# Motion through spacetime

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1. May 20, 2013

### maxverywell

Ok, I understand what the motion through space is, but I have difficulty in understanding what the motion through spacetime is. Spacetime is a 4 dimensional manifold with time being one of the 4 dimensions.

When a particle moves through spacetime with some velocity, it moves through space with the conventional 3-velocity, but it also moves through time. What does it mean to move through time and how is the velocity defined for the motion through the dimension of time?

Last edited: May 20, 2013
2. May 20, 2013

### ghwellsjr

In flat spacetime (the kind we use in Special Relativity where we ignore the effects of gravity), there are two kinds of time to consider. One is the Proper Time of the particle and the other is the Coordinate Time of the Inertial Reference Frame in which the particle is moving at some speed. The ratio of the progression of Coordinate Time to the progression of Proper Time is called Time Dilation which means it takes longer in Coordinate Time for the same amount of Proper Time to progress.

3. May 20, 2013

### WannabeNewton

It is perhaps easiest to understand the time-like component of the 4-velocity, with respect to some orthonormal basis, in SR. Let's say we are in the rest frame $O'$ of an inertial observer, who has setup global inertial coordinates $(t',x^{1'},x^{2'},x^{3'})$ in his frame and has constructed a set of orthonormal basis vectors $\{e_{0'},..,e_{3'}\}$ for the frame, where $(e_{\mu'})^{\alpha'} = \delta^{\alpha'}_{\mu'}$ are the components of each basis vector in this frame; you can imagine $e_{0'}$ as pointing along the "time axis" of the frame setup by the observer and the $e_{i'}$ as pointing along their respective "spatial axes". Note that since the observer is obviously at rest in his own rest frame, his 3-velocity in $O'$ vanishes and his 4-velocity in this frame is simply $\vec{u} = e_{0'}$ i.e. the 4-velocity points along the "time axis" and can be written $\vec{u} = (1,0,0,0)$ with respect to the basis vectors in the frame.

We can in fact take this to be the definition of the 4-velocity of the inertial observer in $O'$. Then, the velocity of this inertial observer as measured by an observer in any other inertial frame $O$ (who has himself set up coordinates $(t,x^1,x^2,x^3)$ and an orthonormal basis $\{e_{0},..,e_{3}\}$ for his frame) is simply given by $u^{\alpha} = \Lambda^{\alpha}_{\beta'}(e_{0'})^{\beta'} = \Lambda^{\alpha}_{0'}$, where $\Lambda$ is the Lorentz transformation. This is nothing more than $\vec{u} = \gamma(1,\mathbf{v})$ where $\mathbf{v}$ is the 3-velocity of $O'$ with respect to $O$. This is of course the usual result.

Now, a particle that is accelerating has no inertial frame in which it is always at rest but what we can do is, for each event on the worldline of the particle, find an inertial frame which is instantaneously co-moving with the accelerating particle and in exactly the same way as above, we define the 4-velocity of the accelerating particle at some event as the $e_{0}$ basis vector of the inertial frame that is instantaneously co-moving with the particle at that event. Now let's tie this in to the definition you are probably more used to. Let $O$ be an inertial frame (with the usual coordinate setup and orthonormal basis setup) instantaneously co-moving with our particle, at a given event, and say that the particle makes an infinitesimal displacement $\vec{d\mathbf{x}}$. In $O$ this is written in components as $\vec{d\mathbf{x}} = (dt,0,0,0)$. We also have that in $O$, the infinitesimal proper time $d\tau$ is simply $d\tau = dt$ so $\frac{\mathrm{d} \vec{\mathbf{x}}}{\mathrm{d} \tau} = (1,0,0,0) = e_{0}$ which by definition is the 4-velocity of the accelerating particle at the given event.

All in all, I guess the important point to take home is that in the rest frame of an inertial observer, or in the instantaneously co-moving frame of an accelerating observer at a given event, the 4-velocity of the observer in that frame simply points along the "time axis". This is one way to interpret the time-like component of the 4-velocity.

4. May 20, 2013

### cosmik debris

Objects "move" through time by simply existing. Everything "moves" through time at 1 sec/sec of proper time. In 4D spacetime objects don't move they have a world line which is a path between events. The distance between these events can be measured in different units, usually seconds or metres.

5. May 21, 2013

### maxverywell

Yes, this is what I'm trying to understand, why an observer can be at rest in space but not in time.

6. May 21, 2013

### WannabeNewton

How can a time-like particle possibly be at rest in time?

7. May 21, 2013

### maxverywell

It can't and this strange because it can be at rest in space. Or in other words, you can move through time and spacetime but not only through space. But space and time should be treated in the same way (the only difference between them is in a minus sign in the metric).

8. May 21, 2013

### WannabeNewton

There is nothing strange here. The time-like direction must be future pointing unless you start getting involved with closed time-like curves. Space and time are not treated the same in every aspect, that is a misconception. Clearly one of the differences is the one you have noted i.e. that in physically realizable cases the time-like direction is future pointing.

9. May 21, 2013

### ghwellsjr

A particle can be at rest in space, that is, in an Inertial Reference Frame (IRF) such that its spatial coordinates don't change with respect to its time coordinate which matches time for it (which we call its Proper Time). But you can transform the events of that stationary particle according to another IRF moving with respect to the original IRF, and you can pick one such that it is moving at a speed approaching that of light in which case it will be Time Dilated so that Proper Time for it will be progressing hardly at all with respect to its spatial coordinate progressing very, very slowly. It's that minus sign that makes this happen.

10. May 21, 2013

### Fredrik

Staff Emeritus
I would say that there's no such thing. But I suppose it depends on your precise definitions. Motion in space is represented by curves in spacetime. That's pretty much all there is too it.

In case you're referring to the "speed through spacetime" and "speed through time" ideas that are discussed here somewhat frequently...

11. May 21, 2013

### maxverywell

Thnx. This is what I wanted to hear.

Edit: wrong...

Last edited: May 21, 2013
12. May 21, 2013

### Fredrik

Staff Emeritus
This doesn't make sense, since there's no such reference frame.

I don't know how meaningful it is to define the reference frame of a tachyon. I have never seen it defined, but the only definition that makes sense to me makes the world line (which is spacelike) the 0 axis, and the 1-3 axes orthogonal to it, with one of them timelike and the other two spacelike. So in its own rest frame, its world line is in the 0 direction, just like for any other particle that has a rest frame.

13. May 21, 2013

### maxverywell

Last edited: May 21, 2013
14. May 21, 2013

### ghwellsjr

What's wrong?
No observer ever sees anything differently just because you (or he) uses a different reference frame to describe what's going on. All that changes are the coordinates assigned to each event.

Here's a diagram for an observer that is stationary in an Inertial Reference Frame (sometimes referred to as his rest frame):

The observer is shown in blue and is stationary at the coordinate location of 0. The progress of his clock marking off his Proper Time is shown by the blue dots. Since he is stationary in this IRF, his Proper Time is not dilated, meaning that it progresses the same as the Coordinate Time marked off by the horizontal grid lines.

Now we transform this IRF into one moving to the left at 0.96c:

Now we see that the observer's location is changing and the progress of his Proper Time is dilated with respect to the Coordinate Time. In other words, it takes more Coordinate Time for his clock to progress through the same amount of time as before which makes his clock tick slower in this frame. All I'm trying to show you is that his clock runs slower and his distance is larger. Isn't that what you were asking about?

However, keep in mind that the observer has absolutely no awareness of anything different in these two frames or any other frame. There is no observation, no measurement, nothing that he can see differently just because we use different frames to describe or analyze what is going on in any scenario. All the changes caused by different frames are 100% coordinate effects.

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15. May 21, 2013

### Fredrik

Staff Emeritus
It looks like he (or she) is saying that you're wrong, but he's actually referring to the things that he edited out of his own post after I replied to them. See the first quote in my reply to his post.

16. May 21, 2013

### ghwellsjr

OK, thanks. I guess you have to be on all the time to capture everything that goes on.

17. May 21, 2013

### maxverywell

Something I wrote was wrong, not your answer. Sorry for the misconception.

18. May 22, 2013

### nitsuj

The idealized "light clock" and SR postulates are great for "seeing" this "transition" between the time & length dimensions via comparative motion; in turn be able to "see" what it means to "move through time".
Light Clock LINK!

At the level I think you are asking about "what it means to travel through the time dimension" is about simple orthogonal geometry. That particular concept is a different and far more simple presentation of the continuum in which we live, if that's your interest then I think GR is what you should learn.

That said understanding dimensions and the geometric presentation of the two types would be important if going to look at a 4D geometric presentation.

for the underlined question, the principle of relativity postulate seems to address "what motion is", it's a comparative.

Last edited: May 23, 2013
19. May 23, 2013

### Stricklandjr

Mr. maxverywell I had some of the same puzzle to study on in my post and Dr. Donis and Mr. Fredrik and others were a big help. They explained the 4 dimensional universe and world tubes and slices for me. The closest Dr. Donis got to answering your question was to say that your world tube means that you are 4 dimensional all there at once but you can only experience a train of 3 dimensional slices, one after another, at the speed of light. But he and Mr. Fredrik said that you can't really know that is really the way the universe works and you can't really know that things are really 4 dimensional all there at once because that's just philosophy and not physics.

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