Integrating an accelerating Worldline

[Dale]

That is what I thought but how are they derived?
The integral equation does not appear to me to include any means of differentiation.

They are derived by differentiating the equation for the worldline, as I showed in the example.

 

[Austin0]

A positive "infinitesimal" is smaller than all positive real numbers. .

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

I would say that the interval between real numbers is zero, not "infinitely small", .

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;
0_{infintesimals} < 0_real-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__

I don't know a way to make sense of "the interval between real numbers" and "the interval between infinitesimals" makes even less sense to me. The distance between two real numbers x and y is defined as |x-y|.
.

I don't understand any of these questions.

.

Georg Cantor: the infinite number of elements of the set of real numbers is greater than the infinite number of elements of the set of integers.
|{R}| =Infinity |{I}|= Infinity... INFINITY _real>INFINITY _Intrg

For any interval I_n =|x-y| there exists an I_n-1 such that I_n-1<I_n where I_n-1=|x-g |... g<x , g>y......Infinite regression.

Math: between any two real numbers there exists an infinite number of other real numbers

....which in Geometry becomes;

Tne line interval AB between any two points A and B contains an infinite number of points P
AB_n-1<AB_n-1 ...infinite regression.

...Which in the Real World becomes:

Zeno's :...the interval between point A [Achilles/the hares / the rabbits current position] and B [the turtle's current position] is infinitely divisable with infinitely regressing shorter intervals which approach zero but never reach it and disappear down the rabbithole.

Make any more sense now?

I am trying to learn Multi Quote and LaTex but can't seem to find the tricks you all use. Is there a source for info?

I tried a search for threads on the subject , no luck.

When I try to use LaTex in a multi reply it goes nuts. Puts things all over the place and doesn't work at all sometimes and latex parts of your quotes which I had deleted magically reappear in strange places.

Also where do you find the real number and integer set symbols?

I found R but it comes out squared.?

Thanks

 

[Dale]

Austin0,

1) do you know how to do differentiation and integration?
2) do you understand the geometric interpretation of differentiation giving you the slope of a curve at a point?
3) do you understand the geometric interpretation of integration giving you the area under a curve?
4) do you understand integration and differentiation as a limit of sums and differences?
5) do you understand integration and differentiation as operators on functions?
6) do you understand integration and differentiation as inverses of each other?

If so, then don't sweat the fine points of infinitesimals, they are not important in this context. If you really want to get into that you should probably start a new thread in the math sub-forum.

 

[Austin0]

Austin0,

1) do you know how to do differentiation and integration?

2) do you understand the geometric interpretation of differentiation giving you the slope of a curve at a point?
Yes
3) do you understand the geometric interpretation of integration giving you the area under a curve?
Yes

4) do you understand integration and differentiation as a limit of sums and differences?
Yes

5) do you understand integration and differentiation as operators on functions?
Yes

6) do you understand integration and differentiation as inverses of each other?
vaguely

If so, then don't sweat the fine points of infinitesimals, they are not important in this context. If you really want to get into that you should probably start a new thread in the math sub-forum.

1) Not sure? In the normal sense no. But;
Last year I wanted to do a table of time and distance figures [between instantaneous velocities] of a constantly accelerating system as measured in an inertial frame.

I took a scale between o and c in increments of .01c

On the assumption that initial acceleration would fall off at a factor of gamma
I used the HyperP calculator to get all 100 gammas.

I then used an initial acceleration function to derive all 100 instantaneous a(v)'s
a(v) =gamma (v) x(a(0))

Applied the normal acceleration math to derive dt's between every.. v and v+(.01 )

From this I got the dx =dt((v+ .01)+v)/2)

I'm sure you will think this was both neolithic and crazy and I would agree , especially after my harddrive went south and I lost it all.

a) Was this a primitive form of differentiation and integration??

b)If I had applied gamma(v)dt and added them all up would the sum = dt'= ds ...?

c) Would this be equivalent to an integrated worldline wrt time?

d)am I crazy?

What really seems like magic and drives me nuts is how your integration works without an acceleration function
LaTex also.
Thanks

 

[Fredrik]

Make any more sense now?

More sense yes, but I still don't see exactly what question I'm supposed to answer. By the way, |x-y| is a real number, not an interval. The interval from x to y is written as [x,y] when the endpoints are included, and as (x,y) when they're not.

I am trying to learn Multi Quote and LaTex but can't seem to find the tricks you all use. Is there a source for info?

This post might help with the quotes. (You will have to copy and paste some quote tags). Regarding the LaTeX, just remember to put each math expression between tex or itex tags, and google "latex symbols" to see how to write the most basic stuff. You can also see how other people here are doing it if you just click the quote button next to a post with LaTeX.

Always use the preview feature to see if it looks OK before you post.

When I try to use LaTex in a multi reply it goes nuts. Puts things all over the place and doesn't work at all sometimes and latex parts of your quotes which I had deleted magically reappear in strange places.

This is a bug. (The last thing at least). It's been that way for a few months. See this thread. The workaround is to refresh and resend after each preview.

 

[Austin0]

More sense yes, but I still don't see exactly what question I'm supposed to answer. By the way, |x-y| is a real number, not an interval. The interval from x to y is written as [x,y] when the endpoints are included, and as (x,y) when they're not.

There was no real question. It was basically a joke or play.

Viewed n the context of my last post you made statements regarding infinitesimals and real numbers [infinitesimals being smaller than real numbers which you said you regarded as zero] which could be taken to imply that the smallest infinitesimal was smaller than zero.

I thought this concept was conceptually similar and intrinsically as amusing as Cantors transfinite sets. One set being "more" infinite than another set.

One 0 being less ,, [less than nothing] than another .

I had assumed you were totally familiar with all of my last post, so was perplexed as to whether you didnt "get the joke" or didnt see the logic behind it or just didn't appreciate my sense of humor.

Also on a more serious side it was consistent with your feelings about infinitesimals as they relate to the real world [physics] and the danger of taking abstract mathematical concepts out of their domain.

With which I totally agree.

Also I questioned [my reading of] your idea that the smallest interval between real numbers was zero.

By the way, |x-y| is a real number, not an interval. The interval from x to y is written as [x,y] when the endpoints are included, and as (x,y) when they're not.


.

Thanks , as you know my command of the formalism is lacking and I just made up the math on the fly, I am sure there are much better and more elegant ways to express it.

On the other hand you stated that the "distance " between real numbers is |x-y|

This seems to indicate that although it may be a real number you also think it indicates an interval.
COuld you explain a little??

Doesn't (x,y) also mean an interval expressed as a real number??

Regarding the LaTeX, just remember to put each math expression between tex or itex tags, and google "latex symbols" to see how to write the most basic stuff. You can also see how other people here are doing it if you just click the quote button next to a post with LaTeX.
Always use the preview feature to see if it looks OK before you post.
.

I have tried to always make sure of correct tags.

I have looked and studied other posts. In the beginning I just copied and pasted from them.

I for the most part haven't made an effort to learn to write LaTeX beyond what I am picking up from that study. I have been using the feature in the reply interface which seems to work fine sometimes but not all. I think you're right ,I will save some headache if I check out the PF LaTeX site.
Thanks for the links on the malfunction ,it has been driving me totally nuts,,,,, so just knowing that it is not just me is a relief.

 

[Fredrik]

I had assumed you were totally familiar with all of my last post, so was perplexed as to whether you didnt "get the joke" or didnt see the logic behind it or just didn't appreciate my sense of humor.

I didn't understand that you were joking.

On the other hand you stated that the "distance " between real numbers is |x-y|

This seems to indicate that although it may be a real number you also think it indicates an interval.
COuld you explain a little??

"The distance between real numbers" is a phrase that doesn't make sense to me. It sounds like a distance between a real number and "the next", but there is of course no "next" real number. The distance d(x,y) between the real numbers x and y is defined by d(x,y)=|x-y|.

A lot of people who aren't me use the term "interval" about something that has more to do with distances than with actual intervals, e.g. the Wikipedia article titled "introduction to special relativity" says "So the spacetime interval between two distinct events is given by $s^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 - c^2 (t_2 - t_1)^2$." That's one of the reasons why I thought you were talking about distances rather than actual intervals.

Doesn't (x,y) also mean an interval expressed as a real number??

The notation can mean many things. The most common are: a) an ordered pair, which can be defined in terms of (unordered) sets as {x,{x,y}}, b) an open interval, which can also be written as $\{t\in\mathbb R\,|\,x<t<y\}$, c) the inner product of the vectors x and y, which is usually written as $\langle x,y\rangle$.

I don't know what you mean by "interval expressed as a real number".

 

[Austin0]

I didn't understand that you were joking.


"The distance between real numbers" is a phrase that doesn't make sense to me.

Me either ,but you said it so i thought maybe you had an idea/

It sounds like a distance between a real number and "the next", but there is of course no "next" real number. The distance d(x,y) between the real numbers x and y is defined by d(x,y)=|x-y|.

A lot of people who aren't me use the term "interval" about something that has more to do with distances than with actual intervals, e.g. the Wikipedia article titled "introduction to special relativity" says "So the spacetime interval between two distinct events is given by $s^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 - c^2 (t_2 - t_1)^2$." That's one of the reasons why I thought you were talking about distances rather than actual intervals.

Now I am not sure of the difference between distance and spatial interval.
I have always assumed that the connection between points A and B was both an interval and the distance.

The notation can mean many things. The most common are: a) an ordered pair, which can be defined in terms of (unordered) sets as {x,{x,y}}, b) an open interval, which can also be written as $\{x\in\mathbb R\,|\,a<x<b\}$, c) the inner product of the vectors x and y, which is usually written as $\langle x,y\rangle$.

I don't know what you mean by "interval expressed as a real number".

1)... By the way, |x-y| is a real number, not an interval. The interval from x to y is written as [x,y] when the endpoints are included, and as (x,y) when they're not.

I assumed that if distance |x-y| was expressed by a real number then... (x,y) which does not include x and y ,,,would also be expressed as a single number not as two more numbers or whatever actual interpretation arises from your post above which is beyond my ability to follow or conceptualize.

Thanks I always learn something

 

[Dale]

1) Not sure? In the normal sense no. But;
Last year I wanted to do a table of time and distance figures [between instantaneous velocities] of a constantly accelerating system as measured in an inertial frame.

I took a scale between o and c in increments of .01c

On the assumption that initial acceleration would fall off at a factor of gamma
I used the HyperP calculator to get all 100 gammas.

I then used an initial acceleration function to derive all 100 instantaneous a(v)'s
a(v) =gamma (v) x(a(0))

Applied the normal acceleration math to derive dt's between every.. v and v+(.01 )

From this I got the dx =dt((v+ .01)+v)/2)

I'm sure you will think this was both neolithic and crazy and I would agree , especially after my harddrive went south and I lost it all.

a) Was this a primitive form of differentiation and integration??

b)If I had applied gamma(v)dt and added them all up would the sum = dt'= ds ...?

c) Would this be equivalent to an integrated worldline wrt time?

d)am I crazy?

What really seems like magic and drives me nuts is how your integration works without an acceleration function

If I understand what you did then in the end you had a worldline which you parameterized by v, i.e. you wound up with the worldline (t(v), x(v)) evaluated at 100 different points. If you wanted to determine the spacetime interval along that worldline then you would set p=v and evaluate the integral I posted earlier:

$$s=\int{ds}=\int{\sqrt{\left(\frac{dt}{dv}\right)^2-\left(\frac{dx}{dv}\right)^2}dv}$$

However, since you only have the function numerically at 100 points you would approximate the integral by a finite sum:

$$s=\int{ds}\simeq \sum_{i=0}^{100}{\sqrt{\left(\frac{\Delta t_i}{\Delta v}\right)^2-\left(\frac{\Delta x_i}{\Delta v}\right)^2}\Delta v}$$

Where $\Delta v = .01 c$ and $\Delta x_i = x_i - x_{i-1}$. This would be a rough numerical approximation to the integral in question.

Do you also know how to (exactly) differentiate and integrate polynomials and simple trig functions by hand? If so, then that is probably sufficient for most purposes. Anything more complicated will generally require looking the answer up in a table or using some math software.

 

[Austin0]

1) Not sure? In the normal sense no. But;
Last year I wanted to do a table of time and distance figures [between instantaneous velocities] of a constantly accelerating system as measured in an inertial frame.

I I then used an initial acceleration function to derive all 100 instantaneous a(v)'s
a(v) =gamma (v) x(a(0))

Applied the normal acceleration math to derive dt's between every.. v and v+(.01 )

From this I got the dx =dt((v+ .01)+v)/2)
I'm sure you will think this was both neolithic and crazy and I would agree , especially after my harddrive went south and I lost it all.

a) Was this a primitive form of differentiation and integration??

b)If I had applied gamma(v)dt and added them all up would the sum = dt'= ds ...? c) Would this be equivalent to an integrated worldline wrt time?

If I understand what you did then in the end you had a worldline which you parameterized by v, i.e. you wound up with the worldline (t(v), x(v)) evaluated at 100 different points. If you wanted to determine the spacetime interval along that worldline then you would set p=v and evaluate the integral I posted earlier:

$$s=\int{ds}=\int{\sqrt{\left(\frac{dt}{dv}\right)^2-\left(\frac{dx}{dv}\right)^2}dv}$$

However, since you only have the function numerically at 100 points you would approximate the integral by a finite sum:

$$s=\int{ds}\simeq \sum_{i=0}^{100}{\sqrt{\left(\frac{\Delta t_i}{\Delta v}\right)^2-\left(\frac{\Delta x_i}{\Delta v}\right)^2}\Delta v}$$

Where $\Delta v = .01 c$ and $\Delta x_i = x_i - x_{i-1}$. This would be a rough numerical approximation to the integral in question.

((((1))))Do you also know how to (exactly) differentiate and integrate polynomials and simple trig functions by hand? If so, then that is probably sufficient for most purposes. Anything more complicated will generally require looking the answer up in a table or using some math software.

Look at b) above. In the process, if instead of using dt to derive dx I had instead used the gamma(v) to derive dt' and then added together all 100 [slightly large] infinitesimals, would I have arrived at a rough approximation of s ,,,,,,,,,,,,,,,?

((((1)))) No, my understanding of differentiation and integration is limited to a rough understanding of the principles and geometric interpretation and original derivation .
That is why I was hoping to see the complete process as applied to a specific worldline.

As it is ,looking at your integral formulas it looks like a computer program without a machine to perform the actual step by step permutations.

My trig is a hazy memory of principles , I don't even actually remember the identities.

My primary immediate motivation is; I want to do some calculations with Rindler and the proper acceleration functions. Both require trig at the onset. SO maybe your software suggestion is the way i should go.
I started this thread with geometry in mind and no real thought it would lead to actual calculus. That it did presented an opportunity I couldn't resist to pose questions.
Thanks for your input.

 

[Fredrik]

I assumed that if distance |x-y| was expressed by a real number then... (x,y) which does not include x and y ,,,would also be expressed as a single number not as two more numbers or whatever actual interpretation arises from your post above which is beyond my ability to follow or conceptualize.

I messed up when I wrote the LaTeX expression because at first I thought we were talking about an interval with endpoints a and b, but then I noticed that you had called them x and y. So I started to change what I had written, but didn't change all of it. I have fixed it now.

What was it that you found hard to "follow or conceptualize"? The notation

$$\{t\in\mathbb R\,|\,x<t<y\}$$

(which is precisely the expression I messed up) just means "the set of all real numbers t such that x<t<y". There's certainly more than two of them.

 

[Dale]

Look at b) above. In the process, if instead of using dt to derive dx I had instead used the gamma(v) to derive dt' and then added together all 100 [slightly large] infinitesimals, would I have arrived at a rough approximation of s ,,,,,,,,,,,,,,,?

Yes, provided dt' refers to the incermental time on the accelerating clock.

((((1)))) No, my understanding of differentiation and integration is limited to a rough understanding of the principles and geometric interpretation and original derivation .
That is why I was hoping to see the complete process as applied to a specific worldline.

As it is ,looking at your integral formulas it looks like a computer program without a machine to perform the actual step by step permutations.

Sorry about that, I didn't realize your calculus background so I thought that I had been providing a complete answer where you could just "fill in the blanks".

My trig is a hazy memory of principles , I don't even actually remember the identities.

My primary immediate motivation is; I want to do some calculations with Rindler and the proper acceleration functions. Both require trig at the onset. SO maybe your software suggestion is the way i should go.
I started this thread with geometry in mind and no real thought it would lead to actual calculus.

What is it that you want to calculate with Rindler coordinates? I am sure that we can walk you through it. I refuse to do anything in Excel, but I am familiar enough with it that I can tell you how to do it. If you do get some math software package then I am sure we can help you solve it in that also.

 

[Austin0]

What is it that you want to calculate with Rindler coordinates? I am sure that we can walk you through it. I refuse to do anything in Excel, but I am familiar enough with it that I can tell you how to do it. If you do get some math software package then I am sure we can help you solve it in that also.

Hi DaleSpam What exactly I am interested in calculating is simple the dilation factor between the front and back of a frame 1km in length at an acceleration of 1000g

10km/sec² and at ...a=1g

I am sure for someone with the skills this is completely trivial but I don't have the trig.

I can't deny I am hoping you might help me out here before giving up on me completely :-)

Thanks

 

[Dale]

What exactly I am interested in calculating is simple the dilation factor between the front and back of a frame 1km in length at an acceleration of 1000g

10km/sec² and at ...a=1g

Sure, no problem. I basically know of two approaches. The first (easier) is to work in the inertial reference frame and use the Doppler shift, and the second (more general) is to work in the Rindler frame and use the metric. We can do it either way, which do you prefer? Also, do you want me to show you the derivation step-by-step or do you want me to walk you through it and have you do it?

 

[Austin0]

Hi DaleSpam What exactly I am interested in calculating is simple the dilation factor between the front and back of a frame 1km in length at an acceleration of 1000g

10km/sec² and at ...a=1g

I Thanks

$$d\tau^2 = dt^2 - dx^2$$
$$d\tau^2 = R^2 \, d\Theta^2 - dR^2$$​

I am assuming a couple of things here. First we measure distance and time in units that make c = 1 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 $\Theta$ equals the observer's own proper time (as measured by his clock), but runs faster or slower than clocks at other fixed R values. The two coordinate systems are related by

$$x = R \cosh \Theta$$
$$t = R \sinh \Theta$$​

Sure, no problem. I basically know of two approaches. The first (easier) is to work in the inertial reference frame and use the Doppler shift, and the second (more general) is to work in the Rindler frame and use the metric. We can do it either way, which do you prefer? Also, do you want me to show you the derivation step-by-step or do you want me to walk you through it and have you do it?

The above was from DrGreg in what he said were non-standard terms.

From this am I correct that the R (x) location is normallized for one g and relative distances are in % of light year??

How is the acceleration factor incorporated??

In the other version I have seen ,the modified Lorentz proper time equation with the negative x factor of the time component; it seems that it will return the same dilation factor no matter what the acceleration magnitude. SO I can see how I could possibly get the dilation for a one g system but not how to extrapolate or what is the metric for x.
Or maybe I am just clueless here.?

Thanks

 

[Dale]

How is the acceleration factor incorporated??

The acceleration depends on R by the formula:
a=c²/R

So for R = 9 Pm (~ 1 light year) you get a = 1 g. If you want a = 1000 g then you would use R = 9 Tm. Of course, you can always adjust your length scale so that R=1. Then your time scale is uniquely determined:
T=c/a

So for a = 1 g we get one unit of time T = .97 year. And for a = 1000 g we get one unit of time T = 8.5 hour.

Will get to the rest in a bit.

 

[yuiop]

Hi DaleSpam What exactly I am interested in calculating is simple the dilation factor between the front and back of a frame 1km in length at an acceleration of 1000g

10km/sec² and at ...a=1g

I am sure for someone with the skills this is completely trivial but I don't have the trig.

I can't deny I am hoping you might help me out here before giving up on me completely :-)

Thanks

Hi Austin,

You need to be clear about something here. If we have points A and B at the back and front of a rocket with length 1km, and the rocket has proper acceleration of 1000g as measured by accelerometers at the back and front, then the rocket will be physically stretched and very quickly torn apart and the length 1km is only valid at one point in time. In this scenario, the length of the rocket appears constant in the original inertial "launch" frame (until it is torn apart) and the length of the rocket is getting longer as measured by the observers onboard the rocket. This is basically related to Bell's rocket paradox and his famous piece of string.

In order for the rocket observers to measure the rocket's proper length as 1km continuously, the front and back ends of the rocket have to accelerate at different rates (and this includes both porper and coordinate acceleration). See Born rigid acceleration for more info. In this case the rocket will appear to be length contracting in the original inertial "launch" frame. If this is what you mean, then you have to specify whether 1000g is applicable to the front or back of the rocket, because it can not apply to both.

 

[Dale]

coordinate time $\Theta$ equals the observer's own proper time (as measured by his clock), but runs faster or slower than clocks at other fixed R values.

Actually, that is not quite correct. The proper time is $\tau$, not $\Theta$. For an observer which is stationary in the Rindler coordinates dR=0 so we get:

$$d\tau^2 = R^2 \, d\Theta^2 - dR^2$$

$$d\tau = R \, d\Theta$$

Which means that for large values of R the proper time advances a lot for a small advance in coordinate time, and for small values of R a lot of coordinate time advances for a small advance in proper time. They are proportional to each other, but not the same except for the observer at R=1.

 

[Austin0]

The acceleration depends on R by the formula:
a=c²/R

So for R = 9 Pm (~ 1 light year) you get a = 1 g. If you want a = 1000 g then you would use R = 9 Tm. Of course, you can always adjust your length scale so that R=1. Then your time scale is uniquely determined:
T=c/a

So for a = 1 g we get one unit of time T = .97 year. And for a = 1000 g we get one unit of time T = 8.5 hour.

Will get to the rest in a bit.

Thanks DaleSpam This may be enough. Let me think it through and se if I can get the results with this before you spend any more time.

 

[Austin0]

Hi Austin,

You need to be clear about something here. If we have points A and B at the back and front of a rocket with length 1km, and the rocket has proper acceleration of 1000g as measured by accelerometers at the back and front, then the rocket will be physically stretched and very quickly torn apart and the length 1km is only valid at one point in time. In this scenario, the length of the rocket appears constant in the original inertial "launch" frame (until it is torn apart) and the length of the rocket is getting longer as measured by the observers onboard the rocket. This is basically related to Bell's rocket paradox and his famous piece of string.

In order for the rocket observers to measure the rocket's proper length as 1km continuously, the front and back ends of the rocket have to accelerate at different rates (and this includes both porper and coordinate acceleration). See Born rigid acceleration for more info. In this case the rocket will appear to be length contracting in the original inertial "launch" frame. If this is what you mean, then you have to specify whether 1000g is applicable to the front or back of the rocket, because it can not apply to both.

Not a problem. I came to this through the Born hypothesis. Yhe assumption is Born rigid acceleration. I simply want to know given this: with 1000g at the back what would be the dilation factor between front and back.
am I correct in thinking this is determined through the Rindler equations without having to factor in relative acceleration between the two?

Thanks

 

[Austin0]

Actually, that is not quite correct. The proper time is $\tau$, not $\Theta$. For an observer which is stationary in the Rindler coordinates dR=0 so we get:

$$d\tau^2 = R^2 \, d\Theta^2 - dR^2$$

$$d\tau = R \, d\Theta$$

Which means that for large values of R the proper time advances a lot for a small advance in coordinate time, and for small values of R a lot of coordinate time advances for a small advance in proper time. They are proportional to each other, but not the same except for the observer at R=1.

Hi DaleSpam I have been diverted to other questions but haven't forgotten this one.

I seem to have run into a wall here.
SO far I have

1 m= 3 *10^-8 ls [rounding c off to 3*10⁵km/s]

8.5 lhrs=3.06*10⁵ ls = 9.18*10¹³m

SO 1 m =3.3333*10^-9 sec and 1000m= 3.3333*10^-6 sec; is this the dilation factor?

How this relates to $$d\tau = R \, d\Theta$$ I am unclear or if I am totally clueless and spinning my wheels.
SO any pointers definitely would be helpful and appreciated. Thanks

 

[Dale]

Hi DaleSpam I have been diverted to other questions but haven't forgotten this one.

I seem to have run into a wall here.
SO far I have

1 m= 3 *10^-8 ls [rounding c off to 3*10⁵km/s]

8.5 lhrs=3.06*10⁵ ls = 9.18*10¹³m

SO 1 m =3.3333*10^-9 sec and 1000m= 3.3333*10^-6 sec; is this the dilation factor?

How this relates to $$d\tau = R \, d\Theta$$ I am unclear or if I am totally clueless and spinning my wheels.
SO any pointers definitely would be helpful and appreciated. Thanks

Hi Austin0,

The first thing that you need to do is to pick your units (make sure that c=1). You have m, lh, ls, for distance and so I am not sure what units you are using. If you have no preference then I would recommend ly for your unit of distance and y for your unit of time. This makes R = 1 ly correspond to an acceleration of .97 g which is close enough.

Then, our 1000g (actually 970 g, but that is close enough) worldline is at R=.001. For this distance we have
$$d\tau = .001 \; d\Theta$$
This means that for every year that elapses on the 1000 g clock theta increases by only .001 radians. By comparison, for every year that elapses on the 1 g clock theta increases by 1 radian.

 

[Austin0]

Hi Austin0,

The first thing that you need to do is to pick your units (make sure that c=1). You have m, lh, ls, for distance and so I am not sure what units you are using. If you have no preference then I would recommend ly for your unit of distance and y for your unit of time. This makes R = 1 ly correspond to an acceleration of .97 g which is close enough.

Then, our 1000g (actually 970 g, but that is close enough) worldline is at R=.001. For this distance we have
$$d\tau = .001 \; d\Theta$$
This means that for every year that elapses on the 1000 g clock theta increases by only .001 radians. By comparison, for every year that elapses on the 1 g clock theta increases by 1 radian.

Hi DaleSpam yes I was trying to pick my units, trying to work from the info you gave me to arrive at meters. I think I see part of the problem. You don't relaize just how terribly limited my math is. I have forgotten all trigonometry beyond basic concepts, let alone hyperbolic functions which I never knew. I don't actually even remember working with radians anyway, only degrees.
SO I assume that the last you gave was a factor of dilation between a 1g system and a 1000g system but as a trig function , is this right?
SO if as you suggest I use 1 ly as basic length unit and one year as basic time unit I would still need some conversion to get down to meters for figuring the dilation factor between R=1 and R=1000 meters , yes?
Thanks for your patience , I realize my lack of math presents a challange.

 

[Dale]

yes I was trying to pick my units, trying to work from the info you gave me to arrive at meters.

No problem. I would recommend doing all of the intermediate calculations in y and ly and then changing to meters and seconds only at the very end. So you will have the following conversion factors at the end:
1 year = 31 556 926 seconds
1 light year = 9.4605284 × 10^15 meters
1 light year/year = 299 792 458 meters / second = 1 c
1 light year/year^2 = 9.50005264 meters / second^2 = 0.96873577 g

SO if as you suggest I use 1 ly as basic length unit and one year as basic time unit I would still need some conversion to get down to meters for figuring the dilation factor between R=1 and R=1000 meters , yes?

I don't think that you want R=1 and R=1000 meters, that would correspond to ~10^16 g and ~10^13 g respectively. You want R = 1 and R = .001 light years, which correspond to ~1 g and ~1000 g respectively.

 

[Austin0]

Hi DaleSpam SO I assume that the last you gave was a factor of dilation between a 1g system and a 1000g system but as a trig function , is this right?
SO if as you suggest I use 1 ly as basic length unit and one year as basic time unit I would still need some conversion to get down to meters for figuring the dilation factor between R=1 and R=1000 meters , yes?
.

No problem. I would recommend doing all of the intermediate calculations in y and ly and then changing to meters and seconds only at the very end. So you will have the following conversion factors at the end:
1 year = 31 556 926 seconds
1 light year = 9.4605284 × 10^15 meters
1 light year/year = 299 792 458 meters / second = 1 c
1 light year/year^2 = 9.50005264 meters / second^2 = 0.96873577 g

I don't think that you want R=1 and R=1000 meters, that would correspond to ~10^16 g and ~10^13 g respectively. You want R = 1 and R = .001 light years, which correspond to ~1 g and ~1000 g respectively.

Hi DaleSpam When I said R=1 and R=1000 meters I was referring to two locations in the same 1000g system, two different parts i.e. front and back of the system and trying to learn how to calculate the dilation factor or differential between them. In essence it should be R=0 and R-1000 but there seems to be some problem or complication with R=0 , yes?

So if R= .001 ly's in the 1000 g system would this mean that R =1,000m would =(.001/9.461*10 ¹⁵)* 1000?
Thaniks again for bearing with me , I have been harried of late but this is getting ridiculous. I feel there is something simple and fundamental I am just not getting

 

[Passionflower]

Austin0 are you trying to get the ratios between the gammas of the front v.s. the back of a 1Km long Born rigid rocket with constant proper acceleration of 1000g?

Assuming my calculations are correct the answer is extremely small, if the front accelerates 9807 m/s² the back will accelerate about 9807.000001 m/s². The ratio of the gammas will be extremely close to 1 and you need a very high precision calculator to even see anything but 1. If you make the spaceship 10,000,000,000 meters the ratio is something like 1.001091311.

Or perhaps that is not exactly what you are looking for?

 

[Dale]

When I said R=1 and R=1000 meters I was referring to two locations in the same 1000g system, two different parts i.e. front and back of the system

In the Rindler coordinate system each location has a different proper acceleration. There are not different locations in a 1000g system, there is a single R coordinate which corresponds to an acceleration of 1000g and any other R coordinate corresponds to a different acceleration.

As Passionflower mentioned, you have to use a high-precision math engine but if you have a 1000 m long rocket and the rear is accelerating at 1000 g then you can calculate the acceleration at the front as follows:
$$r_{rear}=\frac{c^2}{1000 g}=9.165 \; 10^{12}m$$
$$a_{front} = \frac{c^2}{r_{rear}+1000m}= 999.99999989 g$$

 

[Austin0]

Austin0 are you trying to get the ratios between the gammas of the front v.s. the back of a 1Km long Born rigid rocket with constant proper acceleration of 1000g??

Yes. yes,yes !

Assuming my calculations are correct the answer is extremely small, if the front accelerates 9807 m/s² the back will accelerate about 9807.000001 m/s². The ratio of the gammas will be extremely close to 1 and you need a very high precision calculator to even see anything but 1. If you make the spaceship 10,000,000,000 meters the ratio is something like 1.001091311.?

OK this is getting close to exactly what I was wanting to calculate.
But now there is a confusion: I understood that the dilation factor was comparable to some equivalent spatial separation in a gravitational context. So I was expecting a small differential but non-negligable. In the tower experiments in 1g, the distance was not that great but I understood the gamma to be on the order of some factor of 10^-9
I also assumed that an acceleration of 1000g would equate to a Schwarzschild radius in a significantly more massive field with a far greater dilation factor per distance.

Based on what you are telling me here it would appear that in any realistic Born accelerated system of realistic length, the relative dilation would be totally moot. Beyond the ability of instrumentation to detect by measurement of light speed.
That only over an extended course of acceleration would the cumulative effect possibly become significant in the resulting desynchronization of clocks within the system.
Of course I may still be missing something fundamental in this picture.

Or perhaps that is not exactly what you are looking for?

Yes thank you very much this is exactly what I was looking for.
Can I take the figure you gave above for a 10¹⁰ meter system [ 1.001091311]
and simply diminish it by a factor of 10^-7 to get a gamma for 1000m ?

 

[Austin0]

In the Rindler coordinate system each location has a different proper acceleration. There are not different locations in a 1000g system, there is a single R coordinate which corresponds to an acceleration of 1000g and any other R coordinate corresponds to a different acceleration.

As Passionflower mentioned, you have to use a high-precision math engine but if you have a 1000 m long rocket and the rear is accelerating at 1000 g then you can calculate the acceleration at the front as follows:
$$r_{rear}=\frac{c^2}{1000 g}=9.165 \; 10^{12}m$$
$$a_{front} = \frac{c^2}{r_{rear}+1000m}= 999.99999989 g$$

Hi DaleSpam Well I was apparently right obout one thing; I was missing some fundamental understanding of Rindler coordinates.
I thought there was some direct way to calculate the relative gamma directly from position in the Born system without neccessarily first deriving relative acceleration.
I also expected a more significant dilation factor.
Of course I am still in the dark as to how to get the gamma factor from relative acceleration without possibly working out instantaneous velocities at both ends.
Thanks for shedding some light

 

[Passionflower]

That only over an extended course of acceleration would the cumulative effect possibly become significant in the resulting desynchronization of clocks within the system.

The ratio of the gammas would not change over time. But you are right the cumulative effect of the different clock rates over time would increase.

By the way the acceleration differential in the Schwarzschild solution is not exactly the same due to the fact that there are tidal forces but by approximation it is the same.

 

[DrGreg]

I thought there was some direct way to calculate the relative gamma directly from position in the Born system without neccessarily first deriving relative acceleration.

The more general form of the Rindler metric (using a more conventional notation and ignoring y and z) is

$$c^2\,d\tau^2 = \left(\frac{ax}{c}\right)^2 \, dt^2 - dx^2$$​

where a is the proper acceleration of the frame's observer who is located at x = c²/a, and $\tau$ is proper time.

For any stationary point, dx = 0, so

$$d\tau = \frac{ax}{c^2} \, dt$$​

so the time dilation factor between two stationary points at x = c²/a and x = c²/a + L is

$$1 + aL/c^2$$​

c² is a huge number (in (m/s)²) so the dilation effect is very small unless a or L are extremely large.

 

[Passionflower]

But now there is a confusion: I understood that the dilation factor was comparable to some equivalent spatial separation in a gravitational context.

Assuming I get this right, there are basically 3 different kind of situations you would like to consider:

1. A homogenic gravitational field
The inertial acceleration at any height in the field is exactly the same.

To calculate the redshift between two points in such a field you have to use:

$$e^{-gh/c^2}$$
(g and h in meters)​

2. A uniform gravitational field
The inertial acceleration at any height in the field varies by $$c^2/{g + h}$$ (g and h in meters)

To calculate the redshift between two points in such a field you have to use:

$$1-gh/c^2$$​

(g and h in meters)​

Stationary observers in a uniform gravitational field are equivalent with an constantly accelerating frame.

3. A approximate Schwarzschild gravitational field (assume r coordinate is radius)

For a stationary observer the proper acceleration in the field varies by height: $$\frac{M}{r^2}c^2$$ (M and r in meters)

To calculate the redshift between r emitted and r received (both stationary) you have to use:

$$z = \sqrt {\frac {r_R - R_S}{r_E - R_S}} - 1$$​

Source: Gron, Hervik
"Einstein's General Theory Of Relativity With Modern Applications In Cosmology" (Springer, 2007)
(All in meters, R_s is the Schwarzschild radius)​

A Schwarzschild gravitational field is different from the two other fields in that the same acceleration can be found for different masses but at different distances from the r=0.

Here are some numbers:

Acceleration: 9.8 m/s^s
Height: 1 * 10¹³

And only for case 3:
Mass: 1 m (14986661871 kg)
Distance: 957500000 m

Here are the redshifts:

• For a homogenic gravitational field we get: 0.998910197
• For a uniform gravitational field we get: 0.998909603

For the Schwarzschild gravitational field we get:

The redshift is: 1251.154578
g at 1 * 10¹³ is: -8.99 *10¹⁰

Now the same acceleration for a radically different M and distance, namely the Earth:

Mass: 0.004435407 m (66471944.97 kg)
Distance: 6378000.1 m

The redshift is: 1251.154383
g at 1 * 10¹³ is: -3.99 * 10^-12, practically reduced to nothing.

Can others confirm the calculations are correct?

Edited to correct: I forgot the Schwarzschild redshifts and changed the formula and added some clarifying notes. Changed 'constant' to 'uniform'

 

[Austin0]

Assuming I get this right, there are basically 3 different kind of situations you would like to consider:

1. A homogenic gravitational field
The inertial acceleration at any height in the field is exactly the same.

To calculate the redshift between two points in such a field you have to use:

$$e^{-gh/c^2}$$​

(g and h in meters)​

2. A constant gravitational field
The inertial acceleration at any height in the field varies by $$c^2/{g + h}$$ (g and h in meters)

To calculate the redshift between two points in such a field you have to use:

$$1-gh/c^2$$​

(g and h in meters)​

Stationary observers in a constant gravitational fields are equivalent with an constantly accelerating frame.

3. A approximate Schwarzschild gravitational field (assume r coordinate is radius)

The inertial acceleration at any height in the field varies by $$\frac{M}{r^2}c^2$$ (M and r in meters)

To calculate the redshift between two points in such a field you have to use:

$$\sqrt {\frac {1+2M/r}{1+2M/{r+h}}$$​

(M, r and h in meters)​

A Schwarzschild gravitational field is different from the two other fields in that the same acceleration can be found for different masses but at different distances from the r=0.

Here are some numbers:

Acceleration: 9.8 m/s^s
Height: 1 * 10¹³

And only for case 3:
Mass: 1 m (14986661871 kg)
Distance: 957500000 m

Here are the redshifts:

• For a homogenic gravitational field we get: 0.998910197
• For a constant gravitational field we get: 0.998909603

For the Schwarzschild gravitational field we get:
g at 1 * 10¹³ is: -8.99 *10¹⁰

Now the same acceleration for a radically different M and distance, namely the Earth:
Mass: 0.004435407 m (66471944.97 kg)
Distance: 6378000.1 m

g at 1 * 10¹³ is: -3.99 * 10^-12, practically reduced to nothing.

Can others confirm the calculations are correct?

There is a lot here to try and absorb. It seems to me that for a Born rigid accelerated system I would apply #2 is this right??

When you say redshift is that equivalent to time dilation ?
The figure for a constant field looks significant but of course you use a huge distance.

I really want to thank you for your thought and help. This question has been on my mind for some time. Thanks

 

[Passionflower]

By the way this article is very relevant to the topic we are discussing:
http://www.scientificamerican.com/blog/post.cfm?id=atom-interferometer-measures-einste-2010-02-17
and
http://arstechnica.com/science/news...n-highly-accurate-relativity-measurements.ars

"Aside from its gravity-influenced trajectory, the atoms also moved as waves that would interfere constructively or destructively, depending on the phase difference between the waves when the paths realigned. The phase difference was influenced by a number of factors, but the researchers found that the effects all canceled each other out—all except for the redshift from the slightly different gravitational effects of the two different trajectories. Plugging this into the right equations gave a new measure of the gravitational redshift parameter that is 10,000 times more accurate than previously obtained values. "

Regarding my prior posting with the formulas I am doubting whether I have the redshifts for 1 and 2, the Doppler shifts or the inverse. Anybody can take a look?

 

[Dale]

When you say redshift is that equivalent to time dilation ?

Redshift is equivalent to gravitational time dilation only if the two clocks/emitters are stationary wrt each other. If they are moving then there are velocity-induced time dilation and Doppler effects also.

Btw, I don't know if you are trying to derive the time dilation, or just calculate it. If you are just trying to calculate it then we can use the formula:

$$\frac{dt(a)}{dt(b)}=\sqrt{\frac{g_{tt}(a)}{g_{tt}(b)}}$$

where $$g_{tt}(a)$$ is the coefficient of the time² component of the metric equation above evaluated at location a.