I Spacetime interval and basic properties of light

[Ibix]

The right hand side is just $\Delta s^2$, which is invariant in Minkowski space.

...which you can demonstrate by using the Lorentz transforms to express $\Delta t$ etc in terms of $\Delta t'$ etc and seeing that you get an identical formula in terms of the primed quantities.

 

[Nugatory]

However I already read the part about lorenz transformation and a lot of stuff about the basic principles of light several times at different places and a lot of video's

That’s not “working through a suitable textbook”, that’s a haphazard and disjointed activity that is unlikely to ever lead to a coherent understanding.

Taylor and Wheeler’s “Spacetime Physics” is available free online. Start at the first page. We can help you over the hard spots.

 

[Orodruin]

...which you can demonstrate by using the Lorentz transforms to express $\Delta t$ etc in terms of $\Delta t'$ etc and seeing that you get an identical formula in terms of the primed quantities.

I'd rather go the other way. Lorentz transformations are those transformations that keep the form of the Minkowski line element.

 

[Sagittarius A-Star]

So for me easy to accept that the speed of light is constant in all frames so that is not the point.

The point is not, that the speed of light is constant. What you need to understand is, that the speed of light is invariant.

Example: A ship moves with constant velocity $v$ away from the harbor and a person $A$ at rest at the harbor sent a light-pulse to the ship. According to the principle of relativity, a person $B$ on the ship can regard the ship as being at rest and the harbor as moving. Person $B$ will measure, that the light-pulse has velocity $c$ relative to the ship and not velocity $c-v$, as you might intuitively think.

 

[HansH]

The equation of the circle of radius $c \Delta t$ is $(c \Delta t)^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$. Subtract $(c\Delta t)^2$ from both sides and you get
$$
0 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2.
$$
The right hand side is just $\Delta s^2$, which is invariant in Minkowski space.

isn't that exactly the background of what I also had in mind in #25 ?, but seemed to be wrong because there is was conclued that I was not allowed to apply the pythagoras rule but an equivalent rule for (in#30) that was the reason that I got lost. I assume now you apply the pythagoras rule .

 

[Dale]

there he says: all frames agree that the same set of events defines a sphere expanding at c. So that is where the minus sign comes from.
for me the conclusion: 'that is where the minus sign comes from' is still not clear. So there should be some additional thinung stapes in betwen that are logical to Dale but not to me. I see an expanding sphere (or 4 dmentional sphere ok) but then I would like to draw sone lines or whatever to understand the conclusion, but I do not have sufficient information to do that.

I do wish you had actually said this in response to my post. Based on our exchange I thought everything was clear to you after you refreshed your browser.

Ok, let’s go step by step.

1) we start with the equation of a sphere of radius $r$. $$\Delta x^2 + \Delta y^2+\Delta z^2=r^2$$
2) since $c$ is invariant all frames will agree on the events that form a sphere whose radius is expanding at $r=c\Delta t$, this is called the light cone. $$\Delta x^2+\Delta y^2+\Delta z^2=c^2 \Delta t^2$$
3) then we simply do one step of algebra to get: $$-c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 = 0$$
This explains where the minus sign comes from.

The expression on the left seems important so we give it a name: “the spacetime interval”. Not only have we found the reason for the minus sign, we also have shown that all frames agree on the set of events where the interval is 0, or null. In other words, the null interval defines the light cone, which is invariant.

4) we now make the small intuitive leap and ask ourselves “what happens if all spacetime intervals are invariant, not just null ones” $$-c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 = \Delta s^2$$
The answer to that question is that we get all of the experimental predictions of relativity, which we can compare against experimental data.

 

[Orodruin]

isn't that exactly the background of what I also had in mind in #25 ?, but seemed to be wrong because there is was conclued that I was not allowed to apply the pythagoras rule but an equivalent rule for (in#30) that was the reason that I got lost. I assume now you apply the pythagoras rule .

The Pythagorean theorem holds in space. Not in spacetime.

The point is that the quantity
$$
\Delta s^2 = -c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2
$$
is invariant in Minkowski space just as the Pythagorean theorem
$$
a^2 + b^2 = c^2
$$
holds regardless of the coordinate system in Euclidean space.

The $\Delta s^2$ is what corresponds to the squared length of the hypothenuse.

 

[malawi_glenn]

More generally, consider the four quantities $a_0$, $a_x$, $a_y$ and $a_z$.
If $a_0$ transform in the same way as $ct$ and $a_{x,y,z}$ transforms as $x,y,z$, the following combined quantity is invariant $a_0^2 - (a_x^2+a_y^2+a_z^2)$.
Moreover, if you also had four other quantities $b_0$, $b_x$, $b_y$ and $b_z$ which also transforms in that way, the following combined quantities are also invariant $b_0^2 - (b_x^2+b_y^2+b_z^2)$ and $a_0b_0 - (a_xb_x+a_yb_y +a_zb_z)$

 

[Sagittarius A-Star]

... I was not allowed to apply the pythagoras rule but an equivalent rule for (in#30) that was the reason that I got lost. I assume now you apply the pythagoras rule .

The Pythagoras rule is valid, but only for 3D-space, not for 4D-spacetime.

 

[HansH]

regarding #54-59: I think the problem for me is to exactly understand what is meant by the the concept of something being invariant. therefore I am lost at #56 2) already for example. That could probably also explain why I do not understand why pythagoras does not hold in 4d spacetime. so the question is if I read the proposed textbooks of pdf's don't I then run into the same problem? and if so how to solve?

 

[malawi_glenn]

I think the problem for me is to exactly understand what is meant by the the concept of something being invariant.

Post #37 shows what it means.

 

[HansH]

ok perhaps that helps. as said this is what I am digesting at the moment, so I will come back to this later.

 

[Dale]

I think the problem for me is to exactly understand what is meant by the the concept of something being invariant. therefore I am lost at #56 2) already for example.

Invariant just means that all frames agree on something. It doesn't change if you change your reference frame.

The second postulate says that the speed of light is the same in all inertial frames. So if something is going at the speed of light in one frame it is also going at the speed of light in every other frame.

Does that clear up my step-by-step? If not, please quote me directly and explain what specifically is problematic for you.

 

[HansH]

2) since $c$ is invariant all frames will agree on the events that form a sphere whose radius is expanding at $r=c\Delta t$, this is called the light cone. $$\Delta x^2+\Delta y^2+\Delta z^2=c^2 \Delta t^2$$

Here I cannot follow. probsbly I do not fully realize what the expanding sphere exactly means in relation to moving reference frames in relation to each other.

 

[PeroK]

regarding #54-59: I think the problem for me is to exactly understand what is meant by the the concept of something being invariant. therefore I am lost at #56 2) already for example. That could probably also explain why I do not understand why pythagoras does not hold in 4d spacetime. so the question is if I read the proposed textbooks of pdf's don't I then run into the same problem? and if so how to solve?

In my experience most people who try to learn SR haven't studied enough basic physics. Sometimes even concepts like motion, velocity and acceleration are poorly understood. More usually, it is the concept of a reference frame and invariance that are a stumbling block. IMO, these should be studied in classical (Newtonian) physics first before tackling SR. Often, in fact, it's not SR that is the problem, but the concept of studying a kinematic problem from two different reference frames.

Classical physics (more or less) shares the postulate with SR that the laws of physics are the same in every inertial reference frame. If you look at Einstein's original 1905 paper, he actually says "consider a frame where the laws of Newtonian mechanics hold good"!

This is what allows us to do physics on Earth (playing a game of tennis, for example), without having to take into account the motion of the Earth, Sun and Milky Way relative to the fixed stars. Of course, the Earth's surface is not quite an inertial reference frame, hence Foucault's pendulum and large-scale weather systems - and it has gravity - but it's close enough in practical terms in many cases.

In classical physics, lengths and time intervals are invariant. Note that distances, speeds, momentum and kinetic energy are not. In any case, being able to transform a problem from one reference frame to another is a useful ability. E.g. studying a collision of two particles from either the laboratory frame (where one particle may be stationary); or, the centre of momentum frame, where the two particles have equal and opposite momenta is an extremely useful technique.

Once you get to SR, lengths and time intervals are no longer invariant, but the speed of light and spacetime intervals are. It's not, IMO, a question of where a minus sign comes from, but a larger conceptual step to move from the context of classical, Newtonian (Galilean) relativity to Einstein's Special Relavity. The step to General Relativity is a much greater one.

Even to learn the basics of classical physics requires time, focus and effort.

 

[HansH]

Invariant just means that all frames agree on something. It doesn't change if you change your reference frame.

The second postulate says that the speed of light is the same in all inertial frames. So if something is going at the speed of light in one frame it is also going at the speed of light in every other frame.

Does that clear up my step-by-step?

what you say there is not new to me, but probably what it exactly means is still not clear.

 

[Dale]

Here I cannot follow. probsbly I do not fully realize what the expanding sphere exactly means in relation to moving reference frames in relation to each other.

If I have a flash of light, that flash expands in a spherical shape at a speed of $c$. That means that the radius of that sphere is $c\Delta t$. Is that clear?

 

[malawi_glenn]

ok perhaps that helps. as said this is what I am digesting at the moment, so I will come back to this later.

Post #37 shows, with quite many algebraic steps, that the "space-time interval" is invariant, it has the same value in $S$ (using $x$ and $t$ coordinates) as it has in $\tilde S$ (using $\tilde x$ and $\tilde t$ coordinates)

For post #58, use these ($\tilde S$ moves in $S$ with velocity $v$ in the $x$-direction, i.e. a "boost" in the x-direction)
$\tilde a_0 = \gamma \left(a_0 - \dfrac{a_xv}{c}\right) \qquad a_0 = \gamma \left(\tilde a_0 + \dfrac{\tilde a_xv}{c}\right)$
$\tilde a_x = \gamma \left(a_x - \dfrac{a_0v}{c}\right) \qquad a_x = \gamma \left(\tilde a_x + \dfrac{\tilde a_0v}{c}\right)$
$\tilde a_y = a_y \qquad \tilde a_z = a_z$

For instance you would get that $a_0^2 - (a_x^2 + a_y^2+a_z^2) = {\tilde a}_0^2 - ({\tilde a}_x^2 + {\tilde a}_y^2+{\tilde a}_z^2)$ which means that this quantity has the same value as measured in both frames. You should be able to show this with basically the same algebraic steps as for the space-time interval in #37.

 

[HansH]

In my experience most people who try to learn SR haven't studied enough basic physics. Sometimes even concepts like motion, velocity and acceleration are poorly understood. More usually, it is the concept of a reference frame and invariance that are a stumbling block.

the first point should be ok for me (although that is now 40 years ago). I had a final mark 9 at physics at pre-university level, but decided not to do a physics study. So could be the second point of invariance as I already indicated.

 

[Dale]

what you say there is not new to me, but probably what it exactly means is still not clear.

Ok, but you will have to be more explicit. I cannot read your mind. So if you don’t say what specifically is unclear I cannot clarify. Instead of rushing to respond, take some time to read, think, and pin down the actual question

 

[HansH]

If I have a flash of light, that flash expands in a spherical shape at a speed of $c$. That means that the radius of that sphere is $c\Delta t$. Is that clear?

yes. I assume that should be the case for every ovserver that sees that light but moves at a different speed related to the other observer. so gives different points in space crossed at different times for each observer.

 

[HansH]

Ok, but you will have to be more explicit. I cannot read your mind. So if you don’t say what specifically is unclear I cannot clarify. Instead of rushing to respond, take some time to read, think, and pin down the actual question

''Invariant just means that all frames agree on something.''
it is difficult to imagine what they exactly agee on and how to draw that on paper and derive from that the minus sign.

 

[Dale]

yes. I assume that should be the case for every ovserver that sees that light but moves at a different speed related to the other observer. so gives different points in space crossed at different times for each observer.

Yes. And even though they will disagree about the different times and different points, they will all agree that it is a sphere of radius $c\Delta t$. That is what the second postulate means.

Please pause and think a bit. My step by step should now be clear

 

[HansH]

Yes. And even though they will disagree about the different times and different points, they will all agree that it is a sphere of radius $c\Delta t$. That is what the second postulate means.

Please pause and think a bit. My step by step should now be clear

ok thanks. I will first do some cycling that will for sure give the mind some rest to think. I come back later.

 

[malawi_glenn]

it is difficult to imagine what they exactly agee on and how to draw that on paper and derive from that the minus sign.

You are thinking in circles. Forget drawings, just do the calculations.
Minkowski diagrams are very hard to learn from since you will automatically think one should apply Pythagoras theorem for those right triangles. But that is wrong, the "distance" in space time is (ct)²-x² not (ct)²+x².

 

[robphy]

''Invariant just means that all frames agree on something.''
it is difficult to imagine what they exactly agee on and how to draw that on paper and derive from that the minus sign.

Now, try what I suggested in #48.

 

[robphy]

You are thinking in circles. Forget drawings, just do the calculations.
Minkowski diagrams are very hard to learn from since you will automatically think one should apply Pythagoras theorem for those right triangles. But that is wrong, the "distance" in space time is (ct)²-x² not (ct)²+x².

One has to properly learn how to think with Minkowski diagrams... since they have a nonEuclidean geometry... but it's not as bad as you make it sound. (Think trigonometrically... but use hyperbolic-trig.)
(As I have often said, ordinary position-vs-time diagrams in PHY101 also have a nonEuclidean geometry...
but we have learned to sort-of read it and not pay attention to its geometry.)
Both are specific variants of Euclidean geometry:
vary the E-slider in
https://www.desmos.com/calculator/kv8szi3ic8

 

[malawi_glenn]

One has to properly learn how to think with Minkowski diagrams... since they have a nonEuclidean geometry... but it's not as bad as you make it sound. (Think trigonometrically... but use hyperbolic-trig.)
(As I have often said, ordinary position-vs-time diagrams in PHY101 also have a nonEuclidean geometry...
but we have learned to sort-of read it and not pay attention to its geometry.)
Both are specific variants of Euclidean geometry:
vary the E-slider in
https://www.desmos.com/calculator/kv8szi3ic8

It is not impossible no.

 

[PeterDonis]

you mean this?

Yes.

but then it would mean that according to pythagoras

The Pythagorean theorem only holds in Euclidean geometry. The geometry of spacetime is not Euclidean.

 

[PeterDonis]

if you say :In Minkowski space, the equivalent of Pythagoras’ theorem is : then I am back to the openings question of the topic: why, because I still do not understand

As was already pointed out, you could ask the same question about the Pythagorean theorem in Euclidean space: why is that theorem true?

What is your answer to that question? I assume you have one since you seem to have no problem with just accepting that the Pythagorean theorem is true in Euclidean space. Whatever reason that is will work just as well for accepting that the Minkowski version is true in Minkowski space.

 

[PeterDonis]

if you say :In Minkowski space, the equivalent of Pythagoras’ theorem is : then I am back to the openings question of the topic: why, because I still do not understand

Another way of looking at the answer I gave in my previous post just now is this: the Minkowski formula is the metric of Minkowski spacetime just as the Pythagorean formula is the metric of Euclidean space. "Metric" is a general concept and doesn't just apply to Euclidean space, it applies to any geometry. Minkowski spacetime is just a different geometry. (In the older literature it is sometimes referred to as "hyperbolic geometry". One of the key mathematical discoveries of the 19th century was the discovery of non-Euclidean geometries; Minkowski spacetime is just an application of that discovery to physics.)

 

[PeroK]

As was already pointed out, you could ask the same question about the Pythagorean theorem in Euclidean space: why is that theorem true?

What is your answer to that question? I assume you have one since you seem to have no problem with just accepting that the Pythagorean theorem is true in Euclidean space. Whatever reason that is will work just as well for accepting that the Minkowski version is true in Minkowski space.

The book The Pythagorean Proposition has hundreds of proofs of Pythagoras' Theorem!

https://www.goodreads.com/book/show/4651019-the-pythagorean-proposition

 

[robphy]

Minkowski spacetime is just a different geometry. (In the older literature it is sometimes referred to as "hyperbolic geometry". One of the key mathematical discoveries of the 19th century was the discovery of non-Euclidean geometries; Minkowski spacetime is just an application of that discovery to physics.)

While true, "hyperbolic geometry" is a terrible term for Minkowski spacetime geometry...
"hyperbolic trigonometry" might be better.

(Of course, "hyperbolic geometry" is a negatively-curved riemannian-signature geometry [violating Euclid's Parallel Postulate],
whereas "Minkowski spacetime" is a flat lorentz-signature geometry using the hyperbola [hyperboloid] for a circle [which does satisfy Euclid's Parallel Postulate].
Both are nonEuclidean geometries... both are actually Cayley-Klein geometries.)

(Minkowski spacetime violates Euclid's first postulate, suitably reformulated,
as I claim at https://physics.stackexchange.com/q...ids-5-postulates-false-in-minkowski-spacetime )

 

[PeterDonis]

The book The Pythagorean Proposition has hundreds of proofs of Pythagoras' Theorem!

Sure, but all of them assume Euclidean geometry.

 

[PeterDonis]

(Of course, "hyperbolic geometry" is the negatively-curved riemannian-signature geometry [violating Euclid's Parallel Postulate],
whereas "Minkowski spacetime" is flat lorentz-signature geometry using the hyperbola [hyperboloid] for a circle [which does satisfy Euclid's Parallel Postulate].

Yes, you're right, "hyperbolic geometry", strictly speaking, is not the same as Minkowski spacetime. (There is actually a slicing of de Sitter space in which each slice has hyperbolic geometry, I think that's what I was thinking of.)

 

[robphy]

The mass-shell and the "space of 4-velocities", which are certain 3D submanifolds of energy-momentum space and Minkowski spacetime respectively, have hyperbolic geometries.

 

[vanhees71]

Here's my try to explain Minkowski geometry from Einstein's two postulates. Maybe it helps

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[HansH]

at least what I see is that the subject of this topic where the - sign comes from seems to be an already assumed to be known starting point in your equation (1.2.1) at the first page. so based on that it cannot help me I assume.

 

[malawi_glenn]

at least what I see is that the subject of this topic where the - sign comes from seems to be an already assumed to be known starting point in your equation (1.2.1) at the first page. so based on that it cannot help me I assume.

Let's say i stand in the origin and shoot a ray of light in the +x direction. After t = 1s we have that x=ct, after t = 2s then x=2ct, and so on. Therefore (ct)² = x² which is equivalent to 0 = (ct)² - x².
Is this what you are trying to understand? Why there is a minus sign in front of x²?

 

[HansH]

I do wish you had actually said this in response to my post. Based on our exchange I thought everything was clear to you after you refreshed your browser.

Ok, let’s go step by step.4) we now make the small intuitive leap and ask ourselves “what happens if all spacetime intervals are invariant, not just null ones” $$-c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 = \Delta s^2$$
The answer to that question is that we get all of the experimental predictions of relativity, which we can compare against experimental data.

I checked your circle proposal and I think i understand that part. It is basically quite easy: you define a circle wirh a radius r=ct so at t=0 the light starts from the origin and both origins of the stationary frame and moving frame are at the same point at the moment that I fire my lightpulse. at t=1s the light is at a distance c so at a sphere with radius ct. that must be true for both observers because also for the moving observer the light follows from his/her perspective the same spherical expansion.

but then we come to the next point of the same equation with ds at the right side, so not the ''null one'' so for my understanding this means that at t=0 the light already is at a radius ds^2. but the question is then: where are both observers at t=0? are they still both in the same point or not? only when they are in the same point I can understand that they agree on the radius of the light cone being the same for both. But I assume the spacetime interval is generally valid, so also for any initial position of the reference frames. Then for me the reasoning with both circles however does not give a good understanding for me of the situation.

 

[HansH]

Let's say i stand in the origin and shoot a ray of light in the +x direction. After t = 1s we have that x=ct, after t = 2s then x=2ct, and so on. Therefore (ct)² = x² which is equivalent to 0 = (ct)² - x².
Is this what you are trying to understand? Why there is a minus sign in front of x²?

yes I think so. I assume your description with one dimension where the light goes to is same as what Dale showed me with 3 dimensions giving a sphere of light at distance ct. But how should I interpret that in general, so when ds is not 0, especially in relation to both observers.

 

[malawi_glenn]

yes I think so. I assume your description with one dimension where the light goes to is same as what Dale showed me with 3 dimensions giving a sphere of light at distance ct. But how should I interpret that in general, so when ds is not 0, especially in relation to both observers.

Yes it is, the 3D-case follows from that (spherical wavefront)

For other pair of events which do not have "light like" separation, you consult post #37
It follows from the Lorentz transformation that $s^2 = (ct)^2 - x^2$ (1) is invariant (this is the "1D" case since we start with just a boost in the x-direction here).
So just by imposing same speed of light in all inertial frames and the linear transformation, the lorentz-transformation and this result (1) pops out.

Here is an example.
Two flashes of lightning is measured to occur simultanously in frame $S$. They are separated with 250 m along the x-axis. A spaceship is traveling at 0.60c relative $S$ in the x-direction.
In the $S$ frame we have these coordinates $t_1 = t_2$ and $x_2 = x_1+250$ m.
In $S$ the space-time interval (squared) for these two events is $c(t_1-t_2)^2 - (x_1-x_2)^2 = 250^2$ m²
What are the time- and spatial coordinates for these two events in the spaceship frame $\tilde S$?
We consult the Lorentz-tranformation. The gamma factor is 1.25 (I am pretty sure you can calculate it yourself).
$\tilde x_1 = \gamma (x_1 - vt_1) \:,\: \tilde x_2 = \gamma (x_2 - vt_2) \:,\: \tilde t_1 = \gamma (t_1 - x_1v/c^2)\:,\:\tilde t_2 = \gamma (t_2 - x_2v/c^2)$
We can now calculate the space-time interval (squared) as measured in frame $\tilde S$.
Since we already know that this is the same as we had in $S$ we do not need to perform any calculations, we have that $c(\tilde t_1-\tilde t_2)^2 - (\tilde x_1-\tilde x_2)^2 = 250^2$ m² also!
I leave that as an excerise for you to actually compute this :)

Note that we have not mentioned observers yet, just coordinates measured in the different frames. For a stationary observer you also need to account for the time it takes for the information to travel to him/her.

 

[Dale]

so not the ''null one''

OK, so if I am hearing you right, you now understand where the minus sign for the $-c^2\Delta t^2$ comes from in my steps 1)-3), and now you want to move on to point 4). Is that correct?

 

[HansH]

For other pair of events which do not have "light like" separation, you consult post #37
It follows from the Lorentz transformation that $s^2 = (ct)^2 - x^2$ (1) is invariant (this is the "1D" case since we start with just a boost in the x-direction here).

now you introduce a term "light like" separation, which is new to me. The opposite is space like I assume? (at least both terms I recognize in general as going slow or very fast) probably better if I first do a more thorough reading of special relativity, because I see I keep your hole team busy which is probably too much asked and this for me problematic point about the minus sign now seems to be almost clear and after al not difficult. the most difficult part was probably not knowing what was the thought behind. now I think i know. simply an expanding lightcone and the relation between time and position. how simple can it be. small time gives still small expansion of the lightcone, large time gives large expansion so the difference scaled with the proper factor (c) is constant. I will continue with checking #37

 

[HansH]

OK, so if I am hearing you right, you now understand where the minus sign for the $-c^2\Delta t^2$ comes from in my steps 1)-3), and now you want to move on to point 4). Is that correct?

yes, especially what exactly ds means and its consequences for reference frames and lightcones I have the feeling that I stil mis something there but not easy to describe what.

 

[malawi_glenn]

now you introduce a term "light like" separation, which is new to me. The opposite is space like I assume?

Yeah "light-like" means that $\Delta s = 0$. I forgot to add "and time-like" separations.
You have three cases for what $\Delta s^2$ can compute to: zero, greater than zero, smaller than zero.

• Space-like separation means that the spatial separation between two events is greater than that of the distance a light ray would travel, i.e. $(\Delta x)^2> (c\Delta t)^2$ (outside the light-cone)
• Time-like separation means that the spatial separation between two events is smaller than that of the distance a light ray would travel, i.e. $(\Delta x)^2 < (c\Delta t)^2$ (inside the light-cone)
• Light-like separation means that $(\Delta x)^2 = (c\Delta t)^2$ (on the light-cone)

probably better if I first do a more thorough reading of special relativity

Just remember that there are two ways one can start.
1) define the space-time interval to be invariant, then equal speed of light in all frames pops out.
2) define equal speed of light in all frames pops out, then invariant space-time interval pops out.
Both ways leads to equivalent physics.

 

[Dale]

so not the ''null one'' so for my understanding this means that at t=0 the light already is at a radius ds^2

So light is only for the null interval. Any non-null interval refers to something other than light.

The key thing to understand is that $\Delta s^2$ is a sort of “distance”. It is a ”distance” that includes actual standard spatial distances measured by rulers ($\Delta s^2>0$) which are called spacelike, and it includes temporal “distances” measured by clocks ($\Delta s^2<0$) which are called timelike.

So if $\Delta s^2>0$ at $\Delta t=0$ you have a sphere which consists of all of the points that are a ruler-measured distance of $\Delta s$ away from the initial point.

Or if $\Delta s^2<0$ at $\Delta x=\Delta y=\Delta z=0$ then you have the two events that are a clock-measured “distance” of $\Delta \tau =\sqrt{-\Delta s^2/c^2}$ away from the initial event.

the question is then: where are both observers at t=0? are they still both in the same point or not?

The observers are not important here. One of the benefits of this approach is that you can focus on the physics of what is actually physically happening, and focus less on who sees what.

It doesn’t matter where the observers are located or what reference frame they are using, they will agree on $\Delta s^2$. Perhaps a concrete example will help.

Suppose in some reference frame that we have two firecrackers, one at $x=0$ and the other at $x=10$, that explode simultaneously at $t=0$. So $\Delta x=10$ and $\Delta t=0$ and plugging that into the spacetime interval formula gives $\Delta s^2=100$.

Now, in a reference frame moving at $v=0.6c$ we can use the Lorentz transform to get $\Delta x’=12.5$ and $\Delta t’= 7.5$ and plugging that into the spacetime interval formula gives $\Delta s^2=100$. So even though the two frames disagree on the time between the two firecrackers exploding and even though they disagree on the distance between the explosions, they agree on the spacetime interval between the explosions.

 

[PeterDonis]

now you introduce a term "light like" separation, which is new to me.

@malawi_glenn gives a good explanation in post #96. The one point I would add is that this threefold classification of intervals into timelike, lightlike, and spacelike has no counterpart in Euclidean geometry or Newtonian mechanics. In Euclidean geometry $ds^2$ is always positive. So the whole timelike, lightlike, spacelike thing is unlike anything you are used to and you should be very wary of trying to make analogies about it with anything you already know.

 

[HansH]

For other pair of events which do not have "light like" separation, you consult post #37
It follows from the Lorentz transformation that $s^2 = (ct)^2 - x^2$ (1) is invariant (this is the "1D" case since we start with just a boost in the x-direction here).

now you introduce a term "light like" separation, which is new to me. The opposite is space like I assume? (at least both terms I recognize in general as going slow or very fast) probably better if I first do a more thorough reading of special relativity, because I see I keep your hole team busy which is probably too much asked and this for me problematic point about the minus sign now seems to be almost clear and after al not difficult. the most difficult part was probably not knowing what was the thought behind. now I think i know. simply an expanding lightcone and the relation between time and position. how simple can it be. small time gives still small expansion of the lightcone, large time gives large expansion so the difference scaled with the proper factor (c) is constant. I will check #37

So light is only for the null interval. Any non-null interval refers to something other than light.

Perhaps a concrete example will help.

thanks, this really helps.

 

[HansH]

Here are some good notes by Shankar from Yale Open Courses https://oyc.yale.edu/sites/default/files/relativity_notes_2006_5.pdf
video: https://oyc.yale.edu/physics/phys-200/lecture-12Consider .....
Conserved /

Now there are plenty of other invariant quantities in special relativity (such as the invariant mass) which can be proven to be so following a smilar calculation as above. But, it is much nicer to work with four-vector formalism.

good explanation of the concept of invariance. This helps. It looks like magic that ds is invariant. I have the feeling however that the invariance of ds is no coincidence and should follow in a logical way from the general thoughts behind the whole idea.