Rod shortening of General Relativity

[yuiop]

A good approach.

When I spoke about light being slower in a gravitational field, and it gets slower for lower r values, I am talking about the coordinate speed of light, your calculations seem to relate to the speed for a particular observer. However I did mention the word 'measure' and that is definitely a mistake, because different observer will obviously measure this speed to be different. So you are completely right about that.

I also will get back on your calculations in a short while, by the way in the mean time it would be helpful to provide the formulas as well on how you obtained these numbers and when you use ruler distance what the size of that ruler would be if it would be transported our all the way to infinity away from the field.

Yes, I am talking about the speed of light for a particular observer, rather than just the coordinate observer at infinity, but you did say for "all observers".

The ruler distance is the distance measured using very short rods of say 1 meter in length. This rod size is the proper length and is what the length of the rod is when it stationary at r according to a local observer that is also stationary at r and if this ruler is transported to infinity, the observer at infinity will agree that the length of the ruler is 1 meter.

I have had to edit some of the numbers in my last post because I noticed some errors in my calculations when going over the equations again to post them here as requested. Good thing you asked!

The coordinate speed of light according to the observer at infinity is:

\frac{dr}{dt} = (1-2/r)

using units of G=m=c=1 so the Schwarzschild radius is at r=2.

The radar distance from r0 to r1 is obtained by integrating:

t = \int_{r0}^{r1} \frac{1}{(1-2/r)} dr = (r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)

This is the coordinate time it takes light to travel that distance. If the local observer is located at r0, then the radar distance according to that observer (allowing for gravitational time dilation) is:

t = \sqrt{1-2/r0} \left((r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)\right)

This is entirely valid and not an approximation, because the observer at r0 is using a single clock to measure the up and back down time of the radar signal.

The gravitational length contraction factor for a short stationary rod is:

ds = \frac{dr}{\sqrt{1-2/r}}

where ds is the length of the rod according to the local observer and dr is the coordinate difference length of the rod according to the Schwarzschild observer at infinity. By integration, the ruler distance between r0 and r1 is:

s = \int_{r0}^{r1} \frac{dr}{\sqrt{1-2/r}} = \sqrt{r1(r1-2)} - \sqrt{r0(r0-2)} + ln\left(\frac{\sqrt{r0} + \sqrt{r0-2}}{\sqrt{r1} + \sqrt{r1-2}} \right)

O.K. assuming the above is correct, here is another numerical example that demonstrates that the speed of light above an observer can be measured to be greater than c.

The local stationary observer is located at r=4 (again).
The target mirror is located at r=50.

The ruler distance from r=4 to r=50 is 48.9834839.
The radar distance from r=4 to r=50, according to the observer at r=4 is 37.0213588.
The average speed of light in terms of coordinate distance is 46/37.0213588 = 1.24252598c.
The average speed of light in terms of ruler distance is 48.9834839/37.0213588 = 1.32311416c.

In the above example the radar distance is less than both the coordinate and ruler distances and the average speed of light according to the local observer is >c whether measured in terms of coordinate distance or proper ruler distance.

 

[yuiop]

Hi Yuiop,
Well explained. I follow and agree with everything except the last reference to the double transformation that Claire observes. Let's consider the two quotes included. My understanding is that the free falling observer is the "inertial reference frame" and that the Swarzschild observer at infinity (Claire) would see the same a Alice (free falling frame).

The r, dr, dt and angular derivatives of the Schwarzschild metric are defined as measurements made by a stationary observer "at infinity" and only the proper time tau is measured "locally". We have shown that in certain circumstances that the measurements made by a free falling local observer are numerically equivalent to the measurements made by the observer at infinty, but it would be a mistake to assume that their points of view are identical. For example, if a test particle has Schwarzschild parameters r=3 and dr=0, this means that is stationary with respect to the observer at infinity. To the free falling observer falling past r=3, the test particle is not stationary and dr \ne 0.

Infinity is a very difficult concept and in some respects it is better to think of the observer at infinity as just "very far away" where the curvature is almost flat rather than actually being at "infinity" which introduces all sorts of headaches. For example if an observer freefalls from infinity for a finite amount of coordinate or proper time, then by the definition of infinity, they are still at infinity.

 

[Passionflower]

Yes, I am talking about the speed of light for a particular observer, rather than just the coordinate observer at infinity, but you did say for "any observer".

Indeed, so that was clearly a mistake on my part.

The coordinate speed of light according to the observer at infinity is:

\frac{dr}{dt} = (1-2/r)

using units of G=m=c=1 so the Schwarzschild radius is at r=2.

Yes I agree with that. And in this context I hope you would agree that everywhere in the field this speed is < c. However most observers will measure this speed differently due to their changes in relative clock rates. Could we identify an observer for which this is unity? Perhaps a free falling observer going at escape velocity?

The radar distance from r0 to r1 is obtained by integrating:

t = \int_{r0}^{r1} \frac{1}{(1-2/r)} dr = (r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)

Yes full agreement on this.

This is the coordinate time it takes light to travel that distance. If the local observer is located at r0, then the radar distance according to that observer (allowing for gravitational time dilation) is:

t = \sqrt{1-2/r0} \left((r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)\right)

This is entirely valid and not an approximation, because the observer at r0 is using a single clock to measure the up and back down time of the radar signal.

Yes, again full agreement.

The gravitational length contraction factor for a short stationary rod is:

ds = \frac{dr}{\sqrt{1-2/r}}

where ds is the length of the rod according to the local observer and dr is the coordinate difference length of the rod according to the Schwarzschild observer at infinity. By integration, the ruler distance between r0 and r1 is:

s = \int_{r0}^{r1} \frac{dr}{\sqrt{1-2/r}} = \sqrt{r1(r1-2)} - \sqrt{r0(r0-2)} + ln\left(\frac{\sqrt{r0} + \sqrt{r-2}}{\sqrt{r1} + \sqrt{r-2}} \right)

Here it looks we might have some difference of opinion. It appears this hinges on the interpretation of r and if the Schwarzschild solution is 'volume expanding' or 'object reducing'. Perhaps it would help if you could give me an analog of my description of the meaning of r from your perspective and then we can see what the differences really are (if any).

O.K. assuming the above is correct

I did not check the numbers but I fully agree with the formulas, except for what you call 'gravitational length contraction'.

So, it seems, if I am not mistaken, that the gravitational length contraction is the main thing to discuss.

To calculate the physical volume(as opposed to the coordinate volume) between r1 and r2 is:

<br /> \begin{align*}<br /> 4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, <br /> \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M <br /> }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) }+ \\<br /> 20\,\pi \,{M}^{3} <br /> \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ <br /> M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M <br /> \right) }- \\<br /> 10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M <br /> \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M <br /> \right) }- \\<br /> 20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 <br /> \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right) <br /> \end{align}<br />

(Sorry for the alignment, perhaps someone can suggest how to left align. The formula is too long and I had to break it up)

To work out the 'magic bag' metaphor a little bit more I plotted the volume between the event horizon and increasing areas (area * pi) both for an Euclidean space and for a Schwarzschild solution with m=1/2. Note that the plot does not use the r coordinate in any way, the volumes are based on the areas.

[PLAIN]http://img830.imageshack.us/img830/4332/volumeeuclideanvsproper.gif

It is clear the there is more volume between these areas than expected based on Euclidean calculations.

Note that for those who have an objection taking the area of the event horizon, an objection which I do not share by the way, taking the plots between any two larger areas will give the same results, the volume is larger than in the Euclidean calculation.

 

[pervect]

MTW gives a formula for a closely related case on pg 604 which agrees with Passionflower's formula as far as the setup goes (though they don't work out the integral, and I'm not going to either).

It's the volume element for the interior volume for a relativistic star using the interior Schwarzschild solution. Basically it's the same except m(r) isn't a constant in that case, m(r) is a function of r, which can be thought of as the mass within the radius r. (MTW gives more details, but their explanation of how m(r) is computed is a bit on the long side, perhaps a bit hand-wavy, and I don't see a need to go into it. Which may or may not mean it'll turn into a hot topic :-) ).

At any rate the formula is in agreement, i.e.:

dV = 4 pi r^2 dr / sqrt(1 - 2m(r)/r)

Also, they have a sketch of an embedding diagram of the equatorial spatial slice (a lice through the equator of constant Schwarzschild time) on pg 614 - a lot of authors have similar embedding diagrams, it's one of the more popular ones. From the sketch it's pretty clear that the enclosed volume is greater than it would be in Euclidean space.

Of course the volume element depends on the space-time slice, but the obvious spatial slice (orthogonal to Schwarzschild time) is used.

Finding the geometry of a space-slice for a massive object which is not a black hole is only slightly more complicated mathematically, is of great physical interest, and avoids dealing with any need to consider complications that could be introduced by the event horizon.

 

[Mentz114]

<br /> <br /> \begin{align*}<br /> &amp; 4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, <br /> \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M <br /> }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) } \\<br /> &amp; + 20\,\pi \,{M}^{3} <br /> \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ <br /> M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M <br /> \right) } \\<br /> &amp; - 10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M <br /> \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M <br /> \right) } \\<br /> &amp; - 20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 <br /> \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right) <br /> \end{align}<br /> <br />

This is the very integral I mentioned in post#25. I'm glad to see you're paying attention to the crucial facts - i.e. that the volume enclosed by a 2D surface of equal curvature is not dependent on r the way it is in flat space.

[PassionFlower, the alignment in Latex is done with the '&' character.]

 

[Pierre007080]

Hi Yuiop, The maths and terminology being discussed is largely above my head. Passionflower's mention of volume has helped me identify my problem with your "double transformation" Prof Susskind used fluid dynamics to demonstrate how the flow divergence was the mathematical analogy to the gravitational field. The acceleration flow field gave me this image of "space flowing" even though he was quick to mention that space does not actually flow. Using this analogy the speed of "flow" can easily be determined by realting the total work done (GMm/r) to the kinetic energy (1/2mv2). Thus v =√ 2GM/r. This is obviously the speed of a free falling test mass and also the escape velocity at that particular radius. If the free falling reference frame were to be seen as the "instantaneous" inertial frame, the large mass would be approaching the test mass at this speed. If the Lorentz transformations for rod shortening are applied by inserting v =√ 2GM/r, then the gravitational rod shortening factor you gave me earlier √ 1-2GM/r emerges. Is this some sort of circular reference??
Why I mention this "space flowing" instantaneous speed is that if the free falling body had no velocity at infinity, the the "space flow" speed would equal that of the body and only one transformation would apply. If not then of course the second transformation would apply to the difference of speed between "space" and the body. Is this too naive to have intuitive significance?

 

[yuiop]

Hi Yuiop, The maths and terminology being discussed is largely above my head. Passionflower's mention of volume has helped me identify my problem with your "double transformation" Prof Susskind used fluid dynamics to demonstrate how the flow divergence was the mathematical analogy to the gravitational field. The acceleration flow field gave me this image of "space flowing" even though he was quick to mention that space does not actually flow. Using this analogy the speed of "flow" can easily be determined by realting the total work done (GMm/r) to the kinetic energy (1/2mv2). Thus v =√ 2GM/r. This is obviously the speed of a free falling test mass and also the escape velocity at that particular radius. If the free falling reference frame were to be seen as the "instantaneous" inertial frame, the large mass would be approaching the test mass at this speed. If the Lorentz transformations for rod shortening are applied by inserting v =√ 2GM/r, then the gravitational rod shortening factor you gave me earlier √ 1-2GM/r emerges. Is this some sort of circular reference??
Why I mention this "space flowing" instantaneous speed is that if the free falling body had no velocity at infinity, the the "space flow" speed would equal that of the body and only one transformation would apply. If not then of course the second transformation would apply to the difference of speed between "space" and the body. Is this too naive to have intuitive significance?

There is a coordinate system called the "river model" or more formally the Gullstrand-Paineleve coordinates, that is similar to what you are talking about. In an informal description of the river model, "spacetime" is pictured flowing into the black hole and has a velocity equal to the speed of light as it passes the event horizon. In these coordinates everything moves relative to this flowing spacetime and an ingoing photon is moving at 2c as it passes the event horizon and an outgoing photon has a velocity of 0c at the event horizon because it cannot go against the current of the river and cannot escape a bit like trying to walk up a fast escalator. In these informal descriptions, they usually add a disclaimer that nothing is actually flowing because they realize that if something was actually flowing it would be an acknowledgment of an ether. One immediate problem with such a description is that it might lead to the conclusion that the time dilation of a stationary clock in Schwarzschild coordinates is due to the flow of this "substance" past the stationary clock. The stationary clock is effectively moving relative to the flowing river and it turns out 9as you have pointed out) that the time dilation of the stationary clock is equivalent to the time dilation a clock moving relative to an observer in flat SR spacetime with a velocity exactly equal to the falling escape velocity of the river. As you already have pointed out this implies that a clock freefalling with escape velocity is stationary with respect to the river substance and should experience no time dilation, but we know this is not true because we have proven in this forum (jn many different ways) that the time dilation of a falling clock is tau = t*\sqrt{(1-2m/r)}*\sqrt{(1-v^2)} where v is the local velocity measured by a stationary Schwarzschild observer at r. That last equation works whether the clock is moving horizontally or vertically and it does not have to be moving at escape velocity. If the clock does happen to be free falling at escape velocity then the time dilation factor is simply tau = t*(1-2m/r)

 

[yuiop]

<br /> <br /> \begin{align*}<br /> &amp; 4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, <br /> \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M <br /> }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) } \\<br /> &amp; + 20\,\pi \,{M}^{3} <br /> \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ <br /> M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M <br /> \right) } \\<br /> &amp; - 10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M <br /> \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M <br /> \right) } \\<br /> &amp; - 20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 <br /> \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right) <br /> \end{align}<br /> <br />

This is the very integral I mentioned in post#25.

I could not resist shortening that equation a little bit to:

\frac{5}{6}\pi \left[3r_s^3\ln\left(r+\sqrt{rr_s - r_s^2}\right) + 3r_s^2\sqrt{r_2^2-r_2r_s} +2r_s\sqrt{r_2^4-r_2^3r_s} + \frac{8}{5}\sqrt{r_2^6 - r_2^5r_s} \, \right]_{r=r1}^{r=r2}

where rs-2M.

 

[Pierre007080]

After long consideration of the simplistic radial formulae obtained for the theoretical case proposed at the outset of this thread, it appears that the rod shortening and time dilation at a certain gravitational potential are the same as the rod shortening and time dilation that would be obtained by applying Lorentz transformations to the velocity of a body free falling from infinity along the radial. Is there any significance to this fact?

 

[Pierre007080]

Does the modern view of "vacuum" not present sufficient inertial energy in space itself to conform to the flow concept in the "river model"?

 

[Pierre007080]

Hi Guys,
Your previous insights on this thread really helped me to have a naive but practical idea of what GR is about. The next step in my thought process seems to be the EM force.
I would like to pose a question with regard to the radial shortening of a charged object (say negative) being attracted by another "STATIONARY" charge (say positive). Would it be possible to apply the same formulae as the case with gravity as discussed previously in this thread. I realize that the force is much greater than gravity, but the force formula seems to be related (charge in stead of mass).
Perhaps you could use the theoretical example of a proton attracting an electron from infinity to explain?

 

[Dale]

This really should be in a new thread. The title of this thread is not descriptive of your current question.