
#19
Dec112, 04:28 PM

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PF Gold
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I do understand better where you're trying to go with this from your comments. In general what you're doing makes sense; I'm just trying to give some cautions about its limitations. 



#20
Dec112, 04:50 PM

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PF Gold
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In general, the Einstein Field Equation (which the equations on the Wiki page that you refer to are components of) tells how the curvature of spacetime is related to the presence (or absence, in this case) of stressenergy (energy, momentum, pressure, and stress are all components of stressenergy). So in general, the equations you refer to are basically constraints on how the curvature can vary given that the spacetime is vacuum. But the specific form depends on the coordinate chart you adopt. 



#21
Dec112, 04:54 PM

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The way distance works is this.
If you've got a curve, and you use standard definitions, everyone agrees on the length of the curve, the length of the curve is not observer dependent. Nor is the length of the curve coordinate dependent  it doesn't matter what your observer is, or what coordiates you choose, the curve has a length that's inddepenent of all that. So you'll never have a case of a distant observer saying a given curve has a different length than a nearby observer. The length of a curve is observer independent , period. But  we can't say the same about distance. Why is that? Well, the problem is in specifying the curve. If you have a curve, the length of it is observer independent. But the actual curve itself may not be. That's the only tricky issue in measuring distanace. How do you specify this curve whose length you are to measure? Once you've managed to do that you've got the problem solved. Generally, the curve is specified as the shortest curve connecting two points, where all points are on a surface of simultaneity. The surface of simultaneity CAN change with the observer, so in general distance (in SR) can depend on the observer. This is a consequence of the relativity of simultaneity. But the ONLY reason this happens is when you have a disagreement on simultaneity. If you have a nontime varying gravitational field (like the Schwarzschild. case), and you're outside the event horizon, there is a shared notion of simultaneity for all static observers. Thus, in the case of a static observer, IT DOESN"T MATTER where you are when you measure length. Whether you're "on site" or "at infinity" a meter, is a meter, is a meter. Thus it is sufficient to be able to measure the distance onsite with a ruler to determine, say, the height of a building, there's no need to spend a lot of time agonizing about your meter sticks changing length, and your clocks ticking at different rates, and all that. Your local clocks, and local rulers give you the answer you're looking for, the one and only answer. And if the distant observer comes up with a different answer, he's simply wrong. It's necessary that you an the distant observer share a common notion of simultaneity to make this statement  but as long as you restrict yourself to static observers, you hae agreement on length, just as you do in special relativity if you demand that everyone be at rest relative to one another by choosing some particular inertial frame. 



#22
Dec112, 09:19 PM

P: 434

Thank you everyone for your replies so far. :)
(dz / dr) L = m L_t^2 / r^2 which should be true for any arbitrary coordinate system. For instance, in SC, we have z = sqrt(1  2 m / r), L = sqrt(1  2 m / r), L_t = 1 dz / dr = (m / r^2) / sqrt(1  2 m / r) [(m / r^2) / sqrt(1  2 m / r)] sqrt(1  2 m / r) = m (1)^2 / r^2 which works out. In GUC it is z = 1 / sqrt(1 + 2 m / r), L = 1 / sqrt(1 + 2 m / r), L_t = 1 / (1 + 2 m / r) dz / dr = (m / r^2) / (1 + 2 m / r)^(3/2) [(m / r^2) / (1 + 2 m / r)^(3/2)] / sqrt(1 + 2 m / r) = m [1 / (1 + 2 m / r)]^2 / r^2 which also works out, and for EIC it becomes z = (1  m / (2 r)) / (1 + m / (2 r)), L = 1 / (1 + m / (2 r))^2, L_t = 1 / (1 + m / (2 r))^2 dz / dr = (m / r^2) / (1 + m / (2 r))^2 [((m / r^2) / (1 + m / (2 r))^2] / (1 + m / (2 r))^2 = m [1 / (1 + m / (2 r))^2]^2 / r^2 which works out as well. I'm still looking for another relationship that can be put together with this one so that the other two coefficients can be found after making a coordinate choice for one. 



#23
Dec212, 01:00 AM

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#24
Dec212, 09:48 AM

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#25
Dec212, 10:02 AM

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[tex]ds^2 = A(r) dt^2 + B(r) dr^2 + r^2 d\Omega^2[/tex] It's fairly straightforward (though a bit tedious in places) to show that the metric for any static, spherically symmetric spacetime can be written in this form. (MTW lays out the reasoning in some detail.) However, that doesn't mean this form is the best form for looking at all aspects of the physics. Essentially, you *are* starting with SC; any line element written in the above form is just a small variation on SC, and has all of the same restrictions as SC. If all you want to study is static, spherically symmetric spacetimes (i.e., things like stars in stable equilibrium, *not* black holes), these restrictions aren't that much of an issue. But if you want to look at other kinds of spacetimes (e.g., black holes, or the FRW spacetimes used in cosmology), you won't get very far. We've already seen that the SC form of the line element doesn't work well at a BH horizon, because the spacetime at and inside the horizon is not static. An FRW spacetime is even worse: the universe is expanding, so the spacetime isn't static *anywhere*, and you can't even write the metric in the above form at all. The general procedure of looking at the symmetries of a particular spacetime, and trying to match up coordinates with them, is fine as far as it goes; but there's no guarantee that it's going to give you a good handle on all the physics. 



#26
Dec212, 10:57 AM

P: 434





#27
Dec212, 12:26 PM

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There is an invariant associated with the radial part of the metric, but it's not really a local one. To see what it is, go back to the thought experiment I mentioned earlier, with a long, radially oriented rod going between a distant observer with r > infinity and a local observer at finite r. We saw that if the distant observer moves the rod radially by 1 meter, as seen by comparing a local meter stick to marks on the rod made while it was in empty space far from all gravitating bodies, the local observer also sees the rod move radially by 1 meter, as seen by comparing a local meter stick to marks on the rod at his end. However, we could also ask a different question: what is the proper length of the rod between the two observers? I.e., how many marks, each separated by 1 meter (as seen locally), lie between them when the rod is oriented radially? This is where the effect of the dr^2 metric coefficient shows up; it makes the answer *larger* than it would be if spacetime were flat. Another way of expressing this is that the spatial geometry of a slice of constant Schwarzschild time is not Euclidean. In your terms, the geometry of the "flat sheet of paper" on which the distance covered by each coordinate interval dr is the same, is *different* from the geometry of the spatial slice where the distance covered by each coordinate interval dr changes with r. Going by the flat sheet of paper, we would expect r2  r1 marks between the two observers; but going by the actual geometry of the spatial slice, using the SC chart, the number of marks is given by [tex]s = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1  2m / r}}[/tex] This turns out to be *larger* than r2  r1, and that fact is an invariant; it doesn't depend on the particular coordinates you adopt. You can calculate it in GUC and get the same answer. You can also calculate it in EIC, but there you have to be careful, because the labeling of the 2spheres changes; you will have to find *different* values R1 and R2 for the limits of integration, to keep the physical invariants at each limit the same. But the metric coefficients will *not* be among those invariants. One other thing to be careful of: you'll note that in defining this invariant s, I specified the geometry of a slice of constant *Schwarzschild* coordinate time. Charts with different slicings (such as Painleve or EddingtonFinkelstein or Kruskal) will have a *different* "proper distance" between r1 and r2, even if the r values themselves are identical. For example, in the Painleve chart, the proper distance in a slice of constant Painleve time between r1 and r2 *is* just r2  r1; the spatial geometry of a slice of constant Painleve time *is* Euclidean. The slices of constant Painleve time are "cut at an angle" compared to the slices of constant Schwarzschild time. So even in a static spacetime, there are still subtleties. 



#28
Dec212, 12:46 PM

P: 434

Right, the proper length ds of a local radial ruler does not change with the coordinate system, so that is an invariant, and we have
ds = dr / L = dr1 / L1 Coordinate tangent length changes directly with r1 / r, since the circumference of a spherical shell varies directly with r1 / r according to a distant observer, while local observers measure the ruler distance around their shell the same, invariant, so L1_t / L_t = r1 / r L1_t / r1 = L_t / r and z is an invariant, so z1 = z So far I haven't been able to use those to determine anything though. 



#29
Dec212, 01:07 PM

P: 434

(dz / dr) L = m L_t^2 / r^2 will work for any arbitrary coordinate system of this metric form, transforming the variables z, L , and L_t to z1, L1, and L1_t with (dz1 / dr1) L1 = m L1_t^2 / r1^2 dz1 (L1 / dr1) = m (L1_t / r1)^2 whereby dz1 = dz, L1 / dr1 = L / dr, and L1_t / r1 = L_t / r, so that we have dz (L / dr) = m (L_t / r)^2 



#30
Dec312, 07:08 PM

P: 434

Okay well, the relations shown in the last couple of posts are coordinate independent invariants, so if we use those to find some relationship that works out for SC, then it will work for any coordinate system of the same metric form. So far I have happened upon the relationship
[d(L_t / r) / dr] L (r / L_t)^2 =  z or d(L_t / r) (L / dr) / (L_t / r)^2 =  z which is composed entirely of the invariants found in the last couple of posts as can be seen more clearly in the second equation. It works for SC, so it will work for any coordinate transformation from SC. Combining this with the relationship found in the OP, we can also gain m d(L_t / r) =  dz z which is also composed of invariants. In fact we can now see in the OP that those individual invariants shown there for a' are composed of these invariant factors also (and constants), such as a' =  (G M) (L_t / r)^2 / z and a' =  c^2 dz (L / dr) / z We now have a second relationship to combine with the first, which by working through the last two for a', the first relationship was found to be c^2 (dz / dr) L = G M L_t^2 / r^2 or dz (L / dr) = m (L_t / r)^2 so that if we make some arbitrary coordinate choice for one variable, z, L, or L_t in terms of r, we can now directly determine the other two from those two relationships, although it may still be difficult for some with the derivatives involved. I will continue to work through these some more to find more such relationships, but what I'm really looking for is the logical foundations behind them. I'm hoping I might be able to see something in one of them that looks like a reasonable assumption to base the particular resulting equation upon. Does anybody see anything about any of the first three equations shown in this post that looks like something can be determined about what logical foundation that equation might be based upon? 



#31
Dec312, 09:20 PM

P: 434

Here's a rather simple one I have found so far.
z^2 = 1  2 m (L_t / r) So if we make a coordinate choice for z or L_t, we can easily find the other. Then we can find L with one of the other relationships found before, such as L = m (L_t / r)^2 / (dz / dr) I still need the logical foundation for it, though. 



#32
Dec312, 09:45 PM

P: 434

Here's something interesting. We can rearrange that last one to get
(L_t / r) = (1  z^2) / (2 m) and combine it with L = m (L_t / r)^2 / (dz / dr) to get L = (1  z^2)^2 / (4 m (dz / dr)) which gives us the relationship between L and z, but if we then combine that with a' =  c^2 (dz / dr) L / z from the OP, we find a' =  c^2 (1  z^2)^2 / (4 m z) which gives us a direct relationship between the local acceleration and time dilation, independent of r or any other factors other than the mass of the gravitating body M, since m = G M / c^2. So if one only knows the numerical time dilation at some shell (regardless of the value of r for that shell) and the mass of the gravitating body, they can find the local acceleration at that shell by applying this formula which is coordinate independent. Of course that also works vice versely, one can find the time dilation according to whatever the local acceleration is, rather than work with gravitational potential for a specific coordinate dependent r or whatever, but finding the solution for z in terms of a' in that manner appears to be long and messy. 



#33
Dec312, 09:50 PM

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PF Gold
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[tex]a = \frac{M}{r^2 \sqrt{1  2M / r}} = \frac{M}{r^2 z}[/tex] So if I know z and M, I still can't find a without knowing r as well. 



#34
Dec312, 10:15 PM

P: 434

(By the way, just a quick mention that that would actually need to be a' =  G M L_t^2 / (r^2 z) to be coordinate independent, but of course in SC, L_t = 1 so amounts to the same thing for that coordinate system only.) 



#35
Dec312, 11:20 PM

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#36
Dec312, 11:47 PM

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[tex]z = \sqrt{1  \frac{2M}{r}}[/tex] and rearrange to: [tex]r = \frac{2M}{1  z^2}[/tex] and then substitute into the formula for a to obtain: [tex]a = \frac{M}{r^2 z} = \frac{M \left( 1  z^2 \right)^2}{4 M^2 z} = \frac{\left( 1  z^2 \right)^2}{4 M z}[/tex] which is what you wrote a few posts ago. 


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