Spacetime interval and basic properties of light

[Ibix]

thought I was only talking about flat spacetime. At least that was where the question was about. Not sure where I used properties of curved spacetime?

You were explicitly talking about something collapsing into a black hole and the effects thereof. That's a curved spacetime, even if one part of it is flat.

ok, but if you say :In Minkowski space, the equivalent of Pythagoras’ theorem is $ds^2 = c^2 dt^2 - dx^2$ : then I am back to the openings question of the topic: why, because I still do not understand. so it seems difficult to get the basic idea clearly explained at headlines without diving into the books while I hoped this is what the forum could add. I thought I understood but seems to be on the wrong track, so I think I will first check the links in #29

What answer would you give to the question of why Pythagoras' theorem holds?

I think you have two options. First, you can assert that Pythagoras is invariant, derive the implications, and show that they accurately describe the behaviour of rulers and Cartesian coordinate grids on planes. (Other axiomatisations of Euclidean geometry are available.) Second, you can study the behaviour of Cartesian coordinates and rulers on planes and deduce Pythagoras' theorem.

If you can answer that question then we can answer "why the interval" in similar terms.

 

[malawi_glenn]

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-12
Might be of use for you if you want to start from the postulate "speed of light is same in all inertial systems" and then derive the invariance of the space-time interval.

Consider frame $\tilde S$ moving in $S$ with velocity $v$ in the $+x$ direction.
From the galilean transformation the relationship between coordinates $(x,t)$ and $(\tilde x, \tilde t)$ are
$\boxed{t = \tilde t}$
$\boxed{ \tilde x = x - vt \: , \, x = \tilde x + v t}$
But with this transformation, you will get that the speed of light is not the same in $S$ and $\tilde S$ (which Michelson–Morley experiment and Maxwells equations suggested).
Consider instead a more general linear transformation (linear - it is invertible, one coordinate in $\tilde S$ will correspond to exactly one coordinate in $S$) which is called Lorentz transformation:
$\boxed{ \tilde x = \gamma (x-vt) } \: (1) \qquad \boxed{ \tilde t = \gamma \left( t - \dfrac{xv}{c^2} \right) }\qquad \boxed{ x = \gamma (\tilde x + v\tilde t) } \: (2) \qquad \boxed{ t = \gamma \left( \tilde t + \dfrac{\tilde x v}{c^2} \right) }$
where $\gamma$ is a factor which should only depend on $v$ which shall fulfill $\gamma \to 1$ when $v/c \to 0$ which motivates the Galilean transformation as "low speed limit" (I know, there was a recent thread about this which is pretty good read!) and it should give that the speed of light is the same in both $S$ and $\tilde S$.

That speed of light should be the same in both $S$ and $\tilde S$ is implemented as follows:
When the origins of $S$ and $\tilde S$ coincide, a ray of light is emitted. The position of the front of the ray is $x = ct$ in $S$ and $\tilde x = c\tilde t$ in $\tilde S$. Insert $t = x/c$ into the Lorentz-transformation for $\tilde x$ (1) and $\tilde t = \tilde x/c$ into the Lorentz-transformation for $x$ (2).
We get $\tilde x = \gamma \left(x-v\dfrac{x}{c}\right)= \gamma \left(1- \dfrac{v}{c}\right) x$
and
$x = \gamma \left(\tilde x+v\dfrac{\tilde x}{c}\right)= \gamma \left(1+ \dfrac{v}{c}\right) \tilde x$
which means that $\dfrac{\tilde x}{x} = \gamma \left(1- \dfrac{v}{c}\right)$ and $\dfrac{\tilde x}{x} = \dfrac{1}{\gamma \left(1+ \dfrac{v}{c}\right)}$
solve for $\gamma$, the result is $\boxed{ \gamma = \dfrac{1}{\sqrt{1 - \frac{v^2}{c^2}}} }$.

Now consider the quantity $s^2 = (ct)^2 - x^2$. By performing the Lorentz-transformation above, we obtain

$s^2 = (ct)^2 - x^2 = \dfrac{c^2\left(\tilde t + \frac{\tilde x v}{c^2}\right)^2}{1-\frac{v^2}{c^2}} - \dfrac{\left(\tilde x + v\tilde t\right)^2}{1-\frac{v^2}{c^2}} = \dfrac{c^2 \tilde t ^2 + 2 \tilde t \tilde x v+ \frac{\tilde x^2 v^2}{c^2} - \tilde x^2 - 2 \tilde t \tilde x v - v^2 \tilde t^2 }{1-\frac{v^2}{c^2}} =$

$\dfrac{(c^2-v^2)\tilde t^2}{1-\frac{v^2}{c^2}} - \dfrac{\left(1 - \frac{v^2}{c^2}\right)\tilde x^2}{1-\frac{v^2}{c^2}} = \dfrac{c^2\left(1-\frac{v^2}{c^2}\right)\tilde t^2}{1-\frac{v^2}{c^2}} - \tilde x^2 = c^2 \tilde t^2 - \tilde x^2 = (c\tilde t)^2 - \tilde x^2 = \tilde s^2$

Conserved / invariant quantities are very important in physics since they hint that there is some underlying geometrical property, symmetry.

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.

 

[HansH]

The answer was given by @Dale in post #7.

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.

 

[HansH]

You were explicitly talking about something collapsing into a black hole and the effects thereof. That's a curved spacetime, even if one part of it is flat.

What answer would you give to the question of why Pythagoras' theorem holds?

That was another topic and also a reason for me to do first 1 step back and better understand the special relativity. (also because that topic is temporary?? closed this creates room to go back to the basics first) So I assume all questions and answers in this topic should be able to prvent general relativity in order to make it not too complex at this stage.

 

[HansH]

What answer would you give to the question of why Pythagoras' theorem holds?

I think you have two options. First, you can assert that Pythagoras is invariant, derive the implications, and show that they accurately describe the behaviour of rulers and Cartesian coordinate grids on planes. (Other axiomatisations of Euclidean geometry are available.) Second, you can study the behaviour of Cartesian coordinates and rulers on planes and deduce Pythagoras' theorem.

If you can answer that question then we can answer "why the interval" in similar terms.

I think your answer is too general for me to to be able to do anything with it to get better understanding. Be aware that i am not a physics student but someone with a general interest but different background and terms as 'invariant' are long time ago and probably related to different perspective than relativity. So I am looking for more straight to the point derivations that I can try to follow than doing this derivations myself getting in the wrong direction 5 times first.

 

[Sagittarius A-Star]

for me the conclusion: 'that is where the minus sign comes from' is still not clear.

A light pulse moves with velocity $c$ from emitting event $E_1$ to receiving event $E_2$ with spatial distance $\sqrt{\Delta x^2+\Delta y^2 + \Delta z^2}$ between them. The temporal interval $\Delta t$ between those events will be:
$$\Delta t= \frac{1}{c} \sqrt{\Delta x^2+\Delta y^2 + \Delta z^2}$$
$\Rightarrow$
$$-(c\Delta t)^2 + \Delta x^2+\Delta y^2 + \Delta z^2 = 0$$

 

[malawi_glenn]

So I am looking for more straight to the point derivations that I can try to follow than doing this derivations myself getting in the wrong direction 5 times first

What was wrong with my post? #37? You need a book to read, you have been given some suggestions already. Get one of those and read it and fill in the steps.

 

[Orodruin]

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.

No, it is just moving the radius of the sphere (ct) to the same side as the squares of the spatial distances.

 

[HansH]

What was wrong with my post? #37? You need a book to read, you have been given some suggestions already. Get one of those and read it and fill in the steps.

nothing wrong. I am digesting that stuff at the moment. Regarding books: I have a nice example of someone who has a whole bookshelf with books about general relativity and still does not make any progress because it seems that he is running in circles because the books assume pre-knowledge about one part of the topic to be able to understand another part and to be able to understand the first part ypu first have to understand still something else so you will never get there. So I am a bit hesitating and assume that there must be a lot of good information at the internet too to start with first. and especially I rely on a forum like this because that gives the unique opportunity to recognize what the lack of knowledge of someone is and specifically act on that which a book can impossibly do.

 

[malawi_glenn]

I have a nice example of someone who has a whole bookshelf with books about general relativity and still does not make any progress because it seems that he is running in circles because the books assume pre-knowledge about one part of the topic to be able to understand another part and to be able to understand the first part ypu first have to understand still something else so you will never get there. So I am a bit hesitating and assume that there must be a lot of good information at the internet too to start with first. and especially I rey on a forum like this because that gives the unique opportunity to recognize the the lack of knowledge of someone is and specifically act on that which a book can impossibly do.

Wow that was two sentences. Hard to read :)

Poor books, its like collecting sports car but don't know how to drive :(

Sure you need to have the correct pre-knowledge, not going to argue against that. Then wouldn't it be more reasonable to ask "what background knowledge do I need to have, and how can I acquire it, in order to grasp the very basic ideas of SR?" The replies on forums also assumes some background knowledge.
You basically just need some algebra, geometry, calculus and kinematics to do the entire book by Morin for instance. Then as soon as you encounter something you don't understand, you can ask here, or search for an answer online. It is good to stick to one main book all the time to get used to notation and the authors style.

Here is a free online college physics book https://openstax.org/details/books/college-physics-2e the treatment of SR there is very crude, so I would just bother with chapters 1-8.

 

[Nugatory]

especially I rely on a forum like this because that gives the unique opportunity to recognize what the lack of knowledge of someone is and specifically act on that which a book can impossibly do.

Better is to start working through a suitable textbook, and use the forum to help you when you’re stuck at a particular hard spot.

 

[HansH]

No, it is just moving the radius of the sphere (ct) to the same side as the squares of the spatial distances.

I still don't get your point. I have drawn a circle with radius ct and a spatial distance I called d. but then what is next? what do you mean by 'moving the radius of the sphere (ct) to the same side as the squares' ? and how does that lead to the conclusion of the minus sign.

 

[robphy]

Try Bondi, as I suggested in #21.
The “minus sign” shows up in the last step (if you want to take that step).

 

[HansH]

Better is to start working through a suitable textbook, and use the forum to help you when you’re stuck at a particular hard spot.

perhaps better indeed. 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 about that also and that was exactly the hard spot to get clear where the minus sign comes from. So for me easy to accept that the speed of light is constant in all frames so that is not the point.

 

[Orodruin]

I still don't get your point. I have drawn a circle with radius ct and a spatial distance I called d. but then what is next? what do you mean by 'moving the radius of the sphere (ct) to the same side as the squares' ? and how does that lead to the conclusion of the minus sign. View attachment 305666

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.

 

[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