?

Nope. The Earth signals are rays (half-lines) of light-light trajectories sent at periods of a year. You may choose to describe them as rays:
$o_{k}: t - x/c = k \times 1\, \textrm{year} , \; x \geq 0, \; 0 \leq k \leq 10$
Or as pairs of events:
$D_{k}: \left( x = 0, \; t = k \times 1\, \textrm{year} \right) , \; 0 \leq k \leq 10 \\ E_{k}: \left\{ \begin{array}{lll} \left( x = k \times \frac{3}{2} \, c \cdot \textrm{years} , \; t = k \times \frac{5}{2}\, \textrm{years} \right) , & \quad \quad \quad & 0 \leq k \leq 2 \\ \left( x = (10 -k) \times \frac{3}{8} \, c \cdot \textrm{years} , \; t = (6 + k) \times \frac{5}{8}\, \textrm{years} \right) , & & 2 \lt k \leq 10 \end{array} \right.$
The second form is easier to apply the Lorentz transform to.

Indices have nothing to do with it, plenty of them in linear algebra.
A tensor is a generalization of linear algebra that encompasses scalars, vectors, matrices and higher dimensional linear objects.

Look at the sign of the invariant interval in such cases. And I count six cases if you order your events.

• Space-like (C and D_5)
• Same-place, Same-time D_0 and A
• Null-like, forward in time D_3 and E_3
• Null-like, backward in time E_7 and D_7
• Time-like, forward in time A and B
• Time-like, backward in time B and C

11. ### rpennerFully WiredRegistered Senior Member

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You need both 1 time dimension and at least 1 space dimension for movement and light-speed to be sensible things to talk about.

Light signals from Traveller
Just as the family of outbound rays can be written parameterized by the proper time of the Earth clocks at time of sending,
$o_{k}: t - x/c = k \times 1\, \textrm{year} , \; x \geq 0, \; 0 \leq k \leq 10$
the inbound rays can be written parameterized by the proper time of the traveller clock at time of sending. Here we work in coordinates where Earth is at rest.
$i_{k}: t + x/c = \left\{ \begin{array}{lll} 2 k \times 1 \, \textrm{year} , & \quad \quad \quad & 0 \leq k \leq 4 \\ \left( 6 + \frac{1}{2} k \right) \times 1 \, \textrm{year} , & & 4 \lt k \leq 8 \end{array} \right. , x \geq 0$
This gives us start and end events:
$F_{k}: \left\{ \begin{array}{lll} \left( x = k \times \frac{3}{4} \, c \cdot \textrm{years} , \; t = k \times \frac{5}{4}\, \textrm{years} \right) , & \quad \quad \quad & 0 \leq k \leq 4 \\ \left( x = (8 -k) \times \frac{3}{4} \, c \cdot \textrm{years} , \; t = k \times \frac{5}{4}\, \textrm{years} \right) , & & 4 \lt k \leq 8 \end{array} \right. \\ G_{k}: \left\{ \begin{array}{lll} \left( x = 0 , \; t = k \times 2 \, \textrm{years} \right) , & \quad \quad \quad & 0 \leq k \leq 4 \\ \left( x = 0 , \; t = (12 + k) \times \frac{1}{2}\, \textrm{years} \right) , & & 4 \lt k \leq 8 \end{array} \right.$

12. ### rpennerFully WiredRegistered Senior Member

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Relativity of Simultaneity
Now that the events on AB and AC + CB are parameterized by proper time as $D_k$ and $F_k$ respectively, we may create families of all events "simultaneous" with those events in the frames where the trajectory the event is on AB, AC or CB is considered motionless.

$\begin{array}{lcl|l} \textrm{conditions} & & \textrm{equation} & \textrm{equation} \\ \hline \\ t =t_{D_k} & \quad \Rightarrow \quad & t = k \times 1 \, \textrm{year} & t' + \frac{3}{5} \frac{1}{c} x' = k \times \frac{4}{5} \, \textrm{years} \\ t' = t'_{F_k} \; \wedge \; k < 4 & \quad \Rightarrow \quad & t - \frac{3}{5} \frac{1}{c} x = k \times \frac{4}{5} \, \textrm{years} & t' = k \times 1 \, \textrm{year} \\ t'' = t''_{F_k} \; \wedge \; k > 4 & \quad \Rightarrow \quad & t + \frac{3}{5} \frac{1}{c} x = \left( \frac{9}{2} + k \right) \times \frac{4}{5} \, \textrm{years} & t' + \frac{15}{17} \frac{1}{c} x' = \left( \frac{9}{2} + k \right) \times \frac{8}{17} \, \textrm{years} \end{array}$

And in points, we have two new families that are paired DH and IF to form simultaneity slices for Earth and Traveller:

$H_k =\left\{ \begin{array}{lll} \left( x = \frac{3}{5} k \, c \cdot \textrm{years}, \; t = k \, \textrm{years} \right), & \quad \quad \quad & 0 \leq k \leq 5 \\ \left( x = 6 - \frac{3}{5} k \, c \cdot \textrm{years}, \; t = k \, \textrm{years} \right) , & & 5 \lt k \leq 10 \end{array} \right. \\ I_k = \left\{ \begin{array}{lll} \left( x = 0, \; t = \frac{4}{5} k \, \textrm{years} \right), & \quad \quad \quad & 0 \leq k \lt 4 \\ \left( x = 0, \; t = \frac{18}{5} + \frac{4}{5} k \, \textrm{years} \right), & & 4 \lt k \leq 8 \end{array} \right.$

Notice that at x=0 and k near 4, the Earth event simultaneous the with traveller, I, jumps from t=16/5 years to t=34/5 years. That's why relativity of simultaneity is defined between inertial coordinate frames and not generic trajectories, per se.

Summary of event labels

$B = D_{10} = E_{10} = F_8 = G_8 = H_{10} = I_8 \\ \begin{array}{ll} D_{(8,10)} = G_{(4,8)} = I_{(\frac{11}{2}, 8)} \\ D_8 = G_4 = I_{\frac{11}{2}} \\ D_{(\frac{34}{5}, 8)} = G_{(\frac{17}{5}, 4)} = I_{(4, \frac{11}{2})} \\ \lim_{k \to 4 +} I_k = D_{\frac{34}{5}} = G_{\frac{17}{5}} \\ & E_{(2, 10)} = F_{(4, 8)} = H_{(5, 10)} \\ D_{(5, \frac{34}{5})} = G_{(\frac{5}{2}, \frac{17}{5})} \\ D_5 = G_{\frac{5}{2}} & C = E_2 = F_4 = H_5 \\ D_{(\frac{16}{5}, 5} = G_{(\frac{8}{5}, \frac{5}{2})} \\ & E_{(0,2)} = F_{(0, 4)} = H_{(0, 5)} \\ \lim_{k \to 4-} I_k = D_{\frac{16}{5}} = G_{\frac{8}{5}} \\ D_{(0, \frac{16}{5})} = G_{(0, \frac{8}{5})} = I_{(0, 4)} \end{array} \\ A = D_0 = E_0 = F_0 = G_0 = H_0 = I_0$

For all coordinate systems $k \leq \ell \; \Rightarrow \; t_{D_k} \leq t_{E_{\ell}}$ and $k \leq \ell \; \Rightarrow \; t_{F_k} \leq t_{G_{\ell}}$.

Last edited: Mar 21, 2017
13. ### arfa branecall me arfValued Senior Member

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After more reading, I'm confused about why you say this in post #4:
Can you clarify the difference between light-like signals (the red and blue lines in the diagram at 45°), and light-like trajectories given by LMN?

Last edited: Mar 21, 2017
14. ### rpennerFully WiredRegistered Senior Member

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The product LMN is a matrix which represents a linear transform of coordinates and meant to be a gentle introduction into Lorentz transforms.

Like all linear transforms LMN has two special directions which, when we talk about space-time coordinates, means we are talking about the 1) time direction, 2) a space direction or 3) some combination of the two which is the same as talking about a velocity since such a direction is a fixed amount of change in position with change in time. Like all linear transforms, when you apply it to coordinates representing displacement from the origin in one of its special directions, the change is the same as multiplying by a constant. This is a rescaling of the original input or conventionally, a "stretching."

Since it is a linear transform, differences of coordinates get stretched the same way as coordinates.
$LMN \begin{pmatrix} \Delta_{21} x \\ \Delta_{21} t \end{pmatrix} = LMN \begin{pmatrix} x_2 - x_1 \\ t_2 -t_1 \end{pmatrix} = LMN \begin{pmatrix} x_2 \\ t_2 \end{pmatrix} - LMN \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \Delta_{21} \left( LMN \begin{pmatrix} x \\ t \end{pmatrix} \right)$

LMN is special in that the two special directions are given by x = +c t and x = - c t . And the two constants of stretching (eigenvalues) are respectively a and b.
$LMN \begin{pmatrix} + c t \\ t \end{pmatrix} = \begin{pmatrix} + a c t \\ a t \end{pmatrix} = a \begin{pmatrix} + c t \\ t \end{pmatrix} \\ LMN \begin{pmatrix} - c t \\ t \end{pmatrix} = \begin{pmatrix} - b c t \\ b t \end{pmatrix} = b \begin{pmatrix} - c t \\ t \end{pmatrix}$
Since its linear, if the input is the sum of two special directions, the output is the sum of the stretched directions.
$LMN \begin{pmatrix} + c t_1 - c t_2 \\ t_1 + t_2 \end{pmatrix} = a \begin{pmatrix} + c t \\ t \end{pmatrix} + b \begin{pmatrix} - c t \\ t \end{pmatrix} = \begin{pmatrix} + a c t_1 - b c t_2 \\ a t_1 + b t_2 \end{pmatrix}$
And this works in the general case:
$LMN \begin{pmatrix} x \\ t \end{pmatrix} =\frac{t + x/c}{2} LMN \begin{pmatrix} + c \\ 1 \end{pmatrix} + \frac{t - x/c}{2} LMN \begin{pmatrix} - c \\ 1 \end{pmatrix} = \begin{pmatrix} + a c \frac{t + x/c}{2} - b c \frac{t - x/c}{2} \\ a \frac{t + x/c}{2} + b \frac{t - x/c}{2} \end{pmatrix} = \begin{pmatrix} \frac{a - b}{2} c t + \frac{a + b}{2} x \\ \frac{a + b}{2} t + \frac{a - b}{2} x /c \end{pmatrix}$

LMN is not a Lorentz transform unless $a \gt 0, b \gt 0, a \times b = 1$.

So given $a = \sqrt{\frac{c + v}{c -v}} , b = \sqrt{\frac{c - v}{c + v}}$ we see that LMN is the Lorentz transform with
$\frac{a + b}{2} = \frac{c}{\sqrt{c^2 - v^2}} = \frac{1}{\sqrt{1 - v^2/c^2}} \\ \frac{a - b}{2} = \frac{v}{\sqrt{c^2 - v^2}} = \frac{v/c}{\sqrt{1 - v^2/c^2}} \\ LMN \begin{pmatrix} x \\ t \end{pmatrix} = \begin{pmatrix} \frac{v}{\sqrt{1 - v^2/c^2}} t + \frac{1}{\sqrt{1 - v^2/c^2}} x \\ \frac{1}{\sqrt{1 - v^2/c^2}} t + \frac{v/c^2}{\sqrt{1 - v^2/c^2}} x \end{pmatrix}$

Last edited: Mar 22, 2017
15. ### arfa branecall me arfValued Senior Member

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So what you mean by "stretching of lightlike trajectories" can be seen in the diagrams as the different distances between red and blue lines, for instance in the earth frame the red lines are closer together in the left hand diagram than in the right hand one?

16. ### arfa branecall me arfValued Senior Member

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Hmm, or could the "stretching" also be seen as the differences in length? The red and blue lines are different lengths in each diagram.

17. ### rpennerFully WiredRegistered Senior Member

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Yes. But you are using the Euclidean length of separation of the light-like lines in the diagram, not space-time “length” in the sense I originally meant it: as the square root of the invariant interval aka. proportional to the proper time of a time-like interval.
Since in the diagram the red and blue lines are perpendicular, you are looking at the same stretching two different ways. The area of each parallelogram stays the same because the product of a times b is 1.

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Ka-ching!

19. ### arfa branecall me arfValued Senior Member

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Another comment on the two diagrams, regarding the simultaneous hyperplanes (in the left-hand diagram these are all parallel to the x axis for the inertial earth frame, for the outbound frame there is a set of parallel lines meeting the outbound trajectory at 1, 2, 3, 4 years). The 4-year point is also the turnaround point, so there are two hyperplanes 'meeting' at this point, and what appears to be a gap (of several years) in the earth frame's intersection with the travelling frame. These same lines are parallel to the x axis in the right hand diagram (the inertial frame of the outbound traveller).

But the hyperplanes don't represent a causal connection between the different frames--it takes light 3 years to get from earth to the turnaround point where the simultaneous hyperplanes also 'tilt' symmetrically w.r.t. the distance involved. Instead they denote each frame's determination of "now".

20. ### rpennerFully WiredRegistered Senior Member

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First half of post #9.