# Time Dilation Question

1. Dec 31, 2005

I was wondering... If length contraction only contracts in the direction of motion, would we see time slow in other frames if we traveled "perpendicular" to time (what ever that may mean)? perhaphs we would view time moving faster for other frames? Dunno if this even makes since just wondering.

2. Dec 31, 2005

### dicerandom

Well, from a geometrical outlook the time dilation and length contraction due to relative velocities in SR originates from the fact that the two observers world lines (i.e. their time axes) and therefore their lines of simultaneity (x axes) are rotated with respect to one another. If you had a second temporal dimension, and it was possible to have a velocity vector which was only rotated into one spatial and one temporal dimension, then I think you would have a situation where two observers would disagree on their spatial measurements, and on the readings of one set of their clocks, but not on the readings of their other set of clocks.

I not sure that such a situation would be possible though, I would expect that even with two temporal dimensions your motion through space would be connected to both rather than one.

3. Dec 31, 2005

### Mortimer

Perpendicular to time = space (in 4D space-time). So yes, we would see time slow down if we travel through space.

4. Dec 31, 2005

Staff Emeritus
Travelling at right angles to time, as dicerandom said, is travellling in a spacelike direction, and since no time will pass, you would be travelling instantaneously. Needless to say, this is physically impossible.

Generally any speed you may have, relative to some other frame, will have a time component and a space component, and observers in that other frame would see a different sized time and space component, but there is no frame in which the time component shrinks to zero. In your rest frame, for you, the space component shrinks to zero (you aren't aware of moving) but the time component is as big as it gets (you are aware of time passing).

5. Jan 1, 2006

### Mortimer

I'm not sure about your usage of the term "spacelike" here. I assume you probably have the Minkowski light-cone in mind where the spacelike quadrants are indeed drawn in the perpendicular direction to the $ct$ axis. The Minkowski diagram however is just a graphical representation of a mathematical concept and should, in my opinion, not be taken as an actual geometrical representation of a physical reality.
I hold on to my earlier remark, that in 4D space-time, space is perpendicular to time and therefor any spatial velocity component is by definition perpendicular to the time dimension. The maximum velocity $c$, as occurs with massless particles yields a spatial velocity that is fully perpendicular to time, hence no time will pass.

Happy New Year to all,
Rob

Last edited: Jan 1, 2006
6. Jan 1, 2006

### robphy

Of course, the "quadrant" itself isn't Minkowski-perpendicular (henceforth M-perpendicular) to the ct-axis, although those directions in the quadrant are, as you've noted, spacelike-directed [in fact, to all timelike directions... not just the ct-axis]. Rather it is a hyperplane [with normal vector along the ct-axis] that contains the directions M-perpendicular to the ct-axis, i.e. the directions regarded by the ct-axis as "spatial".
You may be interested in the discussions at
http://alcor.concordia.ca/~scol/seminars/conference/
http://www.spacetimesociety.org/conferences/2006/
The conclusion of the last sentence (which I've also seen elsewhere in the context of "reference frames for photons") has always bothered me a little. Of course, you're saying that a lightlike-direction is M-perpendicular to itself, which can be viewed as the limiting case of a timelike-direction with its M-perpendicular direction ["its notion of space"] as the spatial velocity approaches c.

For the timelike case, I agree that "no time will pass" for its M-perpendicular direction ["its notion of space"]. This suggests that events along this "spatial direction" are simultaneous... that they happen all at one instant to this observer.

However, for the lightlike-case, I think there is a problem reaching this same conclusion because... although no proper-time will tick off (probably there is no invariant sense of proper-time for a lightlike direction), two events connected by a lightlike signal still retain the notion of causal precedence. That is, for a future-pointing lightlike vector $$\vec {PQ}$$, event Q is still AFTER event P... even though $$\vec{ PQ}$$ is M-perpendicular to itself. So, in my opinion, the conclusion "no time will pass" for the lightlike case is misleading... and, if used, must be further clarified.

Last edited by a moderator: Apr 21, 2017
7. Jan 2, 2006

### Mortimer

Indeed, the notion of "passing time" for a lightlike case is confusing, in particular because no invariant proper time can be calculated in any reference frame, like you noticed already. I think that causal precedence for the lightlike case can be determined equally well (and perhaps even only) by spatial position (instead of time). After all, there is no way that e.g. a photon can be made to turn and go backwards on its path, so its consecutive spatial positions (or "passing space") always seem to determine causality.
If I try to imagine what the photon would "observe", in analogy to what we observe in our environment, I can only think of a "Flatlander-like" situation where for the photon there is only a 2D space and a third dimension (which is the direction of its travel) that acts as its equivalent of "time". Like suggested above, that third dimension would then represent the basis for causality for the photon (or other massless particles), like time does for us.

8. Feb 2, 2006

### Jesus Rodriguez

Musings of an Ignoramus

Some believe these questions are meaningless because they can not be asked in a mathematical form. That is, attempts to use the velocity equations for an object travelling at light speed result in an attempt to divide by 0, which is "undefined."

However, and correct me if I'm wrong since I'm very unlearned in these matters, I have read that quite some time ago the attempt to merge mathematics and philosophy ended with the proof that the two disciplines do not fully overlap. That is to say, there are valid questions, arguments and answers which can not be framed in the language of mathematics. If this is true, then there are valid questions for discussion which are irreducible to mathematics. Thus an "undefined" operation, such as dividing by zero is not exactly identical to "meaningless." That is, dividing by zero is meaningless in mathematics, but the question that led to the formulation may not be.

Assuming this is true (and I am certainly open to learning I am wrong, as I am rather ignorant in both philosophy, mathematics and physics), this means talking avout the photon's perspective is valid, though incalculable.

This flatland anology is interesting. Like many of you I delight in that little book. The question I posit is wether the photon is like a flatlander (a form of 3D existence which experiences no time, only being imbedded in or) OR is the photon more like the visiting sphere, a 5D object that from our perspective behaves very oddly indeed. Phenomena like entanglement make me wonder if perhaps photons are able to interact in a dimension unachievable below light speed.

I suppose, if you pardon the pun, this could get all entangled in the notion of curled dimensions of which string theorists are so enamored.

Oh well, I know so little perhaps everything above is just giberish. If so, I'd appreciate being told so. Unlike many people, I don't suffer delusions of omniscience, and are quite happy with a response like "go here and read awhile before embarrasing yourself further"

JR

9. Feb 4, 2006

### yogi

J.R - Your certainly do not have to apologize for raising the philosopic issues associated with relativity - these are in fact the most interesting to many of us. There are a lot of individuals on these boards with post graduate credentials in mathematics and physics - so the threads become quickly overloaded with equations where frequently the symbols are not defined nor understood by persons wanting to make sense out of the discussion.
When I was a corporate patent attorney, there was a prevailing attitude that a Patent application loaded with complex equations would snow the examiner, not being able to fully understand, the examiner would be inclined to allow some claims simply because the applicant appeared to have in depth knowledge in the art. In truth, there are very few things that cannot be adequately described using words.

I have always been impressed with Einstein's words in a speach he gave in tribute to Faraday ...some of the persons posting on these boards would do well to read it. I will quote it below in a separate post:

10. Feb 4, 2006

### yogi

JR Here is the excerpt: “To become fully conscious of this change in outlook was a task for a highly original mind whose insight could go straight to essentials, a mind that never got stuck in formulas. Faraday was this favored spirit. ...."

11. Feb 4, 2006

### ZapperZ

Staff Emeritus
You are forgetting that there's a DIFFERENCE between not getting stuck in formulas and not knowing them AT ALL. That quote isn't an excuse to be ignorant of the underlying mathematical description, and it was meant towards people who ALREADY have an underlying understanding of the equation, not to those who have zero clue.

I yearn for the day when you stop playing these quotation game. I'm skeptical that I'll live to see it.

Zz.

12. Feb 4, 2006

### pervect

Staff Emeritus
This was discussed a bit in another thread here recently, in which I screwed up. But I think I've got it right this time around.

If we have an observer moving at a velocity less than c, he has one axis defined by his local clock, which is his time axis, and three perpendicular axes which are Minkowski-perpendiuclar, called his space axes.

If we attempt to explore an observer moving "at c", we have one null axis, defined by the observers motion, which is Minkowski-perpendiuclar to itself, and we also have two space-like axes which are perpendicular to the null vector defined by the observers motion. This gives us only three axes.

But these axes do not span the 4-d space. In order to span the 4-d space, we need to introduce another vector. A convenient choice (though not the only one) is another null 4-vector. This additional null 4-vector is perpendicular to the same two spatial axes that the "velocity" vector of the observer is.

An example helps:

We have coordinates (t,x,y,z)

(1,1,0,0) is the null 4-vector of our observer

(0,0,1,0) is one of our orthogonal space-like vectors (y)

(0,0,0,1) is the other one of our orthogonal space-like vectors (z)

(1,-1,0,0) is a null vector which "fills out" our coordinate system. It is Minkowski-perpendicular to (0,0,1,0) and (0,0,0,1). It is not (and cannot be) Minkowski-perpendiuclar to (1,1,0,0) (that's where I screwed up last time)

Another way of putting this is that we have coordinates

u' = t+x; v'=t-x; y'=y; z'=z

u and v are null coordinates.

Yet another way of looking at this: there is some representation of a photon moving in a "head on collision", and representation of "parallel" photons moving in the same direction. Both of these are null vectors, and hence neither space-like or time-like. In addition, there are two perpendicluar spatial axes.

13. Feb 6, 2006

### pervect

Staff Emeritus
Let's suppress the two spatial dimensions, and consider the problem of representing a 1-space+1-time Minkowski continuum with null vectors.

There will be only 2 null vectors possible (light travelling in the + and - directions), and 2 are required, so there is no problem deciding what vectors to use.

However, there is an issue of how to "scale" the null vectors. Via the mathematical process of an affine parametrization of a light-like geodesic, we can, given a reference interval, mark intervals along a null geodesic equal to the reference interval.

One can think of this process physically as counting the wavelengths of light, once the inital wavelength has been specified. This explains why a reference interval is required.

So what we need to start with is to define two reference intervals, one for the null geodesic in the +x direction, and the other for the null geodesic in the -x direction.

Setting both of these intervals to '1', and an origin point (u=v=0) defines a complete coordinate system.

The metric of the resulting space-time will be of the form k*du*dv, where u and v are the null coordinates, and k is some arbitrary scale factor constant. It will usually be convenient to set k=1.

By the transformation t = (u+v)/2*sqrt(k), x=(u-v)/2*sqrt(k), we can transform the null coordinate system into a traditional space-time coordinate system, with a metric dt^2 - dx^2.

We can see that our choice of initial "equal" null vectors has defined a notion of a "rest frame". This notion is of course not unique - any choice of initial "equal" null vectors will define some notion of a rest frame.

Last edited: Feb 6, 2006
14. Feb 9, 2006

### yogi

Who are you to tell us what Einstein meant?

15. Feb 9, 2006

### ZapperZ

Staff Emeritus
And I could ask the SAME thing to you. Look at the AUDIENCE that that was said to. Are you saying that the CONTEXT of a quote is irrelevant, that it is FAIR to take quotes out of nowhere?

Since when is physics done by a series of quotes, which is something you do all the time? We RESPECT the work done by these people. We don't REVERE them as if their words are the words of GOD that need to be repeated.

Zz.