# Timelike vectors

1. Mar 2, 2006

### Oxymoron

Why is the angle between two timelike vectors given by

$$g(X,Y) = \|X\|\|Y\|\cosh\theta$$

$$g(X,Y) = \|X\|\|Y\|\cos\theta$$?

2. Mar 2, 2006

### robphy

That's the Minkowski metric for you.
The Minkowski-angle (the rapidity) between unit future-timelike vectors is based on the intercepted arc length of the unit hyperbola.

Can you write cosh in terms of tanh alone?

3. Mar 2, 2006

Staff Emeritus
Because in Minkowski geometry the equation satisfied by a generic unit vector (t,x) is $$t^2 - x^2 = 1$$, not $$t^2 + x^2 = 1$$. The first equation is that of a unit hyperbola and it is parametrized by the hyperbolic functions not the circular ones.

4. Mar 2, 2006

### Oxymoron

I will try and do this today and get back to you. Am I correct in assuming that by using some hyperbolic trig identities I can get tanh and is this the relative speed of a timelike, future pointing particle (or an inertial observer)?

Is this due to the metric having minus signs instead of plus? Like $s^2 = t^2-x^2-y^2-z^2$? since dot product is given by

$$u\cdot v = g_{ij}u^iv^i$$

so the 'angle' will be determined on what the metric looks like. In our case it has negatives in it.

5. Mar 2, 2006

### robphy

Yes, but I want you to interpret cosh(theta) with that understanding. If that was too easy, you should work on expressing exp(theta) in terms of tanh(theta), then interpreting.

Yes. In fact, one can draw a picture of the Minkowski metric. It is a hyperbola (generally, hyperboloid)... traced out by the tips of the Minkowski-unit vectors.

6. Mar 6, 2006

### Oxymoron

Im not sure if I have the right answer, I may be complicating it too much.

Consider an observer moving in a straight line in the x-plane. Such motion can be described by a Lorentz transformation which leaves y and z unchanged. As a result the 4x4 Lorentz matrix will look like this

$$[L^{\mu'}{}_{\nu}] = \left(\begin{array}{cccc} L^1{}_1 & 0 & 0 & L^1{}_4 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ L^4{}_1 & 0 & 0 & L^4{}_4 \end{array}\right)$$

The spacetime metric in SR can be written as

$$\Delta s^2 = g_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu}$$

I will assume that the observer (free particle) has a rectilinear path with respect to inertial frames. Transformations preserving the spacetime metric are of the form

$$x'^{\mu'} = L^{\mu'}{}_{nu}x^{\nu} + a^{\mu'}$$

where the coefficients $L^{\mu'}{}_{\nu}$ satisfy

$$g_{\rho\sigma} = g_{\mu'\nu'}L^{\mu'}{}_{\rho}L^{\nu'}{}_{\sigma}\quad\quad[1]$$

which is a Poincare transformation.

Now substituting our Lorentz 4x4 matrix into [1] we have

$$L^1{}_1L^1{}_1 - L^4{}_1L^4{}_1 = g_{11} = 1\quad [2]$$
$$L^1{}_1L^1{}_4 - L^4{}_1L^4{}_4 = g_{14} = g_{41} = 0 \quad [3]$$
$$L^1{}_4L^1{}_4 - L^4{}_4L^4{}_4 = g_{44} = -1 \quad [4]$$

From [4] we have

$$(L^4{}_4)^2 = 1+(L^1{}_4)^2 \geq 1$$

and assuming $L^4{}_4 \geq 1$ we can set

$$L^4{}_4 = \cosh\theta$$

then it simply follows that

$$L^1{}_4 = \pm\sqrt{\cosh^2\theta -1} = \sinh\theta$$

Similarly for [2] we have

$$L^1{}_1 = \cosh\phi$$
$$L^4{}_1 = \sinh\phi$$

and using [3] we have

$$0 = \sinh\theta\cosh\phi - \cosh\theta\sinh\phi = \sinh(\theta-\phi)$$

which implies that $\theta = \phi$.

Now this is where I am stuck. Im trying to express $\cosh\theta$ on its own. I know that

$$\cosh\theta = \gamma[/itex] because I was told, but I still want to show it. I thought by writing everything that I have written above shows me that I should think about $\tanh\theta$ and then use some hyperbolic trig identities, which I will try now... Last edited: Mar 6, 2006 7. Mar 6, 2006 ### Oxymoron Ok, after looking through many appendices I found an identity which may work. Here it goes. A free particle will move with velocity $v = x/t$ along the x-axis, we know this. So [tex]v = \frac{x}{t} = -c\frac{\sinh\theta}{\cosh\theta}$$

Where did the -c come from? Not sure yet but it seems to work atm. Basically I let

$$\tanh\theta = -\frac{v}{c} = \frac{\sinh\theta}{\cosh\theta}$$

Anyway, then I used the identity

$$\cosh^2\theta - \sinh^2\theta = 1$$

and wrote it as

$$\cosh\theta\sqrt{1 - \left(\frac{\sinh^2\theta}{\cosh^2\theta}\right)}=1$$

rearranging...

$$\cosh\theta = \frac{1}{\sqrt{1-\left(\frac{\sinh\theta}{\cosh\theta}\right)^2}}$$

which equals

$$\cosh\theta = \frac{1}{\sqrt{1-\tanh^2\theta}}$$

$$\tanh\theta = -\frac{v}{c}$$

so

$$\tanh^2\theta= \left(\frac{v}{c}\right)^2$$

Therefore

$$\cosh\theta = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma$$

8. Mar 6, 2006

### Oxymoron

If all this is correct, and I think it is, then my assumption that

$$\tanh\theta = -\frac{v}{c} \quad [5]$$

certainly works. Unfortunately I have no justification for setting [5] in the first place. I only obtained it from working backwards.

The problem then is that tanh has profound meaning. In fact $\tanh\theta$ must be the relative speed of the free particle! I wont let this rest. Hopefully someone can help. Is there something which will justify setting [5]???

9. Mar 7, 2006

### Oxymoron

I think I worked it out.

As a particle travels along its worldline in the x-plane, we can look at it at some instant and draw a unit tangent vector at that point. The tangent vector, the t-axis and the x-axis make a triangle, with theta being the angle between the tangent vector and the t axis.

Now, label the tangent vector v. Then from

$$x=v(-ct)$$

we have

$$\frac{x}{t} = -cv$$

But we know that

$$\frac{x}{t} = v =\frac{x/v}{t/v}$$

And we also know from our little triangle on the worldline of the particle that

$$\sinh\theta = \frac{x}{v}$$
$$\cosh\theta = \frac{t}{v}$$

so we have

$$\frac{x}{-ct} = v = \frac{\sinh\theta}{\cosh\theta}$$

which, of course, yields

$$\frac{x}{t} = v = -c\frac{\sinh\theta}{\cosh\theta}$$

After rearranging we have finally

$$-\frac{v}{c} = \tanh\theta$$

How does this look?

Last edited: Mar 7, 2006
10. Mar 7, 2006

### robphy

It's basically correct, but seems inconsistent at times (e.g., $$\frac{x}{t} = -cv$$ ). This relation I pasted should read:
$$\frac{x}{t} = v$$ or $$\frac{x}{ct} = \frac{v}{c}$$... for dimensional consistency.

$$\tanh\theta=v/c [5]$$ describes the SLOPE of the worldline on a spacetime diagram (which is like a position-vs-time graph in which slopes are velocities). Since $$\tanh$$ is dimensionless, the right-hand side must be as well.

Note that
$$\frac{x}{t} = v =\frac{x/v}{t/v}$$
does not algebraically imply
$$\sinh\theta = \frac{x}{v}$$ and $$\cosh\theta = \frac{t}{v}$$ (the units won't work out).
Here's an analogy from Euclidean geometry to show that this method fails algebraically: Does $$\frac{\sin\theta}{\cos\theta}=\frac{4}{5}$$ imply that $$\sin\theta=4$$ and $$\cos\theta=5$$? All you know is that, e.g., $$5\sin\theta=4\cos\theta$$.

11. Mar 7, 2006

### Oxymoron

So how does $\tanh\theta = -v/c$?

12. Mar 7, 2006

### robphy

To describe a particle moving in the forward direction, it's $$\tanh\theta = v/c$$ (no minus sign).

How does such a relation come about in the first place?

One can appeal to the form of the "transformation of velocities" formula, which looks like the identity for the hyperbolic tangent of a sum or difference.

Geometrically, if one draws two rays: x=0 and x=vt (t>0), one cuts an Minkowski arc length of theta on the unit hyperbola. This Minkowski arc-length divided by this radius is the Minkowski-angle [called the rapidity]. Consider the displacement vector along x=vt from the origin to the unit hyperbola. Its spatial component is $$\sinh\theta$$ and its temporal compoent is $$\cosh\theta$$ (the time-dilation $$\gamma$$ factor). The constant velocity is ratio of these two components.

13. Mar 7, 2006

### Oxymoron

Excellent! Thanks robphy. So going back to where I let

$$\tanh\theta = -\frac{v}{c}$$

I now actually have reason for this...because the constant velocity is the ratio of the spatial and temporal components. Which means I could just have written it without the negative sign as you said. I suppose it all comes down to looking at Minkowski space diagrams and seeing how all the angles and vectors interact.

14. Mar 7, 2006

### robphy

That would describe a particle traveling backward [to the left]... just like it would be on an ordinary position vs time graph from introductory physics.

Yes... and finding a physical interpretation. That's the beauty of modelling SR with Minkowski spacetime. The model is faithful... every physical situation can be modeled geometrically. And any geometrical result probably has a physical [though not necessarily interesting] interpretation.

15. Mar 7, 2006

### Oxymoron

I want to use $\tanh\theta$ and rapidity now.

Suppose we have three objects moving parallel with different velocities relative to one another. In particular, A moves with velocity v relative to B and B moves with velocity u relative to C.

u is the velocity of B relative to C. But what is the velocity of A relative to C?

Classically it should be u+v. But what if A,B and C are light beams?

Instead of simply adding the velocities what we need to do is move from the inertial frame of B to the inertial frame of A. In SR this is not as easy as it is for classical physics. Instead there are rules. And these rules are the Lorentz transformations.

So lets move from B to A. We can carry along with us knowledge of the velocity of B relative to C (which is u). But u will change. How much it changes depends on the frame A. Obviously if A is B then u doesnt change. But if A differs from B by any velocity v we need to change u. The way u changes is via

$$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$

Now this is where rapidity comes into play. Robphy mentioned it and now it has intrigued me. If we define relative velocity as

$$v = \tanh\theta$$

then the rapidity $r$ is defined as

$$r = \theta$$

From the identity

$$\tanh(\theta+\phi) = \frac{\tanh\theta + \tanh\phi}{1+\tanh\theta\tanh\phi}$$

can we conclude how this addition rule is consistent with the Lorentz transformation of combined velocities?

Last edited: Mar 7, 2006
16. Mar 7, 2006

### robphy

A Lorentz transformation takes the points of geometric figures and slides them along hyperbolas. These "rotations" are parametrized by the rapidity.

Write the [invariant] relative-rapidity $$\theta_{BA}=\theta_{BO}-\theta_{AO}$$, and define the [invariant] relative-velocity $$v_{BA}=c\tanh\theta_{BA}$$.

By using the identity for hyperbolic tangents, you can write the transformation of velocities formula, relatiing $$v_{BA}$$ in terms of $$v_{BO}$$ and $$v_{AO}$$.

17. Mar 7, 2006

### Oxymoron

Rapidity is additive! So it is by no means like velocity (which is definitely not additive)?

I read that physicists tend to use rapidity over velocity. Is this because it is additive?

18. Mar 7, 2006

### robphy

Unfortunately many physicists (and others) do not use rapidity. Many may make a passing acknowledgment of such a quantity... but they don't use it. [It was de-emphasized in the latest edition of Spacetime Physics because users of the text didn't use it.] But, yes, those who do use it because it is additive, easier to work with, and because it is the natural variable to use to describe the separation between concurrent worldlines.