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This question already has an answer here:

Well, it seems to me that if I move faster in space I move slower in the dimension of time which is orthogonal to the dimension of space.

All speeds are then equal. Is this statement correct?

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marked as duplicate by Yashas, ZeroTheHero, Jon Custer, Qmechanic Mar 29 '17 at 15:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For the velocity four-vector $\eta^{\mu}=\frac{dx^{\mu}}{d\tau}$ [where $d\tau=\frac{dt}{\gamma}$ stands for the infinitesimal proper time, and $x^{\mu}=(ct,x,y,z)$; you can define $\eta^{\mu}$ also as $\frac{dx^{\mu}}{dt}$; both have their advantages and I'm not sure which one is the commonest used but I use the first definition here] we have:

$\eta^{\mu}=\gamma(c,\vec v)$, with $\gamma=\frac 1 {\sqrt{{1-\frac {v^2} {c^2}}}}$.

So even if we stand still ($\vec v=0$, with respect to the Universe) we move with velocity (speed) $c$, which is not a vector (its the time component of the velocity four-vector, but nevertheless, its "direction" is the future).

Now $\eta_{\mu} \eta^{\mu}=c^2$ (you can calculate this yourself if you're familiar with contra- and covariant four-vectors), so the value of the velocity four-vector is always $c$.

If we stand still (but "moving" with speed $c$ through time) and start to move, a lesser portion of ${\eta}^{\mu}$ is moving through time and a $\vec v$ through space, which can have every direction, develops. The faster we move the lesser is the portion of $\eta^{\mu}$ moving through time and the bigger the portion moving through space. If we (almost) reach the velocity (through space) $\vec c$, we are (almost) only moving through space and (almost) no longer through time (as observed by someone who stays behind where we started moving). The photon is a good example of this. It moves through space with velocity $\vec c$, for all observers, while its velocity (speed) through time is zero for all observers (for a photon there is no time ticking away).

Thus every object is or moving with speed $c$ through time (directed to the future) when standing still, as strange as this may seem, or moving with speed ${\gamma}c$ through time and velocity $\gamma{\vec v}$ through space (in whatever direction) with $\eta_{\mu}\eta^{\mu}=c^2$, or with speed $0$ through time and velocity $\vec c$ through space (in whatever direction).

Maybe you can see a difficulty arising when $v=c$ because in that case $\gamma\rightarrow\infty$, giving a problem for $\eta^{\mu}=\gamma(c,\vec{v})$, but $\eta_{\mu}\eta^{\mu}$ always equals $c^2$ (actually it's a bit more complicated than that; in the linked article you can read this is connected to an affine parameter, but let's not go into thát before it's gétting too complicated!).

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    $\begingroup$ +1 It sounds as though your answer is what the OP wanted, but it might be good to point out that not everyone would interpret "speed" in this way (as in John's answer), although of course your reasoning is quite correct. Just to make the OP aware that they will come across the two interpretations (although I thing co-ordinate velocity is likely the commoner one). $\endgroup$ – WetSavannaAnimal Mar 29 '17 at 11:27
  • $\begingroup$ You're absolutely right! It's a sign that most people (including me) think in the Newtonian way when talking about speed (without thinking about the speed through time). Nobody says that the speed of their car is c when the car is standing in front of their house. $\endgroup$ – descheleschilder Mar 29 '17 at 11:35
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    $\begingroup$ Do people really sometimes use the word "speed" to mean the norm of the velocity four-letter? That's certainly not the most common meaning of the word. $\endgroup$ – Tanner Swett Mar 29 '17 at 12:19
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The norm of the four velocity is always $c$, but of course not all four velocities are equal because they can point in different directions and vectors that point in different directions are not identical.

But to claim the word speed means the norm of the four-velocity seems unjustified. By speed we normally mean (the magnitude of) the coordinate velocity, and the coordinate velocity is the derivative of position in our coordinate system with time in our coordinate system. This can of course have any magnitude from zero up to the speed of light.

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My answer incorporates features of the earlier answers, but tries to clearly disentangle the implied uses of "speed" in space and in spacetime [when a phrase like 'slower in the dimension of time' is used].

"Massive objects (like a basketball) all move at constant 'speed' c in spacetime" really means that

  • its 4-momentum vector can be normalized to a "unit" 4-velocity vector with Minkowski-norm c.
  • It has a spatial speed (slope) v < c because the spatial component of its 4-momentum has an absolute-size that is smaller than that of the time-component of its 4-momentum.

"Massless objects (like a light-signal) all move at constant 'speed' zero in spacetime" really means that

  • its 4-momentum vector has Minkowski-norm zero [and thus can't be normalized].

  • It has a spatial speed (slope) c because the spatial component of its 4-momentum has the same absolute-size as that of the time-component of its 4-momentum.

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