Wikipedia: using the four-velocity to understand time dilation (?)

[nonequilibrium]

Hello,

I was wondering if the following quote by Wikipedia (http://en.wikipedia.org/wiki/Four-velocity section "Interpretation") makes sense:

In other words, the norm or magnitude of the four-velocity is always exactly equal to the speed of light. Thus all objects can be thought of as moving through spacetime at the speed of light. This provides a way of understanding time-dilation: as an object like a rocket accelerates from our perspective, it moves faster through space, but slower through time in order to keep the four-velocity constant. Thus to an observer, a clock on the rocket moves slower, as do the clocks in any reference frame that is not comoving with them. Light itself provides a special case- all of its motion is through space, so it does not have any "left over" four-velocity to move through time. Therefore light, and anything else traveling at light speed, does not experience the "flow" of time.

Cause I would think it doesn't make sense: in this reasoning they're acting as if the norm is something like a^2+b^2+c^2+d^2, so that if b^2+c^2+d^2 is big ("rocket [...] moves faster through space"), then to keep the norm constant, a^2^ should be smaller ("slower through time in order to keep the four-velocity constant"). But of course that is not the structure of the norm, so I don't think this reasoning works out. It would even lead to an opposing answer: due to the minus sign in the norm, the "speed through time" should increase along with the spatial speed! (indeed reflected in \eta^0 = \frac{c}{\sqrt{1-u^2/c^2^}}).

I'm not sayig the four velocity is contradicting time dilation, I'm just trying to argue that their reasoning makes no sense. Or if it does, I have something new to learn, so please correct me!

 

[Bill_K]

mr vodka, You're right! But then again you're wrong. In order to maintain the norm of the velocity vector, both the time component and space component must increase. For a particle at rest, V⁰ = c and for a particle that's moving, V⁰ = c γ > c. But since also V⁰ = c dt/dτ where τ is proper time, that means that proper time does run *slower* than coordinate time.

 

[nonequilibrium]

Hm, I understand everything you say, but then how would you support their claim:

In other words, the norm or magnitude of the four-velocity is always exactly equal to the speed of light. Thus all objects can be thought of as moving through spacetime at the speed of light.

I don't see the logical "thus"

 

[Fredrik]

I really dislike the statement you just highlighted. It's been brought up many times before, so I'll just quote myself.

Brian Greene used it in the "The elegant universe". I think that explanation is really bad. I wrote some comments about it in another thread recently. These are the relevant posts: 64, 65. The words "earlier in this thread" in the second one refers to 17.

Edit: Those posts also cover a few things you didn't ask about, so I'll answer the question directly. The "thus" is nonsense. The statement "all objects can be thought of as moving through spacetime at the speed of light" is true because we have chosen to define the "speed through spacetime" as the magnitude of the four-velocity, which by definition of "four-velocity" is always 1. So the highlighted statement really says "thus, for any object, the normalized tangent vector of the world line has length 1". (Since "normalized" implies "length 1", the statement is vacuously true).

(I always use units such that c=1).

 

[nonequilibrium]

Okay, that sounds good :) thanks