Four-vectors and related concepts

  • Thread starter Dale
  • Start date
  • Tags
    Concepts
In summary, the conversation discusses the concept of four-vectors and how they can be used to understand the relationship between energy and momentum in special relativity. The four-vector approach allows for a convenient way to keep track of the things that everyone agrees on and easily determine how any frame sees something. The conversation also touches on the concept of proper time and how it is invariant for all frames of reference, as well as the relationship between energy and momentum being similar to that of time and space. The conversation also mentions that the concept of four-vectors can be understood with just high school algebra and is introduced by Richard Feynman in his series of introductory physics lectures.
  • #1
35,441
13,719
In another thread Naty1 said:
Naty1 said:
I just happened to reread (Richard Feynmann, SIX NOT SO EASY PIECES) that replacing in the Lorentz transformations x with px (for momentum) and replacing t with E (for energy as mc2 yields the four vector momentum.

Is that what you are referring to here? If so, could you elaborate a bit further as I am trying to make some sense of the four vector momentum...whereas time in Lorentz is velocity and distance dependent (u,t) now energy is velocity and momentum dependent...

So it seems like energy and momentum transform/rotate into one another...yes? Is that what your comment implies??
The Wikipedia pages on http://en.wikipedia.org/wiki/Four-vector" ) 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:
Physics news on Phys.org
  • #2


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!

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.

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...

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

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 Feynman in under 20 pages or so...with just high school algebra...it's from his world famous series of introductory physics lectures..
 
  • #3


I just reread you post and noted:
"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..."

yes, Feynman 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


Naty1 said:
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, 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


DaleSpam said:
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.


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


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

[tex]x^{\prime} = \gamma (x - vt)[/tex]

[tex]t^{\prime} = \gamma (t - vx/c^2)[/tex]

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

[tex]x^{\prime} = \gamma (x - \beta ct)[/tex]

[tex]ct^{\prime} = \gamma (ct - \beta x)[/tex]

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

[tex]p_x^{\prime} c = \gamma (p_x c - \beta E)[/tex]

[tex]E^{\prime} = \gamma (E - \beta p_x c)[/tex]

(replacing x with [itex]p_x c[/itex] and ct with E)
 
  • #7
Four-velocity

Let [itex]u=(u^0,u^1,u^2,u^3)=(u^0,\vec u)[/itex] 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 [itex]\vec v[/itex] in that frame. The projection of u onto the hyperplane defined by [itex]x^0=0[/itex] is clearly in the same direction as [itex]\vec v[/itex], so there exists an A>0 such that [itex]\vec u=A\vec v[/itex]. The speed v is completely determined by the angle the tangent vector makes with the 0 axis. We have [itex]v=|\vec u|/u^0[/itex]. But this means that

[tex]u=(u^0,\vec u)=(|\vec u|/v,A\vec v)=(A|\vec v|v,A\vec v)=A(1,\vec v)[/tex]

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 [itex]u^2=-1[/itex].

[tex]-1=A^2(-1+\vec v^2)=-\frac{A^2}{\gamma^2}\implies A=\gamma[/tex]

[tex]u=\gamma(1,\vec v)[/tex]

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 [itex]\vec v[/itex].

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

[tex]u^\mu=\frac{dx^\mu}{d\tau}[/tex]

where [itex]\tau[/itex] 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 ([itex]u^2=-1[/itex]) automatically.
 
Last edited:
  • #8


Fredrik said:
[tex]u=\gamma(1,\vec v)[/tex]

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:
[tex]u=(\cosh\theta,\sinh\theta)=\cosh\theta(1,\tanh\theta)[/tex].
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 [tex]u[/tex].
 
Last edited:
  • #9


Fredrik said:
[tex]u=(u^0,\vec u)=(|\vec u|/v,A\vec v)=(A|\vec v|v,A\vec v)=A(1,\vec v)[/tex]
I'm bumping this because I've been linking to this post, and I just spotted a mistake. The above should be

[tex]u=(u^0,\vec u)=(|\vec u|/v,A\vec v)=(A|\vec v|/v,A\vec v)=A(1,\vec v)[/tex]
 
  • #10


DaleSpam said:
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.

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 necessary 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 necessary 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


Austin0 said:
1) Is it even possible to derive real benefit from the study without a comparable knowledge and study of higher math and other physics?
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.

Austin0 said:
2) Do you consider it necessary to a fundamental understanding and application of SR?
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


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
he writes...
"Minkowski's semi-popular Cologne lecture was an audacious
attempt, seconded by Göttingen mathematicians and their allies, to change
the way scientists understood the principle of relativity. Henceforth,
this principle lent itself to a geometric conception, in terms of the
intersections of world-lines in space-time. Considered as a sales pitch
to mathematicians, Minkowski's speech appears to have been very
effective, in light of the substantial post-1909 increase in mathematical
familiarity with the theory of relativity. Minkowski's lecture was also
instrumental in attracting the attention of physicists to the principle
of relativity...

"
 
Last edited by a moderator:
  • #13


DaleSpam said:
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.

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 there's lots of work ahead. But thanks for your time and advice
 
  • #14


Austin0 said:
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 there's lots of work ahead. But thanks for your time and advice
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


Fredrik said:
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.

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 applied to differing inertial frames but don't 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 necessary 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


Austin0 said:
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.
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 received the same advice early on. (and I agree completely with Fredrik about prioritizing spacetime diagrams)
 
  • #17


Austin0 said:
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?
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 [tex]\sqrt{dt^2-dx^2-dy^2-dz^2}[/tex], 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:

[tex]\int\sqrt{dt^2-dx^2-dy^2-dz^2}=\int dt\sqrt{1-\vec v^2}=\int\frac{dt}{\gamma}[/tex]

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

Austin0 said:
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?
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.

Austin0 said:
4) Hyperbola -------- I can see the ellipsoid as an intrinsic geometric shape as applied to differing inertial frames but don't 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??
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:
  • #18


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?

Fredrik said:
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.

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 isn't 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?

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)./

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:
  • #19


=Fredrik;2307199]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 [tex]\sqrt{dt^2-dx^2-dy^2-dz^2}[/tex], 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:

[tex]\int\sqrt{dt^2-dx^2-dy^2-dz^2}=\int dt\sqrt{1-\vec v^2}=\int\frac{dt}{\gamma}[/tex]
I don't think there's an easier way to think of it than this.

I hope I am not bothering you with all these questions , thanks. The above seems quite clear, but from this, I don't 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


Austin0 said:
it represents a motion in one direction which stops and then reverses direction which leads me to suspect I am not getting it.

Maybe a Newtonian analogy will help. What happens if you throw a baseball straight up?
 
  • #21


Austin0 said:
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.
You probably mean "cartesian product", not "superposition", but your idea still doesn't quite work. It would take some time for me to write down a definition of the term "Lorentzian" that's both accurate and general enough to work even for curved manifolds, and you wouldn't understand it because it requires some differential geometry, so it doesn't seem worth the effort. You should probably focus on special relativity, and in SR, we only care about one Lorentzian manifold. I'm of course talking about Minkowski space, which can be defined as [itex]\mathbb R^4[/itex] equipped with a function g defined by [itex]g(u,v)=-u^0v^0+u^1v^1+
u^2v^2+u^3v^3[/itex]. (Those are component indices, not exponents).

Austin0 said:
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 isn't 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
You seem to think that I would consider the switch from Galilei to Lorentz a coordinate transformation!? I certainly would not. Each Galilei transformation corresponds to a coordinate system on non-relativistic spacetime, and each Lorentz transformation corresponds to a coordinate system on special relativistic spacetime. Rindler coordinates is another type of coordinate system on special relativistic spacetime. So a change from inertial coordinates to Rindler coordinates is a good example of a coordinate change, but the switch from Galilei to Lorentz is a switch to a different mathematical model, and to a different theory of physics.

Austin0 said:
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?
Rindler coordinates is a coordinate system on (a subset of) Minkowski space. You probably meant "transformed to an inertial frame". The description of a set of events will be very different in terms of the inertial coordinates, but it will include the same events as a description in terms of any other type of coordinate system, including Rindler coordinates. E.g. if one coordinate system describes two balls thrown towards each other as missing each other, no coordinate system will describe them as colliding. This is obvious when you understand what a coordinate system is.

Austin0 said:
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?
What curve would that be? Gamma as a function of v? Even if that's a hyperbola, I don't see the relevance, and it seems to me that you have to plot [itex]\gamma[/itex] as a function of [itex]\gamma v[/itex] to get a hyperbola, since [itex]\gamma^2-\gamma^2v^2=1[/itex].

Austin0 said:
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.
Yes, I did. Well, DrGreg sort of did it for me, but I worked through the details and wrote down my version of it in my personal notes.
 
  • #22


Austin0 said:
I hope I am not bothering you with all these questions , thanks. The above seems quite clear, but from this, I don't 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?
Why would [tex]\int\frac{dt}{\gamma}[/tex] be a sum of velocities?

Austin0 said:
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's what constant proper acceleration for all eternity (both past and future) looks like. No matter what inertial frame the diagram represents, the speed will be 0 at some point.

Austin0 said:
That and the light-like asymptotes. What is the source and meaning of this concept?
The meaning is that an object undergoing constant proper acceleration forever will get arbitrarily close to light speed, but never reach it. Isn't that what we should expect, even without doing the math?
 
  • #23


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

I see the relevance of the analogy to my interpretation but the reason I questioned my own interpretation was,, it didnt seem to fit the circumstances.
As far I know, the baseball in Newtonian terms is not in a state of uniform acceleration throughout , whereas the world-lines in the Rindler context are undergoing a constant proper acceleration.
As far as I understand GR, the baseball wouldn't be accelerating in the sense that the points in Rindler coordinates would , through the application of force, except during the initial upward phase.
So I appreciate the help but I am going to need more study and thought on this one.

Thanks
 
  • #24


Austin0 said:
As far I know, the baseball in Newtonian terms is not in a state of uniform acceleration throughout

Yes, in Newtonian terms, it is in a state of uniform acceleration throughout! The acceleration of the baseball is 9.8 m/s^2 in the down direction, both on the way up and on the way down.

On the way up, acceleration (down) and velocity (up) are in opposite directions, so the speed (magnitude of velocity) decreases. On the way down, acceleration (down) and velocity (down) are in the same direction, so speed increases.
Austin0 said:
As far as I understand GR, the baseball wouldn't be accelerating in the sense that the points in Rindler coordinates would , through the application of force, except during the initial upward phase.

I'm not sure what you mean; according to GR, the 4-acceleration of baseball is zero, both going up and coming down. Still, my Newtonian analogy is a good one.
 
  • #25


Originally Posted by Austin0
The above seems quite clear, but from this, I don't 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?

=Fredrik;2307691]Why would [tex]\int\frac{dt}{\gamma}[/tex] be a sum of velocities?
I didnt mean to suggest it was the sum of velocities. I was saying that I didnt see why it would produce a different result from the approach that calculated on the basis of summed infintesimal amounts of dilation resulting from very close samplings of instantaneous velocities. As you know I am making guesses regarding the actual process based purely on hearing the term in this forum. Where it was commented that the end result was different.
I was simply asking why.

That's what constant proper acceleration for all eternity (both past and future) looks like. No matter what inertial frame the diagram represents, the speed will be 0 at some point.
So in actual application where the world -line starts at zero only the top of the hyperbola is relevant?

The meaning is that an object undergoing constant proper acceleration forever will get arbitrarily close to light speed, but never reach it. Isn't that what we should expect, even without doing the math?

DUH ! You definitely have that right. I will take this as a clear sign its bedtime

Thanks
 
  • #26


It might be worth pointing out an analogy with ordinary Euclidean (non-relativistic) 2D space. You can measure 2D space using Cartesian (x,y) coordinates (horizontal and vertical), a square grid. Or you can measure 2D space using polar [itex](r,\theta)[/itex] coordinates (radius and angle), a circular grid. Either way is valid, and they both describe the same 2D space, but geometric equations take a different form in the two coordinate systems.

You measure distances using either of the equations

[tex]ds^2 = dy^2 + dx^2[/tex]
[tex]ds^2 = r^2 \, d\theta^2 + dr^2[/tex]​

For example, if a curve is given by y as a function of x, its curved length is given by

[tex]\int \left( \sqrt{\left( \frac {dy}{dx} \right)^2 + 1} \right)\, dx [/tex]​

If a curve is given by both y and x as functions of [itex]\lambda[/itex], its curved length is given by

[tex]\int \sqrt{\left( \frac {dy}{d\lambda} \right)^2 + \left( \frac {dx}{d\lambda} \right)^2 } \, d\lambda [/tex]​

If a curve is given by both r and [itex]\theta[/itex] as functions of [itex]\lambda[/itex], its curved length is given by

[tex]\int \sqrt{r^2 \left( \frac {d\theta}{d\lambda} \right)^2 + \left( \frac {dr}{d\lambda} \right)^2 } \, d\lambda [/tex]​

The two coordinate systems are related by

[tex]x = r \cos \theta[/tex]
[tex]y = r \sin \theta[/tex]​

So much for 2D Euclidean geometry. Now let's switch to 4D Minkowski geometry, or, rather, to keep it simple, 2D Minkowski geometry. We'll ignore 2 space dimensions and consider just one space dimension and one time dimension.

You can measure this spacetime using Minkowski (x,t) coordinates (distance and time of an inertial observer), a square grid. Or you can measure this spacetime using Rindler [itex](R,\Theta)[/itex] coordinates (distance and coordinate time of an accelerating observer), a hyperbolic grid. Either way is valid, and they both describe the same 2D spacetime, but equations take a different form in the two coordinate systems.

Here I am using deliberately non-standard symbols for Rindler coordinates, to make the analogy clearer.

You measure "proper times" using either of the equations

[tex]d\tau^2 = dt^2 - dx^2[/tex]
[tex]d\tau^2 = R^2 \, d\Theta^2 - dR^2[/tex]​

I am assuming a couple of things here. First we measure distance and time in units that make c = 1 (e.g. 1 light-year per year). Second, that the proper acceleration of the observer is 1 (e.g. 1 light-year per year2, which by an amazing cosmic coincidence is alarmingly close to 1g). The observer is located at R = 1, and coordinate time [itex]\Theta[/itex] equals the observer's own proper time (as measured by his clock), but runs faster or slower than clocks at other fixed R values. Two events that share the same [itex]\Theta[/itex] coordinate are simultaneous according to a co-moving inertial observer i.e. an inertial observer who is momentarily traveling at the same speed as the accelerating observer.

The two coordinate systems are related by

[tex]x = R \cosh \Theta[/tex]
[tex]t = R \sinh \Theta[/tex]​

(If you're not familiar with "cosh" and "sinh", look up Hyperbolic function.)

That's probably enough to think about for now, but consider the similarities between the two coordinate systems for 2D Euclidean space, and the two coordinate systems for 2D Minkowski space.
 
Last edited:
  • #27


George Jones said:
Yes, in Newtonian terms, it is in a state of uniform acceleration throughout! The acceleration of the baseball is 9.8 m/s^2 in the down direction, both on the way up and on the way down.

I was assuming that at the point of equilibrium ,,the momentum wrt the Earth was zero,,,
= maximum potential energy but no motion. so that there two acceleration phases not a single continuous acceleration. SO am I correct that; in this picture , even at rest , without motion, a particle is in a state of unform acceleration due to the force acting on it?


________________________________________
Originally Posted by Austin0
As far as I understand GR, the baseball wouldn't be accelerating in the sense that the points in Rindler coordinates would , through the application of force, except during the initial upward phase.
________________________________________

I'm not sure what you mean; according to GR, the 4-acceleration of baseball is zero, both going up and coming down. Still, my Newtonian analogy is a good one

That is what I meant except I included the initial toss as an acceleration through force.
In GR would this not be viewed as acceleration?
Yes your analogy was good . I just couldn't put it in context because I didnt understand then ,that the hyperbolas were extended into the indefinite past as well as the future so I was looking at it as starting from zero and then continuous and constant from there.
Thanks
 
  • #28


Originally Posted by Austin0
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.
____________________________________________________________________________--

=Fredrik;2307676]You probably mean "cartesian product", not "superposition", but your idea still doesn't quite work. It would take some time for me to write down a definition of the term "Lorentzian" that's both accurate and general enough to work even for curved manifolds, and you wouldn't understand it because it requires some differential geometry, so it doesn't seem worth the effort. You should probably focus on special relativity, and in SR, we only care about one Lorentzian manifold. I'm of course talking about Minkowski space, which can be defined as [itex]\mathbb R^4[/itex] equipped with a function g defined by [itex]g(u,v)=-u^0v^0+u^1v^1+
u^2v^2+u^3v^3[/itex]. (Those are component indices, not exponents).

Agreed that it wouldn't be worth your effort. ANd yes I am so used to thinking of a limited 2-d Minkowski space that I used "superposition" which only makes sense for a 2-d Minkowski diagram and only used the term Lorentzian because you had applied it to Minkowski space.
AM I correct in reading the [g(u,v) =...] expression as a statement of direct one to one correspondence and mapping?Or does it mean something else altogether??
When you mentioned curved manifolds was that applicable to MInkowski space or only to a generalized Lorentzian construct?



Originally Posted by Austin0
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 isn't 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
___________________________________________________________________

You seem to think that I would consider the switch from Galilei to Lorentz a coordinate transformation!? I certainly would not. Each Galilei transformation corresponds to a coordinate system on non-relativistic spacetime, and each Lorentz transformation corresponds to a coordinate system on special relativistic spacetime. Rindler coordinates is another type of coordinate system on special relativistic spacetime. So a change from inertial coordinates to Rindler coordinates is a good example of a coordinate change, but the switch from Galilei to Lorentz is a switch to a different mathematical model, and to a different theory of physics.

I had no thought whatsoever like that. As stated I was simply clarifing context for the question. Having reread my post and your responce I see no disagreement , you simply expanded on exactly the relevant parameters of the question.

Originally Posted by Austin0
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?
__________________________________________________________________

Rindler coordinates is a coordinate system on (a subset of) Minkowski space. You probably meant "transformed to an inertial frame". The description of a set of events will be very different in terms of the inertial coordinates, but it will include the same events as a description in terms of any other type of coordinate system, including Rindler coordinates. E.g. if one coordinate system describes two balls thrown towards each other as missing each other, no coordinate system will describe them as colliding. This is obvious when you understand what a coordinate system is.
__________________________________________________________________
This the understanding of coordinate systems I was operating under.
Is it not true that two baseballs which are traveling parallel in a Euclidian system would never intersect but that in some non-euclidean systems they might very well collide at some point in time?
In this situation, with a series of equidistant points on x' which converge to a degree in Rindler space, the question is would they converge or remain parallel without the inclusion and use of the Rindler system?

What curve would that be? Gamma as a function of v? Even if that's a hyperbola, I don't see the relevance, and it seems to me that you have to plot [itex]\gamma[/itex] as a function of [itex]\gamma v[/itex] to get a hyperbola, since [itex]\gamma^2-\gamma^2v^2=1[/itex].
Yes gamma as a function of v. YOu may be right and even if true having no significance whatever. On the other hand if true it would be interesting and symmetry and correspondence are not always just coincidence
xy=(a^2)/2
Thanks
 
  • #29


DrGreg said:
[tex]ds^2 = dy^2 + dx^2[/tex]
[tex]ds^2 = r^2 \, d\theta^2 + dr^2[/tex]​


The two coordinate systems are related by

[tex]x = r \cos \theta[/tex]
[tex]y = r \sin \theta[/tex]​


You can measure this spacetime using Minkowski (x,t) coordinates (distance and time of an inertial observer), a square grid. Or you can measure this spacetime using Rindler [itex](R,\Theta)[/itex] coordinates (distance and coordinate time of an accelerating observer), a hyperbolic grid. Either way is valid, and they both describe the same 2D spacetime, but equations take a different form in the two coordinate systems.

Here I am using deliberately non-standard symbols for Rindler coordinates, to make the analogy clearer.

You measure "proper times" using either of the equations

[tex]d\tau^2 = dt^2 - dx^2[/tex]
[tex]d\tau^2 = R^2 \, d\Theta^2 - dR^2[/tex]​


The two coordinate systems are related by

[tex]x = R \cosh \Theta[/tex]
[tex]t = R \sinh \Theta[/tex]​

(If you're not familiar with "cosh" and "sinh", look up Hyperbolic function.)

That's probably enough to think about for now, but consider the similarities between the two coordinate systems for 2D Euclidean space, and the two coordinate systems for 2D Minkowski space.

Thats definitely more than enough to think about for now.
I have gotten the definition of h and am in the process of trying give it real geometric meaning in its trig application to hyperbolas. I expect this is going to take a while.
I am learning the formalism, the math language necessary to even communicate effectively on the level that you and Fredrik, et al, operate , let alone comprehend the concepts being discussed. This also is going to take time and I appreciate your patience with my inept communications.
In the meantime I have a couple of more conceptual type questions.

I am assuming a couple of things here. First we measure distance and time in units that make c = 1 (e.g. 1 light-year per year). Second, that the proper acceleration of the observer is 1 (e.g. 1 light-year per year2, which by an amazing cosmic coincidence is alarmingly close to 1g). The observer is located at R = 1, and coordinate time [itex]\Theta[/itex] equals the observer's own proper time (as measured by his clock), but runs faster or slower than clocks at other fixed R values. Two events that share the same [itex]\Theta[/itex] coordinate are simultaneous according to a co-moving inertial observer i.e. an inertial observer who is momentarily traveling at the same speed as the accelerating observer.

On the proper acceleration of an observer; That is a truly amazing and provocative coincidence which has occupied my thoughts since reading it.
1) But having thought about it I can't make the leap from 1 ls/s^2 to 10m/s^2 if the g you are referring to is the local constant. Could you point me towards a reference to how to make this connection?
2) As I understand it g is a local value derived from G which in GR is still a universal constant? so does this mean this applies to G also.
3) Were you possibly being facetious and humerous when you used the term "cosmic coincidence" and that you don't think there is anything coincidental about the relationship whatsoever? That it is actually fully understood.


Just as a check; Am I right in thinking that cartesian, polar and Minkowski coordinate systems all work on the assumption of a uniform time metric within any given frame?
ANd that all three are fundamentally euclidean ?

That the Rindler coordinate system assumes a non-uniform time metric and is different from the other three in this regard??

Am I wrong in thinking that the geometry of space-time is not a static function and that a non-uniform time metric would imply a non-euclidean geometry as described by the motions of points or particles over time?

Or is there something obvious and fundamental I am missing?

Well thanks for your help and you have given me directions and plenty to ponder

.
 
  • #30


Austin0 said:
That is a truly amazing and provocative coincidence which has occupied my thoughts since reading it.
1) But having thought about it I can't make the leap from 1 ls/s^2 to 10m/s^2 if the g you are referring to is the local constant. Could you point me towards a reference to how to make this connection?
2) As I understand it g is a local value derived from G which in GR is still a universal constant? so does this mean this applies to G also.
3) Were you possibly being facetious and humerous when you used the term "cosmic coincidence" and that you don't think there is anything coincidental about the relationship whatsoever? That it is actually fully understood.
I wouldn't read too much into this. It really is just a coincidence. Its only importance is convenience. When working baseball problems it is convenient to round g off to 10 m/s², similarly when working relativistic rocket problems it is convenient to round g off to 1 lightyear/year².

1) 1 lightsecond/second² most definitely does not equal 1 lightyear/year². If you just do the unit conversion you find that 1 lightyear/year² = 9.51 m/s² which is about 3% off from g.
2) no, it also depends on the mass of the earth, the radius of the earth, the mass of the sun, and the distance between the Earth and the sun, none of which are universal constants
3) if it is not coincidental then we would have to be able to derive all of those things above from G and c. Also, it is 3% off, so it is not really that close. It is just close enough for "napkin" calculations.
 
  • #31


Austin0 said:
That is a truly amazing and provocative coincidence which has occupied my thoughts since reading it.
1) But having thought about it I can't make the leap from 1 ls/s^2 to 10m/s^2 if the g you are referring to is the local constant. Could you point me towards a reference to how to make this connection?
2) As I understand it g is a local value derived from G which in GR is still a universal constant? so does this mean this applies to G also.
3) Were you possibly being facetious and humerous when you used the term "cosmic coincidence" and that you don't think there is anything coincidental about the relationship whatsoever? That it is actually fully understood.
It really is just a coincidence, and DaleSpam has said it all on that subject.

To do the calculation yourself: 1 ly/y2 is, by definition, the coordinate acceleration (relative to an inertial frame) that would take you from 0 to c in 1 year of coordinate time. (Of course, such a journey isn't physically possible. As your speed approached c, your proper acceleration would approach infinity, which makes the journey impossible to complete.) Nevertheless, you can use the equation v = at, with v = 3 x 108 m/s and t = 365.25 x 24 x 60 x 60 s to calculate the acceleration in m/s2.

Austin0 said:
Just as a check; Am I right in thinking that cartesian, polar and Minkowski coordinate systems all work on the assumption of a uniform time metric within any given frame?
ANd that all three are fundamentally euclidean ?

That the Rindler coordinate system assumes a non-uniform time metric and is different from the other three in this regard??

In the example I gave, Cartesian and polar coordinates are two different ways to measure static points in 2D space. Time is not involved, we are just looking at the geometry of static points on a flat piece of paper. It's Euclidean geometry whichever coordinate system you use.

The Minkowski and Rindler coordinates are two different ways to measure events in 2D spacetime (in my example, a cut-down version of 4D spacetime). Specifically "flat" spacetime which means gravity is being ignored. It's "Minkowski geometry" whichever coordinate system you use. Some might call it "Lorentzian geometry". The word "geometry" is now being used in an analogous sense as we are thinking of 2D spacetime as being a 2D geometrical entity. One of the dimensions is now time instead of space, but we can still draw spacetime graphs on a flat 2D piece of paper and look and the geometry of the curves. Only we have to use "Minkowski geometry" instead of "Euclidean geometry" and we have to measure our "pseudo-distance" on the graph using ds2 = dt2 - dx2.

The "non-uniform time metric", as you put it, is analogous to the fact that in 2D Euclidean geometry in polar coordinates you have a "non-uniform angle metric"; when an angle changes by one degree, the distance traveled depends on the radius.

Austin0 said:
Am I wrong in thinking that the geometry of space-time is not a static function and that a non-uniform time metric would imply a non-euclidean geometry as described by the motions of points or particles over time?
I'm not quite sure what you mean here, but there's a distinction to be drawn between the geometry of space and the the geometry of spacetime. The physical interpretation of the metric of flat (gravity-free) spacetime is that free-falling objects always move at constant velocity relative to any inertial frame. The trajectories of free-falling objects define what a "straight line" (or "geodesic") is in spacetime which in turn determines the spacetime geometry. When you choose Rindler coordinates instead of Minkowski coordinates, it means the coordinate gridlines that you measure with are no longer all straight lines.

You can see a diagram of Rindler gridlines attached to post #9 of this thread.
 
  • #32


DaleSpam said:
I wouldn't read too much into this. It really is just a coincidence. Its only importance is convenience. When working baseball problems it is convenient to round g off to 10 m/s², similarly when working relativistic rocket problems it is convenient to round g off to 1 lightyear/year².

1) 1 lightsecond/second² most definitely does not equal 1 lightyear/year². If you just do the unit conversion you find that 1 lightyear/year² = 9.51 m/s² which is about 3% off from g.

Oops ,,,so much for snap evaluations. Next time do the math.

2) no, it also depends on the mass of the earth, the radius of the earth, the mass of the sun, and the distance between the Earth and the sun, none of which are universal constants

I understood this, that is why I said local derivation. SInce I didnt understand the significance of 1 lightyear/ year^2 I was unsure if it would apply in some fundamental way to G or not . Isn't the view in GR that; G itself is still a universal constant?

Thanks again Although I am somewhat dissappointed,, it would have been a great coincidence,, I will rest easier now that you have clarified the situation.
 
  • #33


Austin0 said:
I understood this, that is why I said local derivation. SInce I didnt understand the significance of 1 lightyear/ year^2 I was unsure if it would apply in some fundamental way to G or not . Isn't the view in GR that; G itself is still a universal constant?
Yes, c and G are both considered universal constants. However, they are http://math.ucr.edu/home/baez/constants.html" constants, only the dimensionless universal constants are considered fundamental.

Austin0 said:
I will rest easier now that you have clarified the situation.
Excellent! That is very important :smile:
 
Last edited by a moderator:
  • #34


____________________________________________________________________________

Originally Posted by Austin0
Just as a check; Am I right in thinking that cartesian, polar and Minkowski coordinate systems all work on the assumption of a uniform time metric within any given frame?
ANd that all three are fundamentally euclidean ?

That the Rindler coordinate system assumes a non-uniform time metric and is different from the other three in this regard??
__________________________________________________________________-

=DrGreg;2309970]

The Minkowski and Rindler coordinates are two different ways to measure events in 2D spacetime (in my example, a cut-down version of 4D spacetime). Specifically "flat" spacetime which means gravity is being ignored. It's "Minkowski geometry" whichever coordinate system you use. Some might call it "Lorentzian geometry". The word "geometry" is now being used in an analogous sense as we are thinking of 2D spacetime as being a 2D geometrical entity. One of the dimensions is now time instead of space, but we can still draw spacetime graphs on a flat 2D piece of paper and look and the geometry of the curves. Only we have to use "Minkowski geometry" instead of "Euclidean geometry" and we have to measure our "pseudo-distance" on the graph using ds2 = dt2 - dx2.

Would it be incorrect to think that some Euclidean priciples still apply directly in a Minkowski 2D diagram wrt inertial frames??
Eg: The lines of simultaneity and the world lines of points on x ,or x', with a space-like separation, are parallel in the 2D coord space and therefore, will never converge or diverge no matter how extended?
That the triangles delineated by the intersection of the moving systems, lines of simultaneity and the rest frames timeline and x-axis and the triangle formed by the world lines of the moving frame and the restframes world line and x-axis are similar. And the normal correspondences of proportionality that that implies, are valid?

What would be a more correct way to express the idea of "non-uniform time metric" ?
Anisotropic time differential?? Location specific time parameterization?? Other?




____________________________________________________________________________
Originally Posted by Austin0
Am I wrong in thinking that the geometry of space-time is not a static function and that a non-uniform time metric would imply a non-euclidean geometry as described by the motions of points or particles over time?
_____________________________________________________________________________

I'm not quite sure what you mean here, but there's a distinction to be drawn between the geometry of space and the the geometry of spacetime. The physical interpretation of the metric of flat (gravity-free) spacetime is that free-falling objects always move at constant velocity relative to any inertial frame. The trajectories of free-falling objects define what a "straight line" (or "geodesic") is in spacetime which in turn determines the spacetime geometry. When you choose Rindler coordinates instead of Minkowski coordinates, it means the coordinate gridlines that you measure with are no longer all straight lines.

That is pretty much much exactly what I meant there although not expressed as correctly or extensively as you did.
ANd it is getting to the crux of my question.

" the coordinate gridlines that you measure with are no longer all straight lines"

What is your interpretation of this?
Doesnt this neccessarily mean that the spacetime is no longer flat , (Euclidean) ?
That the paths of inertial particles in a system with clocks at different rates ,will inevitably be curved??
Would it be wrong to say this is one functional definition of a non-Euclidean matrix?

See if this makes any sense... In the primary function of providing a chart for the recording of actuated events (measurements etc.) , for assigning precise and unique locations and times to those events , all coordinate systems must be equivalent. Those already transpired ,events in the real world, compose a singular and unique set , so any valid coordinate description of that set must neccessarily be co-related to any other coordinate description by some rational ,consistant transformation ,no matter how complex .
But there is a second important function of these systems; taking sets of events recorded up to the present and extrapolating and projecting them into the future as predictions of events. Or taking hypothetical sets of events or conditions and extrapolating them into a possible future.
In this role all systems are not neccessarily equivalent.
An empirically measured set of events , say particle paths recorded in one system and then transformed into another system could possibly extrapolate into a different future.
A set of hypothetical initial conditions could possibly extrapolate into different final conditions in differing systems depending on those systems implicit geometry and embedded physical assumptions.

So is there any validity to this perception?

Thanks for your patience and help
 
  • #35


Austin0 said:
Would it be incorrect to think that some Euclidean priciples still apply directly in a Minkowski 2D diagram wrt inertial frames??
Eg: The lines of simultaneity and the world lines of points on x ,or x', with a space-like separation, are parallel in the 2D coord space and therefore, will never converge or diverge no matter how extended?
Yes, the concept of parallel lines works just as well in Minkowski spacetime as in does in Euclidean space. And switching from one inertial frame to another doesn't affect parallelism.

Austin0 said:
That the triangles delineated by the intersection of the moving systems, lines of simultaneity and the rest frames timeline and x-axis and the triangle formed by the world lines of the moving frame and the restframes world line and x-axis are similar. And the normal correspondences of proportionality that that implies, are valid?
Look at the attached diagram. OB and OR are the worldlines of two inertial particles. B'R' is the wordline of another particle a constant distance from OB, and it's parallel to OB.

OB' is a line of simultaneity for the B (blue) observer, and so is the parallel line BR. OR' is a line of simultaneity for the R (red) observer.

I think you are asking if triangle OBR is similar to triangle OB'R'? The answer is yes. If you choose scales such that 1 s vertically is equal to 1 light-second horizontally (e.g.) then the similarity is true even in the Euclidean geometry of the paper you draw it on, but (I think) it's also true in some sense in Minkowski geometry too.

Austin0 said:
What would be a more correct way to express the idea of "non-uniform time metric" ?
Anisotropic time differential?? Location specific time parameterization?? Other?
How about "gravitational time dilation"? In relativity we don't distinguish between the "true" gravity due to matter and the "pseudo-gravity" of acceleration, it's all just "gravity".

Austin0 said:
That is pretty much much exactly what I meant there although not expressed as correctly or extensively as you did.
ANd it is getting to the crux of my question.

" the coordinate gridlines that you measure with are no longer all straight lines"

What is your interpretation of this?
Doesnt this neccessarily mean that the spacetime is no longer flat , (Euclidean) ?
In the context of relativity, "flat" and "Euclidean" aren't the same thing at all.

Minkowski geometry isn't Euclidean geometry, but it's still "flat". That's because (sticking to our simplified 2D spacetime) we can draw spacetime diagrams on a flat piece of paper. This is a bit of an oversimplification, but if spacetime is curved, you have to draw the diagram on a curved surface, such as the surface of a sphere. If you use curved gridlines on a flat surface, it's still flat spacetime, and the geometry hasn't changed, just the coordinate system.

Austin0 said:
See if this makes any sense... In the primary function of providing a chart for the recording of actuated events (measurements etc.) , for assigning precise and unique locations and times to those events , all coordinate systems must be equivalent. Those already transpired ,events in the real world, compose a singular and unique set , so any valid coordinate description of that set must neccessarily be co-related to any other coordinate description by some rational ,consistant transformation ,no matter how complex .
But there is a second important function of these systems; taking sets of events recorded up to the present and extrapolating and projecting them into the future as predictions of events. Or taking hypothetical sets of events or conditions and extrapolating them into a possible future.
In this role all systems are not neccessarily equivalent.
An empirically measured set of events , say particle paths recorded in one system and then transformed into another system could possibly extrapolate into a different future.
A set of hypothetical initial conditions could possibly extrapolate into different final conditions in differing systems depending on those systems implicit geometry and embedded physical assumptions.
All valid coordinate systems must make identical predictions about anything that you can measure that isn't just a feature of the coordinates themselves. For example whether or not two particles collide, or how much time will elapse on a single clock transported between two events. The equations that describe the laws of physics may change when you switch between different coordinate systems (but not between two Minkowski coordinate systems in the absence of gravity). And some coordinate systems may not extend to the whole of spacetime, they may only be valid within a restricted region.

For example the Rindler coordinates in post #26 are valid only for R > 0, x > |t|.
 

Attachments

  • Diagram for Four-vectors thread.png
    Diagram for Four-vectors thread.png
    2.5 KB · Views: 440

Similar threads

  • Special and General Relativity
Replies
6
Views
460
  • Special and General Relativity
Replies
4
Views
1K
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
Replies
4
Views
2K
  • Special and General Relativity
Replies
3
Views
965
  • Special and General Relativity
3
Replies
101
Views
3K
  • Special and General Relativity
Replies
20
Views
2K
  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
Replies
6
Views
1K
Back
Top