# With what velocity are we moving along the time dimension?

Does the question make sense? Velocity along time axis means $$v_t=\mathrm dt/\mathrm dt$$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity? Extension of the question is welcomed.

• I've deleted some off-topic or obsolete comments. Everyone please keep in mind that comments are meant for requesting clarification or suggesting improvements to their parent post. – David Z Nov 2 '18 at 8:43
• @DavidZ, two of the comments deleted were precisely a request for clarification (from me) and a helpful answer from Krishna. Can they be restored? – Alfred Centauri Nov 2 '18 at 13:13
• @AlfredCentauri Those were among the obsolete comments I mentioned. It looked like Krishna had already seen your request for clarification and made any edits they would like to make in response to it. Krishna, is that not the case? If you would like to make additional edits to the question to address Alfred's request for clarification, I can bring those comments back for a couple days to give you a chance to do so. – David Z Nov 2 '18 at 19:24
• @DavidZ, the clarification request, IIRC, was to ask whether the question goes to the concept of a temporal velocity in general or whether the question is specifically what is the value of one's temporal velocity is in one's rest frame. I believe the answer was the something like concept of temporal velocity in general with any extensions welcome. Maybe it's all fine just as it is; the debate in the comments regarding this has essentially finished. – Alfred Centauri Nov 2 '18 at 20:08

In non-relativistic mechanics, time $$t$$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $$x(t),y(t),z(t)$$ of this universal parameter $$t$$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $$t$$:

$$\mathbf{v} = \frac{dx}{dt}\hat{\mathbf{x}} + \frac{dy}{dt}\hat{\mathbf{y}} + \frac{dz}{dt}\hat{\mathbf{z}}$$

However, in relativistic mechanics (SR for simplicity), time $$t$$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $$\tau$$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').

The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $$t(\tau),x(\tau),y(\tau),z(\tau)$$ of the particle's proper time $$\tau$$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $$\tau$$:

$$\vec{U} = c\frac{dt}{d\tau}\hat{\mathbf{t}} + \frac{dx}{d\tau}\hat{\mathbf{x}} + \frac{dy}{d\tau}\hat{\mathbf{y}} + \frac{dz}{d\tau}\hat{\mathbf{z}}$$

So, in this coordinate system, the component of the particle's four-velocity in the time direction is

$$U^0 = c\frac{dt}{d\tau}$$

Now, it can be shown that (time dilation)

$$dt = \gamma_v d\tau$$

where

$$\gamma_v \equiv \left(1 - \frac{v^2}{c^2}\right)^{-1/2}$$

and

$$v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}$$

thus

$$U^0 = c\gamma_v$$

This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.

(note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)

• This assumes that we somehow observe ourselves from some other reference frame than the one we are moving in, which is paradoxical. – Ruslan Oct 31 '18 at 19:58
• @Ruslan, this answer makes no such assumption. The exposition refers to a particle that is observed to have relative motion. And as always, when speaking of one's velocity, there is the implied relative to whom?. – Alfred Centauri Oct 31 '18 at 21:00
• @Ruslan Note that this formula also holds for v = 0 (i.e. the rest frame). In that case we get $U^0 = c$. – Graipher Oct 31 '18 at 22:50
• @elliotsvensson, I have no interest in a debate in my comment thread. What is your intention? – Alfred Centauri Nov 1 '18 at 20:26
• Woohoo! I thought my week couldn't get any better but I wake to find that I'm now a Relativity Cleric! Extra cream in the coffee please and donuts for all! – Alfred Centauri Nov 2 '18 at 11:51

There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $$c=1$$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $$+---$$ signature). All of this is coordinate-independent.

If we now specialize to Minkowski coordinates $$(t,x,y,z)$$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $$\tau$$ (not coordinate time $$t$$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $$\gamma$$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $$\gamma$$.

In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $$\gamma=1$$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.

• This assver is incorrect, because it involves the time dilation $\gamma$ in the proper frame self evidently implied in the question (by "we moving"). As such the answer does not correctly emphasize that the local speed of light is always the same. On the conventions, the answer completely ignores the dimensional analysis that ultimately the real life meaning of $c=1$ is one light second per second. Thus the answer wrongly implies that there is no difference between time and space. Finally, using geometrized units here is as unhelpful as suggesting to meet at one billion meters o'clock. – safesphere Oct 31 '18 at 17:38
• @safesphere meters is an element of a vector space, o'clock times are a point in an affine space. I see no such category error in the above answer. "We should meet at 1 billion meters after noon" is closer to what the above answer does. I agree that 1 billion meters after noon is unhelpful, because that is just noon with rounding error (3 seconds). A trillion meters is more reasonable (about an hour). – Yakk Nov 1 '18 at 18:02

At exactly 1 second per second.

• One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity. – safesphere Oct 31 '18 at 15:52
• safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this? – Krishna Deshmukh Oct 31 '18 at 16:09
• @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $\gamma$ as some answers here suggest. In other words, our own $\gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $\gamma=1$ with the speed of $c$. – safesphere Oct 31 '18 at 16:57
• How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant. – Krishna Deshmukh Oct 31 '18 at 17:06
• @KrishnaDeshmukh Please use the @ address. Otherwise people are not notified of your responces. Thanks! The "rate" here is the rate of time moving, not the rate of speed changing. As a simple visual example, if you look at at the second hand of a watch, it's rate of ticking is one tick per secong while it's speed of moving around the dial is always constant. – safesphere Oct 31 '18 at 18:10

In relativity the time coordinate is $$x_o=ct$$ and its time derivative (in the rest frame) is $$c$$. Therefore the time component of four-velocity is the speed of light in vacuum.

• In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame. – Ben Crowell Oct 31 '18 at 15:51
• @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame. – safesphere Oct 31 '18 at 16:07
• @K7PEH "Proper" is self evidently implied in the question by "are we moving". It is not a specialist terminology question, but the meaning is clear. My answer simply answers the question asked. I didn't expect that such an obvious point could be unclear to anyone. – safesphere Oct 31 '18 at 17:06
• @K7PEH Sorry, I do not judge the level of understanding of people I don't know. Also, knowing the technical definition of "proper" is not required for someone talking about his own time. The meaning is still the same. To avoid this confusion I have added the rest frame to the answer. Thanks! – safesphere Oct 31 '18 at 17:49
• I am sorry guys, but if the rest frame is implied then @safesphere is correct, you all have to agree on that. Now, was it implied? This OP just wants know with what speed is HE moving with regard to time coordinate, not what does some other observers see.Why would he care about that? Now, answering this question with rigor like Alfred Centauri did, misses this point although it is a very nice answer. At least we can agree that both answers are correct, but safespheres is more to the point. – Žarko Tomičić Oct 31 '18 at 18:53