Why is the Time component in the Space Time Interval negative?

[hmsmatthew]

Why is the Time component in the Space Time Interval negative?

The space time interval is defined to be:

ds^2=dt^2-dx^2-dy^2-dz^2

or depending on the convention used it may also be:

ds^2=-dt+dx^2+dy^2+dz^2

The equation is defining distance using the pythagorean theorem. This results in the flat spacetime metric being defined as:

-1000
0100
0010
0001

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

and the metric would then be:

1000
0100
0010
0001

By the same logic?

Any help is appreciated.

Thanks

 

[matphysik]

The space time interval is defined to be:

ds^2=dt^2-dx^2-dy^2-dz^2

or depending on the convention used it may also be:

ds^2=-dt+dx^2+dy^2+dz^2

The equation is defining distance using the pythagorean theorem. This results in the flat spacetime metric being defined as:

-1000
0100
0010
0001

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

and the metric would then be:

1000
0100
0010
0001

By the same logic?

Any help is appreciated.

Thanks

A. Einstein took a great interest in Maxwell`s equations. He (apparently) was puzzled as to why the speed of light in vacuum was fixed, regardless of the speed of the source of the light.

In particular, if the source moves with speed `v` then (according to Galileo) the light should move with speed c±v.

So instead of using the motion of material bodies as central, he thought of using the motion of light (in vacuum, as in empty space) as central.

Now whether it was he, or one (or more) of his mathematician friends, the characteristic surfaces of the 3D wave operator were used to define a line element/metric. And that`s why the metric appears as it is, and not how you suggest.

Addendum:

See for example, chapter VIII of "Introduction to partial differential equations with applications" By E. C. Zachmanoglou, Dale W. Thoe (@Google books).

 

[WannabeNewton]

Well I don't know if this is the ONLY reason but consider a photon: a photon moves on null geodesics such that ds^2 = -c^2dt^2 + dx^2 = 0 so dx^2 / dt^2 = c^2 and in this way all observers agree on the speed of light.

 

[atyy]

It isn't based on logic. It's based on experiment. This property makes the speed of light the same in every inertial frame. This is why the metric is called "pseudo-Riemannian".

 

[matphysik]

Well I don't know if this is the ONLY reason but consider a photon: a photon moves on null geodesics such that [itex]ds^2 = -c^2dt^2 + dx^2 = 0[/itex] so dx^2 / dt^2 = c^2[/itex] and in this way all observers agree on the speed of light.

You must mean a worldline (?)

 

[hmsmatthew]

A. the characteristic surfaces of the 3D wave operator were used to define a line element/metric. QUOTE]

What do you mean by "the characteristic surfaces of the 3D wave operator were used to define a line element/metric" ?

Thanks

 

[WannabeNewton]

You must mean a worldline (?)

Pardon? Don't quite understand your question. A photon moves on a null geodesic such that the tangent vector to the geodesic has zero norm, in accordance with events being null - like separated along a photon's trajectory.

 

[hmsmatthew]

Well I don't know if this is the ONLY reason but consider a photon: a photon moves on null geodesics such that ds^2 = -c^2dt^2 + dx^2 = 0 so dx^2 / dt^2 = c^2 and in this way all observers agree on the speed of light.

Would all observers agree on the speed of light because of the invariance of spacetime distance between events as observed in different reference frames? (with the null geodesic as you suggest)

 

[matphysik]

Pardon? Don't quite understand your question. A photon moves on a null geodesic such that the tangent vector to the geodesic has zero norm, in accordance with events being null - like separated along a photon's trajectory.

Isn`t that GR (?)

 

[yuiop]

Look at like this. The spacetime interval using the the +--- convention is ds^2= (cdt)^2-dx^2-dy^2-dz^2. ds^2 is actually c^2 dtau^2 where tau is the proper time. By dividing both sides by c^2 you get dtau^2 = t^2- (dx^2-dy^2-dz^2)/c^2. For a particle traveling at the speed of light, the proper time tau is always zero so if dx^2, dy^2 or dz^2 are non zero positive real numbers, dt has to be equal and opposite in order for dtau to be zero. If you had a ++++ metric any non zero spatial displacement (squared) would require dt to be an imaginary number in order for dtau to be zero. The +--- or -+++ formalisms are not just conventions but dictated by nature. You could also write c^2 = ((cdt)^2-dx^2-dy^2-dz^2)/dtau^2 and note that c is a constant of nature. If we had a +++ metric any change in the spatial displacement would would require imaginary time in order for c to be constant. It is nature that decides c is a constant and that is just the way things are.

Of course you could get around all this by writing the metric as ds^2 = (ic dt)^2+dx^2+dy^2+dz^2 where i is the imaginary number to force a ++++ metric and indeed in the past some people have done this, but it is not generally used these days probably because it would be confusing.

 

[matphysik]

The space time interval is defined to be:

ds^2=dt^2-dx^2-dy^2-dz^2

or depending on the convention used it may also be:

ds^2=-dt+dx^2+dy^2+dz^2

The equation is defining distance using the pythagorean theorem. This results in the flat spacetime metric being defined as:

-1000
0100
0010
0001

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

and the metric would then be:

1000
0100
0010
0001

By the same logic?

Any help is appreciated.

Thanks

A final thought. If the spacetime interval had signature (++++) then what would ds²=0 represent?

 

[JDoolin]

It has something to do with the difference between the trigonometric functions and the hyperbolic trig functions.

Solutions of dy/dx=-x/y lead to x^2 + y^2 = Constant. These are circles.

Solutions of dx/dt = t/x lead to t^2 - x^2 = Constant. These are hyperbolas.

One big difference is that if you have something in front of you, you can turn around and make it to your left, then behind you, then to your right, then in front of you again. It all forms one continuos circle.

On the other hand, you can't take an event that's in your future, and and turn until it is to your left, then in your past, then to your right, then in your future again. They are four separate hyperbolas.

 

[hmsmatthew]

A final thought. If the spacetime interval had signature (++++) then what would ds²=0 represent?

With the ++++ signature I suppose ds^2=0 could only represent no change in space or time coordinates. Would this be the metric without considering special relativity?

I know that we work in units where c=1, so the null interval c^2*dt^2=dx^2+dy^2+dz^2 becomes dt^2=dx^2+dy^2+dz^2 This is the distance traveled for a light particle because it is effectively saying ct=x. So a spacetime interval could be >0 ("timelike") if the particle is slower than light. If the interval is "spacelike" it is <0. Would this imply faster than light travel for the particle? (Because -c^2dt^2+dx^2+dy^2+dz^2)? If so can an interval be less than zero? Perhaps it can but the event must be casually unrelated. It seems the metric signature is this way (-+++) in order to uphold the celestial speed limit of C.

 

[Dale]

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

One of the very important things about these metrics is the class of transformations that leave them unchanged. E.g. using the Cartesian coordinates rotations and translations do not change the distance between two points.

Now, from experiment we know that the speed of light is unchanged by boosts, so we want to find a metric which preserves the speed of light. The equation for a spherical pulse of light is:
$$c^2 dt^2 = dx^2 + dy^2 + dx^2$$
$$0= -c^2 dt^2 + dx^2 + dy^2 + dx^2$$

So any transformation which preserves the Minkowski metric also preserves the speed of light but the same is not true for the Euclidean metric.

 

[bobc2]

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

and the metric would then be:

1000
0100
0010
0001

By the same logic?

Any help is appreciated.

Thanks

The diagram at the left represents a guy in a red rocket moving to the left (with respect to the black rest system) at relativistic speed and another blue guy moving to the right with the same speed. Special relativity tells us that the two will experience different cross-section views of the 4-dimensional universe. Accordingly, when the red guy is at station 9 along his X4 world line his 3-D cross-section view of the 4D universe shows the blue guy to be at station 8. But, when the blue guy is at station 9 his 3-D cross-section view of the 4D universe shows the red guy to be at station 8. Each sees the other guys clock running slow.

Now, using the triangular relation diagrammed on the right, you can derive the metric beginning with the actual Pythagorean Theorem as your intuition seems to prefer. We simply recognize distances layed out in the 4-dimensional space. Look at the right triangle formed by the X4 coordinate of the red guy's system and the X1 and X4 coordinates of the blue guy's system. Just write down the Pythagorean equation directly for the hypotenuse and sides of the triangle. The blue X4 coordinate is the hypotenuse in our example. But, you can now solve the equation for just the red guy's X4 coordinate if you wish. And this is the form for the metric. Nevertheless, the Pythagorean Theorem still works for the spatial distances manifest in the diagram.

 

[JDoolin]

If you know
(1) how to multiply matrices
(2) the Maclaurin series expansion for e^x, sin(x), cos(x), sinh(x), and cosh(x)
then you can verify the following:


$${K1,K2,K3}= \left( \begin{array}{cccc} 0 & \beta_x & 0 & 0 \\ \beta_x & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & \beta_y & 0 \\ 0 & 0 & 0 & 0 \\ \beta_y & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & \beta_z \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \beta_z & 0 & 0 & 0 \end{array} \right)$$

$${S1,S2,S3}= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -\theta_1 \\ 0 & 0 & \theta_1 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \theta_2 \\ 0 & 0 & 0 & 0 \\ 0 & -\theta_2 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta_3 & 0 \\ 0 & \theta_3 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$$


Then $e^{K1}, e^{K2}, e^{K3}$ evaluate to the Lorentz Tranformations; i.e. accelerations in the x, y, z direction. and $e^{S1}, e^{S2}, e^{S3}$ evaluate to rotations in the x, y, z planes.

You notice there is one negative sign in the S1, S2, S3 definitions, but there is no negative sign in the Boost1, Boost2, Boost3 definitions.

As an example note that
$$\left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) ^2 = \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right )= \begin{pmatrix} -\theta^2 & 0\\ 0 & -\theta^2 \end{pmatrix}$$

and
$$\left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) ^3 = \begin{pmatrix} -\theta^2 & 0\\ 0 & -\theta^2 \end{pmatrix} \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) = \begin{pmatrix} 0 & -\theta ^3\\ \theta^3 & 0 \end{pmatrix}$$



Taking e to the power of the matrix

$$e^\left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right )$$

The first seven terms of the Maclaurin series expansion yields:

$$\begin{pmatrix} (1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} ) & (\theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!} ) \\ ( -\theta + \frac{\theta^3}{3!}-\frac{\theta^5}{5!} ) & (1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} ) \end{pmatrix}$$

and you can see this is leading to:


$$\begin{pmatrix} \cos(\theta) & sin(\theta) \\ -\sin(\theta) & cos(\theta) \end{pmatrix}$$

This final matrix is
\begin{pmatrix}
\frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}} &
\frac{\Delta \vec u_{x'}}{\Delta \vec u_{y}} \\
\frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}} &
\frac{\Delta \vec u_{y'}}{\Delta \vec u_{y}}
\end{pmatrix}

Which relates the change in the unit vector in the pre-rotation coordinates, to the unit vectors in the post-rotation coordinates.

For instance, if

$$\vec u_x \overset {def} = \begin{pmatrix} 1\\ 0 \end{pmatrix}$$

and the rotation maps that point

$$\begin{pmatrix} 1\\ 0 \end{pmatrix}\rightarrow \begin{pmatrix} cos(\theta)\\-sin(\theta) \end{pmatrix}$$

then

$$\frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}}=cos(\theta)$$

and

$$\frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}}=-sin(\theta)$$

There are several bits and pieces of this that I haven't worked out yet. For instance, why is it that

$$\begin{pmatrix} \frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}} & \frac{\Delta \vec u_{x'}}{\Delta \vec u_{y}} \\ \frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}} & \frac{\Delta \vec u_{y'}}{\Delta \vec u_{y}} \end{pmatrix} = e^{S1}$$

etc. I mean I can see that it is true, but I can't see why it is true. (Edit: More specifically, what would make someone think to take e to the power of a matrix. I guess it wouldn't have to be any more complicated than somebody saying "hey, I noticed these two things were equal." )

(This discussion extends from https://www.physicsforums.com/showthread.php?t=430956 and specifically Jackson's Classical Electrodynamics.)

 

[matphysik]

With the ++++ signature I suppose ds^2=0 could only represent no change in space or time coordinates. Would this be the metric without considering special relativity?

I know that we work in units where c=1, so the null interval c^2*dt^2=dx^2+dy^2+dz^2 becomes dt^2=dx^2+dy^2+dz^2 This is the distance traveled for a light particle because it is effectively saying ct=x. So a spacetime interval could be >0 ("timelike") if the particle is slower than light. If the interval is "spacelike" it is <0. Would this imply faster than light travel for the particle? (Because -c^2dt^2+dx^2+dy^2+dz^2)? If so can an interval be less than zero? Perhaps it can but the event must be casually unrelated. It seems the metric signature is this way (-+++) in order to uphold the celestial speed limit of C.

YES. The spacelike interval implies c<v. SR does not allow the possibility of a material particle being accelerated from v<c to v>c. However, it does allow the possibility of particles moving at FIXED v≥c.

"It seems the metric signature is this way (-+++) in order to uphold the celestial speed limit of C."

Also, (+---).

 

[GrayGhost]

If the distance between two points in cartesian coordinates is given by:

d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

Then why is the spacetime interval not defined to be:

d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

and the metric would then be:

1000
0100
0010
0001

By the same logic?

Any help is appreciated. Thanks

(ct)² = x²+y²+z²

Let's ignore the z-axis to make it simpler, work in 2d+time ...

(ct)² = x²+y²

Since no length contractions exist wrt axes orthoganal wrt the axis of motion (motion is along +x here), then y=Y where Y is of the other system. But Y=cTau, and so y=cTau. So ...

(ct)² = x²+(cTau)²

and so ...

(cTau)² = (ct)²-x²

or equivalently, after multiplying by -1 ...

-(cTau)² = -(ct)²+x²

Note that this equation has Tau of one system left of the equal sign, and x & t of the other system right of the equal sign. So it defines Tau as a function of the other system x,t.

In imaginary systems, i² = -1. So the above may be rewritten as ...

(icTau)² = -(ct)²+x²

and since the length of the spacetime interval per Minkowski is s = icTau, then since (icTau)²=s², the above becomes ...

s² = -(ct)²+x²

More generically ...

s² = -(ct)²+x²+y²+z²

So s is a length of the X,Y,Z,Tau system (numerically equal to time Tau), but is derived here in terms of the other system x,y,z,t. This is why the metric can vary from the traditional d = ++++ and instead become s = -+++ .

GrayGhost

 

[matphysik]

A. the characteristic surfaces of the 3D wave operator were used to define a line element/metric. QUOTE]

What do you mean by "the characteristic surfaces of the 3D wave operator were used to define a line element/metric" ?

Thanks

Please see pages 274-275 of the book i mentioned.

 

[JDoolin]

Another way of looking at it is that when you are talking $(x,y)$ coordinate distance, you can imagine concentric circles around either particle and ask, what concentric circle the other particle is on. The answer involves a $r=\sqrt{x^2 + y^2}$

With $(x,t)$ 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 $\tau = \sqrt{t^2-x^2}$ or $s=\sqrt{x^2-t^2}$ (depending on which quadrant you're in.)

 

[cosmik debris]

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.

 

[Dale]

you can only travel in one direction in the time dimension of spacetime.

That is primarily due to the fact that there is only one timelike dimension. If there were two or more then you could have closed timelike curves in flat spacetime.

 

[JDoolin]

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.

I think the simplest you can make it, and still get the idea across, is to say the signature is related to the difference between concentric circles (in space-space), and concentric hyperbolas (in space-time).

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 space-time. But all those directions are within the light-cone.

 

[JDoolin]

On further reflection, it occurred to me that if there were only one direction for time in space-time, the metric should be:

$$d \tau^2 = dt^2 + 0 dx^2 + 0 dy^2 + 0 dz^2$$

rather than

$$d \tau^2 = dt^2 - \frac{1}{c} dx^2 - \frac{1}{c} dy^2 - \frac{1}{c} dz^2$$

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 observer-dependent 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.

 

[bcrowell]

A final thought. If the spacetime interval had signature (++++) then what would ds²=0 represent?

Matphysik, please don't quote an entire long post when all you're posting one line. It makes it very difficult to follow the discussion.

 

[JDoolin]

The origin can be traced to Einstein's OEMB paper. Einstein noticed that, when starting from the Lorentz transforms:
(1) $x'=\gamma(x-vt)$
(2) $t'=\gamma(t-vx/c^2)$

by differentiation, one gets:
(3) $dx'=\gamma(dx-vdt)$
(4) $dt'=\gamma(dt-vdx/c^2)$

He further observed that the above results in the discovery of the frame invariance of the expression:
(5) $(cdt)^2-dx^2=(cdt')^2-dx'^2=...$

which prompted Minkowski to develop his 4D formalism where
(6) $ds^2=(cdt)^2-dx^2$

is taken to be the norm of the four-vector $(icdt,dx)$

Welcome to Physics Forums, fft.

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.

$\begin{align*} dx' &=\gamma(dx-vdt) \\ &= \gamma(dx-\frac{v}{c} c dt) \\ &= \gamma dx - \beta \gamma c dt\\ &= \begin{pmatrix} \gamma & -\beta \gamma \end{pmatrix} \begin{pmatrix} dx \\ c dt \end{pmatrix} \end{align*}$

and
$\begin{matrix} dt'= \left (\gamma dt-\frac{v dx}{c^2} \right )\\ \begin{align*} c dt' &=\gamma \left (c dt - \frac{v}{c} \cdot dx \right ) \\ &= \gamma c dt - \beta \gamma dx \\ &=\begin{pmatrix} -\beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} dx \\ c dt \end{pmatrix} \end{align*} \end{matrix}$

hence
$$\begin{pmatrix} dx'\\ c dt' \end{pmatrix} =\begin{pmatrix} \gamma & -\beta \gamma\\ -\beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} dx\\ c dt \end{pmatrix}$$

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:Hyperbolic_functions.svg )

Then

$$\begin{align*} \gamma &= \frac{1}{\sqrt{1-\beta^2}} \\ &= \frac{1}{\sqrt{1-\tanh^2(\varphi)}} \\ &= \frac{1}{\sqrt{1- \frac{ \sinh^2(\varphi) }{ \cosh^2(\varphi)} }} \\ &=\frac{\cosh(\varphi) }{\sqrt{\cosh^2(\varphi) -\sinh^2(\varphi)}}\\ &=\cosh(\varphi) \end{align*}$$

and similarly

$$\begin{align*} \beta \gamma &= \frac{\beta}{\sqrt{1-\beta^2}} \\ &= \frac{ \tanh(\varphi) }{\sqrt{1-\tanh^2(\varphi) }} \\ &= \frac{\tanh(\varphi)}{\sqrt{1- \frac{ \sinh^2(\varphi) }{ \cosh^2(\varphi)} }} \\ &=\frac{\cosh(\varphi) \tanh(\varphi)}{\sqrt{\cosh^2(\varphi) -\sinh^2(\varphi)}}\\ &=\sinh(\varphi) \end{align*}$$

Now,
$$\begin{align*} (c dt')^2 &= (-\sinh(\varphi )dx +\cosh(\varphi)c dt)^2\\ &=\sinh^2 (\varphi)(dx)^2 -2 \cosh(\varphi)c dx dt +\cosh^2(\varphi)(c dt)^2 \end{align*}$$

and

$$\begin{align*} (dx')^2 &=\left( \cosh(\varphi )dx -\sinh(\varphi)c dt\right)^2\\ &=\cosh^2 (\varphi)(dx)^2 -2 \cosh(\varphi)c dx dt +\sinh^2(\varphi)(c dt)^2 \end{align*}$$

Subtracting the two gives

$$\begin{align*} (c dt')^2 - (dx)^2 &=\left[\cosh^2(\varphi) - \sinh^2(\varphi) \right] (c dt)^2 +\left[\sinh^2(\varphi) - \cosh^2(\varphi) \right] (dx)^2 \\ &= (c dt)^2 -(dx)^2 \end{align*}$$

So this quantity will always be the same.

 

[GrayGhost]

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.

While it is true that we all seem to travel thru 1 time dimension, I don't see how what you say here supports an answer to the question of a -+++ metric?

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, c-v, and therefore ct-vt. 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 ...

-s² = (ct)²-(vt)²​

Or ...

s² = -(ct)²+(vt)²​

where (vt)² = x² + y² + z²

GrayGhost