# Moving throuh Spacetime

1. Jun 28, 2007

### Futobingoro

I have a quick question:

Does the arrangement of spacetime directly dictate rates like velocity?

For example, if it takes a man 10 minutes to run 2 kilometers, does that mean that there were 10 minutes of time and 2 kilometers of space between his starting and ending position?

2. Jun 28, 2007

### StatusX

No, different observers partition his trip into different amounts of space travel and time travel. The underlying principle is that, however you do this, the combination:

$$(c \Delta t)^2 - (\Delta x)^2$$

is the same for all (inertial) observers. So, for example, someone might be moving relative to the runner so that he appears to be moving faster and so cover more distance. That is, $\Delta x$ is larger that 10 km for them. But then in order to maintain the value of the above expression, $\Delta t$ would also have to increase, ie, they would observe the runner's watch to be running slow.

Last edited: Jun 28, 2007
3. Jun 28, 2007

### neutrino

Not really. But the spacetime interval between the two events (the start and finish of the run) is the same in all frames. The space-part and time-part will differ based on who is making the measurements.

EDIT:Yes, what StatusX said. ;-)

4. Jun 28, 2007

### robphy

(minor typo corrected)

5. Jun 28, 2007

### pervect

Staff Emeritus
If the 2 km course is marked out by a stationary observer, and the time it takes the man to run the 2 km is measured by a watch that he carries around with him, the ratio distance / time is known as a "rapidity" rather than a velocity.

The difference between rapidity and velocity is that the time for the man to run the distance would be measured by clocks in the lab frame. Typically one would have one clock at the start of the course, one clock at the finish of the course, and the issue of synchronizing clocks arises. This could get rather detailed, so I'll just skip ahead a lot and say that the runner's time measured on the clock he carries with him according to relativity be shorter than the time reading measured in the lab frame that goes into the velocity equation.

6. Jun 28, 2007

### robphy

The standard definition of "rapidity" is that is the Minkowski-angle $$\theta$$ (the spacelike-arclength along the Minkowski-unit-circle), between two future unit-timelike vectors (along the "t"-axis of the stationary inertial observer and the "s"-axis of the inertial runner). Its hyperbolic-tangent yields the ordinary [spatial-]velocity $$v=c\tanh\theta$$ (the ratio of [instantaneous] spatial-displacement and temporal-displacement, i.e. ["opposite over adjacent"]).

Maybe the term you mean is "celerity" (coined by Levy-Leblond? see http://arxiv.org/abs/physics/0608040 )
which is "the spatial-part of the 4-velocity $$v^a$$"
$$\gamma v=(\cosh\theta) (c\tanh\theta)= c\sinh\theta$$, i.e. ["opposite over hypotenuse"] . (A spacetime diagram would probably clarify this.)

7. Jun 28, 2007

### pervect

Staff Emeritus
You're right, I double-checked with
http://www.eftaylor.com/pub/spacetime/STP1stEdP41to59.pdf

A good reference, which I'd recommend to the OP, and some calculations show me the difference is important. Sorry for any confusion.

A celerity of 2 (covering 2 light years (lab frame) in 1 year (proper time)) is equivalent to a rapidity of 1.4436 via the Taylor & Wheeler definition, and a velocity of .8944 c, i.e. 2 light years (lab frame) in 2.236 years (lab time).

8. Jun 28, 2007

### robphy

Here is the Levy-Leblond article: