# Relativistic velocity in the time dimension

1. Apr 29, 2005

### Mortimer

I would be interested to hear what others think of the general concept of relativistic velocity in the time dimension, analogous to spatial velocity.
To get going, consider the Minkowski velocity 4-vector:
$$\gamma(c,v_1, v_2, v_3)$$
The time component of this 4-velocity is $$\gamma c$$. With increasing spatial velocity, this component goes to infinity. With spatial velocity zero, it equals c.
Obvious questions are:
- how to give (physical) meaning to this component; seconds per second is not very usable.
- why does the time component go to infinity, while it is clear that proper time goes to zero.

Last edited: Apr 29, 2005
2. Apr 29, 2005

### Meir Achuz

Not all 4-vectors have simple physical interpretations.
Velocity is an awkward variable for modern (SR or QM) physics.
As you indicate, you can easily confuse yourself working with velocity.
The momentum 4-vector has good physical properties.

3. Apr 29, 2005

### robphy

In relativity, measurements done by an observer can be described in terms of dot-products of that observer's (normalized) 4-velocity with the quantity whose components are being measured.

Consider two inertial observers, one with 4-velocity $$\tilde u$$ and one with 4-velocity $$\tilde v$$. The quantity $$\tilde u \cdot \tilde v$$ is (up to factors of c) simply $$\gamma$$, the [dimensionless] time-dilation factor [as you've written]. In fact,
one can decompose $$\tilde v$$ into a piece parallel to $$\tilde u$$ and the remainder, which is necessarily (Minkowski-)perpendicular to $$\tilde u$$. That remainder is $$\gamma$$ times the relative-velocity vector.

For another example, the electric field observed by an inertial observer with 4-velocity $$\tilde u=u^a$$ is (up to signature conventions and factors of c) $$u^aF_{ab}$$, where $$F_{ab}$$ is the electromagnetic field tensor.

In summary, the 4-velocity is essential for physical measurements.

4. Apr 30, 2005

### Mortimer

Hello RobPhy,

Isn't this $$\gamma (u)\gamma (v)\{c^2-u_1v_1-u_2v_2-u_3v_3\}$$? If $$\tilde u=\tilde v$$ the dot product equals $$c^2$$, or am I overlooking something with regard to doing dot products with Minkowski vectors?

Although this particular expression is new to me, I can see how that works. The effect is the same as doing a Lorentz transformation on $$F_{ab}$$ to the restframe of $$\tilde u$$, right?

These two examples use the full velocity 4-vector. This was not exactly what I was trying to trigger. I would like to focus in particular on the time component of the Minkowski velocity 4-vector and its properties c.q. physical interpretation.

5. Apr 30, 2005

### robphy

You are correct, although you might not see the implication. Let me clarify our notations.
$$\gamma_{uv}$$ means $$1/\sqrt{1-(V_{uv}/c)^2}$$, where $$V_{uv}$$ is the relative-velocity of v according to u.

Let "our frame" be $$\tilde t$$. In our frame, $$\tilde t$$ has coordinates $$(c,0,0,0)$$.
In our frame, $$\tilde u$$ has coordinates $$\gamma(u)(c,u_1,u_2,u_3)$$, which I write as $$\gamma_{tu}(c,u_1,u_2,u_3)$$,
and simliarly for $$\tilde v$$.

So,
$$\tilde u \cdot \tilde v=\gamma_{uv}=\gamma_{tu}\gamma_{tv}\{c^2-\vec u \cdot \vec v\}$$.
For simplicity, assume all motions are "in the x-direction" (and let's absorb the speed of light factor), so we can write
\begin{align*} \tilde u \cdot \tilde v=\gamma_{uv} &=\gamma_{tu}\gamma_{tv}\{1- uv\}\\ \frac{1}{\sqrt{1-V_{uv}^2}} &=\frac{1}{\sqrt{1-u^2}}\frac{1}{\sqrt{1-v^2}}(1- uv) \end{align*}.

If you do a little algebra and solve for $$V_{uv}$$, you'll find the relative-velocity formula:
$$V_{uv}=\frac{u-v}{1-uv}$$.

Yes.

The physical interpretation is that it is the time-dilation "gamma factor" relating the observer with 4-velocity u with the observer with 4-velocity v.

6. Apr 30, 2005

### Mortimer

Interesting. I've seen a number of different derivations of the velocity addition formula but not this one. Most I've seen use what I call for myself the "division of derivatives of the classical transformation equation" (e.g. "University Physics" uses it). Another common approach uses hyperbolic functions (so called pseudo-angle method; see e.g. Foster&Nightingale). Feynman uses yet another approach with a little math and a little reasoning in his lectures on physics. And yet even Einstein himself uses again a different method in his original paper although I'm not sure if that one could be related to the hyperbolic functions. Kind of curious. There seems not to be much consensus on the proper method. One of the textbooks even changed the method between two succesive editions.

But I'm drifting away from the original topic.
I'm not sure if I'm happy with this. I think what you are saying is that gamma is a tool that relates and controls the individual components in and amongst the 4-vectors. Perhaps that is all there is to it and I am trying to find meanings for the time component in the velocity 4-vector that do not exist. If we purely look at the time-velocity component in the observer's rest frame it equals c. That seems like a very concrete thing. The observer moves with velocity c in the time dimension. Things go wrong if the observers calculates the time-velocity component of a moving object. It increases above c! It may be just a mathematical (Minkowski) curiosity but I'm not convinced. You would expect it to decrease, as shows from the clock slowing down.
Am I looking for ghosts?

Last edited: Apr 30, 2005
7. Apr 30, 2005

### robphy

Since $$\tilde u$$ is timelike, $$0 < \tilde u \cdot \tilde u =\gamma^2(c^2-\vec u \cdot \vec u)$$. So, it follows that $$\gamma^2\vec u\cdot \vec u < (\gamma c)^2$$. With the speed of light as limit of the spatial-speed of an observer (i.e., the magnitude of the spatial-component of $$\tilde u$$), we must expect that the magnitude of the temporal-component must be at least $$c$$. Geometrically, it simply says that: for the 4-velocity to be inside the light cone, the magnitude of the temporal component must be larger than that of the spatial component.

8. Apr 30, 2005

### Mortimer

I agree that this makes (mathematical) sense in Minskowski space. The temporal component can have no physical meaning in the sense that we are accustomed to with spatial velocity. Even the spatial component in the Minkoswki 4-vector is not the velocity $$\vec u$$ but $$\gamma \vec u$$ and $$\gamma \vec u > \vec u$$. So the whole "velocity" 4-vector is not like any real velocity at all, nor are any of its components. They are even not constructed like we construct velocities in every day life. I mean: normal velocity would be $$dx_\mu/dt$$. The 4-vector components are $$dx_\mu/d\tau$$. One could wonder why we call it a velocity vector at all....

9. Apr 30, 2005

### robphy

Consider yourself with 4-velocity $\tilde t$ and "moving observer" $\tilde v$ meeting at a common event O.

After a time of 1 second on each observer's clock, the event "1 second later along each worldline" is given by the 4-velocity multiplied by a time s="1 second".
The temporal displacement of $s\tilde v$ is $\gamma s$.
The spatial displacement of $s\tilde v$ is $\gamma V s$.
The ratio of spatial part to temporal part is $$\displaystyle\frac{\gamma V s}{\gamma s}=V$$
which one might call the "spatial velocity" of observer v, which is a spacelike vector orthogonal to observer t.

Symbolically, $$\tilde v= \gamma \tilde t + \gamma \tilde V$$.

In addition, in the "classical limit" this 4-velocity becomes
$$\tilde v_{classical} = \tilde t + \tilde V$$,
which, when multiplied by s, corresponds to a temporal displacement of s for any classical 4-velocity and a spatial displacement of Vs. In other words, in the classical limit, the spatial component is the ordinary Newtonian velocity vector and the temporal component reflects the absolute-time structure of Galilean relativity. (A similar thing happens for the 4-momentum.)

Hopefully, these comments help to justify the use of the term 4-velocity vector.

10. May 1, 2005

### Mortimer

So the spatial velocity $$V$$ is the ratio of the spatial and temporal components of the relativistic displacement of observer v. That indeed sounds very familiar in terms of ordinary velocity calculations (i.e., $$dx/dt$$).
It is obviously a result of the fact that the gamma's cancel in the ratio, which in fact hides the effects of $$\tau$$ on the individual components of $$dx_\mu/d\tau$$.
Still it puzzles me that the displacement 4-vector $$\tilde x$$ is $$(ct, x_1, x_2, x_3)$$ but it requires $$d\tilde x/d\tau$$ to get the velocity 4-vector. From the perspective of the observer t that doesn't seem to make sense. After all, any velocity calculations he does should be using $$dt$$ instead of $$d\tau$$. $$\tau$$ is simply not part of his "environment" or "tools". But OK, I don't want to push it too far. It is just the way Minkowski 4-vectors have to be defined in order to yield invariants. Thanks anyway for your patience and efforts!

11. May 8, 2005

### pmb_phy

Hi rob

Nice to see you posting again. Either I've been very absent lately or you have.

If I have the 4-momentum P of a particle then the 4-velocity of an observer will get you to the time component of P. However the spatial components are coordinate dependant and as such one requires more than the observers 4-velocity. One requires the coordinate basis vectors of the particular coordinate system used. After all when one measures a physical object one is required to measure components. The spatial components of a tensor require the basis vectors of the particular coordinate system used.

However, that said, if one is seeking only the 3-vector associated with the spatial component then I agree. In fact MTW has a list of some definitions somewhere (lack of reference due to laziness as usual).

..and is a scalar.
.. and is a 4-vector. This is known as the electric field 4-vector. Not to be confused with the electric field 3-vector. There is a magnetic field 4-vector too which is similar to this (there is another tensor, the Levi-Civita tensor, required to define it). Its as wise to make this distinction as it is wise to distinguish 3-momentum from 4-momentum.

Nice to have you back rob. Its rare to find people who know what the electric field 4-vector is.

Pete

Last edited: May 8, 2005
12. May 8, 2005

### Chronos

It also makes sense in FRW space.

13. May 8, 2005

### pmb_phy

This is the position 4-vector, not the displacement 4-vector. The infinitesimal displacement 4-vector is (c dt, dx_1, dx_2, dx_3). The position 4-vector is only defined in flat spacetime whereas the infinitesimal displacement 4-vector is always defined. The position 4-vector is defined as a finite displacement from the event which you choose to call the "origin" of the coordinate system. The position 4-vector is a Lorentz 4-vector (not a generalized 4-vector) whereas the infinitesimal displacement 4-vector is not; its a generalized 4-vector.

In component languagte - A Lorentz 4-vector is any object whose components transform in the same way as the components of the position 4-vector which serves as a prototype.

In geometric language - A Lorentz 4-vector is a linear map from position 1-forms to scalars.

The reason for dividing the dx by d(tau) is that the result will be a 4-vector. Had you divided by dt then the result would not be a 4-vector.

Pete

Last edited: May 8, 2005
14. May 9, 2005

### robphy

Hi Pete.
Yes, I've been busy.

I agree.
Of course, two non-proportional timelike 4-vectors (say the particle's 4-momentum P and observer's 4-velocity u) define a plane. While the Minkowski scalar-product yields the time-component of P, it also yields the spatial-component along a "natural" spatial direction, namely the spacelike relative-velocity 4-vector [along the other direction in this plane]. In other words, these two non-proportional timelike 4-vectors already define two of the (n+1) basis-vectors in an (n+1)-Minkowski space. Of course, one can choose any basis one wants. However, this construction already provides two, which may be all that is needed for the discussion.

Yes, although I would be careful to always clarify that this $$E_b=u^aF_{ab}$$ is "the electric field 4-vector observed by u".
There is of course the electromagnetic field $$F_{ab}$$, which needs no clarification.

As you know, 3-vectors are observer-dependent projections of 4-vectors, which [in my opinion] should be avoided for as long as possible until the end of a calculation, if they are needed at all. I feel this avoidance is not unlike "waiting until the end to plug in numerical values". Of course, one may have a lot of possibly-unfamiliar and possibly-ugly algebra to deal with... but at least one is confident that the calculation is coordinate-independent. The 4-vector result is "something that may be interpreted by introducing an observer's basis". I prefer this to a temporal-and-3-vector result "that needs to be transformed from one observer's basis to another".

15. May 9, 2005

### pmb_phy

Since "electric field 4-vector" can have no other meaning it can be assumed that its what one means. That's why I've chosen to ignore it. This was the convention that Thorne and Blanchard chose after they defined it and then used it. This text is online at
http://www.pma.caltech.edu/Courses/ph136/yr2002/index.html

See chapter 1. Its nice for those who are unfamiliar with this stuff and don't have literature at home.

Pete

16. May 9, 2005

### robphy

On page 39 of http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf ,
they say
which is in agreement in spirit with what I suggested earlier
Of course, if it is clear to you, you may omit the clarification.

17. May 9, 2005

### pmb_phy

Thank you.

Pete

18. May 11, 2005

### pmb_phy

This was never a problem for me in the past. Even as an undergrad. This also smacks of something specific, i.e. it seems that in such a case you're only looking to get an end tensor relation. Suppose its the end tensor relation that you're starting with?

Consider a charged nucleus that has a finite diameter. The charge is being accelerating in the x direction where E(x) is not constant. What is the energy if you don't neglect the size and you want to write an expression which is theortically accurate as possible while remaining within the classical domain? Hint: stress contributes to the energy as well as the mass m and in this case theory says that E does not equal mc^2.

Pete