# Four-vectors and related concepts

1. Jan 31, 2009

### Staff: Mentor

The Wikipedia pages on http://en.wikipedia.org/wiki/Four-vector" [Broken]) are not wonderful, but they are a good starting point. I don't know exactly how far your background in these concepts extends, so I apologize if I go over stuff you already know.

Basically, starting with the Lorentz transform we can notice two important things:
1) time and space are not entirely separate entities but one frame's time gets split into another frame's space and vice versa.
2) there is a notion of "distance" called the http://en.wikipedia.org/wiki/Spacetime#Space-time_intervals" which also mixes space and time together and is agreed upon by all reference frames (i.e. is invariant under the Lorentz transform).

What we would like is a convenient way to keep things organized so that we can easily keep track of the things that everyone agrees on and easily determine how any frame sees something. This is exactly what the four-vector approach accomplishes. If we take our normal space coordinates that we have all seen since introductory physics, (x,y,z), and we add a time coordinate, (ct,x,y,z), then we have four-vectors and spacetime. Now we can write the http://en.wikipedia.org/wiki/Lorentz_transformations#Matrix_form" and easily switch between reference frames. This is very useful in itself for figuring out how different frames look at times and distances.

We can also consider the spacetime interval (aka Minkowski norm) as the length of the four-vector. One important use of this approach is that in an object's rest frame all of the space coordinates are 0, so the spacetime interval is immediately seen to be the time in an object's rest frame or its "proper time". Since the interval is invariant then we see that all frames agree on the proper time along any worldline. This is important because we can take any four-vector, use the derivative wrt this proper time, and come out with another four-vector. That new four-vector will transform according to the same Lorentz transform matrix as above, and the norm of the new four-vector will also be invariant.

So, in Newtonian physics velocity is the time derivative of position. Similarly, if we start with the position four-vector, (ct,x,y,z), and take the derivative wrt proper time then we get the four-velocity. Note, the norm of the four-velocity is always c, and for an object at rest the four-velocity is (c,0,0,0). In other words, a stationary object is still "moving" through time.

Now, if we multiply the four-velocity by the rest mass we get the four-momentum. For an object at rest we have p=(mc,0,0,0). Using the famous E=mc² equation you can re-write the above as, p=(E/c,0,0,0) for an object at rest. Thus energy is seen as "momentum" through time, and energy and momentum are seen to have the same relationship to each other as time and space have. Energy and momentum are not entirely separate entites and one observer's energy gets split into another observer's momentum in the same way as space and time above, and using the same Lorentz transform matrix. The norm of this four-momentum is the invariant rest mass, and the timelike component is the energy or the relativistic mass.

I apologize for the length. I am sure that my rambling explanation left more than one question unanswered, so please don't hesitate. This is an important subject and my discovery of four-vectors is what finally made SR "click" for me, so I am certainly willing to try and help.

Last edited by a moderator: May 4, 2017
2. Jan 31, 2009

### Naty1

Re: Four-vectors

Dalespam..thanks several of your comments help...my Matrix math is from many years ago so I can sometimes follow a mathematical development/derivation with explanation but Wiki is just too notational dependent for me at the moment,,,and four vector algebra a bit more so...I'm a bit "rusty" in other words!!!

yet,I have gotten prostestations on this forum when I have said the same thing....it IS a simple way to view maximum speed in this universe...That's a slick way to look at it...and I knew it was a valid viewpoint...

Yes....that's what I was questioning in my original post....as time and space become mixed so do energy and momentum....I believe that was your point in the other thread....

For anyone interested in this discussion, an introduction to Dalespams's post here is insightfully presented by Richard Feynmann in under 20 pages or so....with just high school algebra...it's from his world famous series of introductory physics lectures..

3. Jan 31, 2009

### Naty1

Re: Four-vectors

I just reread you post and noted:
yes, Feynmann notes (pg 109) that E2 - p2 = m0^2 (where c=1) is invariant in all coordinate systems...so in a coordinate system moving with the particle (in which the particle is standing still) it would have no momentum so all it's energy is rest energy mo...

This stuff is subtle enough it's no wonder we have so much trouble communicating ideas/concepts...it's tough without being able to rephrase to confirm....math is clear in that regard but the implications/interpretations are not!!!

4. Jan 31, 2009

### Staff: Mentor

Re: Four-vectors

Yes, a lot of people don't like the four-velocity. It does have some mathematical problems. For instance, you can add two position four-vectors and get another valid one, same with four-momenta. But since the four-velocity must always have a norm of c the sum of two four-velocities is not in general a four-velocity. I think it is an ok stepping-stone to the four-momentum but you do have to be more careful with the four-velocity than some other concepts.

5. Jan 31, 2009

### robphy

Re: Four-vectors

That "the sum of two 4-velocities is not a 4-velocity"
is analogous to the Euclidean fact that
"the sum of two unit-vectors is generally not a unit-vector".

For me, the 4-velocity of an observer
is the key to neatly expressing tensorial quantities
in terms of components according to that observer.

6. Jan 31, 2009

### Staff: Mentor

Re: Four-vectors

To see explicitly what the Lorentz transformation for energy and momentum looks like:

The Lorentz transformation for x and t is commonly stated in the form

$$x^{\prime} = \gamma (x - vt)$$

$$t^{\prime} = \gamma (t - vx/c^2)$$

Rewrite it a bit by defining $\beta = v/c$ and using ct instead of t. This gives us the Lorentz transformation for the position 4-vector (ct, x, 0, 0).

$$x^{\prime} = \gamma (x - \beta ct)$$

$$ct^{\prime} = \gamma (ct - \beta x)$$

The 4-momentum is usually written with the components $(E/c, p_x, p_y, p_z)$, but I'll multiply it through by c to get rid of the division, and because pc fits naturally into the equation $E^2 = (pc)^2 + (m_0 c^2)^2$. In parallel with the position/time Lorentz transformation shown above, the momentum/energy Lorentz transformation for $(E, p_x c, 0, 0)$ is:

$$p_x^{\prime} c = \gamma (p_x c - \beta E)$$

$$E^{\prime} = \gamma (E - \beta p_x c)$$

(replacing x with $p_x c$ and ct with E)

7. Feb 1, 2009

### Fredrik

Staff Emeritus
Four-velocity

Let $u=(u^0,u^1,u^2,u^3)=(u^0,\vec u)$ be the components of a tangent vector of the world line of a massive particle in some inertial frame S. It's easy to see that this four-vector contains all information about the particle's velocity $\vec v$ in that frame. The projection of u onto the hyperplane defined by $x^0=0$ is clearly in the same direction as $\vec v$, so there exists an A>0 such that $\vec u=A\vec v$. The speed v is completely determined by the angle the tangent vector makes with the 0 axis. We have $v=|\vec u|/u^0$. But this means that

$$u=(u^0,\vec u)=(|\vec u|/v,A\vec v)=(A|\vec v|v,A\vec v)=A(1,\vec v)$$

The magnitude of this four-vector is irrelevant (it has nothing to do with the velocity), so we can define the four-velocity of the particle in frame S at that point on the world line to be the u that satisfies $u^2=-1$.

$$-1=A^2(-1+\vec v^2)=-\frac{A^2}{\gamma^2}\implies A=\gamma$$

$$u=\gamma(1,\vec v)$$

This is equivalent to simply defining the components of u to be (1,0,0,0) in the co-moving inertial frame, and then Lorentz transforming to the frame where the velocity of the particle is $\vec v$.

It's also equivalent to defining the four-velocity to be the tangent vector we get when the world line is parametrized by proper time. What this means is that if we define the components of u by

$$u^\mu=\frac{dx^\mu}{d\tau}$$

where $\tau$ is the proper time of the particle's world line (from some irrelevant starting point to the point we're interested in), we get the standard normalization ($u^2=-1$) automatically.

Last edited: Feb 1, 2009
8. Feb 1, 2009

### robphy

Re: Four-velocity

In a (1+1)-spacetime, that neat but possibly cryptic-to-beginners expression can
be expressed in terms of rapidities (the Minkowskian-angle between future-timelike vectors) as:
$$u=(\cosh\theta,\sinh\theta)=\cosh\theta(1,\tanh\theta)$$.
This allows one to use a variant of one's Euclidean intuition in doing calculations and interpreting geometrically.
For instance, one can see that this is a future-timelike unit-vector in Minkowski spacetime.
One can also more easily interpret the projections of an arbitrary 4-vector into spatial- and temporal-parts according to that observer with 4-velocity $$u$$.

Last edited: Feb 1, 2009
9. Aug 4, 2009

### Fredrik

Staff Emeritus
Re: Four-velocity

I'm bumping this because I've been linking to this post, and I just spotted a mistake. The above should be

$$u=(u^0,\vec u)=(|\vec u|/v,A\vec v)=(A|\vec v|/v,A\vec v)=A(1,\vec v)$$

10. Aug 7, 2009

### Austin0

Re: Four-vectors

Hi When I initially looked at the basics of the four-vector it was quite obvious that it would be indispencable for any real world 4-D application like electrodynamics etc. within the context of SR. But as I have neither the tools nor the intention of any of these applications it didnt seem like it was neccessary for the fundamental understanding of SR , where it is mainly 2 -d coordinate measurments limited to the congruent ( x , x' ) axis and time. It appeared that in this situation it reduced to the fundamental formulations of the Lorentz math and my time would be better spent in other areas.
Since then I have repeatedly encoutered it in many threads, among them your recurrent mentions of their importance and utility and have begun to wonder if my initial impressions may have led me astray.
AS someone who actually understands the use and application I hope you can help me.
1) Is it even possible to derive real benefit from the study without a comparable knowledge and study of higher math and other physics????
2) Do you consider it neccessary to a fundamental understanding and application of SR???

My time, like for most, is limited but I would definitely make the time to pursue this if it was crucial or offered greater basic understanding. SO your knowledgeable advice would be welcome. Thanks

11. Aug 8, 2009

### Staff: Mentor

Re: Four-vectors

I don't know exactly what you would classify as "higher math", but I think that you need a solid understanding of Euclidean geometry and algebra in order to understand Minkowski geometry (4-vectors). This, together with Newtonian mechanics, is a sufficient background for learning problems with inertial objects or with objects undergoing instantaneous accelerations or collisions. If you want to do more realistic physics problems then you should also know calculus.

Yes. I can only give you advice based on my personal experience. I struggled with SR off and on for about 7 years using the algebraic approach. Then I quite literally stumbled on the Minkowski geometric approach. Suddenly all of the things that I couldn't understand previously just "clicked" into place in my mind. I especially liked the clean unification of different concepts that I had previously seen as separate (space and time, energy and momentum, etc.). So, at least in my case, yes, an understanding of Minkowski geometry was necessary to a fundamental understanding and application of SR.

12. Aug 8, 2009

### robphy

Re: Four-vectors

In the concluding remarks of Scott Walter's
"Minkowski, Mathematicians and the Mathematical Theory of Relativity"
http://www.univ-nancy2.fr/DepPhilo/walter/papers/mmm.xml [Broken]
he writes...

Last edited by a moderator: May 4, 2017
13. Aug 10, 2009

### Austin0

Re: Four-vectors

I was afraid that this is what you would say. It is not that I am lazy [at least not totally] but my time is tight. So I was hoping you would tell me I could pass on this one.
To say my math is rusty and largely forgotten is overly charitable so theres lots of work ahead. But thanks for your time and advice

14. Aug 10, 2009

### Fredrik

Staff Emeritus
Re: Four-vectors

If you're studying it just for fun, and want to gain as much understanding as possible with as little effort as possible, then you should focus on spacetime diagrams and ignore most of the algebraic stuff for now.

15. Aug 12, 2009

### Austin0

Re: Four-vectors

Hi Fredrik I have been meaning to ask you a couple of questions relevant to this .

When I first encountered your reference to the intergrated path length of an accelerating world-line I immediately pictured a basic euclidean geometric form based on partitioning orthoganal to a projected linear world-line between the endpoints.
Since encountering Rindler coordinates I have realized that might have been naive.

1) If the integration technique you used was non-euclidean is there any simple geometric or topological conception you can verbally describe?

2) From what I have been able to grasp so far , it seems that the Rindler system creates a local non-eucliean sub-sector of Minkowski space. Is this at all accurate??

3) I am unclear on whether this means:
A) Rindler coordinates are simply a convention ,tranforming Minkowski coordinates and placing them in a different frame as a mathematical convenience for simplified application of maths???
OR
B) They themselves are an expession of physics and lead to different predictions of the behavior of particles and their interactions over time, which are then transformed back into Minkowski coordinates???

4) Hyperbola -------- I can see the ellipsoid as an intrinsic geometric shape as appled to differing inertial frames but dont understand the relation of the hyperbola as a fundamental shape. Does it arise solely in the context of lines of simultaneity or is there a more fundamental relationship???

Also ; is the curve derived from the gamma function hyperbolic??

I have a fair understanding of the fundamental concepts and principles of calculus , the how and why it works , but am in the process of learning the notation and specific operations neccessary for even the most simple application.

So any direct mathematical explanations would be wasted on me at this time.

SO my interest is "for fun" in that nobodie's paying me, but quite serious in terms of the time and mental effort I am putting into it. Having virtually totally forgotten trig I am also into a crash refresher course there too.

Thanks for your time and help

16. Aug 12, 2009

### Staff: Mentor

Re: Four-vectors

If your time is tight then I would definitely further emphasize the value of Minkowski geometry. I would have literally saved myself years if I had recieved the same advice early on. (and I agree completely with Fredrik about prioritizing spacetime diagrams)

17. Aug 13, 2009

### Fredrik

Staff Emeritus
Re: Four-vectors

The definition of the proper time integral is what you expect it to be, i.e. you pick N points on the curve and imagine a curve consisting of N+1 straight lines, each one going from one of those points to the next. Then you calculate the integral over that curve instead, and finally you take the limit N→∞ in a way that ensures that the maximum (euclidean) length of the set of straight line segments goes to 0.

The contribution to the integral from each straight line segment is $$\sqrt{dt^2-dx^2-dy^2-dz^2}$$, where the "d-somethings" represent how much the relevant coordinate (t,x,y or z) changes from one endpoint to the other. Note that we can take "dt" outside the square root:

$$\int\sqrt{dt^2-dx^2-dy^2-dz^2}=\int dt\sqrt{1-\vec v^2}=\int\frac{dt}{\gamma}$$

I don't think there's an easier way to think of it than this.

I would say A, but I'm not 100% clear on what B means. One thing you should realize is that a coordinate system is just a mathematical function that assigns four numbers (coordinates) to each event in some region of spacetime. So a coordinate change can't change a region of a manifold (e.g. spacetime in SR) from being flat to being curved for example, or from being Euclidean to non-Euclidean. Minkowski space isn't Euclidean by the way. It's Lorentzian. (I don't know if that term is recognized by mathematicians). However, if you consider a set of points in 1+1-dimensional Minkowski space, that all have the same value of the Rindler "x"-coordinate, what you get is a hyperbola, and that's a curved (one-dimensional) manifold. So the answer to your question 2 is "I would say no, but I guess we could say yes", since a coordinate change won't change the properties of spactime, but it gives you an easy way to specify a submanifold that does have some new properties, like non-zero curvature.

The hyperbolas you see in a diagram of Rindler coordinates represent constant proper acceleration (with a different constant associated with each hyperbola). Proper acceleration is the coordinate acceleration (i.e. the second derivative of the spatial coordinates with respect to the time coordinate) in a co-moving inertial frame. For a derivation of how that gives us a hyperbola, I suggest you check out DrGreg's posts in this thread. (Math is unavoidable here I'm afraid).

Last edited: Aug 13, 2009
18. Aug 13, 2009

### Austin0

Re: Four-vectors

Originally Posted by Austin0
2) From what I have been able to grasp so far , it seems that the Rindler system creates a local non-eucliean sub-sector of Minkowski space. Is this at all accurate??

3) I am unclear on whether this means:
A) Rindler coordinates are simply a convention ,tranforming Minkowski coordinates and placing them in a different frame as a mathematical convenience for simplified application of maths???
OR
B) They themselves are an expession of physics and lead to different predictions of the behavior of particles and their interactions over time, which are then transformed back into Minkowski coordinates???

Couldn't you say Lorentzian space is a superposition of two Euclidean ,cartesian coordinate spaces having different metrics?
Both having uniform metrics ,there are no mapping problems,, but it makes them less directly readable, ie: some information requires conversion.

Question B above which I didnt make clear.
In another context. The Lorentz transformation does not simply assign different values to Galilean coordinates , it also assigns different relationships between those assigned locations and times. Galilean transforms have zero physical implications outside of the invariance of Newtonian mechanics. Lorentzian maths produce predictions of a different physics. Eg: extended muon lifetimes. I understand the mindset that this can be looked on as a coordinate transformation. But isnt that really semantics?? Scientists on an isolated space lab, with no concept of other inertial frames, would still inevitably discover the Lorentz math through electrodynamics etc. They would just consider it physics. IMO

In any case that was my question. Does Rindler coordinates simply assign different coordinates values and context or does it also predict different relationships???
Different world-lines over time, that after calculation are then transformed back into Minkowski space-time locations???

Thanks for the link. It was very interesting and helpful. Without having a chance to really look at it , it seems to me that the curve described by the lorentz gamma function is an equilateral hyperbola. That rotated 45deg it would be congruent with a hyperbola having the same rectangular dimensions. It would also seem to fit the definition regarding asyptotes.
I will have to wait till I get home to really look at the math but is this basically accurate???
Did you ever complete your workup of an accelerating system from an inertial frame or were you only interested in the math principals involved???
Thanks again , you have been a real help.

Last edited by a moderator: Aug 13, 2009
19. Aug 13, 2009

### Austin0

Re: Four-vectors

I hope I am not bothering you with all these questions , thanks. The above seems quite clear, but from this, I dont see why the end result would be different from a sum of instantaneous relative velocities. It seems like just a different approach to the same end.
What am I missing???

One problem I am having with the hyperbola is ; in the drawings I have seem it appears that, interpreted in terms of Lorentzian space , it represents a motion in one direction which stops and then reverses direction which leads me to suspect I am not getting it.
That and the light-like asymptotes. What is the source and meaning of this concept???
Thanks again

20. Aug 13, 2009

### George Jones

Staff Emeritus
Re: Four-vectors

Maybe a Newtonian analogy will help. What happens if you throw a baseball straight up?