# Minkowski space

1. May 9, 2013

### ash64449

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..

2. May 9, 2013

### WannabeNewton

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 $\Delta s^{2} = 0$.

3. May 9, 2013

4. May 9, 2013

### ghwellsjr

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.

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. May 9, 2013

### ash64449

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. May 9, 2013

### micromass

Staff Emeritus
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. May 9, 2013

### ghwellsjr

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. May 9, 2013

yes.i know

9. May 9, 2013

### Ibix

Minkowski was the guy who realised 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:
$$d^2=x^2+y^2+z^2$$
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:
$$s^2=c^2t^2-x^2-y^2-z^2$$
Warning: some people put the signs the other way around!

The funny thing about the interval, $s^2$, compared to the square of the distance you get from the familiar Pythagoras formula, is that it can be negative ($d^2$ can never be negative). This leads to a classification system for the intervals between events. If $s^2$ 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
$$s^2=c^2t^2-x^2-y^2-z^2=c^2t^2-d^2$$
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 $s^2$). So if $s^2$ 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 $s^2$ 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 $s^2$ 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. May 9, 2013

### ash64449

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. May 9, 2013

### ash64449

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. May 9, 2013

### ash64449

oh,, so you think lorentz transformation is enough to understand minkowski space?

13. May 9, 2013

### ghwellsjr

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. May 9, 2013

### robphy

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...

15. May 9, 2013

### ash64449

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. May 9, 2013

### ash64449

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

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?

I didn't quite understood this point...

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. May 9, 2013

### Staff: Mentor

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]

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. May 9, 2013

### ash64449

Sorry,what is the meaning of this statement?

19. May 9, 2013

### ash64449

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?

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: May 9, 2013
20. May 9, 2013

### ash64449

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?