# Integrating an accelerating Worldline

1. Jul 7, 2010

### Austin0

Geometrically:
Is it correct of think that a line integration in effect converts a curvilinear segment
into an equivalent straight line??

If this is basically applicable then it would mean converting the accelerating worldline into a straght line in Minkowski space.
My question is how is the slope of this line determined ?

Thanks

2. Jul 7, 2010

### Passionflower

No.

Perhaps you are mixing up integration and differentiation?

3. Jul 7, 2010

### Austin0

Originally Posted by Austin0
Geometrically:
Is it correct of think that a line integration in effect converts a curvilinear segment
into an equivalent straight line??

So I am wrong in thinking that the integration derives a linear length [obviously longer that the linear length between the end points of the curve] which could be then drawn as a straight line on the euclidean plane of the curve???

You could very well be right as my knowledge of calculus is self evidently limited that I would be asking this question.

As I understood the explanations regarding worldlines in the Minkowski 2D plane it is a matter of of a large number of infinitesimally small ( virtually straight) intervals which are summed to derive a linear length. Is this integration or differentiation?
Or is my basic understanding of the process flawed??
I have worked trying to get an understanding regarding this question for some time so any insights or help is appreciated.

4. Jul 7, 2010

### Staff: Mentor

A line integration is just like any other integration: it adds up an infinite number of infinitesimal values. In the case of a line integral those infinitesimal values are some finite function which is defined along the line times a differential change of some variable which parameterizes the line.

5. Jul 7, 2010

### yossell

Might this help:

your original question it sounded as if you thought a curved line was somehow transformed into a straight line by integration. As Passionflower said, that's not true.

What we can do, though, is *approximate* a curve by breaking it up into n distinct pieces - say by putting n crosses on that curve - and drawing straight lines between those curves, to get a kind of zig zag shape which roughly follows the curve. Each straight line will have a length', and, for any finite number of lines, we can add the finite number of lengths' to find the length' of the whole one.

The more pieces we break the curve into, the smaller the straight lines, though the greater the number - and the better the zigzag approximates the curve, and the closer the addition of the finite number of lengths of the little segments approximates the length' of the curve. The result of integration is the number we approach when we add up the lengths' of finitely many, finitely sized little zigzags as the number of crosses approaches infinity and the size of the individual zigzags tends to zero. In well behaved spaces, the result of this process gives us the length' of the curve.

This is a way of thinking about integration in this area without having to think there really are infinitesimals - I prefer it, as infinitesimals, and the whole dx dy notation, make me weep.

I have put length in inverted commas for the following reason: if we're imagining the curve drawn in space, then it's really length we're talking about. But we can extend the basic ideas to other quantities. In the relativistic case, we don't have curves in space, but curves in space time. Here, the relevant quantity is separation rather than length - it can be thought of as a kind of space-time distance, but the analogy isn't perfect. However, separation between two space-time points is independent of coordinate system, just as length in Euclidean space, is independent of coordinate system.

6. Jul 7, 2010

### Austin0

If I am understanding you correctly; my concept of integration was basically correct but my idea of its application was off.
The length in Minkowski space has a geometric relationship to the value of the total interval. With an inverse time value , yes?

The geometric length is converted through the gamma function to a value.

I was thinking that the process simply integrated the geometric length of the worldline and then applied the function to the derived linear length.

I got this impression because Fredrik
had used the analogy of a straight line being the shortest distance etc. Explaining why the curved worldline must be longer than the straight distance ,,between any two points on the worldline.

SO it sounds like the gamma is derived as a function of the slope of the the infinitesimal intervals of the integration ,,in the process , so it returns an already converted value rather than a geometric length. Maybe??

Thanks

7. Jul 7, 2010

### Austin0

That is exactly how I was looking at it. The only difference was I was assuming the this "length" , which is simply a value, could be drawn on the 2D space of the curve and " in effect" be considered equivalent to the curve straightened out .

That was the way I was thinking about integration but I thought it was infinitesimals. SO I dont get your distinction. But thats not surprising .
But in Minkowski spacetime the curves are really in a 2D space arent they?

8. Jul 7, 2010

### Fredrik

Staff Emeritus
A positive "infinitesimal" is smaller than all positive real numbers. When physics books are using the term infinitesimal, they're not really talking about infinitesimals. You should think of that word as a warning that lets you know that the next equality you see is only accurate to order n in some variable(s) (and in at least 90% of those cases, n=1). For example, to say that when x is infinitesimal, sin x=x, is just a sloppy way of saying that for some positive integer n (in this case n=1), there exists a positive real number M such that

$$\left|\frac{\sin x-x}{x^n}\right|\leq M$$

for all x in some interval that contains 0.

Minkowski spacetime is 4-dimensional, but we often consider the 2-dimensional version of it because that's sufficient for a discussion of most features of SR, like length contraction and time dilation.

9. Jul 7, 2010

### Staff: Mentor

Yes, although I don't know what you mean by "inverse time value".

In standard vector calculus in Euclidean space we have the metric (Pythagorean theorem):
$$ds^2=dx^2+dy^2+dz^2$$
which gives the length of a differential segment of a line. If we have a line parameterized by some arbitrary parameter p then we can find the length of the line by integrating as follows:
$$s=\int{ds}=\int{\frac{ds}{dp}dp}=\int{\sqrt{\left(\frac{dx}{dp}\right)^2+\left(\frac{dy}{dp}\right)^2+\left(\frac{dz}{dp}\right)^2}dp}$$

Similarly in Minkowski space we have the metric (units where c=1):
$$ds^2=dt^2-dx^2-dy^2-dz^2$$
which gives the proper time along a differential segment of a worldline. If we have a worldline parameterized by some arbitrary parameter p then we can find the proper time of the worldline by integrating as follows:
$$s=\int{ds}=\int{\frac{ds}{dp}dp}=\int{\sqrt{\left(\frac{dt}{dp}\right)^2-\left(\frac{dx}{dp}\right)^2-\left(\frac{dy}{dp}\right)^2-\left(\frac{dz}{dp}\right)^2}dp}$$
In the case where the line is parameterized by time we have p=t.
$$s=\int{\sqrt{\left(\frac{dt}{dt}\right)^2-\left(\frac{dx}{dt}\right)^2-\left(\frac{dy}{dt}\right)^2-\left(\frac{dz}{dt}\right)^2}dt}=\int{\sqrt{1-v^2}dt}=\int{\gamma^{-1}dt}$$

10. Jul 8, 2010

### Austin0

Poorly stated I meant that the time value was inversely proportional to the integrated length.
Isn't that what is indicated by the -1 gamma exponent in

$$\int{\gamma^{-1}dt}$$

I think I am getting a dim grasp of the basics but why is the z component factored by dp or dt???

It seems to me like this is a basic formulation or blue print for the integration process,
but in actuall application to a worldline how are the values for dx/dp derived??

From the acceleration function for the curve???

Or is my grasp dimmer than i think???

Thanks for your help I think it did??

11. Jul 8, 2010

### Austin0

When you say ((1)) above are you meaning smaller than the infinitely small interval between real numbers . Cantoresque relative infinities?

Understood, but I was talking about the physical 2-D plane of the drawing surface.
I thought that a Minkoski diagram was an accurate representation of the geometry.
That in the drawing itself the physical distances were a geometrically true expression of the values. So it seemed that; just like the integrated length of a curve drawn on paper could then be drawn on the same sheet as a linear line ,,,an integrated worldline should be able to be drawn on the same paper as a straight line and retain its geometric validity???
Is this not true then???
Thanks

12. Jul 8, 2010

### matheinste

No it isn't.

Leaving aside geometrical complications and the fact that the result of an integration is just a number, think of what a worldline in Minkowski spacetime is. Roughly,it is the line passing through every event that an object, or strictly speaking for a worldline, a point particle, is present at in its lifetime. If this worldine is curved it is not possible, on the same diagram, to draw a straight worldline that passes through the same events as those visited by the curved worldline. That is the physical reality of the situation which the geometry must respect if it is to model the real world

Matheinste.

13. Jul 8, 2010

### yossell

I'm not sure what you have in mind by the physical 2-d plane of the drawing surface.' If you literally mean the piece of paper, this is just (approximately) Euclidean and purely spatial. The physical drawing is a representation of a space-*time*. It's an accurate *representation* - but you have to make sure you're reading it properly. For instance, two points that are very far apart on the diagram might have a very small spatio-temporal distance'. For instance, any two points on a light cone have *zero* separation, even though, on the piece of paper, they may be as far apart as you like.

I'm not sure what you have in mind by an integrated world line. There are just curves in space-time. These curves can be approximated by a series of zigzag lines of the same space. By integrating, we can calculate the length' of the curved line from the lengths' of the straight line segments that make up the line. But the integration is not naturally thought of as a map from a curve to a particular straight line.

Of course, it is true, that for any length' you calculate for a curve, there are many straight segments you can draw in space-time that have the same length. But, as far as I know, this isn't closely connected with Minkowski space-time.

Hope this helps

14. Jul 8, 2010

### Staff: Mentor

No, the integrated length is directly proportional to the proper time. In units where c=1 they are equal. Look at the units here, gamma is unitless so the integral has units of time, not inverse time. Since gamma is unitless the exponent has no significance for units.

When you have a curve you can parameterize it by some arbitrary parameter p. That means that you write the curve as e.g. (x(p),y(p),z(p)). Then, in order to calculate its length you differentiate by p to get dx/dp, dy/dp, and dz/dp, and you substitute into the metric. This is just standard vector calculus.

If the parameter p is equal to time then dx/dp=dx/dt= the x component of the velocity. If the parameter p is not equal to time then it is a little more complicated, but it is still related to velocity as follows:
$$\frac{dx}{dp}=\frac{dx}{dt}\frac{dt}{dp}$$

15. Jul 8, 2010

### Austin0

$$\int{\sqrt{1-v^2}dt}=\int{\gamma^{-1}dt}$$ I understand that gamma is unitless. But here $$\gamma$$-1 is applied to coordinate time (which is in units) derived from the integrated length. This would seem to mean that the resulting proper time is inversly proportional to the coordinate dt and inversely proportional to the intgrated length. Greater length means less proper time.
If I am off here what is the meaning of the -1 exponent??
Or my basic view??

I understand this . I was confused by thinking of the curve as it is in Minkowski space with a non-uniform time metric. Which would seem to mean a non-uniform p .
Oops
Ok I understood this but in this case isn't the velocity itself non-uniform, changing with dt???
And the dt here is coodinate time which is uniform , right??

( dz/dp)dp
WHy is this??

Last edited: Jul 8, 2010
16. Jul 8, 2010

### Austin0

((1)) What exactly do you mean by this??
That two points on a light cone can have zero separation in real world spacetime??

((2)) WHy would this not apply in a Minkowski spacetime daigram plane which is essentially Euclidean both on paper as well as in fact. Isn't the xy plane flat not only on paper but in spacetime??
But to be geometrically meaningful the integrated straight line would have to have the proper slope??

17. Jul 8, 2010

### starthaus

No, not at all.

Through a parametrization of the trajectory described by the travelling clock.

18. Jul 8, 2010

### yossell

(1) Yes - it sounds very strange if one thinks that the separation between two points of space time is just the Minkowskian four-dimensional version of our 3-dimensional distance in space. Well, there are analogies between the two notions, but there are quite significant differences too.

Like distance in Euclidean space, separation is frame invariant - the spacetime separation between two points does not depend upon whatever Lorentz frame you work in. Like distance, the separation can be expressed as a function of coordinates of the points. In Cartesian coordinates the square of the distance from a point at the origin is: root{x^2 + y^2 + z^2}. But in Relativity, the square of the separation from the origin of a Lorentz coordinate system is: root{-t^2 + x^2 + c^2 + z^2}. (working in units where speed of light equals 1 - sometimes, you'll see the time coordinate multiplied by c^2). Clearly, since time squared is subtracted rather than added, it's possible for this quantity to be zero. It's even possible for this quantity to be negative. In these respects, separation is very unlike our notion of distance in Euclidean space.

However, the quantities can be given physical interpretation: If s^2 squared between two points is negative, the separation is called timelike, and a clock that travelled freely in a straight line between those points will measure s units of time as passing. If s^2 is positive between two points then the separation between the two is spacelike, and in a frame where the two points are simultaneous, and a rod in such a frame will measure the (spatial) distance as s. If the separation is null, then the straight line between them represents the path a light beam could travel.

All this shows that one has to be fairly careful in how you read the Minkowski diagram and interpret the invariant separation as a kind of space-time length. The analogy can be very helpful, but it can also be quite misleading.

19. Jul 8, 2010

### yossell

Austin0
(2) Well, it would apply - what I'm trying to convey is that it's not clear to me what you have in mind when you talk about the integrated straight line. There's no need to consider any straight line that has the same length' as a curve when you do the integration. What line's slope are you talking about?

The letter s has a certain length. I calculate by splitting into finer and finer segments and adding up the lengths of the little bits. Eventually, I come up with a number. I can then draw \ and / and --- and | as straight lines which have this length and thus have the same length as s. But this last part involves an arbitrary choice and isn't really germane or relevant to the integration.

Last edited: Jul 8, 2010
20. Jul 8, 2010

### Fredrik

Staff Emeritus
I would say that the interval between real numbers is zero, not "infinitely small", but then I know almost nothing about infinitesimals. There might be a way to make sense of that too. I'm talking about an x that satifies 0 < x < 1/n for all n=1,2,... There's no need to ever consider infinitesimals in physics, but for some reason physicists like to use the word "infinitesimal" as soon as a Taylor expansion is involved.

1+1-dimensional Minkowski spacetime is just the vector space $$\mathbb R^2$$ with a bilinear form that isn't positive definite, instead of an inner product. The diagram is just a picture of $$\mathbb R^2$$. I don't think I understand what you're asking towards the end of the quote. We can draw a world line, but its proper time (the result of the integration) is a number.

21. Jul 8, 2010

### Staff: Mentor

Yes, you are very much off here. If a and b are proportional to each other then you can write:
$$a=k\;b$$
You can also write:
$$a\;k^{-1}=b$$
They mean the same thing, a and b are proportional to each other, not inversely proportional regardless of the exponent of k.

There is not an additional factor for z. Note that the dp you are looking at is outside the square root. It is the variable of integration.

Last edited: Jul 8, 2010
22. Jul 9, 2010

### Austin0

Originally Posted by Austin0

an integrated worldline should be able to be drawn on the same paper as a straight line and retain its geometric validity???
Is this not true then???
Hi
Couldn't you say that a 2-d M diagram has two different aspects?
Hypothetically it is derived from actual observations and as such is exactly as you described . A graphing of a series of events.
But isn't there a second interpretaion which is purely geometric?

With two worldlines that are both inertial:--- The relationship of the values of those lines, regarding space and time , is geometrically correct.
Couldn't you could take a line interval from one line and using normal pythagorean relationships applied to the intersecting lines of time and space in the other frame, derive a Euclidean length from which you would then get the relative values by applying the gamma transformation ???
Or knowing the slope get the equivalent length through normal trig and apply gamma?

Of course I am not suggesting that anyone in practice would want to do this. ANd I was not thinking that an equivalent straight line of an accelerating frame was equivalent to the curved worldline as a history but only regarding geometric value.

I know this all must seem somewhat silly and moot but I tend to visualize things geometrically and am trying to fully grasp M-space in all its aspects.

Thanks

23. Jul 9, 2010

### Austin0

Originally Posted by Austin0
When you say ((1)) above are you meaning smaller than the infinitely small interval between real numbers . Cantoresque relative infinities?

If the interval between real numbers is zero and the interval between infinitesimals is smaller yet would this imply that;
0infintesimals < 0real-n's

Could you perhaps say that the interval of real numbers decreases to the limit 0 at infinity (the end of the universe???)
or that it decreases through infinite regression down Zeno's rabbithole??

COuld both of the above also be apply to infinitesimals?

In any case I take your message to be;steer clear of both and that in physics infinitesimals means some incredibly small finite interval ,yes ???

Thanks
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24. Jul 9, 2010

### Austin0

tex]s=\int{\sqrt{\left(\frac{dt}{dt}\right)^2-\left(\frac{dx}{dt}\right)^2-\left(\frac{dy}{dt}\right)^2-\left(\frac{dz}{dt}\right)^2}dt}=\int{\sqrt{1-v^2}dt}=\int{\gamma^{-1}dt}[/tex][

Originally Posted by Austin0
I understand that gamma is unitless. But here -1 is applied to coordinate time (which is in units) derived from the integrated length. This would seem to mean that the resulting proper time is inversly proportional to the coordinate dt and inversely proportional to the intgrated length. Greater length means less proper time.
If I am off here what is the meaning of the -1 exponent??
Or my basic view??

((1)) $$\int{\gamma^{-1}dt}$$

((2)) $$gamma^{-1} \int{\dt}$$

dx'= $$gamma$$dx
and it is commonly written dt'=$$gamma$$dt but shouldn't that be correctly written dt'= $$gamma^{-1}$$ dt ???? and could be said to be inversely proportional relative to $$gamma$$dx ????

Is this correct???
$$\int{\gamma^{-1}dt}$$=$$gamma^{-1} \int{\dt}$$

Oops optical illusion. It looked to me that the dp was under the z exponent and included in the square root.

Thanks

25. Jul 9, 2010

### starthaus

You can't do that since $$\gamma$$ is not constant along the integration interval. Remember that the motion is accelerated (see the title of your OP) so $$v=v(t)$$. You are missing the basics.