Straight line through spacetime.

[Rotormaster]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

 

[Passionflower]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

The path length of a photon in spacetime is zero, it is a point.

 

[Fredrik]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times. Does this mean that the only entity to ever have and that ever will achieve a straight line through space time is light?

Any particle can move as described by a "straight line" in spacetime. The technical term is "geodesic". There are three different types of geodesics: Timelike, null and spacelike. A timelike gedesic represents the motion of a massive particle in free fall. A null geodesic represents the motion of a massless particle. (Massless particles are always in free fall if they're moving through a vacuum).

If you meant to ask specifically about null geodesics, then the answer is that any massless particle in free fall would move as described by a null geodesic. The only massless particles in the standard models are photons and gluons, and I don't think gluons are ever in a state that can be described as free fall. So I guess photons are the only particles that will move this way, at least over long distances. But there could exist other (massless) particles that do the same (if they are very rare or interact very weakly with other kinds of matter...otherwise they would have been discovered by now).

 

[Rotormaster]

But aren't photons the only particle with constant velocity? would this not mean that since everything started they would be the only thing to maintain absolute synchronization with time's arrow ie achieve a straight line through space time?

 

[Ich]

Since to travel in a straight line through space time requires the subject in question to have zero change to its state of motion, ie maintain constant velocity at all times.

Moving "in a straight line" in curved spacetime means moving "along" a geodesic. From the moving particles' point of view, there is no change of their "state of motion" as they are alwasy at rest wrt themselves and feel no forces acting on them. (Photons don't have a point of view, btw.)
Measured in some coordinate system, their speed and direction may change. That's true for light also.

 

[Fredrik]

But aren't photons the only particle with constant velocity?

No. Einstein's equation tells us which curves in spacetime are geodesics. We then define zero acceleration as "motion described by a geodesic", and acceleration as a specific measure of much a particle's motion deviates from geodesic motion. So if there are any timelike geodesics at all in spacetime (and there always are), then massive particles can have zero acceleration as well.

"Constant velocity" isn't a very precise term. We can make the velocity anything we want it to be, and depend on the time coordinate in any way we want it to, just by choosing the coordinate system carefully. That's why I'm talking about zero acceleration instead. This is a coordinate-independent concept.

 

[Dale]

The path length of a photon in spacetime is zero, it is a point.

Careful here. The Minkowski norm is degenerate, so the fact that two points have a separation of 0 does not imply that they are the same point like it would with the Euclidean norm.

 

[stevmg]

Careful here. The Minkowski norm is degenerate, so the fact that two points have a separation of 0 does not imply that they are the same point like it would with the Euclidean norm.

Is your statement due to the so-called "null" line or surface of the space-time coordinate system? (In other words, if something is traveling at c or light speed, wouldn't the interval be \Deltax²/c² - \Deltat² = 0) Neither spacelike nor timelike.

Am I on the right track here?

Remember, I am one of those uneducated ones...

 

[stevmg]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

 

[Dale]

Is your statement due to the so-called "null" line or surface of the space-time coordinate system? (In other words, if something is traveling at c or light speed, wouldn't the interval be \Deltax²/c² - \Deltat² = 0) Neither spacelike nor timelike.

Am I on the right track here?

Yes. In Euclidean geometry the set of all points with 0 distance from the origin is a single point, but in Minkowski geometry the set of all points with 0 interval from the origin is a cone called a light cone.

 

[Passionflower]

Yes. In Euclidean geometry the set of all points with 0 distance from the origin is a single point, but in Minkowski geometry the set of all points with 0 interval from the origin is a cone called a light cone.

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

 

[stevmg]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

Help...

Someone answer this one...

 

[matheinste]

Can anyone out there explain the difference between "Aristotilean" and "Galilean" spacetime viewpoints? They look pretty similar to me.

There is a lengthy description of both viewpoints in the early chapters of Geroch's General Relativity from A to B.

Matheinste.

 

[stevmg]

There is a lengthy description of both viewpoints in the early chapters of Geroch's General Relativity from A to B.

Matheinste.

Thanks Matheinste, I read Geroch's book and that's why I asked the question. All I know is there are straight lines up, bevelled decks of cards, etc., etc. I don't know if I am coming or going.

Maybe, if I phrase the question this way, at slow non-relativistic speeds as we are in the "real world" what is the most utilitarian approach? Forget about\sqrt{(1 - v^2/c^2)}

 

[Dale]

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

None of that statement is true in general. For two arbitrary points inside or even on the light cone the interval may be timelike, spacelike, or null. Also, the interval between two events is purely geometric and has nothing to do with photons nor the arrow of time.

 

[Passionflower]

None of that statement is true in general. For two arbitrary points inside or even on the light cone the interval may be timelike, spacelike, or null. Also, the interval between two events is purely geometric and has nothing to do with photons nor the arrow of time.

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

 

[matheinste]

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

If we are discussing the spacetime interval why is it not true that a zig zag path can give a null value if we sum along the path, even if we restrict the motion of the photon to forward pointing in time.

Matheinste.

 

[Passionflower]

If we are discussing the spacetime interval why is it not true that a zig zag path can give a null value if we sum along the path, even if we restrict the motion of the photon to forward pointing in time.

Matheinste.

If we restrict it we could obviously not "zig zag" from a future to a passed event inside the light cone.

 

[matheinste]

If we restrict it we could obviously not "zig zag" from a future to a passed event inside the light cone.

If we are referring to any two events in any time order that may be true. But we can still have a zig zag null path going forward in time between events joined by a timelilke vector within a light cone without either part of the zig zag path being on the light cone.

Matheinste.

 

[matheinste]

Thanks Matheinste, I read Geroch's book and that's why I asked the question. All I know is there are straight lines up, bevelled decks of cards, etc., etc. I don't know if I am coming or going.

I think that in the Aristotlean view there is a priviledged state of absolute rest. In the Galilean view there is no such absolute state, all motion is relative.

Matheinste.

 

[DrGreg]

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

So what? The "interval" between two events inside the cone is the longest "distance" (or rather, time) between the events, not the shortest.

 

[stevmg]

I think that in the Aristotlean view there is a priviledged state of absolute rest. In the Galilean view there is no such absolute state, all motion is relative.

Matheinste.

That makes sense in accordance of what I read.

Interesting that relativity is tacked onto Galilean.

 

[stevmg]

Using the equation:

(curvature of space-time geometry) = G * (Mass density of matter in space-time)

How did Einstein know what the mass-density of the Sun was? Was the curvature measured in degrees or radians or whatever?
Does G = 6.67259 x 10^-11 N m²/kg²?

Or did he do it backwards... He measured the curvature angle, used Cavenish's G and calculated the mass-density of the Sun by back substitution.

 

[stevmg]

If we are referring to any two events in any time order that may be true. But we can still have a zig zag null path going forward in time between events joined by a timelilke vector within a light cone without either part of the zig zag path being on the light cone.

Matheinste.

But within each leg of the zig-zag path, wouldn't the path be on the "local" light cone eminating from each "vertex?"

Damn it, I wish I could draw what I'm talking about. It would be so easy with a pencil and paper.

 

[matheinste]

But within each leg of the zig-zag path, wouldn't the path be on the "local" light cone eminating from each "vertex?"

Damn it, I wish I could draw what I'm talking about. It would be so easy with a pencil and paper.

From the start of each leg yes. And all events along each leg.

An event in spacetime is the vertex of a light cone. The set of events on a particles worldine have associated with them a set of light cones.

I was considering a pair of events linked by a timelike vector within the future light cone of some other event in the past of those two events.

Matheinste.

 

[Dale]

Sorry but you are wrong, provided we do not restrict photons to go only forward in time we can always "zig zag" inside the cone to make the distance 0.

This has nothing to do with it. The distance between two points is not the same thing as the length of an arbitrary path between the two points. This is analogous in both Euclidean and Minkowski geometry. When we say that my door is 5 m from my window we do not mean that every possible path from my door to my window is 5 m long. Again, these are purely geometric statements and have no relationship whatsoever with photons or the arrow of time.

 

[Passionflower]

This has nothing to do with it. The distance between two points is not the same thing as the length of an arbitrary path between the two points. This is analogous in both Euclidean and Minkowski geometry. When we say that my door is 5 m from my window we do not mean that every possible path from my door to my window is 5 m long. Again, these are purely geometric statements and have no relationship whatsoever with photons or the arrow of time.

My original statement which you described as "None of that statement is true in general" was:

However, the shortest possible distance between any two points inside the cone is also 0, it is only when we restrict photons to go only forward in time that the points are only on the cone.

You were wrong and rather than admitting that you now seem to cover it up by twisting what I was writing.

Using two arbitrary points A and B inside a light cone we can construct a set of light cones that connect us from A to B. The total length of such a connection is 0, which is the shortest possible path. That is simply a feature of a Minkowski spacetime.

 

[Dale]

Using two arbitrary points A and B inside a light cone we can construct a set of light cones that connect us from A to B. The total length of such a connection is 0, which is the shortest possible path.

This is correct, but the length of such a path is not the distance between A and B. Therefore this statement is simply not true.

the shortest possible distance between any two points inside the cone is also 0

When you measure a distance between two points you do not use an arbitrary path, you use a geodesic. The path you describe is not a geodesic. And photons going forward in time is not relevant to any of this.

Let's say that we have a light cone at the origin, and we take the events A = (t_A,x_A,y_A,z_A) = (1,0,0,0) and B = (t_B,x_B,y_B,z_B) = (3,0,0,0) which are two points inside the cone. Then the distance AB between them is 2 despite the fact that there is a path ACB with C = (t_C,x_C,y_C,z_C) = (2,1,0,0) where AC+CB = 0 and a path ADB with D = (t_D,x_D,y_D,z_D) = (2,1.41,0,0) where AD+DB = 0+2i. Those other paths are completely irrelevant to what the distance AB is.

 

[Passionflower]

When you measure a distance between two points you do not use an arbitrary path, you use a geodesic. The path you describe is not a geodesic.

I really see no point in continuing this "discussion". Hopefully other people can benefit from my initial statement, it is clear it is lost on you.

 

[Dale]

Do you really not understand the difference between the distance between A and B:
\sqrt{c^2(t_B-t_A)^2 - (x_B-x_A)^2 - (y_B-y_A)^2 - (z_B-z_A)^2}

and the length of a path from A to B:
\int_A^B \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2 } d\lambda