# Does relativity work

## Main Question or Discussion Point

Does relativity work for a coordinate system traveling with a light photon?

Lets see, if the observer is traveling with a photon then the universe is Lorenz contracted into a two dimensional plane perpendicular to the direction of the photon velocity vector.

All mass objects would have infinite relativistic mass, however photons have only relativistic mass which does not exist perpendicular to the direction of motion so the universal plane would have no gravitational affect on a rest massless photon.

All matter at rest in our coordinate system must exist as photons as that is the only thing known to travel at the speed of light.

Other photons could move toward and away from the coordinate photon in the universal plane as the photon crosses the universe but no time passes for the coordinate photon.

So other photons must exist at all positions it has in time in the universal plane simultaneously.

Or other photons, however, must travel at the speed of light, regardless?

That’s it!! my brain is fried.

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JesseM
Does relativity work for a coordinate system traveling with a light photon?
Nope, you can't construct a coordinate system where a photon is at rest and yet all the same laws of physics apply as in sublight frames, so it's understood that the "inertial frames" of special relativity are only those of objects moving at sublight speeds.

Nope, you can't construct a coordinate system where a photon is at rest and yet all the same laws of physics apply as in sublight frames, so it's understood that the "inertial frames" of special relativity are only those of objects moving at sublight speeds.
I think hes talking about the limit as v -> c. However even in that "frame" light would appear to travel at c.

JesseM
I think hes talking about the limit as v -> c. However even in that "frame" light would appear to travel at c.
I think the limit just isn't well-defined. For example, consider a particle A moving at some v < c, and a photon B moving at c. In the limit as v approaches c, both A and B are moving at c, and B is still moving at c in the frame of A. On the other hand, consider two particles moving at the same speed v, at rest relative to one another. In the limit as v approaches c, both A and B are moving at c, and B is at rest in the frame of A.

2D universe

Yes, I am considering the limit as V approaches C.

The part that confuses me is the pancake universe.

So light and other fattened masses are moving around in a plane?

Does the universe really turn 2D and I louse time and one dimension?

Why isn't the "inertial frame" of a photon simply not the entire universe at a single point standing still? The entire universe is a sigularity in the rest frame of a photon. No?

JesseM
Yes, I am considering the limit as V approaches C.

The part that confuses me is the pancake universe.

So light and other fattened masses are moving around in a plane?

Does the universe really turn 2D and I louse time and one dimension?
Well, if we assume matter goes on forever, then for every v smaller than c (relative to average rest frame of all the matter), it still goes on forever, even if individual galaxies and the distances between them are highly flattened. This is another case where the limit isn't really well-defined, in figuring out the "width" of the universe in the limit you're effectively multiplying infinity by zero.

Another point is that if I'm moving at some very large fraction of c relative to the galaxy, not only do I measure the galaxy to be highly compressed in the direction I'm going, but I also see clocks on either end of the galaxy as wildly out-of-sync...if the galaxy is 100,000 light years long in its own frame, and I'm moving at speed v relative to it, then two clocks at either end of the galaxy which are synchronized in the galaxy's frame will be out of sync by (100,000 ly)*(v)/c^2 in my frame. So, in the limit as v approaches c, clocks on either end of the galaxy are out-of-sync by 100,000 years, so at the same moment that it's 2008 A.D. on the leading edge, it's 102,008 A.D. on the trailing edge. And yet in the limit as v approaches c, the distance between clocks along the direction of motion is compressed to zero. So in the limit, perhaps you could say that the photon's entire history is traversed instantly, since it's going zero distance and all the different clock-readings it passes are squashed together on this zero-length path. But again, there are a lot of aspects of the limit that aren't well-defined, and it's definitely not correct in SR to talk about a photon having its own rest frame.

I think the limit just isn't well-defined. For example, consider a particle A moving at some v < c, and a photon B moving at c. In the limit as v approaches c, both A and B are moving at c, and B is still moving at c in the frame of A. On the other hand, consider two particles moving at the same speed v, at rest relative to one another. In the limit as v approaches c, both A and B are moving at c, and B is at rest in the frame of A.
The limit is perfectly defined it just involves a lot of weird zeros and infinities. Everything is infinitely contracted and infinitesimally dilated. Photons still move at c, but so does everything else.

JesseM
The limit is perfectly defined it just involves a lot of weird zeros and infinities. Everything is infinitely contracted and infinitesimally dilated. Photons still move at c, but so does everything else.
Why isn't it valid to treat the case of two photons by taking the limit as v -> c for two particles that are both traveling at v together? In this limit each particle is at rest in the frame of the other.

Why isn't it valid to treat the case of two photons by taking the limit as v -> c for two particles that are both traveling at v together? In this limit each particle is at rest in the frame of the other.
From what I know, this would be because photons are not travelling at the limit v->c, but are travelling at c. The Lorenz Transform equations break down, and no longer apply, when v=c, so I would assume many of the basics of Special Relativity do too.

JesseM
From what I know, this would be because photons are not travelling at the limit v->c, but are travelling at c. The Lorenz Transform equations break down, and no longer apply, when v=c, so I would assume many of the basics of Special Relativity do too.
Yes, this was what I said before. But I was responding to the idea that even if photons don't have a valid rest frame in SR, we could still create a pseudo-frame for them by looking at the limit as v -> c; my point was that even this notion of a limit isn't sufficiently well-defined.

Applying the Lorentz transformations for a massless particle is incorrect.

Applying the Lorentz transformations for a massless particle is incorrect.
Why?

The Lorentz transformations speak to the way bodies with mass move through and interact with space-time.

As JesseM said, the limit for V = C is ill-defined.

A massless particle, or more specifically, a particle with no rest mass, by definition will move at v=c. At that point, distance covered equates exactly to time, unless this has been shown incorrect somewhere and I missed it?

For a beam of light, any change in spatial coordinates is accompanied by a corresponding change in time coordinates.

x=t

One second of time for a massless particle will see it cross one light-second of distance.

If you were to apply the lorentz transformations to light, then you would have to assume either it interacted with spacetime in a way similar to mass, and then add in a whole other factor in the lorentz calculations to show that... or you'd have to essentially rewrite them entirely.

JesseM
For a beam of light, any change in spatial coordinates is accompanied by a corresponding change in time coordinates.

x=t
That isn't a meaningful coordinate transformation. In a coordinate transformation you label one set of coordinates differently than the other--often using primed and unprimed coordinates like (x,t) for the first coordinate system and (x',t') for the second--and then you show how to determine the coordinates of an event in the second system as a function of the coordinates of the event in the first system. For example, the Lorentz transformation can be written as:

x' = gamma*(x - v*t)
t' = gamma*(t - v*x/c^2)

On the other hand, x=t seems to relate the space and time coordinates of the same system, so it's really just giving you a path through spacetime x(t) = t. Perhaps you meant to write t' = x? If so, what is x' as a function of x and t?

The other problem is that you really never answered the question about what the physical meaning of this coordinate system is. Nothing forces you to consider the inertial rest frame of a slower-than-light observer to be his "perspective" (it doesn't reflect what he sees visually, for example), but it's useful to do so because the coordinates reflect what he would measure on a system of rulers and clocks at rest relative to himself, and with the clocks synchronized by the Einstein synchronization convention (which ensures that the laws of physics will work the same way in every inertial frame). You haven't provided any explanation whatsoever about why you think the coordinate system you're thinking of should be called the "perspective" of a light beam.

The Lorentz transformations speak to the way bodies with mass move through and interact with space-time.
Lorentz transforms don't speak, they transform. Don't even get me started. Postmodernist pretense and it's kin--my thesis.

That isn't a meaningful coordinate transformation. In a coordinate transformation you label one set of coordinates differently than the other--often using primed and unprimed coordinates like (x,t) for the first coordinate system and (x',t') for the second--and then you show how to determine the coordinates of an event in the second system as a function of the coordinates of the event in the first system. For example, the Lorentz transformation can be written as:

x' = gamma*(x - v*t)
t' = gamma*(t - v*x/c^2)

On the other hand, x=t seems to relate the space and time coordinates of the same system, so it's really just giving you a path through spacetime x(t) = t. Perhaps you meant to write t' = x? If so, what is x' as a function of x and t?

The other problem is that you really never answered the question about what the physical meaning of this coordinate system is. Nothing forces you to consider the inertial rest frame of a slower-than-light observer to be his "perspective" (it doesn't reflect what he sees visually, for example), but it's useful to do so because the coordinates reflect what he would measure on a system of rulers and clocks at rest relative to himself, and with the clocks synchronized by the Einstein synchronization convention (which ensures that the laws of physics will work the same way in every inertial frame). You haven't provided any explanation whatsoever about why you think the coordinate system you're thinking of should be called the "perspective" of a light beam.

Sorry, been mixing up arguments from an anti-relativist board I bumped into, so converting from trying to figure out why they are arguing against x' = x-vt based on x' having dual abscissal values in S (huh?), to actual Relativistic concepts is a pain sometimes.

I just called it perspective because it is convention, same with someones rest frame being called their perspective.

As for x=t, I'm just thinking about the spacetime interval as it relates to a beam of light.

The more that I wish to increase my dx'^2, the more I have to decrease my dt'^2 to keep the same spacetime interval as the light beam.