Everything travels at the speed-of-light ?

[trewsx7]

I've been reading various articles about the speed-of-light and I came across a statement that intrigued me: Everything travels at the speed of light.

Now, I realize that most of this motion for everyday objects is diverted in both the time and space directions, so that theoretically an object at rest is traveling through spacetime at light-speed through the time direction while light travels at this speed through space.

But my question is this: "objects at rest" that travel at the speed-of-light through time DO NOT apply to me sitting here at my desk, correct, since I am moving with the orbit of the earth, correct? These "objects at rest" traveling at 100% lightspeed though the TIME DIMENSION ONLY only apply to hypothetical objects in a vacuum (like empty space), and not anything here on Earth?

Please tell if I am right or wrong on this. And if I am wrong, please elaborate objects at perfect rest and give me an example (all travel is done through time and none through space).

Thanks!

 

[robphy]

Here's my comment on this "Everything travels at the speed of light" idea
https://www.physicsforums.com/showthread.php?p=565681#post565681
(The short story is that the idea doesn't seem useful to me... and does not deserve the emphasis it is given.)

 

[Dale]

The direction in spacetime that corresponds to "TIME DIMENSION ONLY" is completely arbitrary, just like the direction in space that corresponds to "X AXIS ONLY". So any inertially moving objest is moving through time only in some reference frame.

 

[A.T.]

But my question is this: "objects at rest" that travel at the speed-of-light through time DO NOT apply to me sitting here at my desk, correct, since I am moving with the orbit of the earth, correct? These "objects at rest" traveling at 100% lightspeed though the TIME DIMENSION ONLY only apply to hypothetical objects in a vacuum (like empty space), and not anything here on Earth?

It all depends how you define "travel". When you observe an inertially moving clock and compare it to a clock at rest by measuring

$$dt$$ : time passed on the clock at rest
$$d\tau$$ : time passed on the moving clock (it's proper time)
$$dx$$ : the spatial displacement of the moving clock

you will find that

$$(c\,dt)^2$$ = $$(c\,d\tau)^2$$ + $$dx^2$$

for every velocity of the moving clock. And since this looks very much like Pythagoras, one could interpret it geometrically by assuming $$c\,d\tau$$ and $$dx$$ to be orthogonal dimensions of something called "space-(proper)time". So that $$c\,dt$$ would be the distance "traveled" by every object through this space-(proper)time during the time $$dt$$. In simple words: Everything travels with c trough this space-(proper)time.

Here a visualization:
http://www.adamtoons.de/physics/relativity.swf

 

[robphy]

In simple words: Everything travels with c trough this space-(proper)time.

That would only be for "things with non-zero mass"
... whose spacetime-tangent-vector is timelike (and can be normalized).

For "things with zero-mass" (like light), it would [in analogy] "travel with speed 0"
... since a light-signal's spacetime-tangent-vector is null (and can't be normalized).

 

[A.T.]

For "things with zero-mass" (like light), it would [in analogy] "travel with speed 0"
... since a light-signal's spacetime-tangent-vector is null (and can't be normalized).

In space-propertime light is not a special, but the most trival case. If you assume $$d\tau$$ for light to be zero, then it simply moves only along the spatial dimensions of space proper-time, and $$(c\,dt)^2$$ = $$(c\,d\tau)^2$$ + $$dx^2$$ simplifies to the obvious $$c\,dt = dx$$

As I said, it all depends how you define "travel speed". If you define it as displacement in space-propertime divided by coordinate time (dt), then everything (including light) travels at c.

 

[robphy]

In space-propertime light is not a special, but the most trival case. If you assume $$d\tau$$ for light to be zero, then it simply moves only along the spatial dimensions of space proper-time, and $$(c\,dt)^2$$ = $$(c\,d\tau)^2$$ + $$dx^2$$ simplifies to the obvious $$c\,dt = dx$$

As I said, it all depends how you define "travel speed". If you define it as displacement in space-propertime divided by coordinate time (dt), then everything (including light) travels at c.

That is why Minkowski's [i.e., the original] spacetime formulation is much more natural, more geometrical, and more consistent... with fewer ambiguities than any other "space and time" formulation. Its reduction to the Galilean case follows easily... and its analogue with Euclidean analytic geometry and trigonometry is also natural. All of these claims can be made precisely.

 

[A.T.]

That is why Minkowski's [i.e., the original] spacetime formulation is much more natural, more geometrical, and more consistent...

Most laymen find Minkowski's space-time rather counter intuitive. Mainly due to it's non-Euclidean signature, which makes it different from what they know as "natural geometry".

 

[robphy]

Most laymen find Minkowski's space-time rather counter intuitive. Mainly due to it's non-Euclidean signature, which makes it different from what they know as "natural geometry".

Yes, that's unfortunate... but because it naturally generalizes to general relativity, it is necessary for anyone who wants to really understand relativity [and not merely one aspect of it]. The required mathematics (and physics) relies on this non-Euclidean signature.

"Anyone who studies relativity without understanding how to use simple space-time diagrams is as much inhibited as a student of functions of a complex variable who does not understand the Argand diagram."

J.L. Synge in Relativity: The Special Theory (1956), p. 63

In my opinion, one way to make Minkowski's spacetime less counter-intuitive is to make more use of it [which few introductory textbooks do]... and to find better ways to explain it [as I am trying to do]... hopefully tying it to one's more familiar Euclidean-geometric and Galilean-kinematic intuition.