Exploring Minkowski Space: Understanding Special Theory of Relativity

In summary, Minkowski space refers to flat spacetime in Special Relativity. The spacetime interval between two events in Minkowski space can be classified as either time-like, space-like, or null-like. This classification depends on the possibility of measuring the interval using an inertial clock or ruler. If the measurements are possible, the interval is either time-like or space-like. If not, it is null-like. Transforming coordinates in an Inertial Reference Frame can help determine the classification of the spacetime interval.
  • #1
ash64449
356
15
Hello friends,

I am reading Einstein's special theory of relativity and came across this subject:

Minkowski space.

I cannot understand exactly what it is when i read that book and went to Wikipedia for understanding more about it. But i didn't understand much.

What i understand about minkowski space is that it setup helps us to understand how Special Theory actually 'works'. It adds time as being the 4th dimension and introduces a new manifold which is called Space-time.

From wikipedia,they said that :

The spacetime interval between two events in Minkowski space is either:
space-like,
light-like ('null') or
time-like.

The thing that i cannot understand is what is the meaning of "Space like", "Light-like" and "Time-like"

What is the difference between these three invariant space-time interval?

I hope people can understand what i am saying. Just tell me what minkowski space is. I think that would help me a bit..
 
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  • #2
Hi ash! Consider the usual vector space ##\mathbb{R}^{4}##. Recall that usually, we pick the standard basis ##\left \{ e_1,e_2,e_3,e_4 \right \}## and write any vector ##v\in \mathbb{R}^{4}## as a linear combination ##v = \sum v^{i}e_i = v^{1}e_{1} + v^{2}e_{2} + v^{3}e_{3} + v^{4}e_{4}##. In Newtonian physics, we use the Euclidean inner product ##< ,>:\mathbb{R}^{4}\times \mathbb{R}^{4}\rightarrow \mathbb{R}## given by ##<v,w> = \sum v^{i}w^{i} = v^{1}w^{1} + v^{2}w^{2} + v^{3}w^{3} + v^{4}w^{4}## which is just the usual, familiar inner product.

Consider now a different map on ##\mathbb{R}^{4}## which I will denote ##\eta:\mathbb{R}^{4}\times \mathbb{R}^{4}\rightarrow \mathbb{R}##. ##\eta## will be almost like an inner product except for a key difference that will end up giving rise to the notions of "null-like", "space-like", and "time-like" vectors. Again using the standard basis, ##\eta## is given by ##\eta(v,w) = -v^{1}w^{1} + v^{2}w^{2} + v^{3}w^{3} + v^{4}w^{4}##; let's call ##\eta## the Minkowski metric. Notice that the only difference between ##\eta(v,w)## and ##<v ,w>## is that in ##\eta(v,w)##, the first term has a negative sign in front; this has a huge effect. The pair ##(\mathbb{R}^{4},\eta)## is called Minkowski space-time.

Consider for example the vector ##v = (1,1,0,0)##. Let's calculate its length squared using the Euclidean inner product (recall that the length squared of a vector is given by ##<v,v>##) so that ##<v,v> = 1 + 1 = 2##. Now I define the length squared of the vector under the Minkowski metric in a similar manner as ##\eta(v,v)## so that ##\eta(v,v) = -1 + 1 = 0##. So under ##\eta##, vectors in ##\mathbb{R}^{4}## can have zero length squared without even being the zero vector! What's more, consider the vector ##w = (1,0,0,0)## and note that its length squared under ##\eta## is ##\eta(w,w) = -1 ##. So now vectors can even have negative length squared! This is totally different from what we're used to under the Euclidean inner product where for any ##v\in \mathbb{R}^{4}##, we have that ##<v,v>\geq 0## and ##<v,v>= 0\Leftrightarrow v=0##. We call this condition positive-definite and say that ##\eta## is indefinite. We define a vector ##v\in \mathbb{R}^{4}## to be space-like if ##\eta(v,v) > 0##, time-like if ##\eta(v,v) < 0##, and null-like (light-like) if ##\eta(v,v) = 0##.

Finally, note that using ##\eta## I can define a "distance" function on Minkowski space-time (sort of like how inner products on vector spaces naturally induce a metric in the sense of analysis). Given any two points ##(t_1,x_1,y_1,z_1),(t_2,x_2,y_2,z_2)\in \mathbb{R}^{4}##, define ##\Delta s ^{2} = -(t_2 - t_1)^{2} + (x_2 - x_1)^{2} + (y_2 - y_1)^{2} + (z_2 - z_1)^{2}##. We call these points events in Minkowski space-time and call ##\Delta s^{2}## the space-time interval. Similar to above, we say the interval is time-like if ##\Delta s^{2} < 0##, space-like if ##\Delta s^{2} > 0##, and null-like if ##\Delta s^{2} = 0##.
 
  • #4
ash64449 said:
Hello friends,

I am reading Einstein's special theory of relativity and came across this subject:

Minkowski space.

I cannot understand exactly what it is when i read that book and went to Wikipedia for understanding more about it. But i didn't understand much.

What i understand about minkowski space is that it setup helps us to understand how Special Theory actually 'works'. It adds time as being the 4th dimension and introduces a new manifold which is called Space-time.
The term "Minkowski space" merely means "flat spacetime" as applied to Special Relativity in contrast to "curved spacetime" as applied to General Relativity.
ash64449 said:
From wikipedia,they said that :

The spacetime interval between two events in Minkowski space is either:
space-like,
light-like ('null') or
time-like.

The thing that i cannot understand is what is the meaning of "Space like", "Light-like" and "Time-like"

What is the difference between these three invariant space-time interval?

I hope people can understand what i am saying. Just tell me what minkowski space is. I think that would help me a bit..
Pick any two events in an Inertial Reference Frame (IRF). If it is possible for an inertial (non-accelerating) clock to be present at those two events, then the spacetime interval between those two events is time-like and the delta time between the readings on the clock between those two events is the spacetime interval.

If it is possible to measure the spacetime interval between those two events at the same time with an inertial ruler then the spacetime interval is space-like.

If neither measurement is possible, then the spacetime interval is null, because only a flash of light (photon) can be present at both events.

Another way of saying all this is that for any two events defined with coordinates in an IRF if you can transform the two events into another IRF (if necessary) so that they either have the same spatial coordinates then the difference in the time coordinates is a time-like spacetime interval OR so that they have the same time coordinates then the difference in the spatial coordinates is a space-like spacetime interval. Otherwise the spacetime interval is null.

Can you work out some examples? Do you know how to transform the coordinates of events using the Lorentz Transformation process?
 
  • #5
WannabeNewton said:
Hi ash! Consider the usual vector space ##\mathbb{R}^{4}##. Recall that usually, we pick the standard basis ##\left \{ e_1,e_2,e_3,e_4 \right \}## and write any vector ##v\in \mathbb{R}^{4}## as a linear combination ##v = \sum v^{i}e_i = v^{1}e_{1} + v^{2}e_{2} + v^{3}e_{3} + v^{4}e_{4}##. In Newtonian physics, we use the Euclidean inner product ##< ,>:\mathbb{R}^{4}\times \mathbb{R}^{4}\rightarrow \mathbb{R}## given by ##<v,w> = \sum v^{i}w^{i} = v^{1}w^{1} + v^{2}w^{2} + v^{3}w^{3} + v^{4}w^{4}## which is just the usual, familiar inner product.

Consider now a different map on ##\mathbb{R}^{4}## which I will denote ##\eta:\mathbb{R}^{4}\times \mathbb{R}^{4}\rightarrow \mathbb{R}##. ##\eta## will be almost like an inner product except for a key difference that will end up giving rise to the notions of "null-like", "space-like", and "time-like" vectors. Again using the standard basis, ##\eta## is given by ##\eta(v,w) = -v^{1}w^{1} + v^{2}w^{2} + v^{3}w^{3} + v^{4}w^{4}##; let's call ##\eta## the Minkowski metric. Notice that the only difference between ##\eta(v,w)## and ##<v ,w>## is that in ##\eta(v,w)##, the first term has a negative sign in front; this has a huge effect. The pair ##(\mathbb{R}^{4},\eta)## is called Minkowski space-time.

Consider for example the vector ##v = (1,1,0,0)##. Let's calculate its length squared using the Euclidean inner product (recall that the length squared of a vector is given by ##<v,v>##) so that ##<v,v> = 1 + 1 = 2##. Now I define the length squared of the vector under the Minkowski metric in a similar manner as ##\eta(v,v)## so that ##\eta(v,v) = -1 + 1 = 0##. So under ##\eta##, vectors in ##\mathbb{R}^{4}## can have zero length squared without even being the zero vector! What's more, consider the vector ##w = (1,0,0,0)## and note that its length squared under ##\eta## is ##\eta(w,w) = -1 ##. So now vectors can even have negative length squared! This is totally different from what we're used to under the Euclidean inner product where for any ##v\in \mathbb{R}^{4}##, we have that ##<v,v>\geq 0## and ##<v,v>= 0\Leftrightarrow v=0##. We call this condition positive-definite and say that ##\eta## is indefinite. We define a vector ##v\in \mathbb{R}^{4}## to be space-like if ##\eta(v,v) > 0##, time-like if ##\eta(v,v) < 0##, and null-like (light-like) if ##\eta(v,v) = 0##.

Finally, note that using ##\eta## I can define a "distance" function on Minkowski space-time (sort of like how inner products on vector spaces naturally induce a metric in the sense of analysis). Given any two points ##(t_1,x_1,y_1,z_1),(t_2,x_2,y_2,z_2)\in \mathbb{R}^{4}##, define ##\Delta s ^{2} = -(t_2 - t_1)^{2} + (x_2 - x_1)^{2} + (y_2 - y_1)^{2} + (z_2 - z_1)^{2}##. We call these points events in Minkowski space-time and call ##\Delta s^{2}## the space-time interval. Similar to above, we say the interval is time-like if ##\Delta s^{2} < 0##, space-like if ##\Delta s^{2} > 0##, and null-like if ##\Delta s^{2} = 0##.

Sorry WannabeNewton, i am way below your standard in order to understand your mathematics.

Can you try to explain it without mathematics? You can use Simple algebra too..
 
  • #6
ash64449 said:
Sorry WannabeNewton, i am way below your standard in order to understand your mathematics.

Can you try to explain it without mathematics? You can use Simple algebra too..

He used very basic mathematics though. In a sense he only did use algebra. If you really want to understand SR, then I recommend that you brush up your mathematics a bit. Otherwise, you'll have a very incomplete understanding what's going on.

There are many nice Linear Algebra books available. For example Lang's "A first course in linear algebra" is very nice and not too difficult. I recommend that you pick it up.

Of course, you can also specifically say what you don't understand. We can try to explain it.
 
  • #7
ash64449 said:
You can use Simple algebra too..
The Lorentz Transformation is simple algebra. And that's all you need to be able to understand the answer to your question. I asked you if you know how to do Lorentz Transformations. What is your answer?
 
  • #8
ghwellsjr said:
The Lorentz Transformation is simple algebra. And that's all you need to be able to understand the answer to your question. I asked you if you know how to do Lorentz Transformations. What is your answer?

yes.i know
 
  • #9
Minkowski was the guy who realized that Einstein's maths in Special Relativity was equivalent to saying that space and time were one four-dimensional whole. Einstein later took that idea and extended it to describe gravity as four dimensional structures with curved geometry - but that's not relevant here.

You are probably used to Euclidean space, where the distance between two points is given by Pythagoras' Theorem:
[tex]d^2=x^2+y^2+z^2[/tex]
You and I might measure different x, y, and z, but we'll agree on d. For example, place a pencil on the table in front of you, pointing straight away from you. You'll say that the x-coordinates of each end of the pencil are the same, but the y-coordinates are different. If I'm sitting round the corner of the table from you, I'll say the x-coordinates are different but the y-coordinates are the same. However, we'll agree on the length.

There is a similar expression for the distance between two points in Minkowski space. We tend to use the word "event" to describe a point in four-dimensional spacetime, to make clear that we mean a particular point in space at a particular point in time. Again, you and I might disagree about the x, y, z, and t-coordinates of two events, but we will agree about the so-called "interval" between them:
[tex]s^2=c^2t^2-x^2-y^2-z^2[/tex]
Warning: some people put the signs the other way around!

The funny thing about the interval, [itex]s^2[/itex], compared to the square of the distance you get from the familiar Pythagoras formula, is that it can be negative ([itex]d^2[/itex] can never be negative). This leads to a classification system for the intervals between events. If [itex]s^2[/itex] is positive then the separation is said to be "time-like". If it is zero, the separation is said to be "null" or "light-like", and if the interval is negative, the separation is said to be "space-like". Why?

Well - note that
[tex]s^2=c^2t^2-x^2-y^2-z^2=c^2t^2-d^2[/tex]
ct is the distance that light travels in time t. d is the three-dimensional distance between the two events (in the frame in which the measurements were made - not everyone will agree on that, only on [itex]s^2[/itex]). So if [itex]s^2[/itex] is positive then ct>d, which means that light can travel further than the distance between the events in the time between the events. That means it's possible for someone to be at both events. For that person, they appear at the same place at different times - hence "time-like separation".

If [itex]s^2[/itex] is negative then ct<d, which means that not even light can get from one event to the other in the time between them. It is impossible for one person to be at both events - hence "space-like separation".

The edge case where [itex]s^2[/itex] is zero means that ct=d. Light, and only light, can get from one event to the other in the time between them. Nobody can see them as being the same time or the same place.

In summary:
  • Time-like: Everyone agrees that the events happened at different times; some people say that they happened at the same place.
  • Space-like: Everyone agrees that the events happened at different places; some people say that they happened at the same time.
  • Light-like: Everyone agrees that the events happened at different times and places.

Does that help?
 
  • #10
micromass said:
He used very basic mathematics though. In a sense he only did use algebra. If you really want to understand SR, then I recommend that you brush up your mathematics a bit. Otherwise, you'll have a very incomplete understanding what's going on.

There are many nice Linear Algebra books available. For example Lang's "A first course in linear algebra" is very nice and not too difficult. I recommend that you pick it up.

Of course, you can also specifically say what you don't understand. We can try to explain it.

I didn't understand what WannabeNewton said. And i don't think he used simple algebra.No where in my textbook i studied inner product and all and i haven't seen anything like what WannaBeNEwton wrote in my textbook.

I am just 15 year old and i only know what a 15 year old knows. Now i am going to study what 16 year old one would study.

Ok... My first chapter of my mathematics is sets and then relations and functions. I hope now you can understand my mathematical background.
 
  • #11
ghwellsjr said:
The Lorentz Transformation is simple algebra. And that's all you need to be able to understand the answer to your question. I asked you if you know how to do Lorentz Transformations. What is your answer?

sorry george for i kept you waiting. I was reading the article that the 2nd replier gave. I was reading responces step by step... Well,i have read some of your responce but i not in the system that helps me quote your responce..
 
  • #12
ghwellsjr said:
The Lorentz Transformation is simple algebra. And that's all you need to be able to understand the answer to your question. I asked you if you know how to do Lorentz Transformations. What is your answer?

oh,, so you think lorentz transformation is enough to understand minkowski space?
 
  • #13
ash64449 said:
oh,, so you think lorentz transformation is enough to understand minkowski space?
It's the only kind of calculation you need to do to answer your question mathematically. I don't think you will have a clear understanding of Minkowski space unless you draw spacetime diagrams for different Inertial Reference Frames (IRF's) and that's what I have been doing for you numerous times in your other threads. Does any of it sink in?
 
  • #14
While linear algebra (and, along that line of thinking, group theory and differential geometry) would help more fully understand special relativity, the essence of your question about timelike, spacelike, and lightlike only requires an enlightening diagram that emphasizes the physics (causality) and a minimal amount of math (intro algebra) to define the square-interval.

http://books.google.com/books?id=rwPDssnbHPEC&q=spacelike#v=snippet&q=spacelike&f=false (Geroch, General Relativity from A to B, see... p89 [Important Diagrams] and onward )



There's a great quote from Mermin in a letter to the editor in the AJP... in bold. I included the surrounding sentences for context...
Mermin said:
( "Lapses in Relativistic Pedagogy ", AJP 62, 11 (1994) http://dx.doi.org/10.1119/1.17728 )

"I couldn't agree more with Mallinckrodt that the Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it."
 
  • #15
ghwellsjr said:
It's the only kind of calculation you need to do to answer your question mathematically. I don't think you will have a clear understanding of Minkowski space unless you draw spacetime diagrams for different Inertial Reference Frames (IRF's) and that's what I have been doing for you numerous times in your other threads. Does any of it sink in?

oh,so do you think if i keep practicing space-time diagrams for different inertial reference frame,i will be able to understand minkowski space?

I think you have explained me about how to draw space-time diagrams and how can i understand from them. Clearly, they are very important to me...(understood this from your reply!)
 
  • #16
ghwellsjr said:
The term "Minkowski space" merely means "flat spacetime" as applied to Special Relativity in contrast to "curved spacetime" as applied to General Relativity.

Pick any two events in an Inertial Reference Frame (IRF). If it is possible for an inertial (non-accelerating) clock to be present at those two events, then the spacetime interval between those two events is time-like and the delta time between the readings on the clock between those two events is the spacetime interval.

inother words, Finite time has passed between them and both the events are non-simultaneous.Correct?

ghwellsjr said:
If it is possible to measure the spacetime interval between those two events at the same time with an inertial ruler then the spacetime interval is space-like.
Events are simultaneous in this case and those two events are separated by space intervel.

I don't know whether i am correct or not,well,(i think) in spacetime,if events are simultaneous,they occupy the same position and if they are spatially separated, then space-time intervel between them is behaving like Space.. Am i correct?

ghwellsjr said:
If neither measurement is possible, then the spacetime interval is null, because only a flash of light (photon) can be present at both events.

I didn't quite understood this point...

ghwellsjr said:
Another way of saying all this is that for any two events defined with coordinates in an IRF if you can transform the two events into another IRF (if necessary) so that they either have the same spatial coordinates then the difference in the time coordinates is a time-like spacetime interval OR so that they have the same time coordinates then the difference in the spatial coordinates is a space-like spacetime interval. Otherwise the spacetime interval is null.

Can you work out some examples? Do you know how to transform the coordinates of events using the Lorentz Transformation process?

Now i understood why you recommend space-time diagrams,because i can represent many situations in space-time, and some situations are like time-like,some space-like and some light-like?
Correct?
 
  • #17
ash64449 said:
oh,so do you think if i keep practicing space-time diagrams for different inertial reference frame,i will be able to understand minkowski space?

I think you have explained me about how to draw space-time diagrams and how can i understand from them. Clearly, they are very important to me...(understood this from your reply!)

When you draw a space-time diagram with x and t axes, you're drawing a diagram of two-dimensional minkowski space. There is just one seriously important thing that you have to remember: The distance between points in the diagram should be calculated using:
[tex]
\Delta{s}^2 = \Delta{x}^2 - c^2\Delta{t}^2[itex]

instead of the Pythagorean formula.

If you pick two points and calculate the distance between them this way, then Lorentz-transform their x and t values and calculate the distance again... It will come out the same. It is this property that makes Minkowski space so useful; things that are true in Minkowski space are true regardless of the velocity of the various observers.
 
  • #18
ghwellsjr said:
. Does any of it sink in?

Sorry,what is the meaning of this statement?
 
  • #19
Nugatory said:
When you draw a space-time diagram with x and t axes, you're drawing a diagram of two-dimensional minkowski space. There is just one seriously important thing that you have to remember: The distance between points in the diagram should be calculated using:
[tex]
\Delta{s}^2 = \Delta{x}^2 - c^2\Delta{t}^2[itex]

The equation that you said is still a Pythagorean theorem!
\Delta{s}^2 + c^2\Delta{t}^2 = \Delta{x}^2
Doesn't this look like Pythagorean theorem?

Nugatory said:
If you pick two points and calculate the distance between them this way, then Lorentz-transform their x and t values and calculate the distance again... It will come out the same. It is this property that makes Minkowski space so useful; things that are true in Minkowski space are true regardless of the velocity of the various observers.

I think i will use Lorentz transform to transform their values.. Anyway they are same,, carrying out some examples will help to understand space-like,time-like and light-like intervels of space-time...
 
Last edited:
  • #20
Ibix said:
Minkowski was the guy who realized that Einstein's maths in Special Relativity was equivalent to saying that space and time were one four-dimensional whole. Einstein later took that idea and extended it to describe gravity as four dimensional structures with curved geometry - but that's not relevant here.

You are probably used to Euclidean space, where the distance between two points is given by Pythagoras' Theorem:
[tex]d^2=x^2+y^2+z^2[/tex]
You and I might measure different x, y, and z, but we'll agree on d. For example, place a pencil on the table in front of you, pointing straight away from you. You'll say that the x-coordinates of each end of the pencil are the same, but the y-coordinates are different. If I'm sitting round the corner of the table from you, I'll say the x-coordinates are different but the y-coordinates are the same. However, we'll agree on the length.

There is a similar expression for the distance between two points in Minkowski space. We tend to use the word "event" to describe a point in four-dimensional spacetime, to make clear that we mean a particular point in space at a particular point in time. Again, you and I might disagree about the x, y, z, and t-coordinates of two events, but we will agree about the so-called "interval" between them:
[tex]s^2=c^2t^2-x^2-y^2-z^2[/tex]
Warning: some people put the signs the other way around!

The funny thing about the interval, [itex]s^2[/itex], compared to the square of the distance you get from the familiar Pythagoras formula, is that it can be negative ([itex]d^2[/itex] can never be negative). This leads to a classification system for the intervals between events. If [itex]s^2[/itex] is positive then the separation is said to be "time-like". If it is zero, the separation is said to be "null" or "light-like", and if the interval is negative, the separation is said to be "space-like". Why?

Well - note that
[tex]s^2=c^2t^2-x^2-y^2-z^2=c^2t^2-d^2[/tex]
ct is the distance that light travels in time t. d is the three-dimensional distance between the two events (in the frame in which the measurements were made - not everyone will agree on that, only on [itex]s^2[/itex]). So if [itex]s^2[/itex] is positive then ct>d, which means that light can travel further than the distance between the events in the time between the events. That means it's possible for someone to be at both events. For that person, they appear at the same place at different times - hence "time-like separation".

If [itex]s^2[/itex] is negative then ct<d, which means that not even light can get from one event to the other in the time between them. It is impossible for one person to be at both events - hence "space-like separation".

The edge case where [itex]s^2[/itex] is zero means that ct=d. Light, and only light, can get from one event to the other in the time between them. Nobody can see them as being the same time or the same place.

In summary:
  • Time-like: Everyone agrees that the events happened at different times; some people say that they happened at the same place.
  • Space-like: Everyone agrees that the events happened at different places; some people say that they happened at the same time.
  • Light-like: Everyone agrees that the events happened at different times and places.

Does that help?

i didn't quite understand space-like intervel from your post. If light can itself cannot reach between the time between the two events,then i agree that it is space-like separated but how can people judge when the events happened if light itself hasn't reached at the other space-like event?
 
  • #21
ash64449 said:
The equation that you said is still a Pythagorean theorem!
\Delta{s}^2 + c^2\Delta{t}^2 = \Delta{x}^2
Doesn't this look like Pythagorean theorem?

Draw it, and then tell us what you think... Remember that the right angle is between the x and t axes so the s side is the hypotenuse.

Also important to remember that the in any frame of reference you have the x and t coordinates of two events; it's the interval between the two events that is the unknown. And we always want the unknown to show up all by itself on one side of the equal sign; so your rearrangement is just turning [itex]unknown = known1 - known2[/itex] into [itex]known1 = unknown + known2[/itex]; and that doesn't help us calculate the unknown.
 
  • #22
ash64449 said:
i didn't quite understand space-like intervel from your post. If light can itself cannot reach between the time between the two events,then i agree that it is space-like separated but how can people judge when the events happened if light itself hasn't reached at the other space-like event?

Let's say that I'm sitting at the origin of my coordinate system (my frame of reference) at rest, at time zero. I snap my fingers. That defines event 1: Fingers snapped at x=0, t=0. Four years later I'm looking in the direction of the star alpha centauri, which I know is four light-years away, and I see a flash of light. That defines event 2: Light hit my eyes at x=0, t=4. I know that the star is four light years away so I know that the flash left the star four years ago; that defines event 3: Flash left star at x=4, t=0.
(I hope you're plotting these three points on your space-time diagram).

Events 1 and 2 are time-like separated.
Events 2 and 3 are light-like separated.
Events 1 and 3 are space-like separated. And you'll notice that although I did eventually learn about event 3, I couldn't until my path through spacetime had reached event 2.

[Edit: And you might try using the Lorentz transform to transform the coordinates of all three events into the coordinates of an observer in spaceship traveling at .6c; the x and t distances will be different (length contraction and relativity of simultaneity) but the intervals will remain the same]
 
  • #23
Nugatory said:
Draw it, and then tell us what you think... Remember that the right angle is between the x and t axes so the s side is the hypotenuse.

Also important to remember that the in any frame of reference you have the x and t coordinates of two events; it's the interval between the two events that is the unknown. And we always want the unknown to show up all by itself on one side of the equal sign; so your rearrangement is just turning [itex]unknown = known1 - known2[/itex] into [itex]known1 = unknown + known2[/itex]; and that doesn't help us calculate the unknown.

i can tell at the first sight of your comment that it cannot form phytagorean theorem. As 's' is considered as hypotenuse.. You made my point clear. Thank you!
 
  • #24
Nugatory said:
Let's say that I'm sitting at the origin of my coordinate system (my frame of reference) at rest, at time zero. I snap my fingers. That defines event 1: Fingers snapped at x=0, t=0. Four years later I'm looking in the direction of the star alpha centauri, which I know is four light-years away, and I see a flash of light. That defines event 2: Light hit my eyes at x=0, t=4. I know that the star is four light years away so I know that the flash left the star four years ago; that defines event 3: Flash left star at x=4, t=0.
(I hope you're plotting these three points on your space-time diagram).

Events 1 and 2 are time-like separated.
Events 2 and 3 are light-like separated.
Events 1 and 3 are space-like separated. And you'll notice that although I did eventually learn about event 3, I couldn't until my path through spacetime had reached event 2.

[Edit: And you might try using the Lorentz transform to transform the coordinates of all three events into the coordinates of an observer in spaceship traveling at .6c; the x and t distances will be different (length contraction and relativity of simultaneity) but the intervals will remain the same]

thank you nugatory, your post cleared space-like separated events. Thank you!
 
  • #25
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  • #26
ConformalGrpOp said:
Hi ash, you seem to be doing pretty well with this topic. At the risk of being too simplistic, the thing to understand about "minkowski space" is that, like euclidean space where you have x, y and z coordinates and triangles obey the pythagorean theorem, minkowski space is the same, that is, everything within the space is related to everything else in the same "orthogonal" way as in euclidian space, except that there is this additional axis "t" to account for time.
It's not that you are being too simplistic, but you are teaching wrong concepts in what follows. You have a lot to learn yourself.

ConformalGrpOp said:
In the three dimensional euclidean space, we are used to thinking about a "point" in space which can be identified by the unique x,y and z coordinates that identify that location. In Minkowski space, everything that relates to a unique locus of x,y,z and t coordinates is considered to be not just a "point" location (a place), but rather an "event", because the coordinates include a "place" for time. The difference between the two is that if you want to describe the motion of the Earth around the sun in euclidian space, then the path of the orbit is described by an elipse and the Earth just keeps going around and around on the path traced by the elipse. In spacetime, the path is described by a helix. This is because you are describing the orbit where the Earth's location at any point in time as time progresses. Thus, in spacetime, the orbit of the Earth will never retrace itself because, as it orbits the sun, it continues goes around in space in an eliptical rotation that moves along the "time axis". So, what looks like an elipse in euclidian space looks like a helix in Minkowski space.
This is not the significant difference between Euclidean and Minkowski space. You should also add the time coordinate in Euclidean space to show the orbit of Earth as a helix that never traces out that same points.

ConformalGrpOp said:
Special relativity describes how different observers at different places (in space and time), relate to each other and to other "events" that they may observe that happen at other places in space and time, and especially where they are in relative motion with respect to each other and/or to such events. So, when people ask about what is the meaning of Minkowski space, the key thing to understand is that everything located in Minkowski space is an event, not just a place. This characteristic has lead to the use of the term "space-time" to describe what Minkowski realized about the meaning of Einstein's theory of Special Relativity and lead him to develop "Minkowski space".
The significant difference between Euclidean and Minkowski space is in the transforms that get you from one frame to another. In Euclidean space, time is not in the transform whereas in Minkowski space, it is.

ConformalGrpOp said:
Others here have posted adequate explanations of the questions you had about what the terms "space like", "time-like" and "light like" refer to. In thinking about these terms, the most important thing to remember is that the speed of light is constant...everywhere, for everybody no matter how fast they are moving toward or away from the source of the light.
That is the significant difference between Minkowski space which is flat and curved space where the speed of light is not the same everywhere.

ConformalGrpOp said:
This fact has important consequences for any set of observers of events happening in space-time (Minkowski space). First, depending on how far away you are from an event...let say, and exploding star, you could observe that event way before someone else who is located much further away from the event than you are. But, on the other hand, he could be much closer to another exploding star than you are. Now if you think about it, if both exploding stars blew up at exactly the same time, but you see the explosion of Star A before you see the explosion of Star B (but you know they exploded at the same time), you can conclude that you are closer to Star A than to Star B. But, if someone else sees Star B explode before Star A, and they also know that both stars exploded at the same time, then they will know that they are closer to Star B than Star A. But, if neither of you know when either star exploded, and the other guy tells you he just saw Star B explode, and you tell him , heck, I just saw Star A explode, and he looks up in the sky and says...Star A is still there, and you look up in the sky and say, well, Star B is still there, you can say that there is a problem that you both are not able to simultaneous observe those two events at the same time from different locations even though both events actually happened simultaneously.
Your last comment, "actually happened simultaneously", is a strong indication that you do not understand relativity. In fact, all these same characteristics that you describe apply in Euclicean space with a finite speed of light.

Another thing that reveals that you do not understand relativity is your statement, "the other guy tells you he just saw Star B explode", as if instant communication over a long distance is possible.

ConformalGrpOp said:
The ability to observe an event in space-time is intimately bound up with the time it takes for light coming from the source of the event to reach the observer. It is this fact that gives rise to the concept of the light cone in space-time. In order to understand it, you have to understand that in Minkowski space, the time axis, "t", is to be treated conceptually as orthogonal to each of the other space axes.
There's more to it than that. This would also be true in Euclidean space with a finite speed of light.

ConformalGrpOp said:
So, when you look at an image of a light cone, you are actually looking at a "surface" which relates the speed of light traveling through space to locations where that light can be observed. If you are located in the "path" of light coming from the source at the origin of the light cone, you have to be located somewhere on that surface. But, you have to always remember, that every observable event that happens in space time has its own unique light cone. The only events you can observe happening are those events where the surface of the light cone intersects the space-time coordinates of where you are "located" in space-time...that is...the "event" of "where/when" you are making your observation. Now, if you are in motion with respect to events that are happening (or did happen), you are changing the where/when of your observation with respect to the light cones of those events. Depending on how your motion relates to the light cones of those events, you may not be able to observe those events where if you were not moving, you would have been able to observe them.
No, events do not have unique light cones. And your statement that you can only observe certain events is totally wrong. Everyone can observe all events no matter where they are if they just wait long enough.

ConformalGrpOp said:
With this understanding, it is possible to gain some insight into what is meant by space-like, time-like and light like events. Let's go back to the situation of two observers, only, instead of thinking about two different exploding stars, let's think about two comets. On Sunday night, at 8:00 pm, you look up in the sky and you note that you see Comet A passing exactly across the path of the North Star, or better, at celestial coordinates x(a), y(a), and z(a), and they see Comet B, way over on the horizon. Someone else located far away (like, on a planet in orbit around another star far away), walks out on the very same Sunday at 8:00 pm and looks at the exact same celestial location (x(a), y(a), and z(a) where you are looking, but they see Comet B at the same place where you are seeing Comet A! When they look to find Comet A, it is way off at a completely different place in the sky!
Again, a total lack of understanding of relativity. Your idea that there is a "very same Sunday" betrays your lack of understanding.

ConformalGrpOp said:
Now the fact is, that Comet A and Comet B never come close to one another even though they trace paths through Space which intersect at x(a), y(a), and z(a) at completely different times. So, how is it possible that both you and the other guy look up at exactly the same time, and see two different comets at exactly the same place?? Well, clearly, one of you is watching something that the other one won't see for a long time...that is, they are watching an event that the other one will only be able to observe in their spacetime future, while the other one is watching an event that happened long ago in the other one's spacetime past, even though you both are making your observations at exactly the same time and observing completely different events happening at exactly the same place. But, at the time that both of you are making your observations, the fact is that neither comet is actually at x(a), y(a), and z(a). We'll call this situation the Comet example and come back to it later.
The fact that different observers see events in different orders is also true in Euclidean space with a finite speed of light but does not justify your continued statement "at exactly the same time".

ConformalGrpOp said:
Now, let's consider a different example of two exploding stars. In this example, the two exploding stars are exactly the same distance from you, but, there is a gigantic gas cloud that lies closer to exploding star A than exploding star B and lies between the two stars such that it would take 3 months for light to get from star B to the gas cloud, but only about 1/3 that time for the light from star A to get there. However, the light from both exploding stars will take at least a year to reach you. Now, if you observe the gas cloud lighting up exactly one month after you saw both exploding stars blow up, can you determine which exploding star caused the gas cloud to light up?
This has nothing to do with relativity. If one exploding star is responsible for lighting up gas cloud, then all explanations would have to conform to this fact.

ConformalGrpOp said:
I have to run now, but will come back to continue this if it seems to be helpful, but think about which situations described in these examples might be space like, time like or light like.
Please, study Special Relativity yourself. You are in no condition to be teaching it.
 
  • #27
@ghwells; Good grief! I appreciate the energy you devoted to evaluating my post. The idea of over simplifying the concepts ash was wrestling with was concededly probably ill-conceived and, as you so thoroughly pointed out, probably more troublesome than helpful. I deleted the post in its entirety and leave it to you guys to help him gain an exacting appreciation of the subject matter, which is no doubt what he was looking for. BTW, I do accept private messages if you have any concerns about my posts.
 
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  • #28
ash64449 said:
The equation that you said is still a Pythagorean theorem!
\Delta{s}^2 + c^2\Delta{t}^2 = \Delta{x}^2
Doesn't this look like Pythagorean theorem?
It isn't very useful to think of it as a triangle, it is better to think of it as the radius of a circle. If you start at the origin of a Euclidean space and trace out all of the points that are a distance ds away then you use the usual formula and the result is a circle. If you start at the origin of a Minkowski space and trace out all of the events that are a timelike interval ds away then you use the relativistic formula and the result is a hyperbola.
 
  • #29
ConformalGrpOp said:
@ghwells; Good grief! I appreciate the energy you devoted to evaluating my post. The idea of over simplifying the concepts ash was wrestling with was concededly probably ill-conceived and, as you so thoroughly pointed out, probably more troublesome than helpful. I deleted the post in its entirety and leave it to you guys to help him gain an exacting appreciation of the subject matter, which is no doubt what he was looking for. BTW, I do accept private messages if you have any concerns about my posts.
Well, like Einstein is credited with the quote: "Everything should be made as simple as possible--but no simpler", being simple is not an asset if it is wrong. Your post is similar to a video that was presented in a recent thread especially at the two-minute mark:

https://www.youtube.com/watch?v=lsCbaXn6Jew

This might be simple, but it's wrong and that needs to be pointed out. Do you see how the video is explaining simultaneity in a similar manner to what you were explaining? And can you see how it needs to be pointed out that it is wrong, even if it is easy to understand?
 

1. What is Minkowski space?

Minkowski space is a mathematical construct developed by Hermann Minkowski in 1908 to describe the physical world in terms of time and space. It is a four-dimensional space-time continuum that combines the three dimensions of space with one dimension of time.

2. How does Minkowski space relate to special theory of relativity?

Minkowski space is the mathematical framework used to describe the concepts of special theory of relativity, which states that the laws of physics are the same for all observers in uniform motion. It helps us understand the effects of time dilation, length contraction, and the relativity of simultaneity.

3. What are the key concepts of Minkowski space?

The key concepts of Minkowski space are space-time, events, world lines, and the space-time interval. Space-time is the four-dimensional continuum in which all events take place. Events are points in space-time that can be described by their coordinates of time and position. World lines are the paths of objects moving through space-time. The space-time interval is a measure of the distance between two events in space-time.

4. How is Minkowski space represented mathematically?

Minkowski space is represented mathematically using the Minkowski metric, which is a tensor that defines the space-time interval between two events. It is also represented using the Lorentz transformations, which are a set of equations that describe how coordinates and measurements change between different frames of reference in Minkowski space.

5. What are some real-world applications of Minkowski space and special theory of relativity?

Minkowski space and special theory of relativity have many applications in modern physics, including particle accelerators, GPS technology, and the study of black holes. They also have practical applications in the fields of telecommunications, navigation, and satellite communications. Understanding Minkowski space and special theory of relativity is crucial for many technological advancements in our modern world.

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