Spacetime interval and basic properties of light

[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.