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I'm just starting to learn about relativity, and I just can't wrap my head around this one question. I must be imagining something incorrectly. My understanding is that everything is traveling at "the speed of light," be it through space or time. So if light is traveling along the space axis alone, it should appear to someone only moving through time to appear for one moment and be gone the next.

I have some guesses about what I'm imagining wrong, but I hope someone can help me get a better handle on it. I'm realizing that in my conception, light would experience everything else as totally frozen in time, but there would be no "space compression" like I know is supposed to exist.

I assume that the picture in my mind is incorrect because I'm making the speed of light infinite, but I don't have a feel for what the picture would look like if this were not true.

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  • $\begingroup$ You are exactly right: If light were traveling along your space axis, it would appear to you to appear for one moment and be gone the next. But that's not what actually happens. Conclusion: Light cannot travel along your space axis. $\endgroup$ – WillO Apr 21 '17 at 17:59
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A good way of understanding what is going on in special relativity is using the four-vector notation. An event in space-time is described by a four-vector $(ct,x)$ where $t$ is the time and x is the position of the event.

If we consider a moving particle, we can write its trajectory as $(ct,x(t))$ where now the position $x$ depends on the time $t$, so for each $t$ we get an event in space-time.

If a particle is not moving, its spacial coordinate is constant: $x=x_0$. We then call its time coordinate the eigentime $\tau$. The trajectory then reads $(c\tau,x_0)$.

The four-velocity $u$ of a particle is defined as the derivative of the trajectory with respect to the eigentime. For a particle that is not moving, the four-velocity is $(c,0)$. So it is not moving through space but moving along the time axis with velocity $c$.

If a particle is moving, then: $$u=\tfrac{d}{d\tau}(ct, x(t))=(c\tfrac{dt}{d\tau},\tfrac{dx(t)}{dt}\tfrac{dt}{d\tau} )=\tfrac{dt}{d\tau}(c,v)$$ where $v=dx/dt$ is the usual three-velocity.

One can show that $\tfrac{dt}{d\tau}=\gamma=1/\sqrt{1-v^2/c^2}$ (ask if you want to see the proof). Therefore, $u=\gamma(c,v)$.

As you might know, distances in space-time are calculated using the Minkowski metric. This also holds for any four-vector. So we can calculate the absolute value of the four-velocity:

$$|u|^2=u_0^2-u_1^2=(\gamma c)^2-(\gamma v)^2=\gamma^2(c^2-v^2)=c^2$$

So, for any $v$, the four-velocity is $|u|=c$.

What happens if we choose $v=c$? It still holds that $|u|=c$. The components of $u$ then are $u=\gamma(c,c)$. So light travels through time and space with the same velocity. However, for $v=c$ we have $\gamma=\infty$. So the two (equal) components of $u$ are both infinite.

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  • $\begingroup$ Ahh got it thank you. I think I solved what was going wrong, but it's just introducing a lot more questions. I'll post the questions that remain after I think about it some more. $\endgroup$ – Jeff Bass Apr 21 '17 at 19:25

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