# Is time a fourth dimension?

1. Oct 17, 2004

### ramcg1

Spacetime raises many questions.

Three-dimensional space is easy to visualise. We live in it. We can denote orthogonal axes as x, y and z and have oodles of fun with three-dimensional geometry.

One wonders, is there really a fourth dimension and is that dimension time?

Newtonian physics has a time parameter. All of space exists at the same unique point in time and moves, as a unit, forward through time at a constant rate.

Spacetime tells us that that an event occurs at a unique point in spacetime with a set of coordinates such as (t, x, y, z). From there we can we can use a lot of advanced mathematics to have oodles of fun with four-dimensional space.

Being inquisitive and sceptical by nature, I ask myself what this fourth dimension "t" really represents.

When I boil down the introductory explanations to spacetime, I find that the "t" represents how long it takes electromagnetic radiation (light) to travel from one spacetime event to another. This velocity is taken because it leaves the spacetime interval invariant under changes of coordinates. Thus, one unit along the "t" axis represents one second of flight time for electromagnetic radiation. That's it, end of story, now get on with the mathematics of spacetime.

Take an imaginary empty sphere with a shell that emits no electromagnetic radiation. Inside the sphere, there is nothing, no matter and no electromagnetic radiation. This sphere has volume so it has the three Cartesian dimensions. Allow the outer surface of the sphere to reflect light. I can see the sphere and it exists in the linear Newtonian time that I am used to. If there has been and never will be any electromagnetic radiation within the sphere, how can I assign the speed of light to the conversion factor between space and time? Does spacetime have any meaning if there is no electromagnetic radiation? Should I now assign zero the conversion c reducing four-dimensions to three-dimension? However, a division by zero would render the spacetime transformations meaningless.

To be able to have flight times (where the velocity is c - the speed of light) between imaginary clocks there must be electromagnetic radiation. If there is no electromagnetic radiation then there are no flight times (where the velocity is c) between imaginary clocks.

All electromagnetic radiation must have a source. From a point source, an electromagnetic wave front would be spherical. Does spacetime have any meaning before the formation of the electromagnetic radiation? Is negative time only relevant to the current position of an existing wavefront or photon?

When I turn on my bedroom lamp (assume it a point source that illuminates in all directions at once) the wave front of the light is a sphere moving a velocity c. I can represent the position of any photon on that sphere by the four coordinates (t, x, y, z) where t is the time of flight. Things to note:
• This t has no meaning before the lamp is turned on
• Before the lamp was turned on the light had no existence and no velocity

My neighbour can turn on his bedroom lamp and achieve the same result.
I can extend this to an infinite number of neighbours and for every one of these point sources I can either have an expanding sphere of light or nothing.

Now normally turning on and off point sources would be a random event and each point source would be independent from all the others. I could allocate these events to positions on a linear time scale. I could explore the maths of flight times between positions on the expanding spheres and between expanding spheres, but that is not a good model of spacetime. I would have positions in my four-dimensional space that do not have expanding spheres of light.

Alternatively, I could switch them all on. Now I have an infinite number of expanding spheres. I can measure the distances between two point sources by the linear three-dimensional distance between them and by the time, it would take the light to reach the other point source. Thus, I can construct a four-dimensional vector for light travelling from one point source to the anther point source.

Now that I have my four-dimensional vector, a model of spacetime, I can go on to play with it mathematically and do all the wondrous things of spacetime. Does this mean that Minkowski space is a model of the surface area of an infinite number of electromagnetic expanding wave fronts originating from an infinite number of point sources?

Not a good model of time as a fourth dimension though, is it, not in the same ilk as the Cartesian dimensions, quite Newtonian actually.

2. Oct 17, 2004

### rcgldr

Since you can't travel backwards in time, it's not considered a dimension. You should do a web search on Steven Hawkins ideas on this.

3. Oct 18, 2004

### HallsofIvy

It would help if you stated what definition of "dimension" you are using. Generally, the number of "dimensions" simply tells how many numbers are required to specify one "point" or one instance of whatever it is you are talking about.

Physics deals with "events"- things that happen at a specific point at a specific time. You need 3 numbers to specify the point and one number to specify the time: 4 numbers= 4 dimensions. That was what Albert Einstein meant when he said "a 4-dimensional space-time continuum".

4. Oct 18, 2004

### spacetime

discrimination/special treatment for time

Imagine people of a world who can visualize only two dimensions similar to the way we human beings are "comfortable" with three dimensions. But they are living in the same world as ours.

Albert Einstein(II) living there is looking at a square block. He moves a bit and still can see just the two dimensions. So what he sees is that the dimensions of the block have changed because of his motion. You can easily visualize that will happen. Whoa, he says. When I move, dimensions of bodies which are not in my frame change. He conjectures that the world is possibly three dimensional. ( the way Albert Einstein (I) told us that the world is 4-D)

Beleive me here. I cannot feel what he feels for that third dimension. You cannot do that. But I am sure he doesn't feel the way we do. To us its all the same for the three dimensions. He will have a different idea of the "third dimension". You can get an idea of that form the way you and I feel for time - which is no different.

How do you think about the three dimensions. They are so similar. They are just the same. I hope someday you and I are able to feel the same way for all the four dimensions.

spacetime ( I really like my username )
www.geocities.com/physics_all/index.html

5. Oct 18, 2004

### pervect

Staff Emeritus
One of the reasons for calling time "the fourth dimension" is that time and space intermix under the Lorentz transform. Using units where c=1

x' = gamma*(x-v*t) t' = gamma*(t-v*x)

where gamma = 1/sqrt(1-v^2)

An interval that is purely spacelike or purely timelike for one observer is a mixture of both for a second observer.

6. Oct 18, 2004

### MiGUi

When Einstein and Lorentz worked out the special relativity, Minkowski tried to give the equations more treatable mathematically, and that manner was creating a 4D space where the coordinates of a vector where (x, y, z, ict) and its geometry was not like our 3d common space.

7. Oct 18, 2004

### ramcg1

Sorry People,

I forgot how literal we can be and that most of us would not read such a long posting.

The thread title “Is time a fourth dimension?” was meant to be rhetorical.

My real issue is this.

If I am to accept Spacetime and all that follows from it. I need to have a clear understanding of the variables that I am using. I have no problem with the normal Cartesian dimensions but quite some difficulty determining what the time variable (dimension, axis) really represents, especially as it seems to be derived from a velocity.

Introductory use abstractions based on clocks and flight times. This may be sufficient information to base the mathematics upon but it seems a poor foundation for a branch of physics. If you do not have a clear definition of time at the start how can your derived equations have any clear meaning? That is, if you start with time derived from a velocity then your times is your derived equations is also derived from velocity and is different from the idea of the passage of time that we base our daily lives upon.

What I posted was a discussion of what the time could mean.

I thought someone would provide me with a definition that would provide clarity to these equations.

8. Oct 18, 2004

### pervect

Staff Emeritus
I'm not sure how much this is going to help, but a reasonably good defintion of time is what you measure with a clock.

The current NIST definition of time might also be helpful:

http://physics.nist.gov/cuu/Units/current.html

Current atomic clocks are the end result of a long sequence of empirical development of what makes a good clock.

The very first accurate clocks were inspired by a practical problem - the desire to find the longitude of ships, so that they could navigate the seas. (See the PBS program - "Longitude", for instance).

The lesson was learned, even at this early stage of development, that good clocks tended to be small.

Atoms are the smallest unit of matter, and atoms of the same type are all identical, so it is not surprising that they make excellent clocks, and that our current time standard is based on them.

9. Oct 18, 2004

### ramcg1

Thanks for the definition, but it does not answer my question.

10. Oct 18, 2004

### pervect

Staff Emeritus
You really aren't being very clear about what you want. Since you seemed to be asking for a defintion of time in your response

that's what I gave you, a definition of time.

I'll take the opportunity to expand on my first remarks, though.

This is a more complicated exposition on the theme of "Lorentz transformations mix together space and time" which I gave in my first response (that you didn't like), with the additional note that "this mixing together is much like the mixing together of horizontal and vertical axis in a plane with respect to rotation of the plane".

These ideas are more fully developed in standard relativity texts, such as Taylor & Wheeler's "Spacetime Physics".

For a rotation of the plane, we write

x' = cos(theta)*x - sin(theta)*y
y' = sin(theta)*x + cos(theta)*y

And we note that distances are unchaged by rotation, which means that

x^2+y^2 = x'^2 + y'^2

For the lorentz transform (with c=1), we write this with hyperbolic trig functions instead of the usual trig functions

x' = cosh(alpha)*x - sinh(alpha)*t
t' = cosh(alpha)*t - sinh(alpha)*t

Here velocity = v = tanh(alpha)

And we note that the Lorentz interval is unchanged by rotation, which jmeans that

x^2 - t^2 = x'^2 - t'^2

The full importance of this geometrical view is more apparent in general relativity, where one considers space-time to be a 4 dimensional manifold.

The simple version of this detailed explanation is that space-time is a 4 dimensional entity because space and time can be "intermixed" by the Lorentz transformation. The comparison of the Lorentz transform to rotation in a plane simply illustrates the point in more detail.

Last edited: Oct 18, 2004
11. Oct 19, 2004

### NoTime

ct is as good as it gets for current knowledge

12. Oct 19, 2004

### franznietzsche

Yes

Actually, its a four dimensional cone, the set of all points seperated from the original emission of light by a spacetime interval of 0.

No.

13. Oct 20, 2004

### ramcg1

I am attempting to obtain a clear understanding of the foundations of Spacetime. It is fun to play with mathematics but how can I fully understand the implications of Spacetime if I do not fully understand the foundations that it is built upon.

If, within an inertial frame of reference, you determine a straight-line path fight time by comparing the clocks at the beginning and the end of the straight-line path then you are using three dimensional geometry to determine the time, distance and velocity of that flight. Thus the conversion factor between space and time is based upon a three dimensional concept.

I took a hypothetical point source as being a very small source that irradiated in all directions at once. I could have considered a single atom releasing a single photon in a random direction and plotting the possible positions with time. In either case the three dimensional wavefront would be a sphere. As I am examining the foundations of Spacetime I am still within my inertial frame of reference and observing things three dimensionally.

If I can explain something three dimensionally then the only reason for me to move to four dimensions is mathematical convenience.

14. Oct 20, 2004

### ramcg1

I appreciate any response and I thanked you for yours. You are right. I have not been very clear possibly because my original posting contained several questions and I had not thought this one through enough.

Perhaps the following makes more sense:

Suppose a 100 km length of straight and flat road.
I travel, in my motor car, along this length of road at 100 km per hour.
I apply t = d/v and determine that t = 1 hour.

If my car is parked at one end of this length of road then v = 0 so d/v is undefined, hence t is undefined.

When the car moved along the road is determined by calendar time. So the hour that the car moved for can be compared to calendar time but is not the same as calendar time because it is undefined when the car is not moving. Hence I am looking at two types of time.

The question then becomes is the time of Spacetime only valid for moving electromagnetic radiation, and as a consequence is the mathematics of Spacetime only valid for simultaneously moving electromagnetic radiation?

If I do not have a clear understanding of the time used in spacetime then how can spacetime have any real meaning to me except as a mathematical exercise. I made my original posting hoping clear the confusion.

15. Oct 20, 2004

### MiGUi

Speed depends on space and time, but space is one thing, time is another thing. Both two are independant on speed. You can stay with no speed but you can set your position and another observer in movement respect you can do it as well.

So your proposition of t = d/v is not correct since t is the independent variable as d, meanwhile v is dependant.

The structure of spacetime is a property inherent to the spacetime. It don't distinguish between EM radiation or something else. I don't understand how you get from spacetime structure to electromagnetic radiation.

The example of the lamp at #1. We must define a (0,0,0,0) origin to do this experiment due to the marvelous statement that the laws of physics don't depend on position (x,y,z,t). So, we can't ask about "before" since it has no sense. You may not ask about the x coordinate of any photon before turn on the lamp. With the time occurs exactly the same. The filament of the lamp is very heat so it radiates so much and this radiation occupies a portion of the visible spectra. So, the filament begins to emit photons which speed is exactly c at the moment of its creation.

16. Oct 20, 2004

### pervect

Staff Emeritus
OK, I think I may be able to offer some comments that will help.

The definition of time that I offered earlier was that of time as an interval, one you measure with a clock. This sort of time isn't quite sufficient, however, to measure the velocity of the car.

Using the notion of time interval, one could have a clock onboard the car. One would note the time, according to the car's clock, when one started the trip, and the time, according to the car's clock, when one reached the end of the road.

However, from relativity, we know that the car's clock is not running at the same rate as the Earth's clock. When we want to measure the car's velocity in the Earth frame, we want to use Earth clocks, as well as Earth rulers.

In order to accomplish this, we need more than one clock. We need two clocks, one at the start of the trip, and one at the end. And, we need a way to synchronize them.

We then note the time reading on the first (Earth) clock when we pass it at the start of our trip, and then we observe the time reading on the second (Earth) clock at the end of the trip, then we subtract the two readings to get the total amount of time the trip took.

This defines the notion of coordinate time, which is the combination of the idea of time as something that can be measured with a clock, plus a notion of how to synchronize two clocks at different locations.

One can envision (in principle at least) an infinite array of clocks everywhere in space, all synchronized by some method, setting up a coordinate system for time as well as space.

So, how do we synchronize clocks at two different locations? The standard answer is to use Einstein's method, and it's a good method. But let's talk about why it's a good method some more.

Imagine that we have two clocks, and that they are not synchronized. In fact, let's assume that they are way, way, way off. Suppose that the clocks are in fact an hour off, because one clock is in a different "time zone" than the other.

What happens if we take this bad definition of synchronization at face value? What happens is that we come up with some unphysical results. If the car travels at 100 km/hour, we find that it travels the distance in no time at all going one way, having therefore an infinite velocity, while it takes 2 hours to go the other way, thus travelling only at 50 km/h.

So, we demand that our clocks be "fairly" synchronized, that the clock synchronization be "isotropic", so that the speed of the automobile is the same in both directions.

This is easy enough to say, but it's not very rigorous. What exactly do we mean by an automobile running "at the same speed"? Can we make this definition more rigorous? There are at least two alternate ways we can accomplish this.

One way to solve the problem does not require any reference at all to light, or light speed, and requires only the knowledge and acceptance of basic Newtonian physics. This is the notion that two cars of the same mass have equal and opposite velocities when their total momentum is zero. A collision test can be done (in principle, at least) to determine if the total momentum is zero or not - if the cars are on a frictionless surface, and collide inelastically, they will come to rest only if their total momentum is zero.

By making use of this physical property of matter, momentum, the notion of fair and/or "isotropic" clock synchronization is well-defined. There is one and only one way of synchronizing clocks such that the momentum, a physically measured quantity, will be the same for a car (or other moving body) of some defined mass M making the trip in either direction with the same measured velocity.

A more convenient method of synchronizing clocks relies on the fact that the speed of light is constant for all observers. This is a much more practical method, and it's known as the Einstein clock synchronization method. Here one sets off a light signal midway along the route. This method relies on the fact that the speed of light is constant for all observers to synchronize the clocks.

The Einstein method works only because the speed of light turn out to be a constant for all observers. The notion of how to synchronize clocks can, however, be defined independently of this special property of light, via the notion of isotropy.

In fact, isotropy was what prompted Einstein to propose his method of synchronizing clocks, as can be seen from reading his original paper. (I don't have a URL handy for this paper, but his reference to isotropy is really quite brief anyway).

17. Oct 20, 2004

### manzaloros

the Universe has eleven dimensions, plus spacetime (including). Time is not constant and the only benchmark for time is the observer, or gravity.

18. Oct 21, 2004

### ramcg1

You have given me a lot of material to consider.
I will go away, digest it and get back if I have any futher questions.

19. Oct 24, 2004

### ramcg1

As your hypothetical being has no concept of thickness, but a concept of change, he is going to believe that time is the third dimension.

I did not like your example for the following reasons:

An observer in 3D can look down onto the 2D world and perceive 2D shapes.
An observer in 2D has no thickness (no volume), no mass etc hence no brains.
The block as it moves through has a cross-section that the 3D observer can see.
The 2D plane through the block has no thickness so what is the 2D observer going to perceive?

I think a better example would be:

An intelligent computer is used to analyses 2D animation. You can allow the computer to view the animation from a point with in the animation and to move this point of observations. As the computer would be capable of analysing all the 2D array of information for each frame it would be able to perceive 2D objects and, being intelligent, wonder if there is another dimension.

20. Oct 24, 2004

Pervect