# Having problems understanding Minskowski space.

1. Jan 6, 2009

### Peon666

I've been studying special Relativity on my own for the last few days and I'm understanding basics of Time dilation, Length Contraction, Relativity of Simultaneity etc. quite nicely, as far as their theoretical and mathemmatical aspects are concerned but I'm having problems understanding Minskowski space/diagrams.

Now, is there any online resource providing a good introduction of Minskowski diagrams regarding SR?

Thanks.

2. Jan 6, 2009

3. Jan 6, 2009

4. Jan 7, 2009

### granpa

no. there arent. at least not any that I have found. just make one yourself. its not as hard as it looks.start with a stationary object and draw in its time and space coordinate system (trivial). and draw a line representing the path that a photon passing through the origin would take. then draw in a second coordinate system from the point of view of an object moving through the origin.

5. Jan 7, 2009

### Fredrik

Staff Emeritus
You can read the relevant pages of "A first course in general relativity" by Bernard Schutz at Google books. Link. I really like that book for its presentation of special relativity and its introduction to tensors.

6. Jan 8, 2009

### granpa

the time coordinate line (y axis for the stationary object coordinate system) and the space coordinate line (x axis for the stationary object coordinate system) are symmetric around the photon path line. if that helps.

7. Jan 8, 2009

### Peon666

I've read the article at wikipedia and it was good. One thing is still not fully clear to me: Why the axis are not taken to be orthogonal?

8. Jan 8, 2009

### jambaugh

The page on which they are written is Euclidean. What is orthogonal in Euclidean space will not be orthogonal in Minkowski space and vis versa. The two oblique looking axes x' and t' are orthogonal under the Minkowski metric of Special Relativistic space-time.

9. Jan 8, 2009

### robphy

The t-and-x axes are orthogonal...according to the Minkowski metric. Same for t'-and-x'.
They aren't (generally) Euclidean-orthogonal because this is a Minkowski spacetime diagram.
(There have been attempts to use Euclidean-orthogonal axes on a Minkowski spacetime diagram... While initially more pleasant in interpretation of some problems, it often leads to unpleasant consequences in other problems.)

10. Jan 8, 2009

### jambaugh

Let me add something that helped me a great deal in conceptualizing this.

Remember that rotations preserve orthogonality and length in Euclidean space.
You can express rotated coordinates using a matrix with trigonometric entries:

$$\left(\begin{array}{c} x' \\ y' \end{array}\right) = \left(\begin{array}{cc} cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right)\left(\begin{array}{c} x \\ y \end{array}\right)$$

Note that this matrix is an exponential of a generating matrix:

$$\left(\begin{array}{cc} cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right) = \exp\left[ \theta \cdot \left(\begin{array}{cc} 0 & -1 \\ +1 & 0\end{array}\right) \right]$$

Now Minkowski space-time has a hyperbolic geometry and you can use those hyperbolic trig functions to express special relativistic pseudo-rotations (a.k.a. Lorentz boosts)

$$\left(\begin{array}{c} x' \\ ct' \end{array}\right) = \left(\begin{array}{cc} cosh(\psi) & \sinh(\psi) \\ \sinh(\psi) & \cosh(\psi) \end{array}\right)\left(\begin{array}{c} x \\ ct \end{array}\right)$$

These Lorentz transformations preserve proper-length and proper-duration as defined in the Minkowski space-time of SR. They also preserve orthogonality in that geometry.

Again this transformation matrix is generated exponentially:

$$\left(\begin{array}{cc} cosh(\psi) & \sinh(\psi) \\ \sinh(\psi) & \cosh(\psi) \end{array}\right) = \exp\left[ \psi \cdot \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) \right]$$

I gave the exponential forms to help you see why the signs are as they are. To relate this to the usual component form of a Lorentz transformation use the fact that the slope of the new x' axis is the velocity/c and this is the hyperbolic tangent of the pseudo-angle $\psi$

$$\psi = \tanh^{-1}(v/c) = \tanh^{-1}(\beta)$$

Also $\gamma = \cosh(\psi)$ and $\gamma\beta=\cosh(\psi)\tanh(\psi)=\sinh(\psi)$.

This is all found on the wikipedia article http://en.wikipedia.org/wiki/Lorentz_transformation" [Broken].

Note that the Wikipedia article http://en.wikipedia.org/wiki/Minkowski_diagram" [Broken]
uses regular trig functions but the angle used in these is not frame independent. You can use this hyperbolic form to work out the addition of velocities rather easily. You just add the pseudo-angles and figure the new velocity:

$$v_{1+2}/c = \tanh\left(\tanh^{-1}(v_1/c) + \tanh^{-1}(v_2/c) \right)$$

Where I find this helps understand the structure of Minkowski space-time is when you carry out larger and larger boosts and see what happens to the t and x axes. You'll in mapping these onto paper see them both swing toward the line x=ct. They appear to be getting closer to each other but this is an artifact of trying to draw them on Euclidean paper.

You can in fact diagonalize the 2x2 boost matrix by choosing null axes x=ct and x=-ct.
You then see the Lorentz boost is stretching one null direction and shrinking the other. Since they are null, lengths in those directions are preserved $e^{\psi}0 = 0$.

One other "weird" property of null vectors is that they are orthogonal to themselves. Once you wrap your head around that (and it may take some time) you'll begin to be comfortable working with Mink. space.

Last edited by a moderator: May 3, 2017
11. Jan 9, 2009

### Jack3145

The Schwarzschild solution at

$$ds^2 = 1/(1-2*m/R)*dR^2 + R^2*d\theta^2 + R^2*(sin(\theta))^2*d\phi - (1-2*m/R)*dt^2$$

What if

R --> infinity

then the object would be a fixed point in the sky never waivering so

d$$\theta$$ --> 0
d$$\phi$$ --> 0

giving you the Minkowski Metric.

12. Jan 9, 2009

### Phrak

This is actually a good question. Unit vectors of the boosted coordinates don't look orthongonal. But are they?

Orthogonality would require the inner product, <u,v> of two vectors to be equal to zero.

In the case of comparing coordinate systems, we would want to know if the unit vectors
$$\hat{e}_{(x')}$$​
and
$$\hat{e}_{(t')}$$ ​
have an inner product of zero in the unprimed coordinates.

I suspect it the inner product must be zero, but I haven't actually done it. You're very proficiant at this, so I thought maybe could prove it/disprove it...

Last edited: Jan 9, 2009
13. Jan 9, 2009

### Mentz114

Phrak,
boosted unit vectors are orthogonal

boosting e_1=(0,1,0,0) gives (cosh(b), sinh(b), 0, 0) and boosting e_0 gives (sinh(b), cosh(b), 0, 0). so forming the inner product with the metric of these two gives
cosh(a)sinh(a)-cosh(a)sinh(a)=0.

14. Jan 9, 2009

### Fredrik

Staff Emeritus
Yes, the t' and x' axes are orthogonal with respect to the "scalar product" (actually "metric tensor") used in Minkowski space, but more importantly: The t' axis is the world line* of the second observer in the first observer's spacetime diagram, and the x' axis must make the same angle with the x axis as the t' axis makes with the t axis, because otherwise the speed of light won't be c in the primed coordinates. Another way of looking at it is that the x' axis is the set of events that the second observer considers simultaneous with the event at the origin, and the obvious definition of simultaneity (see e.g. the Google books link I posted earlier) forces us to draw x' that way.

*) A "world line" is a curve in spacetime that represents the motion of an object.

15. Jan 9, 2009

### Phrak

Thanks, Mentz. I think you transposed the cosh's and sinh's, but I get it. Because it's zero valued, things are a little supspect, but a little work shows that all inner products are conserved.