# B Moving in the 4th Dimension -- Does Time have a velocity?

1. Aug 5, 2017

### eaglechief

Hello all,

i read several threads concerning the a.m. topic, but are still not sure if i got it right.

Is "passing time" or moving in the 4th dimension in relativistic means a real changing of position in grade 4 spacetime or is it just a mathematical effect ?

Why asking: If i understood correctly, a resting massobject with no velocity in space S3 has a worldline parallel to the time-axis. If we compare two points in time on the worldline of this object, it has moved in the coordinate system from f.i. A to B in direction of D4, but not in space S3.

If so, we do not recognice this movement (or at least, i don't do it). Is this comparable to other movements in space (and spacetime), that we do not recognice, f.i. the turning of planet earth or the movement of earth around the sun, which do really happen ?

Swen

2. Aug 5, 2017

### Staff: Mentor

How would you tell the difference. I mean, what kind of experiment could you do to determine which is right?

3. Aug 5, 2017

### Ibix

We don't know. The maths of relativity can be interpreted as describing a 4d spacetime, but it can be interpreted otherwise. Certainly the 4d spacetime is the most popular interpretation, and you often see people writing as if it were the only option. But we don't know for certain.
S3 means something different from what you're using it for here - I'd advise against that notation.
How would you recognise that you were in motion through space?

4. Aug 5, 2017

### stevendaryl

Staff Emeritus
Yes. For instance, the Sun (and the planets along with it) is traveling at over 200 kilometers per second through the Milky Way galaxy. This motion is certainly not detectable without using astronomy. Motion is not detectable, only relative motion (the distances between different objects changes) and acceleration (changes in velocity). Motion through the 4th dimension can be thought of as undetectable inertial motion.

5. Aug 5, 2017

### eaglechief

Thanks for the answers and hints so far.

So that means, that while i am sitting on my couch writing this post, my couch and me are travelling with a certain rate of change along the 4th dimension through spacetime. An observer in a higher grade inertial system (4 up) could therefore measure my changing position in way of the 4th dimension from A to B. Is this correct ?

Is it valid to call this rate of change through the 4th dimension "velocity" ? I would assume that velocity only is valid for movements inside 3-dimensional space.

6. Aug 5, 2017

### Ibix

There's no evidence such a thing exists, so this isn't really answerable.
It depends. You can take the derivative of your path through spacetime with respect to your proper time, and the result of this is a 4-vector usually called a "4-velocity". However, 4-velocities all have magnitude 1, so are a measure of direction in spacetime rather than anything else. I would tend to say that, if we're adopting the 4-d spacetime model, nothing moves in 4-d spacetime. Movement is what you get when you take two 3-d slices through spacetime, declare them to be "the universe, now" and "the universe, a little later" and compare them.

7. Aug 5, 2017

### Yukterez

It is valid to call it (a component of the) 4-velocity (in contrast to the 3-velocity, which it isn't part of). Since the coordinate time component is multiplied with c to give a length, and then differentiated by proper time just like the other space-components, you could imagine the object travelling through your coordinate system t,x,y,z. The faster its velocity on your x,y,z-axes, the slower its velocity on your t-axis. In this 4D coordinate system t is just as orthogonal to x,y,z as they are to each other. Therefore the euqations of motion always give not only x',y',z' or r',θ',φ' but also t' (meaning dt/dτ), so if not "velocity" you can at least call it "motion" with good conscience.

Last edited: Aug 5, 2017
8. Aug 5, 2017

### Staff: Mentor

No, this is not correct. The "velocity on the t-axis" is $dt / d\tau$, the $t$ component of the 4-velocity. If you compare two objects, one at rest and one moving, $dt / d\tau$ for the second object--the moving one--will be larger than for the first, not smaller.

9. Aug 6, 2017

### Orodruin

Staff Emeritus
Just to mention that we even have a name for this: time dilation. The faster an object moves in the spatial directions, the more coordinate time elapses per proper time of the object.

10. Aug 11, 2017

### eaglechief

thanks for the answers. using dt/dτ as expression for the t-component of the 4-velocity, we get a non-dimensional number. But don't we talk about a "velocity". Am i forced to use c = 1 or could i also use "real" velocity by inserting SI-units instead ?

11. Aug 11, 2017

### Orodruin

Staff Emeritus
The usual way to do things if you really want to obscure the physics by inserting an arbitrary conversion constant into your equations is to work with the rescaled time coordinate $x^0 = ct$. The components of the 4-velocity are then $V^\mu = dx^\mu/d\tau$. In particular, the 0-component is $V^0 = c\, dt/d\tau$.

12. Aug 11, 2017

### sweet springs

From four velocity
$(u^0,u^1,u^2,u^3)$ where $u^0u_0-u^1u_1-u^2u_2-u^3u_3=1$
you can make
$$c(1,\frac{u^1}{u^0},\frac{u^2}{u^0},\frac{u^3}{u^0})$$ where 1,2 and 3 components are ordinary 3d your "real" velocity. It is not a vector any more in the sense of Relativity theory.

Last edited: Aug 11, 2017
13. Aug 11, 2017

### vanhees71

Here the minus-signs must go, because $u_1=-u^1$ etc. The Minkowski product is
$$u_{\mu} u^{\mu}=(u^0)^2-(u^1)^2-(u^2)^2 - (u^3)^2=u_0 u^0 +u_1 u^1+u_2 u^2 +u_3 u^3.$$
That's because, by definition
$$u_{\mu}=\eta_{\mu \nu} u^{\nu}$$
with $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$.

14. Aug 11, 2017

### sweet springs

15. Aug 12, 2017

### eaglechief

Thanks for the answers so far.

Another question concerning ibix's quote and the SRT resp. GRT in general.

Do the SRT / GRT equations lead to a blockuniverse as a "must" or are there other interpretations possible with a dynamic universe, as well ?

16. Aug 12, 2017

### tophatphysicist

eaglechief, you sure bring some provocative inquiries. There are physicists who subscribe to the Block Universe concept. And in that context, and along with your original question, we have the picture of an observer moving along his world line at the speed of light. However, note that the observer's physical body would be a 4D object, frozen in spacetime so to speak, in that view--no motion for the physical body. So, it raises the question, "What is doing the moving?" Perhaps you have some thoughts about that.

17. Aug 12, 2017

### Ibix

It's certainly possible in SR to just "pick a frame" and declare that it's the "real" frame and the universe only exists at its "now". Anyone using any other frame is, in some sense, wrong. It's not a claim that is testable, though, since there's no consequence to your choice of "real" frame. You can pick a non-inertial one if you like. Or you can pick the block universe.

It's really up to you.
Nothing is moving in the block universe picture. The four-velocity is simply the tangent vector to the worldline in this model.

18. Aug 12, 2017

### Staff: Mentor

19. Aug 12, 2017

### Staff: Mentor

A fairly long thread hijack has been cleaned up ant the thread is reopened

20. Aug 13, 2017

### PAllen

There is a related GR interpretation, evolving block universe, pursued by, among others, Ellis of Hawking and Ellis fame. MTW refers to this approach as many fingered time. Ellis puts forth arguments based on QM + GR to prefer EBU models as philosophically preferable.

Point is, the model of a complete manifold cannot say anything about what part of the model is 'real', nor can this be determined by experiment, so all choices are metaphysics rather than physics.