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Belated Time travel .. ?


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#52 modest

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Posted 04 January 2008 - 05:38 AM

I apologize if my example was confusing or unhelpful Drum. By using the example of a black hole I was trying to eliminate the property of g-force and in doing so demonstrate a point. That probably just overcomplicated things needlessly.

I will stipulate that gravitational time dilation can be mathematically equivalent to special relativity’s time dilation of relative velocities via the equivalence principle so long as the actual physical nature of the differences are understood. As long as we stay away from words like g-force and free-fall and the funny definitions they’ve taken on, I think we can stop in agreement here. Since this is all tangent to your original post I think that may be best.

If you are looking for insightful differences and similarities between the two, there are people here well-enough versed in SR and GR to explain it - but I’m afraid it’s not me :shrug:

- modest

#53 snoopy

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Posted 04 January 2008 - 06:45 AM

I don't really understand black holes, but the fact that our bodies (and everything else) are stretched as we cross the event threshold implies that gravity is measurably different between our ankles and our ears once we descend into it. This would cause an exponential squeezing effect.

Still trying to understand all of your post.

cool bananas ... Drum



No as you cross the event horizon you wouldnt feel anything much especially if the black hole was especially large as Modest has already explained......

Only when you approached the singularity would you feel the gravitational tidal forces that would rip you apart. This may take a very long time depending on how big the black hole is.

Peace
:shrug:

#54 Drum

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Posted 04 January 2008 - 06:30 PM

Hey Modest ... I re-read Craigs post and I am in agreement with you.

In practical human terms, gravitational time dilation is more significant than velocity time dilation. Although many people familiar with them are aware that GPS satellites have their clocks adjusted to compensate for time dilation, many of these people believe that, due to their velocity time dilation, a standard clock on a GPS satellite ticks slower than one on Earth’s surface, when in actuality it ticks faster, because its gravitational time dilation is greater (about [math]+45.9 times 10^{-6} ,mbox{s/day}[/math] vs. [math]-7.2 times 10^{-6} ,mbox{s/day}[/math]). Rather than aging minisculely less, present day astronauts, such as those on the ISS, age minisculely more than non-astronauts. Science types do tend to be detail (tree) oriented ;)

However, the distinction between gravitational and velocity time dilation is an important “big picture” (forest) one. Though theoretically related, it’s not such a simple relationship as “gravity causes acceleration, acceleration relative velocity, relative velocity time dilation.” As I describe above, gravity causes time dilation even when it causes no acceleration.


But now I am really puzzled ... Velocity time dilation would certainly make time run slower for a GPS satellite from our perspective. But, according to Craig this is balanced, even exceeded, in the opposite direction by Gravity time dilation. I would not have thought that the height of the satellite above surface earth would have contributed such a difference.

No as you cross the event horizon you wouldnt feel anything much especially if the black hole was especially large as Modest has already explained......

Only when you approached the singularity would you feel the gravitational tidal forces that would rip you apart. This may take a very long time depending on how big the black hole is.


I don't really know much about black holes. I always get stuck whenever the apparent horizon is meeting the event horizon. My logic tells me that up to the point where they merge the Apparent Horizon should have been 'outside' the Event Horizon', not within it.

Snoop ... If the gravitational tidal forces rip you apart deep inside the hole, would they still reach to the Event Horizon, no matter how feeble. Logic would dictate that they must, as they (gravity) are the cause of the Event Horizon Boundary ???

cool bananas ... Drum :evil:

#55 CraigD

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Posted 04 January 2008 - 06:54 PM

No as you cross the event horizon you wouldnt feel anything much especially if the black hole was especially large as Modest has already explained......

Only when you approached the singularity would you feel the gravitational tidal forces that would rip you apart. This may take a very long time depending on how big the black hole is.

As best I can figure, conditions inside a very massive black hole need not have singularities or extreme gravitational tidal forces. This is because the radius of the event horizon of a black hole is proportional to its mass, while the volume within the event horizon is proportional to the cube of its radius, so the average density within the event horizon can be very small.

Here are a few numbers I calculated to get an intuitive sense of this.
Body describes a typical object of mass M, “M SM” means “million solar masses”, R is the radius of its event horizon, F the tidal force (tension) experienced by a 1 meter cable connecting a pair of 50 kg bodies (my idealization of a human body), and D the average density within a sphere of radius R.
Body                      M (kg)  R (m)     F (N)     D (kg/m^3)
Planet                      1e25  1.48e-2   1.52e20   7.33e29
Star                        1e31  1.48e4    2.05e10   7.33e17
Globular Cluster (5M SM)    1e37  1.48e10   2.05e-2   7.33e5
40 M SM                     8e37  1.19e11   3.20e-4   1.15e4
50 M SM                     1e38  1.48e11   2.05e-4   7.33e3
130M SM                   2.6e38  3.85e11   3.03e-5   1.08e3
                            8e39  1.19e13   3.20e-8   1.15e0
Galaxy                      1e42  1.48e15   2.05e-12  7.33e-5
Universe                    1e53  1.48e26   ~1e-34    7.33e-27
Note that in the vicinity of 40 million solar masses, average density is about the same as water, while at about 8e39 ([math]8 \times 10^{39}[/math]) kg, average density is about the same as sea level Earth atmosphere. For a galaxy-mass black hole, the density can approach a “soft” vacuum, while for a universe-mass one, the density is about 4 hydrogen atoms / m^3 – a very “hard” vacuum.

I can’t imagine any natural process that would allow such a black hole to form, but if one could somehow entice about 50 million solar masses into a sphere about the radius of the Earth’s orbit without them coalescing into something super-dense or singularity-ish, you’d have a black hole with no terrible tidal forces anywhere, and an average density less than that of water. At its center, like the center of a star or planet, gravitational force would be zero, and clocks would tick at the same rate as very distant ones.

#56 modest

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Posted 04 January 2008 - 08:03 PM

As best I can figure, conditions inside a very massive black hole need not have singularities or extreme gravitational tidal forces.


While I do not question what you have shown I am a bit confused. I believe in all your examples a singularity must exist in the example's future. Even the universe example. I agree absolutely that any density material can create a black hole so long as the radius is large enough - but I think anything within this radius must be moving toward its center or collapsing - even photons. Have I misunderstood?

I do understand the point of the post describing less tidal stress for larger black holes and agree completely.

- modest

#57 freeztar

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Posted 05 January 2008 - 10:15 AM

Like Modest, I found your analysis very enlightening, Craig. I had never considered density within a black hole before. One thing that confuses me is your last statement:

At its center, like the center of a star or planet, gravitational force would be zero, and clocks would tick at the same rate as very distant ones.


Can you elaborate on this, Craig. I assumed it would be the opposite.
If the gravitational force is zero at the center, then how can there be a singularity at all? Why would this not cause a clock to run at a different speed than say a clock at 1/4r?

#58 modest

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Posted 12 January 2008 - 05:08 PM

Hey Modest ... I re-read Craigs post and I am in agreement with you.

But now I am really puzzled ... Velocity time dilation would certainly make time run slower for a GPS satellite from our perspective. But, according to Craig this is balanced, even exceeded, in the opposite direction by Gravity time dilation. I would not have thought that the height of the satellite above surface earth would have contributed such a difference.


I put the formulas for

[math]T'=T\sqrt{1-v^2/c^2}[/math] Velocity time dilation

[math]T'=T\sqrt{1-\frac{2GM}{rc^2}}[/math] and gravitational time dilation

in excel and got some some results:
all are relative to a stationary observer on earth and all orbits were considered circular

International Space Station:
Altitude: 340 Km
Velocity Time Dilation: -25.19 µS/Day (microseconds (millionths) per day)
Gravitational Time Dilation: +3.05 µS/Day
Total Time Dilation: -22.13 µS/Day

GPS satellites:
Altitude: 20200 Km
Velocity Time Dilation: -5.55 µS/Day
Gravitational Time Dilation: +45.77 µS/Day
Total Time Dilation: 40.22 µS/Day

Geosynchronous satellites:
Altitude: 36000 Km
Velocity Time Dilation: null
Gravitational Time Dilation: +51.14 µS/Day
Total Time Dilation: +51.14 µS/Day

Moon's orbital distance:
Altitude: 378640 Km
Velocity Time Dilation: -0.139 µS/Day
Gravitational Time Dilation: +59.19 µS/Day
Total Time Dilation: +59.05 µS/Day

A couple interesting things - the velocity time dilation is zero with something orbiting at 36,000 Km because it has no relative motion. Geosynchronous satellites are only gravitationally time dilated from our perspective.

Also, gravitational and velocity dilation are an equal +/- 18.17 µS/Day at 2750 Km. Meaning there is no net time dilation of a satellite at that altitude.

- modest

#59 Drum

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Posted 14 January 2008 - 04:09 PM

Hi Modest .... could you PM me your figures if possible ... I'm a dumbarse and would like to understand them better ... ie: what values for V and G and M

cool bananas if you can ... Drum :doh:

#60 CraigD

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Posted 14 January 2008 - 06:24 PM

I put the formulas for … in excel and got some some results: …

Cool data! :shrug:

I duplicated modest's input with this little MUMPS program:
s c=299792458,u=398600441800000,r0=6378100 s V=r0*2*/86400,T0V=V/c**2*-1+1**.5,T0G=-2*u/r0/c/c+1**.5 f  r A q:A=""  s R=(A," ")*1000+r0,V=u/R**.5,T1V=V/c**2*-1+1**.5,T1G=-2*u/R/c/c+1**.5 w (R,10)," ",(V,9,2)," ",(T1V/T0V*86400-86400,0,14)," ",(T1G/T0G*86400-86400,0,14)," ",(T1V/T0V*T1G/T0G*86400-86400,0,14),!
getting very similar, but not identical results (headers added and manually formatted a little for prettiness):
Height    Radius    Speed    VT dilation       GT dilation       Net dilation
  (km)       (m)    (m/s)    (s/day)           (s/day)           (s/day)
   340   6718100    7702.75 -0.00002841554917  0.00000304054361 -0.00002537500556
         9550711.05 6460.28 -0.00001995721399  0.00001995721399  0
 20200  26578100    3872.64 -0.00000710527853  0.00004566108491  0.00003855580638
 36000  42378100    3066.89 -0.00000441763425  0.00005103637356  0.00004661873931
378640 385018100    1017.49 -0.00000039421261  0.00005908321682  0.00005868900420
A key feature is that the surface stationary observer not only has a gravitational time dilation (T0G=.99999999930464746 = -0.00006007845946 s/day), but a velocity time dilation (T0V=.99999999999880314 = -0.0000001034087 s/day)

all are relative to a stationary observer on earth and all orbits were considered circular

I also used a circular orbit, but had my earthbound observer stationary relative to the surface of Earth at the equator at sea level, rather than relative to its center. I use a standard 86400 s day, and multiplied the two time dilations rather than adding them

A couple interesting things - the velocity time dilation is zero with something orbiting at 36,000 Km because it has no relative motion. Geosynchronous satellites are only gravitationally time dilated from our perspective.

My calculation differs. As I understand Relativity, you only have zero motion relative to something when both you and it are not accelerating. Both a clock stationary to a point on Earth and one in Geostationary orbit are accelerating centripetally.

Also, gravitational and velocity dilation are an equal +/- 18.17 µS/Day at 2750 Km. Meaning there is no net time dilation of a satellite at that altitude.

I get a slightly different altitude of about 3172611.05 m, radius from center of the Earth 9550711.05 m.

Both my net time dilation of 0.00003855580638 s/day and modest’s of 0.00004022 s/day disagree slightly with the “about 38 microseconds per day” (0.000038 s/day) given by the Wikipedia article “GPS”. I suspect they’re using the sidereal time, where the day is just a bit longer than 86164 s, in which case my result, rounded to the nearest microsecond (µs) matches wikipedia’s.

#61 modest

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Posted 14 January 2008 - 07:20 PM

Hi Modest .... could you PM me your figures if
possible ... I'm a dumbarse and would like to understand them better
... ie: what values for V and G and M

cool bananas if you can ... Drum :confused:


Ok, for gravitational time dilation, the equation is:

[math]T'=T\sqrt{1-\frac{2GM}{rc^2}}[/math]

where:
T' = how much time passes for the observer in the gravitational field
T = how much time passes for an observer outside the gravitational field (or an infinite distance from it)
G = the gravitational constant = [imath]6.67428 \times{10^{-11}} m^3
kg^{-1} s^{-2}[/imath]
M = mass of object causing gravitational field = earth kg = [imath]5.9736 \times{10^{24}} kg[/imath]
c = speed of light = [imath]3\times{10^8}m/s[/imath]
r [imath]\approx[/imath] distance between earth's center of mass and observer

Be sure to solve for both the satellite and the earth-bound observer. When you do, the ratio is the gravitational time dilation. Tell me if you have any trouble and I'll work through it w/ ya.

- modest

#62 modest

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Posted 14 January 2008 - 07:51 PM

As I understand Relativity, you only have zero motion relative to something when both you and it are not accelerating. Both a clock stationary to a point on Earth and one in Geostationary orbit are accelerating centripetally.


I can't believe I did that. Talk about 2D thinking. Thank you for pointing this out.

My input values were grossly rounded - earth's mass and radius, etc. (bad and lazy modest). What equation did you use for an orbiting speed? V = (GM/R)^1/2?

I feel challenged now and think I will graph these parameters in Mathematica.

Thank you for the review Craig.

-modest

#63 modest

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Posted 15 January 2008 - 03:10 PM

Posted Image

The x axis is radius from center of earth.
The y axis is proper time added to satellite (second) per earth (equator) second from earth-equator's reference frame.
Green is gravitational time dilation
Blue is velocity time dilation
Red is total time dilation
And Brown is r=earth's radius

I think a better intuitive sense of what's happening can be gotten when looking at a visual representation.

-modest

#64 EWright

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Posted 14 March 2014 - 05:26 AM

Interesting.