Exploring Light Cones & Causality Diagrams in Relativity

[Grimble]

There is something that has been puzzling me about the way that light cones are drawn, when we consider causality diagrams.
Lines representing a substantial point (as Minkowski referred to them) that is traveling as close to the speed of light, that is virtually at the speed of light, is drawn at an angle of 45°, that is that it will have traveled one light second along the x-axis in a time of one second, that is measured as 1 light second on the ct axis. All very simple and logical, and correct using Newtonian mechanics.
Yet it seems to me that in Special Relativity this cannot be supported as drawn this way the speed of that substantial point would be 1.414 light seconds per second. Which is of course the problem with Newtonian Mechanics.

Complying with the Postulates of Relativity, however, we have a very different diagram where we measure from the known speed of the particle. For we know how far the particle will have traveled in the time measured along the rotated time axis of the moving particle measured from the stationary Frame.

We see this in the following Figure using two light clocks with clock B moving at 0.6c relative to clock A; where the time in the moving clock, measured from the stationary clock, is measured along the rotated time axis of that moving clock.

In the first diagram we see Newtonian mechanics, where clock A is stationary and clock B is moving and its time axis is rotated through angle α (where tan α = v/c). it is easy to see here that when clock B has traveled at 0.6c for 1 second, as measured from A, the speed has to be 1.16 c.

Whereas, in the second diagram obeying the postulates of relativity, we see that clock A, seen as moving from the stationary clock B, has traveled 0.6 light seconds in the time that the light has traveled 1 light second - in one second at the speed of c.
This also is in accordance with the time from the clocks passing event being an expanding circle (as light would travel in every direction from a flash of light at that event, the movement of light being the best clock there is).
In this case the time axis for the moving clock A is rotated through angle β (where sin β = v/c) and the time measured in Frame B is 0.8 seconds.

It is also worth noting that when 1 second has passed in frame B, the time in the moving clock A, measured from Clock B is 1.25 seconds, and a body moving at 0.6c would have a Lorentz factor of 1.25 and its time dilation after 1 second would be 1.25 seconds.

It seems to me, therefore that applying the special relativity postulates to the drawing of the light cone would, rather than a cone result in a hemisphere and that the cone of causality would include all within that hemisphere as for a body moving at hear the speed of light the time dilation would be approaching ∞.

https://ac0077b2-a-62cb3a1a-s-sites.googlegroups.com/site/specialrelativitysimplified/home-1/minkowski-diagrams/Mechanics%20compared.png?attachauth=ANoY7crGxhWFd0ch06oJfptUpnkbzusP2f5ViLUigLEnjsR7_HCrWyZPBAkoi07gtycp2uDeWilO-KLM7phtStO5N_nZlwR2PjvkzMxgFgzZjyW4rs7u-3d7SVo17lm8gQpiGxSzDCruxMj4LLGHeKxQCwk_v1deozl9zuPDQWB5-lkMcw4pcroQT_1t8HIaaJwsOquR0N8mzwnojA622xtpHaKtyhvUFzy6ec7Bf5kW2dKwMikuokNo13wAbWblANn88dAaErIrfyyqvsOJqqVOqK-mt-Sc6A%3D%3D&attredirects=0

Can someone explain why this is not the case?

 

[Dale]

For we know how far the particle will have traveled in the time measured along the rotated time axis of the moving particle measured from the stationary Frame.

It is true, we do know this. But that is not what is plotted in a spacetime diagram. Only use the coordinate time in the same inertial frame as distance is measured.

The spacetime interval from the origin to an event $(t,x,y,z)$ is $s^2=-c^2 t^2+x^2+y^2+z^2$. Set $s=0$ and you have the equation of a cone.

 

[vanhees71]

It's of course a cone in 4D spacetime, i.e., a 3D hypersurface.

 

[Grimble]

It is true, we do know this. But that is not what is plotted in a spacetime diagram. Only use the coordinate time in the same inertial frame as distance is measured.

The spacetime interval from the origin to an event $(t,x,y,z)$ is $s^2=-c^2 t^2+x^2+y^2+z^2$. Set $s=0$ and you have the equation of a cone.

Yes. and to simplify into 2 dimensions, with y and z = 0 we have c²t² = x² which will be a 45° line which is the result of plotting ct against x.
This is when using light as a visible measure of time is so illuminating (pardon the pun ).
For when the light has traveled equal distances along each axis (ct, x) Its total distance traveled (as using ct for time and light makes it a real distance), its total distance traveled will be 1.414 x the distance on either axis.

Look at it this way: in Newtonian mechanics we use vector addition. The path of the light in a light clock moving at 0.6c laterally from a stationary observer would be represented as the vector of the light traveling 1 light second vertically (the orientation of the path to the mirror in the light clock) plus the vector of the movement of the clock, 0.6 light seconds horizontally giving the diagonal vector of the light, as measured by the stationary observer as 1.166 light seconds.
Simple vector addition in Newtonian mechanics where any speed is viable.

But in Relativity that doesn't work because of the second postulate.
No, surely in relativity we know that the resultant path of the light will be measured as 1 light second after one second of travel.
What we know in Relativity is that when the clock has traveled 0.6light seconds the light will have traveled 1 light second. If we draw a circle centred on the null point the light could travel to any (indeed every) point on that circle.

So if we plot where the 0.6 light second coordinate cuts the 1 ls arc we have the path of the light as measured by the observer. This will be at the point (0.6,0.8).
Extend that line to where the light hits the mirror at point (0.75,1.0) and the time on the clock's time axis is 1.25 light seconds; in exact agreement with time dilation - 1 second measured by the stationary observer in his Frame of Reference is dilated to 1.25 seconds in the Frame of Reference of the moving clock, which transformed by the Lorentz factor of 1.25 for a speed of 0.6c.
Similarly the 0.75 light seconds transformed by the Lorentz factor of 1.25 to 0.6 light seconds length contraction for the Stationary observer.

It seems to me that we cannot just add vectors willy nilly in Relativity as we have the limiting factor of c; but by starting from the known result we are led to the time dilation and length contraction.

I wish I could have added a diagram to illustrate this but although the diagrams in my last post looked fine in the preview they haven't displayed in the post.

 

[Grimble]

I suppose the way to look at it is that in the Newtonian mechanics time is a separate absolute entity independent of the measurements. So plotting one second against 1 light second still gives distance/time the speed.
Whereas in a Space/time diagram in relativity it is one distance plotted against another, with the resultant line being a combined distance rather than just a speed?

 

[Ibix]

You need to decide if you are talking about distance through space (in which case the distance measure is $\Delta x=1$) or through spacetime. In the latter case the appropriate distance measure is the interval, s, where $s^2=(c\Delta t)^2-\Delta x^2=0$. Trying to measure Euclidean distance on a Minkowski diagram is rather missing the point that Euclidean geometry does not apply.

 

[Grimble]

But that is exactly what I am saying, that euclidean geometry - or Newtonian Mechanics as I put it doesn't work for Relativity!

But when one is drawing a lightcone that is euclidean geometry, as you put it, because you are ignoring the second postulate, because plotting 1 unit along the x-axis against one unit on the time (ct?) vertical axis will give a distance for a particle, or light to have traveled 1.414 units in one second! How can it not?

I suppose that what I am saying is that one must not bring in old fashioned Euclidean concepts into the drawing of Minkowski diagrams.

 

[Ibix]

Then I don't understand. You seem to be saying that Minkowski geometry is not the same as Euclidean geometry, which is kind of a trivial statement.

Light cones are called light cones because they look like cones on a Euclidean map. The terminology may not be perfect, but that's what the maths is for. You can call them "null surfaces" if you prefer.

 

[Grimble]

Yes, but the light cones are used to represent the relativity relationship of cause and effect, yet they are using Euclidean geometry. Simple Euclidean geometry. Why on Earth use Euclidean Geometry to diagram Relativity?
If they are representing Relativity where is the time dilation and length contraction?
The light cone is effectively the Frame or Reference of a stationary observer depicting the paths of moving objects - if they weren't moving they would all be along the time axis!
Light can only travel at 45° when plotted against time and distance in Euclidean geometry.
Light plotted on a relativity diagram, such as the vertical light in a light clock is plotted along the ct axis, because it travels 1 light second per second in that direction; if the light clock is moving there is another vector involved -the clock's travel and the light in the clock travels in two directions - at the speed of light within the clock and at the speed of the clock away from the observer. Two separate velocities that have to be added together. One of them being c. Giving a total speed greater than c, which cannot be.
In relativity we must measure where the light has traveled to after 1 second, which will be, on the one hand, 1 light second from the emission of the light; and on the other hand the distance it has traveled with the clock - which gives us a line rotated through angle β, where sin β = v/c - the distance traveled by the clock over the distance traveled by the light in the clock from point (0,0)! This is not Euclidean Geometry. Light in the 'cone' diagram would be plotted along the x-axis as v = c in that case.

 

[Dale]

For when the light has traveled equal distances along each axis (ct, x) Its total distance traveled (as using ct for time and light makes it a real distance), its total distance traveled will be 1.414 x the distance on either axis.

No, this is untrue. The spatial distance is x. The spacetime interval is 0. There is no relevant measure whose value is 1.414 x.

Please stop just making stuff up.