# The usage of ct instead of t

1. Apr 1, 2008

### NanakiXIII

I'm looking for an explanation as to why a lot of results of Special Relativity are described using $$ct$$ as a fourth dimension instead of just $$t$$. Now, I understand that using $$ct$$ in a Minkowski diagram with identically scaled axes will cause worldlines of light to angle at a nice 45 degrees, and I've seen how $$x-ct$$, if I remember correctly, is Lorentz invariant, as well as some other cases where it does seem useful to use ct, but the general nature of this usefulness eludes me. Could anyone tell me more about this?

2. Apr 1, 2008

### MeJennifer

Each component in the decomposition of spacetime in "space" and "time" components must have an identical dimensionality. Both the "space" and "time" components use a measure of "distance". Hence: ct = m/s * s = m

By the way the decomposition in "space" and "time" is observer dependent in relativity.

Last edited: Apr 1, 2008
3. Apr 1, 2008

### Staff: Mentor

Also note that with a suitable choice of units, you can use simply t instead of ct in calculations. For example, measure time in seconds and distance in light-seconds (the distance light travels in one second).

4. Apr 1, 2008

### Staff: Mentor

Yeah, but (ct,x,y,z) is easier to write and read than (t,x/c,y/c,z/c)

5. Apr 1, 2008

### MeJennifer

Interestingly, in a way we already do that since the meter is defined as the length of the path traveled by light in vacuum during a time interval of 1/299,792,458 of a second. :)

6. Apr 1, 2008

### robphy

Conceptually, I prefer the latter.
In some sense, measuring a time seems more fundamental than measuring a length. (Radar measurements use a clock.)
In addition, the latter form lends itself more easily to the analyzing the Galilean limit.

7. Apr 3, 2008

### Antenna Guy

Why?

I don't see the (general) need for any such requirement - but scaling time by c does provide a convenient notation within Minkowski space.

Regards,

Bill

8. Apr 3, 2008

### MeJennifer

One would not be able to construct a Minkowski metric or more advanced metrics in curved spacetimes when the dimensions do not match.

Time is not scaled by c, since both dimensions are different, e.g. s * m/s.

9. Apr 3, 2008

### shoehorn

Eh? I think you've got that precisely the wrong way round, imo. I can't think of a single example of how one could measure an interval of time without *first* having to measure a spatial length.

10. Apr 3, 2008

### robphy

Refer to "The Interval" chapter (around p.70) in Geroch's General Relativity from A to B
(see p.72)

The key to a clock is a periodicity.
To perform radar measurements, one makes use of light rays and clocks.

11. Apr 4, 2008

### shoehorn

Care to give me an example of a measurement of an interval of time that can be made without measuring a spatial interval then?

12. Apr 4, 2008

### robphy

In a light-clock, one can declare a given separation of mirrors [without "measuring it"] to define a standard tick of this clock. (Allow two mirrors to travel inertially in the same direction. Changing the separation effectively changes the resolution of the clock.) Again, the key is [regular] periodicity of clock. Instead of the light-clock, one could crudely use your heartbeat to measure time [a la Galileo].

The point is that a "wristwatch" time measurement, unlike a "meterstick" measurement that must be extended out, is local to the observer's worldline. This has conceptual advantages in relativity.

From a more practical point of view, distances to remote objects are not done by extending a ruler out to meet it. Radar using light-rays is often used.

13. Apr 4, 2008

### shalayka

As strange as it sounds... If t was used instead of ct, then the maximum velocity would be 1 instead of c. Of course, I think this is actually done when the speed of light is arbitrarily set to 1 in geometrized units. This follows from the fact that the magnitude of the 4-velocity vector must equal c (or 1) at all times. Where v = 0, t = 1, and so the entire magnitude is due to motion through time (ct = c). At v = c, t = 0, and so the entire magnitude is due to motion through space (ct = 0).

Last edited: Apr 4, 2008
14. Apr 4, 2008

### phyti

Watch a flashing light.
Record the time of flash one.
Record the time of flash two.
You just recorded a time interval with no spatial measurements.

This also gets to the essence of time measurement.
It's matching world events to clock (reference) events.

Like matching an object to a ruler to measure length.

15. Apr 4, 2008

### Antenna Guy

I think the duration of a photon (single complete wave [clarification: passing through a plane of reference]) would constitute both a time and space interval measured by a clock. f=1/t, l=c*t

Conversely, the spatial measurement of one period of a standing wave (segment of a continuous stream of photons) can be used to calculate t.

Regards,

Bill

Last edited: Apr 4, 2008