Integrating an accelerating Worldline

[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.