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If you are moving at $c$ in 3D space and $c$ in time axis too, What would be your total speed?

Edit: Since question has been voted to be closed, I shall make an Edit. In 4D world all objects move with speed $c$. This implies there should be a relation to compare speeds in two realms space and time.

Qmechanic
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Anubhav Goel
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    This "moving on the time axis" meme is not good physics. I know that it's all over the internet and on tv, but that is not what is happening. The world has three dimensions and there is a local clock attached to each observer. We are not "moving" trough some sort of four dimensional continuum with one global time variable in addition to three global spatial variables. That is exactly the opposite of what relativity is about. – CuriousOne May 23 '16 at 02:08
  • @CuriousOne Is this only stupidity then?, A photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any velocity component in the the fourth dimension, its velocity would be different from c, which is not the case. – Anubhav Goel May 23 '16 at 02:15
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    Yes, it's all lies to kids and we shall not repeat those here. A velocity is something that happens globally in affine space, while Minkowski space is really just a tangent space construct. Will a layman understand the difference? Unlikely, if they didn't understand the difference when it was taught to them for Euclidean space in high school while learning vectors and linear algebra. – CuriousOne May 23 '16 at 02:21
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    It isn't clear what you are asking. I'd guess you're asking about the norm of the four-velocity, which is always $c$. – John Rennie May 23 '16 at 06:19
  • @JohnRennie Yes, you are correct. The question was inspired from http://physics.stackexchange.com/q/257427/ – Anubhav Goel May 23 '16 at 06:29
  • Have a look at motion across space-time here http://goo.gl/fz4R0I – Sean May 24 '16 at 14:49

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The spacetime interval is a relativistic invariant, and is proportional to the travelers proper time. So in a since you are traveling one second per second, per your own wrist-watch. Every other measurement would be the speed of some other inertial reference system, measured with your clock.

Let $s^2 = x^2 + y^2 +z^2- (ct)^2$, where $x$, $y$, $z$ are coordinates in some inertial reference frame, and $t$ is the clock from that same frame. Then $s$ is the spacetime interval for the object that began at location $(0,0,0,0)$, and is now located at $(x,y,z,t)$. Perhaps you set off a firecracker at the beginning, and again at the end: two spacetime events. While different inertial observers will observe different coordinates and times, based on their relative velocity, they will all agree on $s^2$ for the spacetime interval between the two events.

You, the person setting off the firecrackers, can also measure $s^2$. Perhaps you are standing still -- then $s^2 = (ct)^2$ according to your wristwatch. Now calculate your speed: $ds/dt = c$, and in units where $c=1$, it is just one tick per tock.

Peter Diehr
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