Integrating an accelerating Worldline

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SUMMARY

The discussion centers on the concept of line integration and its application to accelerating worldlines in Minkowski space. Participants clarify that line integration does not convert a curvilinear segment into a straight line but rather approximates the length of a curve by summing infinitesimal segments. The slope of the resulting line is determined through the integration of the proper time along the worldline, using metrics defined in Minkowski space. The gamma function is highlighted as a critical component in relating geometric length to the proper time of the worldline.

PREREQUISITES
  • Understanding of line integrals in calculus
  • Familiarity with Minkowski space and its metrics
  • Knowledge of proper time and worldlines in special relativity
  • Basic grasp of the gamma function and its applications
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  • Study the derivation and application of line integrals in physics
  • Learn about the metrics of Minkowski space and their implications for spacetime
  • Explore the concept of proper time and its calculation along worldlines
  • Investigate the role of the gamma function in relativistic physics
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  • #61
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?
 
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  • #62
Austin0 said:
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
 
  • #63
Passionflower said:
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 !

Passionflower said:
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.

Passionflower said:
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 ?
 
  • #64
DaleSpam said:
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
 
  • #65
Austin0 said:
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.
 
  • #66
Austin0 said:
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.
 
  • #67
Austin0 said:
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, Rs 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'
 
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  • #68
Passionflower said:
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
 
  • #69
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?
 
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  • #70
Austin0 said:
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.
 

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