Spacetime interval and basic properties of light

[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.

 

[HansH]

@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.

you say Euclidean geometry, but if I understand that well it means not curved. but special relativity is also not curved. and this topic is about special relativity. So do I mis something here? or do you mean 4D flat spacetime not being part of Euclidean geometry?

 

[PeterDonis]

you say Euclidean geometry, but if I understand that well it means not curved.

"Not curved" is one of the properties of Euclidean geometry, yes, but it's hardly the only one. Another one is "positive definite metric", which means what I said, that $ds^2$ is always positive.

special relativity is also not curved

That's correct, Minkowski spacetime is also not curved. It happens to share that property with Euclidean geometry. But it does not share other properties with Euclidean geometry; in particular, $ds^2$ is not always positive in Minkowski spacetime.

 

[robphy]

@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 general, starting with "[future]-timelike" (along inertial worldlines [which are arguable more physical than spatial lines])
I define "spacelike" to be orthogonal to "timelike"
and "null" to be eigenvectors of the boost.
(Using the metric (a figure representing the unit "circle": circle, "$\tau=1$" hyperbola, or "t=1-hyperplane"),
the spacelike direction is the tangent to the "circle" where the timelike radius vector meets the "circle".)With this scheme, the classification still holds for the Newtonian/Galilean case, with "spacelike" and "null" coinciding (as seen in the "opening up the light-cone" construction of the Newtonian/Galilean limit).
This seems consistent with the Cayley-Klein classification of geometries,
as seen in the diagram in my answer to
https://physics.stackexchange.com/q...ids-5-postulates-false-in-minkowski-spacetime
that was linked in #83 above.
(Yaglom 's "ordinary lines" or "lines of the first kind" are "timelike",
those of the "second kind" are "spacelike", and
the "special lines" are "null" (or "lightlike" in special relativity).
Ref: p. 184 of A Simple Non-Euclidean Geometry and Its Physical Basis: An Elementary Account of Galilean Geometry and the Galilean Principle of Relativity
https://www.google.com/search?q="lines+of+the+second+kind"+yaglom ).

 

[PeterDonis]

I define "spacelike" to be orthogonal to "timelike"

This is much too restrictive for MInkowski spacetime. Most spacelike vectors are not orthogonal to most timelike vectors.

and "null" to be eigenvectors of the boost.

This part is fine, and is in fact a good way of looking at how the action of boosts is fundamentally different for null vectors as compared to spacelike and timelike vectors; boosts rotate (hyperbolically) the latter but dilate the former.

With this scheme, the classification still holds for the Newtonian/Galilean case, with "spacelike" and "null" coinciding

I'm not sure how this works for the Newtonian/Galilean case, since AFAIK there is not a single vector space that includes all of the relevant vectors and has a meaningful definition of orthogonality.

 

[robphy]

This is much too restrictive for MInkowski spacetime. Most spacelike vectors are not orthogonal to most timelike vectors.

If a vector is spacelike (which is an invariant notion), there is a timelike vector orthogonal to it.
All one needs is one timelike vector.

For clarity, maybe I should rephrase:​

a "spacelike" vector is defined as a vector that is orthogonal to some timelike vector.​

(Geometrically, fix a point-event...​

a "timelike" vector from that point-event is a radius vector to a "circle" centered at that point-event ;​

a "spacelike" vector from that point-event is a vector parallel to the tangent to a "circle" centered at that point-event)​

I'm not sure how this works for the Newtonian/Galilean case, since AFAIK there is not even a single vector space that includes all of the relevant vectors.

I'm not sure what you mean here.
A position-vs-time graph (with its Newtonian/Galilean structure), a Minkowski spacetime diagram, and the usual Euclidean plane
all have the structure of an affine space.