Proper distance, Area and Volume given a Metric

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SUMMARY

The discussion focuses on calculating the proper distance, area, volume, and four-volume using a specific metric defined by the equation involving a constant A. The proper distance is derived as \( \Delta s = R \left(1 - \frac{AR}{2} \right) \). The proper area is calculated as \( A = 4\pi R^2 \), while the proper volume is expressed as \( V = \frac{4}{3} \pi R^3 \left( 1 - \frac{3}{5} AR^2 \right) \). The four-volume is determined using the formula \( V_4 = \frac{4}{3} \pi R^3 \cdot cT \cdot \left( 1 - \frac{6}{5}AR^2 + \frac{3}{7}A^2R^4 \right) \), indicating a deeper exploration of the metric's implications in future studies.

PREREQUISITES
  • Understanding of differential geometry concepts
  • Familiarity with integration techniques in multivariable calculus
  • Knowledge of metrics in general relativity
  • Basic grasp of spherical coordinates
NEXT STEPS
  • Study the implications of metrics in general relativity
  • Learn advanced integration techniques for multivariable functions
  • Explore the concept of curvature in differential geometry
  • Investigate the physical significance of the constant A in various metrics
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Students and researchers in physics and mathematics, particularly those focusing on general relativity, differential geometry, and multivariable calculus.

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Homework Statement


[/B]
(a) Find the proper distance
(b) Find the proper area
(c) Find the proper volume
(d) Find the four-volume

ricci2.png


Homework Equations

The Attempt at a Solution



Part (a)

Letting ##d\theta = dt = d\phi = 0##:

\Delta s = \int_0^R \left( 1-Ar^2 \right) dr = R \left(1 - \frac{AR}{2} \right)

Part (b)

At ##r=R##:

A = R^2 \int_0^{2\pi} d\phi \int_0^{\pi} sin \theta d\theta
A = 4\pi R^2

Part (c)

V = \int_0^R r^2 \left( 1 - Ar^2 \right) dr \int_0^{\pi} sin \theta d\theta \int_0 ^{2\pi} d\phi
V = \frac{4}{3} \pi R^3 \left( 1 - \frac{3}{5} AR^2 \right)

Part (d)

V_4 = c\int_0^R r^2 \left( 1 - Ar^2\right)^2 dr \int_0^T dt \int_0^{\pi} sin \theta d\theta \int_0^{2\pi} d\phi

V_4 = \frac{4}{3} \pi R^3 \cdot cT \cdot \left( 1 - \frac{6}{5}AR^2 + \frac{3}{7}A^2R^4 \right)Is this the correct method? This question seems a little too straightforward..
 
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I would guess that, if you look forward in your text, you will find this metric gets a lot of work. For example, you may be doing some interesting work studying the place where r^2 = 1/A.

So quite likely this is just a warm-up getting some interesting features of the metric.
 

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