# Geometric Description of SR

1. Aug 26, 2010

### telegramsam1

I'm learning about special relativity in its differential geometry formulation. I don't understand how special relativistic effects can be derived from the Minkowski metric. It isn't obvious to me where relative velocity comes in, or why this makes things look different. Can somebody explain how this geometry manifests itself into physics? Deriving the lorentz transform from the metric would be helpful.

2. Aug 27, 2010

### Mentz114

The Lorentz transformation leaves the proper interval unchanged, and so has a special relationship with the Minkowski spacetime. I suppose it can be derived from that requirement alone. It's a postulate of SR that the proper time is the time measured by a clock on the worldkine.

It's easy enough to get time dilation by working out $d\tau/dt$ directly from the metric.

3. Aug 27, 2010

### Hurkyl

Staff Emeritus
The connection has to do with coordinate systems. The coordinate charts you're used to using are orthonormal ones.

Here's a fun, purely Euclidean example. Draw a long, thin (but not too thin), vertical rectangle on a sheet of paper. Draw in coordinate axes, with the x-axis horizontal and y-axis vertical. Draw in a cross section of the rectangle parallel to the x-axis. How long is it?

Now, draw a different pair of coordinate axes (still perpendicular. Or not, if you like). Draw in a cross section of the rectangle parallel to the new x-axis. How long is it?

4. Aug 27, 2010

### Daverz

The first chapter of Spacetime Physics by Taylor & Wheeler should be illuminating. (Section 1.8 in the original edition covers the LT.)

5. Aug 27, 2010

### Fredrik

Staff Emeritus
This is done by showing that the isometry group of Minkowski spacetime is isomorphic to the Poincaré group. I'll quote myself for the definition of an isometry:
I'll copy-and-paste from my personal notes for the proof. A few comments before I do that: Recall that Minkowski spacetime is the set $\mathbb R^4$ with a manifold structure and a specific metric. This means that the identity map on $\mathbb R^4$ can be interpreted as a coordinate system.

The usual definition of the Poincaré group goes like this: For each linear function $\Lambda:\mathbb R^4\rightarrow\mathbb R^4$, and each $a\in\mathbb R^4$, define $T(\Lambda,a):\mathbb R^4\rightarrow\mathbb R^4$ by $T(\Lambda,a)(x)=\Lambda x+a$. Define

$$P=\{T(\Lambda,a)|\Lambda^T\eta\Lambda=\eta\}$$

where

$$\eta=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

Note that in the definition of P above, the $\Lambda$ that appears to the right of the | is the matrix of components of the linear operator $\Lambda$ in the standard basis. (If you need to refresh your memory about the relationship between linear operators and matrices, see this post). The group structure on P is defined by taking the binary operation to be composition of functions.

I'm getting database errors, so I'll try putting the stuff from my notes in a separate post. (Wait a few minutes....I need to edit it a little).

6. Aug 27, 2010

### Fredrik

Staff Emeritus
OK, here we go. Note that I'm not really distinguishing between a linear operator and its component matrix here either. That's why I seem to be getting the result that P=G rather than "P is isomorphic to G".

First let $$\phi$$ be an arbitrary member of G. This means that it's an isometry, so for all x, and all tangent vectors u,v at x, its pullback $$\phi^*$$ satisfies

$$(\phi^*g)_x(u,v)=g_x(u,v)$$

The components of the right-hand side in the identity map coordinate system are

$$=\eta_{\mu\nu}u^\mu v^\nu$$

for all x. The left-hand side is

$$=\phi^*g_{\phi^{-1}(x)}(u,v)=g_{\phi^{-1}(x)}(\phi_*u,\phi_*v)$$

$$=(\phi_*u)^\mu(\phi^{-1}(x))\ (\phi_*v)^\nu(\phi^{-1}(x))\ g_{\phi^{-1}(x)}\big(\partial_\mu\big|_{\phi^{-1}(x)},\partial_\mu\big|_{\phi^{-1}(x)}\big)$$

$$=(\phi_*u)^\mu(\phi^{-1}(x))\ (\phi_*v)^\nu(\phi^{-1}(x))\ \eta_{\mu\nu}$$

We can simplify this. Let's call the identity map I.

$$(\phi_*u)^\mu(\phi^{-1}(x))=(\phi_*u)_{\phi^{-1}(x)}(I^\mu)=u(I^\mu\circ\phi) =u(\phi^\mu)=u^\rho\partial_\rho\big|_x(\phi^\mu)$$

$$=u^\rho(\phi^\mu\circ I^{-1}),_\rho(I(x))=u^\rho{\phi^\mu},_\rho(x)$$

So the left-hand side of the equality we started with is

$$=u^\rho{\phi^\mu},_\rho(x) v^\sigma{\phi^\nu},_\sigma(x) \eta_{\mu\nu}$$

This is equal to the right-hand side for all u,v and x if and only if

$${\phi^\mu},_\rho(x) \eta_{\mu\nu} {\phi^\nu},_\sigma(x)=\eta_{\rho\sigma}$$

for all x. This is equivalent to the matrix equation

$$J_\phi(x)^T \eta J_\phi(x) = \eta$$

where $$J_\phi(x)$$ is the Jacobian matrix of $$\phi$$ at x. This equation implies that $$\phi$$ must be a first-degree polynomial in x. (The details of that part of the proof aren't interesting at all, so I'm not including them here).

$$\phi(x)=\Lambda x+a$$

The partial derivatives of $$\phi$$ at x are

$${\phi^\mu},_\nu(x)=\partial_\nu({\Lambda^\mu}_\rho x^\rho+a^\mu) ={\Lambda^\mu}_\rho\delta^\rho_\nu ={\Lambda^\mu}_\nu$$

so

$$J_\phi(x)=\Lambda$$

This implies that the matrix equation above is equivalent to

$$\Lambda^T \eta \Lambda = \eta$$

so $$\phi$$ must be a member of $$P$$.

Now let $$\phi$$ be an arbitrary member of P. We can see that its Jacobian satisfies

$$J_\phi(x)^T \eta J_\phi(x) = \eta$$

by examining the calculations above. Those calculations also tell us that this condition is equivalent to saying that $$\phi$$ is an isometry of the Minkowski metric, so $$\phi$$ must be a member of G.

Last edited: Aug 27, 2010
7. Aug 27, 2010

### dx

The affine Minkowski space with the minkowski metric does contain all the mathematical content/structure of special relativity. Relative velocity comes in when we want to represent the 4-velocity in a coordinate sytem: Let {t, x} be the instantaneous comoving reference frame with observer I and v' the tangent vector of the world line of observer II. Then v' = γ(v)∂t + vγ(v)∂x, where v is the relative velocity of II with respect to I. For example, we could find the velocity addition formula as follows: Let B move with a speed u with respect to A, and let C move with a speed v with respect to B, and C move with a speed w with respect to A. Then the 4-vectors will have the following representations:

$$v_{CA} = \gamma(w)\partial_t + w\gamma(w)\partial_x$$
$$v_{BA} = \gamma(u)\partial_t + u\gamma(u)\partial_x$$
$$v_{CB} = \gamma(v)\partial_{t'} + v\gamma(v)\partial_{x'}$$
$$v_B = \partial_{t'}$$

To find the relationship between u, v and w, we note that g(vCA,vBA) = g(vCB,vB), so γ(v) = γ(w)γ(u) - wvγ(w)γ(u), which gives w = (u + v)/(1 - uv).

Lorentz transformations:

Let {t, x} and {t', x'} be two observers, with II moving with a speed v with respect to I. Thus, ∂t' = γ(v)∂t + vγ(v)∂x.

$$dt' = \mathcal{G} \cdot \partial_{t'} = (dt^2 - dx^2) \cdot (\gamma(v)\partial_t + v \gamma(v) \partial_x )$$

so dt' = γ(v)dt - vγ(v)dx, which implies t' = γ(v)t - vγ(v)x. The formula for x' can be derived in a similar way.

Last edited: Aug 27, 2010
8. Aug 27, 2010

### JDoolin

I have no idea about the "differential formulation" of the geometry, but I can tell you, geometrically, what you are doing to the space-time diagram is scaling by a factor of s on the x=ct axis, and scaling by a factor of 1/s on the x=-ct axis (or vice-versa).

$$\begin{pmatrix} ct' \\ x'\ \end{pmatrix}= \begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix} \begin{pmatrix} t \\ x\ \end{pmatrix} = \begin{pmatrix} \cosh(\theta) & -sinh(\theta) \\ -sinh(\theta) & \cosh\theta \end{pmatrix} \begin{pmatrix} t \\ x\ \end{pmatrix}= \begin{pmatrix} \frac {1+s}{2} & \frac {1-s}{2} \\ \frac {1-s}{2}& \frac {1+s}{2} \end{pmatrix} \begin{pmatrix} \frac {s^{-1}+1}{2} & \frac {s^{-1} -1}{2} \\ \frac {s^{-1}-1}{2}& \frac {s^{-1}+1}{2} \end{pmatrix} \begin{pmatrix} t \\ x\ \end{pmatrix}$$

9. Aug 27, 2010

### starthaus

The last equality seems total nonsense, where did you copy it from?

The first two are incorrect as well, here is the correct one:

$$\begin{pmatrix} ct' \\ x'\ \end{pmatrix}= \begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix} \begin{pmatrix} ct \\ x\ \end{pmatrix} = \begin{pmatrix} \cosh(\theta) & -sinh(\theta) \\ -sinh(\theta) & \cosh\theta \end{pmatrix} \begin{pmatrix} ct \\ x\ \end{pmatrix}$$

10. Aug 27, 2010

### DrGreg

Actually (apart from the typo with c), if you multiply the matrices you will see that what JDoolin wrote is correct, with $s=\exp(-\theta)$. But the equation isn't all that obviously connected with the words that precede it. A better way of putting it would be to say

$$\begin{pmatrix} ct' + x'\\ ct' - x'\ \end{pmatrix} = \begin{pmatrix} e^{\theta} & 0 \\ 0 & e^{-\theta} \end{pmatrix} \begin{pmatrix} ct + x\\ ct - x\ \end{pmatrix}$$​

11. Aug 27, 2010

### Fredrik

Staff Emeritus
It's not "differential formulation". It's a formulation based on differential geometry. There are several different versions of special relativity, each with a different mathematical structure representing space and time. The simplest structure that works is the vector space $\mathbb R^4$ with a bilinear form g defined by $g(u,v)=u^T\eta v$. Another option is to use a 4-dimensional smooth manifold with a specific metric tensor field that I'm too lazy to define properly here. This thread is about the version of SR that's based on the latter choice.

12. Aug 27, 2010

### bcrowell

Staff Emeritus
A minor comment: the OP asked about SR "in its differential geometry formulation." I think you might as well say "in its geometry formulation," because you don't need differential geometry to describe a space that has uniform geometrical properties.

13. Aug 27, 2010

### JDoolin

I didn't copy the last portion of the equation, but I did verify it using Mathematica. (of course, Dr Greg is right, there is a typo with the t's instead of ct's.)

My idea was that I realized since the speed of light has to be preserved, the directions x=ct and x=-ct have to be eigenvectors in the transformation. (Sorry if I'm not using the math lingo exactly correctly.) But those directions have to be preserved but the scale has to change.

Then, to produce the actual matrices, what I did was rotate 45 degrees, scale (by s) along x axis, and rotate back, and for the other matrix, I rotated, scaled (by 1/s) along y axis, and rotated back. (That's essentially what I did--if you're a stickler, I might have some detail out of order.)

14. Aug 29, 2010

### telegramsam1

Thank you for all your responses. A further question: (Excuse the matlab notation, I don't know latex.)

Transformations of the form {cosh a, -sinh a; -sinh a, cosh a} leaves the Minkowski metric invariant. How are we to determine 'a' from this knowledge? How would I know that 'a' is a function of relative velocity, let alone 'a = tanh^-1(v/c).

Also @dx. I don't understand your notation. You have terms which are gamma(v) * d/dt. It seems you have equated a differential operator with a velocity. Can you explain what is meant by this notation?

15. Aug 29, 2010

### dx

It's the usual notation in differential geometry. The differential operators (∂t) are to be interpreted as tangent vectors, and objects like dt' and dx' are to be interpreted as 1-forms.

For example, in a reference frame {t, x} which is comoving with some inertial observer, his world-line is simply φ(τ) = (τ, 0), and the tengent vector (4-velocity) to his world-line is ∂t.

The vectors ∂μ and the 1-forms dxμ form bases for the tangent and cotangent spaces respectively, at each point of Minkowski space. The metric tensor can be represented in terms of them:

$$\mathcal{G} = dt \otimes dt - dx \otimes dx$$

$$\mathcal{G}^{-1} = \partial_t \otimes \partial_t - \partial_x \otimes \partial_x$$

Last edited: Aug 29, 2010
16. Oct 5, 2010

### JDoolin

$$\begin{pmatrix} ct' \\ x'\ \end{pmatrix}= \begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix} \begin{pmatrix} c t \\ x\ \end{pmatrix} = \begin{pmatrix} \cosh(\theta) & -sinh(\theta) \\ -sinh(\theta) & \cosh\theta \end{pmatrix} \begin{pmatrix} c t \\ x\ \end{pmatrix}= \begin{pmatrix} \frac {1+s}{2} & \frac {1-s}{2} \\ \frac {1-s}{2}& \frac {1+s}{2} \end{pmatrix} \begin{pmatrix} \frac {s^{-1}+1}{2} & \frac {s^{-1} -1}{2} \\ \frac {s^{-1}-1}{2}& \frac {s^{-1}+1}{2} \end{pmatrix} \begin{pmatrix} c t \\ x\ \end{pmatrix}$$​

It's not immediately clear that the last two matrices represent scaling on the x=c t axis and the x=-c t axis. The http://ysfine.com/articles/dircone.pdf" [Broken]has made the tranformation much more elegant (though I may have a sign or two wrong somewhere)

$$\left( \begin{array}{c} \text{ct}' \\ z' \end{array} \right) = \left( \begin{array}{cc} \cos (45) & -\sin (45) \\ \sin (45) & \cos (45) \end{array} \right) \left( \begin{array}{cc} e^{\eta } & 0 \\ 0 & e^{-\eta } \end{array} \right) \left( \begin{array}{cc} \cos (45) & \sin (45) \\ -\sin (45) & \cos (45) \end{array} \right) \left( \begin{array}{c} \text{ct} \\ z \end{array} \right)$$​

Last edited by a moderator: May 5, 2017