Skip to content

Frames and light spheres Cont.

October 6, 2011

…. continued

Light Emitters

Rather than talk about abstract centres it is better to talk about specific objects. We can alter the scenario slightly, by having 2 light emitters involved, E and E’. E’ moves at speed v relative to E. E is stationary in Frame S and thus E’ is stationary in E. We have sync’d the frames such that (t,x) = (0,0) is the same point in space-time as (t’,x’) = 0. Therefore the emitters pass through one another at that moment. The emitter velocities are then vectors defined at (0,0). If an emitter is moving with speed v in a frame then its velocity vector is (1,v). At (0,0) the light sphere is emitted. So which emitter emits it?

Someone in Frame S will say “Emitter E is at the centre of the sphere, it did!” but someone in Frame S’ will say “No, E’ is at the centre, it did!”. Add more emitters who all meet at the same time and who are all moving in different directions and someone in each of their rest frames will say “NO, you’re all wrong! Mine did!”. Jack_ and Motor Daddy of SciForums consider to be a problem. They claim it means relativity says there’s multiple emission events. So how is this resolved?

Light Cones

The problem is that the light sphere is not the whole story. In relativity it’s common to draw space-time diagrams, where you draw the line an object moves along through space-time. So a photon moving through space sweeps out the line x = t (or x = -t). In 1 dimension a light ‘sphere’ is just a pair of points, at time T (t,x) = (T,\pm T). In higher dimensions they make circles or spheres (obviously). But when you consider all the spheres together, growing linearly in time, you get a light cone. In 1 dimension it’s an upside down triangle with its apex at (0,0).

Suppose we now ask the different frames where in space-time the apex of the light cone is. They will all agree, pointing (metaphorically, as they can’t point through time) to the emission event when all the emitters are at the same location. When asked at any time after that where the centre of the sphere is they will each point to the emitter which is at rest in their respective frames. But what they are doing is saying where the centre of  a slice through the light cone is. Just the equation y=6 defines a line in the (x,y) plane of a graph setting time to some value defines a line through space-time. It is that frame’s notion of ‘now’. In Newtonian physics everyone, no matter how they are moving, agrees on this slice but in relativity different frames take different slices.

In 1 dimension each slice is a line. In Frame S the ‘now’ slice for E is a horizontal line at t=T. All x values are allowed, so we can parameterise this line by (t,x) = (T,s) \equiv L(s) for parameter s. Note that when s = \pm T the point is on the light cone and that \Vert L(-s)- L(0) \Vert = \Vert L(+s)- L(0) \Vert. However, if you asked E what the equation for the ‘now’ slice of E’ it would be something different. To work it out you just apply the Lorentz transform the line from the Frame S’ . In S’ the line is (t',x') = (T,s) = L'(s).

\Lambda^{-1}\cdot L'(s) = \gamma T (1,v) + \gamma s (v,1) \equiv \tilde{L}(s)

Written in this form makes it clear that the old intercept on the t’ axis, when s=0, maps to a point off the t axis, \gamma T (1,v). What about the s = \pm T points? They map to  the following,

\gamma T (1\pm v,v \pm 1) = \gamma T (1\pm v,\mp (1 \pm v)) .

So we see they too satisfy x = \pm t and thus are on the light cone. As expected, points on the light cone map to points on the light cone, even though they are no longer occurring at the same time. Furthermore Frame S sees that \Vert \tilde{L}(-s)- \tilde{L}(0) \Vert = \Vert \tilde{L}(+s)- \tilde{L}(0) \Vert. So it seems the different frames agree about the layout of points along slices too. Below is the set of lines in Frame S. The red is the light cone, the blue the Frame S’ axes, the horizontal grey is ‘now’ to Frame S and the diagonal grey is ‘now’ to Frame S’.

If we reverse the analysis we obtain the point of view of Frame S’ ; the light cone, the axes of Frame S and the ‘now’ of each frame.

What remains now is to address whether or not, to use Jack_’s phrasing, relativity says there are multiple emission points. The true emission point is the apex of the light cone, a point in space-time, not a point in space. Previously we asked which of the emitters did the emitting, so which is it? Well we’ve just seen that it doesn’t matter, they all would produce the same light cone. All that defines a light cone is the location of the emitter, not it’s velocity. We’ve seen all the different frames agree on the space-time structure of the light cone. They also agree on what is inside the cone and what is outside.

What they disagree on is what points lie on a spatial slice of the cone. This is equivalent to disagreeing on what the velocity of the emitter was because the slice is defined by the choice of frame and the emitter which is stationary in a given frame is the one which appears to be in the middle of the sphere from that frame’s perspective. But as we’ve seen, it doesn’t alter the structure of events or create contradictions about the light cone. For each ‘now’ slice in Frame S’ there is a line slice through Frame S with the same layout, it’s just at an angle. This shows how all frames agree on the space-time structure and causality, which is the underlying physically important thing. Having a different ‘now’ slice might sound odd but it is no different in terms of mathematical consistency to doing a rotation.

There are ways to make this much more formal, talking about how Lorentz transforms strictly act on vectors in tangent spaces, not coordinates and then are upgraded by the flat properties of Minkowski space-time to acting on coordinates. The lack of dependence on the velocity can be phrased in terms of tangent bundles. I went over it with Jack_ on SciForums ages ago here so if anyone is interested have a read of that.

No comments yet

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: