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Frames and light spheres

October 5, 2011

Light spheres seem to be a sticking problem for people who don’t like special relativity. Let’s have a look at them. Throughout we use units where c=1 and \gamma = \frac{1}{\sqrt{1-v^{2}}}

Lorentz Boosts

We could work in general N dimensional space but if we’re considering Lorentz boosts then we can just use time and one space dimension. We have 2 frames, S and S’, with coordinates (t,x) and (t’,x’) respectively. A Lorentz boost by speed v in the x direction is then written as

t' = \gamma(t-vx) \quad,\quad x' = \gamma(x-vt) .

The inverse of this is just a boost in the opposite direction so we just swap the coordinates and put a minus sign on the v,

t = \gamma(t'+vx') \quad,\quad x = \gamma(x'+vt') .

Light Spheres

A light sphere centred on the origin in S is then written as x^{2} = t^{2}. What does this become after a Lorentz transform? First we rearrange it to 0 = -t^{2}+x^{2} and so

- \gamma^{2}(t'+vx')^{2} + \gamma^{2}(x'+vt')^{2} = \gamma^{2}\Big( -(1-v^{2})(t')^{2} + (1-v^{2})(x')^{2} Big) = -(t')^{2} + (x')^{2}.

So this is the same form as the original light sphere, the Lorentz boost maps a light sphere centred on the origin to a light sphere centred on the origin. We have done this with a specific Lorentz transform but its actually true of any Lorentz transform. This is because all Lorentz transforms, in any number of dimensions, preserve the Minkowski metric and the light sphere in question can be written in terms of the Minkowski metric \eta. In Cartesian coordinates \eta is a diagonal matrix with -1 as the first entry on the diagonal and +1 on all the others. For our case we have

\eta = \left( \begin{array}{rr} -1 & 0 \\ 0 & +1 \end{array}\right) .

Let’s do this explicitly. We write \mathbf{X} = (t,x) as the vector of coordinates of S and \mathbf{X}' for those of S’. From that it follows

-t^{2} + x^{2} = \mathbf{X}^{\top} \cdot \eta \cdot \mathbf{X} \equiv \eta(\mathbf{X},\mathbf{X}) .

So we now want to show that \eta(\mathbf{X},\mathbf{X}) = \eta(\mathbf{X}',\mathbf{X}'). To do this we need to relate $latex \mathbf{X}$ and \mathbf{X}'. We define a matrix representation for the Lorentz transform via $\mathbf{X}’ = \Lambda \cdot X$. For the above explicit examples we get

\Lambda = \left( \begin{array}{rr} \gamma & - \gamma\,v \\ - \gamma\,v & \gamma \end{array}\right) \quad,\quad \Lambda^{-1} = \left( \begin{array}{rr} \gamma & \gamma\,v \\ \gamma\,v & \gamma \end{array}\right) .

Note that since \Lambda is a Lorentz transform for any value of v then so is its inverse, as it is obtained by changing v to -v. We can now state in a simple way one of the defining and essential properties of Lorentz transforms, that they preserve \eta, easily,

\eta = \Lambda^{\top} \cdot \eta \cdot \Lambda.

For those at home you can see this explicitly for yourself for the above specific case by just putting in the numbers. Now we’re ready to prove the result in general. We start by considering \eta(\mathbf{X}',\mathbf{X}'),

\eta(\mathbf{X}',\mathbf{X}') = \eta(\Lambda \cdot \mathbf{X},\Lambda \cdot \mathbf{X}) = \mathbf{X}^{\top} \cdot(\Lambda^{\top} \cdot \eta \cdot \Lambda ) \cdot \mathbf{X} .

The term in the bracket is, by definition of the Lorentz transforms, just the metric and so we arrive at the result

\eta(\mathbf{X}',\mathbf{X}') = \mathbf{X}^{\top} \cdot \eta \cdot \mathbf{X} = \eta(\mathbf{X},\mathbf{X}) .

Sphere Centres

Both frames see a sphere centred on the origin, their origin. In just 1 spatial dimension this corresponds to x=0 in S and x’=0 in S’. When S has seen time T pass the light sphere has radius T (remember c=1). So the origin is then the space-time location \mathbf{X}_{\textrm{or}} = (T,0). So where does this map to under the Lorentz boost? Easy, \Lambda \cdot \mathbf{X}_{\textrm{or}} = \gamma(T,-vT). This is obviously not \mathbf{X}'_{\textrm{or}} = (0,0). Similarly, the origin in S’ maps to \Lambda^{-1} \cdot \mathbf{X}'_{\textrm{or}} = \gamma (T,v T), which is not \mathbf{X}_{\textrm{or}}.


So here’s the issue, if everything is consistent how can two frames disagree? After all if you put an object like a ball into the space-time then if relativity is consistent when you ask two people in different frames to point at the ball they should agree. So how is it that if they are asked to point at the centre of the sphere they will disagree? This is the crux of the issues had by, among other people, Jack_ on SciForums. An example thread can be found here. I’ll go over that next….


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