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Please don't mark as duplicate. I am asking specifically for the 4vektor formula to see at speed c in space, I will get 0 for the time dim.

I was reading this question:

4 dimensional interpretation

specifically the #2 question

"If a body is at rest in the classical 3 dimension, it would travel through time at cc, but if traveling at cc in space, it would be resting in the "time" dimension.."

I understand the first part. But specifically my question is about this part:"but if traveling at cc in space, it would be resting in the "time" dimension"

And John Rennie's answer where he says

"As it happens, you are absolutely correct."

v⃗ =(c(dt/dτ),dx/dτ,dy/dτ,dz/dτ)

I understand that if I am at rest in space then I am still moving at speed c in time. The magnitude is always c. But I don't understand if I am moving at speed c in space, how can the four-velocity explain that I am at rest in time, so my time component c(dt/dτ) is 0?

Question:

  1. How can you explain with the four-velocity that when something is moving at speed $c$ in space, then it is at rest in time?
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Árpád Szendrei
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    Possible duplicate: http://physics.stackexchange.com/q/29082/50583 – ACuriousMind Oct 28 '16 at 20:20
  • This is backwards. As observers travel faster through space (i.e. covering more space per tick of their proper time), they travel faster through time, not slower. More time passes for every tick of their clock. – knzhou Oct 28 '16 at 20:23
  • I must disagree. this formula of the four-vektor implies that Δτ^2=Δt^2−1/c^2(Δx^2+Δy^2+Δz^2) if you move in space at speed c, Δt^2 will be equal to 1/c^2(Δx^2+Δy^2+Δz^2). So Δτ^2 will be 0? I was asking here for specifically this formula explanation if it is true? – Árpád Szendrei Oct 28 '16 at 20:32
  • Hi Árpád your questions are very interesting to me, as I need to revise the points you raise, but could you think of spending 30 minutes learning mathjax? It is easy and quick to learn and will make your questions stand out much better. The best way to learn it is open my answer to you on your last question, by using edit, and copy and paste the mathjax, then try them out in an answer box, (but DON'T change my answer:). You can find the instructions at http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference thank you. –  Oct 28 '16 at 21:07
  • How can you explain with the four-velocity that when something is moving at speed $c$ in space, then it is at rest in time - An entity with speed $c$ in an inertial reference frame (and thus speed $c$ in all inertial reference frames) has no four-velocity, i.e, the four-velocity is not defined. Note that, from an IRF, the entity with speed $c$ has a world line and thus, 'moves' through the 'time dimension'. However, since the world line is light-like (null), the change in proper time, $d\tau$, along the world-line is zero. – Alfred Centauri Oct 29 '16 at 01:24

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