# Thread: Special Relativity: Spacetime Diagrams

1. ## Special Relativity: Spacetime Diagrams

I'm having some confusion understanding the basic idea of a spacetime diagram and what it really shows. Here is (hopefully) a diagram representing two frames, S and S', where S' moves at relative velocity u to S.

Now, I'm fine with tan(theta) = u/c, since S' moves at velocity u, x = ut gives the blue ct' line.

P happens when x'=0, t'=T'

Inverse Lorentz transform says t = T = gamma[u](t' + vx'/(c^2)) = gamma[u]*T'

I'm taking this to mean that the t-coord in S, of event P, is given by this relation.

tan(theta) = u/c => gamma[u] = [1 - tan^2(theta)]^-1/2

Hence T = T'*[1 - tan^2(theta)]^-1/2

However, using basic trig on the right-angled triangle:

T = T'*cos(theta)

cos(theta) does not equal [1 - tan^2(theta)]^-1/2 !

2. Originally Posted by narg
tan(theta) = u/c => gamma[u] = [1 - tan^2(theta)]^-1/2
Your problem is right here. You are missing a letter:
$\beta = \frac{u}{c} = tanh(\theta)$

This is the hyperbolic tangent of $\theta$:
$tanh(\theta) = \frac{e^{\theta} - e^{-\theta}}{e^{\theta} + e^{-\theta}}$

Just to mention one difference between tan and tanh: $\theta$ does not behave like an angle. There is no mapping between 0 and $2\pi$.

-Dan

3. Are you sure? Wikipedia uses tan, and I'm sure tan works when you consider the triangle.

Minkowski diagram - Wikipedia, the free encyclopedia

4. Originally Posted by narg
Are you sure? Wikipedia uses tan, and I'm sure tan works when you consider the triangle.

Minkowski diagram - Wikipedia, the free encyclopedia
Interesting! I've never seen this. (And after looking at it I hasten to say that what I am calling $\beta$ is not what the Wiki article mentions.)

I think I have a handle on it now. You must be very careful in which diagram you choose to do the problem. Using the Minkowski diagram (which is what you used) we do not have that
$tan(\theta) = \frac{u}{c}$
Angle $\theta$ is between the time axes and angle $\alpha$ is between the space axes. In this diagram we do not generally (and I think never, actually) have that $\theta = \alpha$. The main problem is that we don't know the angle between the x' and ct' axes as measured with respect to the x-ct coordinate system.

Frankly I find that these diagrams are confusing. If I had to use one I'd use the Loedel diagram since it is more in the spirit of SR symmetry. (Note that you can do your problem using the Loedel diagram without any trouble at all because, again, we don't know the angle between the x' and ct' axes. This is clearly indicated by the fact that the right triangle that is giving you problems is no longer a right triangle. The lack of this information is a bit clearer, to my mind, than it is for the Minkowski diagram.) But then again, my SR training is "non-classical" in the sense that I have dealt with the coordinate transformations and associated group symmetry work than seeing what things look like graphically.

-Dan

5. I'm fairly certain they are the same angle... the Wiki article says

"If ct instead of t is assigned on the time axes, the angle α between both path axes results to be identical with that between both time axes. This follows from the second postulate of the special relativity, saying that the speed of light is the same for all observers, regardless of their relative motion (see below)."

In my lecture notes we have theta being the time axes and tan(theta)=u/c also.

6. Originally Posted by narg
I'm fairly certain they are the same angle... the Wiki article says

"If ct instead of t is assigned on the time axes, the angle α between both path axes results to be identical with that between both time axes. This follows from the second postulate of the special relativity, saying that the speed of light is the same for all observers, regardless of their relative motion (see below)."

In my lecture notes we have theta being the time axes and tan(theta)=u/c also.
Between both path axes the angle is the same as between the time axes, yes. The path axes are not the same as the time axes. There is no way to displace an object along the time axis alone.

The article goes on to say:
For the graphical translation it has been taken into account that the scales on the inclined axes are different from the Newtonian case described above. To avoid this problem it is recommended that the whole diagram be deformed in such a way that the scales become identical for all axes, eliminating any need to stretch or compress either axis. This can be done by a compression in the direction of 45° or an expansion in the direction of 135° until the angle between the time axes becomes equal to the angle between the path axes. The angle β between both time and path axes is given by $sin(\beta) = v / c$.
It is only in this system, the Loedel diagram not the Minkowski, that the angle between the time axes is the same as the angle between the space axes.

-Dan

7. Ah ok, so if the coordinates along the time axis in S' are somehow distorted, I can't use trigonometry on the diagram. Which raises the questions in my mind of what the diagram is really useful for, and how we can say tan(theta)=u/c.

No worries though, thanks for the help!

8. Originally Posted by narg
Ah ok, so if the coordinates along the time axis in S' are somehow distorted, I can't use trigonometry on the diagram. Which raises the questions in my mind of what the diagram is really useful for, and how we can say tan(theta)=u/c.

No worries though, thanks for the help!
I'm with you. I think the Loedel diagram is much more useful.

-Dan