# Time as a fourth dimension

1. Aug 18, 2010

### maxverywell

In Minkowski 4-D spacetime the cooridnates of an point are (ct,x,y,z).
My question is why we multiply time with speed of light c and not some other speed v (speed of object moving in three spatial dimantions)? If we treat the time as a fourth dimension it would be more reasonable to say that (t,x,y,z) and not (ct,x,y,z).

2. Aug 18, 2010

### Fredrik

Staff Emeritus
Because -c2t2+x2+y2+z2 is invariant under Lorentz tranformations (functions that change coordinates from one inertial frame to another), and -v2t2+x2+y2+z2 isn't for any v≠c.

3. Aug 18, 2010

### maxverywell

It's a spacetime interval?
Ok, i know it, but what is the physical meaning of $ct$? Can we say that all objects in the real world are moving in a four dimensional space time at a constant velocity as that of light? So objects which are at rest are moving only in the time dimension?

4. Aug 18, 2010

### JesseM

I don't understand, the coordinates of a point are just t,x,y,z, it's only when calculating the spacetime interval that you must use ct, i.e. the square of the interval is giving by ds^2 = x^2 + y^2 + z^2 - (ct)^2

5. Aug 18, 2010

### Staff: Mentor

The only purpose of ct is to make the units dimensionally consistent. In the end, it doesn't really matter, we can put the factors of c into the coordinates or into the metric:

$$r=(ct,x,y,z)$$ and $$g=\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

or

$$r=(t,x,y,z)$$ and $$g=\left( \begin{array}{cccc} -c & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

As long as you are consistent you will get the correct result either way. I first learned the former, but I now prefer the latter. And, of course, they are the same in units where c=1.

6. Aug 18, 2010

### bcrowell

Staff Emeritus
On topographical maps in the U.S., they typically use miles for the horizontal coordinates x and y, and feet for vertical coordinate z. If you were going to start doing a whole bunch of three-dimensional calculations with vectors based on data from such a map, you definitely wouldn't want to use vectors that looked like (x in miles,y in miles,z in feet). You would want to get everything in the same units. If you liked, you could write this is (x,y,z/5280). Someone could then object and say, "Hey, z is just the third dimension. Why are we treating it differently than x and y, and throwing in a factor that only applies to z?" Well, the extra factor of 1/5280 is actually there so that you can treat z the *same* as x and y. It's just that we initially made a poor choice by using different units to define the symbols x, y, and z. We should have just defined x, y, and z all in the same units, and then we wouldn't need the extra factor.

Same thing with c. In relativity, it makes more sense to measure x, y, z, and t all in the same units, and then you *do* simply write the four-vector as (t,x,y,z). The extra factor of c is only needed if you inappropriately choose different units for t than for x, y, and z. In these units, c=1, so you never need to write it.

7. Aug 18, 2010

### Halcyon-on

We physicists like to compare apples with apples and bananas with bananas. When you write (t,x,y,z) you are putting apples (an object with dimensionality of a time) with bananas (an object with dimensionality of a length). This prevents the following nonsense: t + x (apple + banana = ?).

Last edited: Aug 18, 2010
8. Aug 18, 2010

### JesseM

The only issue with this analogy is that once you choose the same units for z, you see that z really is just another spatial dimension, and distances can by calculated with the simple pythagorean formula $$ds = \sqrt{dx^2 + dy^2 + dz^2}$$, whereas even if you choose units where c=1, you still find that time is treated differently in the metric than the other dimensions, i.e. $$ds = \sqrt{dx^2 + dy^2 + dz^2 - dt^2}$$ (and one consequence of this is that geodesics in spacetime maximize the absolute value of ds^2, as illustrated by the twin paradox where the inertial twin ages more than the twin who takes a non-inertial/non-geodesic path, whereas geodesics in space are the paths which minimize the distance between points). So although time is a fourth dimension in relativity, it is not equivalent to a fourth spatial dimension.

9. Aug 18, 2010

### bcrowell

Staff Emeritus
Sure, there are two completely separate issues. The factor of c is a notational issue that can be eliminated. The fact that the metric is not ++++ is a real physics fact with observable consequences.

I wouldn't completely agree that "time is treated differently in the metric than the other dimensions." Under a Lorentz boost, time and space mix together, and therefore there isn't really a separate dimension called time. I think you and I agree on what we mean when we talk about the signature of the metric not being ++++, but this thread is partly about how pop-sci sayings like "time is the fourth dimension" can be misconstrued, and I think "time is treated differently" is also a statement that can be misconstrued.

10. Aug 18, 2010

### JesseM

I don't understand that argument--what would it mean for there to "really" be "a separate dimension called time"? Would you say that there isn't a separate dimension called time in Newtonian physics either, since time and space "mix together" in the sense that the space coordinate in one inertial frame is a function of both the space and time coordinates in another, according to the Galilei transformation? (though unlike with the Lorentz transformation, under the Galilei transformation the time coordinate in one frame does not depend on the space coordinate in another frame, only on the time coordinate in the other frame)

In any case, you can define the separateness of time in a coordinate-independent way--for any pair of points in spacetime, the separation between them is either time-like, space-like or light-like. With spatial dimensions, there's no way you can say the separation between points is "x-like" or "y-like" or "z-like" in a coordinate-independent way.
In what specific way do you think people would misconstrue it? (particularly given my accompanying explanation for what I meant by this in post #8)

Last edited: Aug 18, 2010
11. Aug 18, 2010

### yuiop

Can we not say the time dimension is different to the spatial dimensions because outside a Schwarzschild event horizon we are free to move forwards or backwards spatially but we are constrained to move only forward in time. This makes time special. Below the event horizon, we are constrained to move in only one radial direction (towards the singularity) but we are free to move backwards and forwards in time. Viewed like this there is a clear distinction between the time and space dimensions. Outside the event horizon, time is the unique dimension that you do have a choice about the direction you move in.

12. Aug 18, 2010

### JesseM

That's just because Schwarzschild coordinates behave "badly", with the Schwarzschild t coordinate becoming spacelike inside the horizon, and the r coordinate becoming timelike. Kruskal-Szekeres coordinates don't have this problem, for example. You'd be free to come up with a "bad" non-inertial coordinate system in ordinary flat SR spacetime where, beyond some arbitrary "horizon", a coordinate that was formerly spacelike becomes timelike, and a coordinate that was formerly timelike becomes spacelike, just like what happens with Schwarzschild coordinates at the event horizon of a Schwarzschild black hole.

Last edited: Aug 18, 2010
13. Aug 18, 2010

### DrGreg

Below the event horizon, we are constrained to move in only one radial direction (towards the singularity) but we are free to move backwards and forwards in the t coordinate, but the t coordinate is space not time. The r coordinate is time.

14. Aug 18, 2010

### yuiop

O.K. by this definition, time is the unique dimension?/coordinate? that constrains us to move in one direction, above or below the event horizon. That still makes time special and different to space.

15. Aug 19, 2010

### Fredrik

Staff Emeritus
The description of motion that you're asking about in the first sentence has been brought up here many times, because Brian Greene used it in the "The elegant universe". I think that explanation is really bad. I wrote some comments about it in another thread recently. These are the relevant posts: 64, 65. The words "earlier in this thread" in the second one refers to 17.

Your second sentence is fine. Just keep in mind that no object is objectively at rest, and that all objects can be said to be at rest in some coordinate system. Such a coordinate system maps the curve that represent's the object's motion to a curve that's parallel to its time axis, so your second sentence is true by definition of "coordinate system" and "at rest".