Understanding Metrics & Forces in Extreme Situations

[oldman]

I need help, please, in understanding two extreme but quite different situations, say 1 and 2 below:

1. Suppose that an observer falls freely and radially towards a neutron star. As he approaches the star he will begin to detect, by observing test particles in his local inertial frame (set up far from the star), increasingly more apparent tidal phenomena. For instance test particles he releases from rest along a line through the centre of the star will be measured to accelerate and separate from each other.

In fact this observation must lead him to conclude that his inertial frame is getting too big for its boots, as it were, and that he must restrict it to a volume in which such tidal phenomena remain imperceptible. The extent of a local inertial frame is of course subjective and depends on circumstances.

Setting aside this caveat, the observer will find that if he ties the particles together with string before releasing them, the string will eventually break. This he will attribute to a tidal force, if he adopts a Newtonian perspective instead of explaining such phenomena in terms of the Schwartzchild metric.

2. Consider the same observer (somehow surviving) in an inflating flat FRW universe that begins to expand exponentially rapidly after he has set up his local inertial frame. Suppose he again releases two test particles from rest in this frame. What happens as the scale factor, and its derivatives with respect to time, change exponentially with time?

Am I correct in assuming that nothing at all happens, and that his local inertial frame, with its test particles at rest, remains undisturbed despite the extreme "stretching of space" that takes place everywhere as his universe inflates? (I believe that this is the view taken by cosmologists.)

Or am I wrong, and will a string connecting these particles break?

 

[MeJennifer]

I need help, please, in understanding two extreme but quite different situations, say 1 and 2 below:

1. Suppose that an observer falls freely and radially towards a neutron star. As he approaches the star he will begin to detect, by observing test particles in his local inertial frame (set up far from the star), increasingly more apparent tidal phenomena. For instance test particles he releases from rest along a line through the centre of the star will be measured to accelerate and separate from each other.

In fact this observation must lead him to conclude that his inertial frame is getting too big for its boots, as it were, and that he must restrict it to a volume in which such tidal phenomena remain imperceptible. The extent of a local inertial frame is of course subjective and depends on circumstances.

Setting aside this caveat, the observer will find that if he ties the particles together with string before releasing them, the string will eventually break. This he will attribute to a tidal force, if he adopts a Newtonian perspective instead of explaining such phenomena in terms of the Schwartzchild metric.

2. Consider the same observer (somehow surviving) in an inflating flat FRW universe that begins to expand exponentially rapidly after he has set up his local inertial frame. Suppose he again releases two test particles from rest in this frame. What happens as the scale factor, and its derivatives with respect to time, change exponentially with time?

Am I correct in assuming that nothing at all happens, and that his local inertial frame, with its test particles at rest, remains undisturbed despite the extreme "stretching of space" that takes place everywhere as his universe inflates? (I believe that this is the view taken by cosmologists.)

Or am I wrong, and will a string connecting these particles break?

I just want to point out that the two situations are not exactly opposites.

While "time" in a gravitational field has a non-linear relationship with distance from the center, in a FRW model the expansion is linear with regards to time.
If this impacts the "coupling" of space-time with regards to EM-forces I do not know.

 

[pervect]

I need help, please, in understanding two extreme but quite different situations, say 1 and 2 below:

1. Suppose that an observer falls freely and radially towards a neutron star. As he approaches the star he will begin to detect, by observing test particles in his local inertial frame (set up far from the star), increasingly more apparent tidal phenomena. For instance test particles he releases from rest along a line through the centre of the star will be measured to accelerate and separate from each other.

In fact this observation must lead him to conclude that his inertial frame is getting too big for its boots, as it were, and that he must restrict it to a volume in which such tidal phenomena remain imperceptible. The extent of a local inertial frame is of course subjective and depends on circumstances.

Setting aside this caveat, the observer will find that if he ties the particles together with string before releasing them, the string will eventually break. This he will attribute to a tidal force, if he adopts a Newtonian perspective instead of explaining such phenomena in terms of the Schwartzchild metric.

In GR, the tidal force is one of the components of the Riemann tensor.

The tidal force in the radial direction works out to be just 2GM/r^3 in a GR "frame field", so the result is very similar to the Newtonian result. (There area compressive tidal forces as well, just like the Newtonian case).

frame field:
http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

I think the author (probably Chris Hillman from the way it reads) addreses the GR "tidal tensor" and mentions that it's the same as the Newtonian version.

This blows up as one approaches the singularity (r->0). That's why singularities are singular.

2. Consider the same observer (somehow surviving) in an inflating flat FRW universe that begins to expand exponentially rapidly after he has set up his local inertial frame. Suppose he again releases two test particles from rest in this frame. What happens as the scale factor, and its derivatives with respect to time, change exponentially with time?

Am I correct in assuming that nothing at all happens, and that his local inertial frame, with its test particles at rest, remains undisturbed despite the extreme "stretching of space" that takes place everywhere as his universe inflates? (I believe that this is the view taken by cosmologists.)

Or am I wrong, and will a string connecting these particles break?

If the matter density is low enough not to affect the expansion, (case A) the expansion will be exponential and the tidal force per unit length in a "frame field" will be constant for a "De-sitter" universe.

http://en.wikipedia.org/wiki/De_Sitter_universe

If the matter density is high enough, it will attempt to "fight" the expansion for a while, the expansion will be slower than exponential. As the matter density thins, this will eventually become an exponential expansion like case A.

So the tidal force won't increase indefinitely in a "De-sitter" universe with a cosmological constnat, the amount of tidal force / unit length will basically be set by the value of the cosmological constant.

I hope this is isn't too unclear

 

[oldman]

I just want to point out that the two situations are not exactly opposites.

While "time" in a gravitational field has a non-linear relationship with distance from the center, in a FRW model the expansion is linear with regards to time.
If this impacts the "coupling" of space-time with regards to EM-forces I do not know.

Yes, the point I am interested in is whether in the case of the FRW model there is or there isn't "coupling" of space-time with regards to EM-forces, as you put it.

In cosmology folk may have mixed views, from what I've read. Some seem to assume there is no coupling at all, while others talk of expansion exerting (in the case of the less-extreme expansion of the Hubble flow) a weak "force" that tends to disrupt EM and/or gravitationally-bound objects.

I'm hoping that the experts in this specialised forum will be able to clarify the matter. Thanks for your help, MJ.

 

[oldman]

If the matter density is low enough not to affect the expansion... the expansion will be exponential and the tidal force per unit length in a "frame field" will be constant for a "De-sitter" universe.

If the matter density is high enough, it will attempt to "fight" the expansion for a while, the expansion will be slower than exponential. As the matter density thins, this will eventually become an exponential expansion like case A.

So the tidal force won't increase indefinitely in a "De-sitter" universe with a cosmological constnat, the amount of tidal force / unit length will ... be set by the value of the cosmological constant.

I follow your reply to the first toy situation I described. Thanks very much for the explanation, which I follow and agree with.

In your discussion of the second situation (the de Sitter FRW case), though, I didn't quite understand some of the points you made.

First, you seem to accept that there is indeed a "tidal force" that can "fight the expansion". Is there really and truly a "force" of this kind in the de Sitter case? If so, exactly how does it arise from a dilation or expansion that is isotropic, in a homogeneous universe? This is actually the central question I'm seeking an answer for -- the string-breaking question, as it were!

Second, I didn't understand what you meant by "tidal force/unit length ", which sounds a bit like surface tension in a liquid. But this is rather a minor point.

I hope I'm not just being stupid about the whole thing.

 

[pervect]

In your discussion of the second situation (the de Sitter FRW case), though, I didn't quite understand some of the points you made.

First, you seem to accept that there is indeed a "tidal force" that can "fight the expansion".

Sorry, when I was writing it I could tell I wasn't being very clear, but I wasn't sure what to do to fix it.

Consider two masses.

A----------------------B

Without a cosmological constant, and with matter present, B always accelerates towards A.

This is why the expansion of a universe slows down (decelerates) in cosmologies without a cosmological constant.

With a cosmological constant, without matter present, B always accelerates away from A. (Well, it depends on the sign of the constant, actually. I should say, with a "dark energy" sort of cosmological constant, B always accelerates away from A).

When you have both (matter, and "dark energy") the two effects fight each other.

Note that the directon of expansion doesn't have anything to do with the direction of acceleration. While all FRW models without a cosmological constant have B acclerating towards A, the universe still expands in these models. (B moves away from A, but acclerates towards it, i.e. B slows down as it gets further away from A).

Is there really and truly a "force" of this kind in the de Sitter case? If so, exactly how does it arise from a dilation or expansion that is isotropic, in a homogeneous universe? This is actually the central question I'm seeking an answer for -- the string-breaking question, as it were!

If you stick a big meter stick in a matter-free de-Sitter universe (well, it still has the meter stick in it! But no other matter) the meter stick will experience a tidal force. You can calculate the magnitude if you know the Riemann curvature tensor from the geodesic deviation equation. The force will be a "pull it apart" sort of force as per the diagram. In formal language, B and A both follow geodesics, but the geodesics accelrate away from each other (geodesic deviation).

Second, I didn't understand what you meant by "tidal force/unit length ", which sounds a bit like surface tension in a liquid. But this is rather a minor point.

I hope I'm not just being stupid about the whole thing.

The accleration between A and B is proportional to the distance. Thus, force / distance.

If A and B are twice as far apart, the relative acceleration (aka "force") is twice as great.

 

[oldman]

If you stick a big meter stick in a matter-free de-Sitter universe (well, it still has the meter stick in it! But no other matter) the meter stick will experience a tidal force. You can calculate the magnitude if you know the Riemann curvature tensor from the geodesic deviation equation. The force will be a "pull it apart" sort of force as per the diagram. In formal language, B and A both follow geodesics, but the geodesics accelrate away from each other (geodesic deviation).
The accleration between A and B is proportional to the distance. Thus, force / distance.
If A and B are twice as far apart, the relative acceleration (aka "force") is twice as great.

It's all starting to make sense to me, thanks to you. Give me a short while, and I may yet end up understanding expansion properly.

One further point: since a meter stick experiences a "tidal force" in an (otherwise) matter-free expanding universe, which as you say is a "pull it apart" sort of force, what happens during the extreme expansion of inflation?

In situations (such as near a neutron star or sun-sized black hole) where tidal forces are extreme, material objects would literally be pulled apart. Objects even as small as nuclei would eventually be disrupted if they approached close enough to the singularity of a a black hole.

How do elementary entitities (electrons? quarks?) survive inflation? Is it perhaps the high initial mass/energy density that mitigates inflation's extreme disruptive "tidal" effect, as you said:

When you have both (matter, and "dark energy") the two effects fight each other.

 

[oldman]

How do elementary entitities (electrons? quarks?) survive inflation? Is it perhaps the high initial mass/energy density that mitigates inflation's extreme disruptive "tidal" effect, as you said:

Upon reflection I see that the solution I suggested above for this question is sheer nonsense. My apologies.

The helpful analysis you gave, Pervect, should I think be viewed from a perspective where the word "force" is banned. In a sense general relativity describes gravitational effects kinematically, i.e. without invoking the concept of "force". This is why discussing the other forces of nature and gravity in the same context makes one speak with a forked tongue, as it were!

So, when it comes to inflation, one should regard the universe's expansion by many orders of magnitude in an infinitesimal instant as a kinematic given. Of course this involves enormous kinematic relative accelerations between separated objects which, as you say, increases with their separation.

But I can't then understand the survival of even elementary bits of matter !

 

[Jorrie]

... So, when it comes to inflation, one should regard the universe's expansion by many orders of magnitude in an infinitesimal instant as a kinematic given. Of course this involves enormous kinematic relative accelerations between separated objects which, as you say, increases with their separation.

But I can't then understand the survival of even elementary bits of matter !

It helps ease the headache if one thinks that space between particles expanded and the particles felt no force or acceleration. That is, if there were any material particles during inflation proper - perhaps just photons; I don't know!

 

[oldman]

It helps ease the headache if one thinks that space between particles expanded and the particles felt no force or acceleration. That is, if there were any material particles during inflation proper - perhaps just photons; I don't know!

Yes, I do need something like Grandpa Headache Powders for all this!

But the idea you express that "space between particles" expands and that "the particles felt no force or acceleration" is exactly the delusion I'm trying to unravel! Cosmology is full of the nonsense notion that "space expands", when in fact nobody even knows what "space" is. The better texts (e.g. Peebles' Principles of Physical Cosmology ) carefully avoid this trap.

The particles you mention, like the meter stick mentioned in the quote below will, if they are finite in size, "try to stretch" with expansion against whatever cohesive forces they are endowed with.

If you stick a big meter stick in a matter-free de-Sitter universe (well, it still has the meter stick in it! But no other matter) the meter stick will experience a tidal force. You can calculate the magnitude if you know the Riemann curvature tensor from the geodesic deviation equation. The force will be a "pull it apart" sort of force as per the diagram. In formal language, B and A both follow geodesics, but the geodesics accelrate away from each other (geodesic deviation).

I disagree here only with the use of the word "force".

But you are right that during inflation there may not even be such things as material particles --- who knows. The question I asked may therefore be moot.

 

[pervect]

Been a busy day, and then PF crashed.

I probably should fix a few glitches in my explanation :-(. If you have two particles A and B, and they are both following geodesics in a DeSitter space, or near the Earth, or near a black hole, tidal forces will tend to make them accelerate relative to each other.

It is only when external forces (such as the electromagnetic forces in a meterstick) act on A and B that they maintain a constant separation. It turns out that the force needed to keep them a constant distance apart is just their mass times the acceleraton they would have if you let them follow geodesics. So when you put a meter stick in space, the ends of the meter-stick are not following geodesics, but we can still use the geodesic equation to calculate the force being applied to keep the two particles at a constant distance.

The tidal forces in De-sitter space are conceptually the very same tidal forces as those due to a black hole, or due to the Earth.

In GR, we just calculate the magnitude of the components of the Riemann tensor (easy with software, not so easy to do by hand) in a frame-field, and we have the "tidal force" resulting from a metric, regardless of whether that metric is a black hole, De-sitter space, or whatever.

One thing that you may not appreciate is how tiny they are. The magnitude of the force in geometric units turns out to be H^2, where H is the Hubble constant.

The MKS units are (meters / sec^2 ) / meter = 1/sec^2. So there isn't any big unit conversion issue, you just need to express H in units of inverse seconds.

To get a number for the tidal force for two objects 1 km apart in m/s^2 I'd need an estimate of what H was during the inflation era, and I don't have a clue. (But I can tell you that it would be 1000/H^2, where H was expressed in seconds).

Space Tiger might (or might not) know the value of the Hubble constant during inflation (Ned Wright says that it's constant, but gives no value) - or have some insight into whether or not it's even knowable.

There's also the issue of whether we need to correct the formula I mentioned (for empty De-sitter space) for the presence of matter - however since expansion during inflation is usually described as being exponential, I think that implies that the gravitational effects of matter were totally overwhelmed during inflation and can be safely ignored.

 

[oldman]

One thing that you may not appreciate is how tiny they are.

Well, I certainly don't feel stretched by the present expansion rate of expansion! But had I been around during inflation, I might:

Although H (say (71 Km per sec) per MPc, or about 3 x 10 ^ (-19) per sec), is now very small, it must have been vastly bigger during inflation.

Inflation caused a fractional increase in linear scale of the Universe of about 10 ^ 43 and lasted for about 10 ^ (-34) sec (Andrew Liddle, Introduction to Modern Cosmology, p. 106). Averaging H out (to get a rough estimate) then estimates H at about 10 ^ 77 per sec during inflation, which is 10 ^ 96 times bigger than today's value, unless I'm being too simple minded and have got it all wrong.

Big enough accelerations to disrupt a meter stick and perhaps much else!

You have sorted out my muddled thinking (the quote below) very nicely, thanks Pervect, and I hope the points you raised do get clarified in this thread.

The magnitude of the force in geometric units turns out to be H^2, where H is the Hubble constant... To get a number for the tidal force for two objects 1 km apart in m/s^2 I'd need an estimate of what H was during the inflation era, and I don't have a clue. (But I can tell you that it would be 1000/H^2, where H was expressed in seconds).

Space Tiger might (or might not) know the value of the Hubble constant during inflation (Ned Wright says that it's constant, but gives no value) - or have some insight into whether or not it's even knowable.

There's also the issue of whether we need to correct the formula I mentioned (for empty De-sitter space) for the presence of matter - however since expansion during inflation is usually described as being exponential, I think that implies that the gravitational effects of matter were totally overwhelmed during inflation and can be safely ignored.

 

[MeJennifer]

I probably should fix a few glitches in my explanation :-(. If you have two particles A and B, and they are both following geodesics in a DeSitter space, or near the Earth, or near a black hole, tidal forces will tend to make them accelerate relative to each other.

It is only when external forces (such as the electromagnetic forces in a meterstick) act on A and B that they maintain a constant separation. It turns out that the force needed to keep them a constant distance apart is just their mass times the acceleraton they would have if you let them follow geodesics. So when you put a meter stick in space, the ends of the meter-stick are not following geodesics, but we can still use the geodesic equation to calculate the force being applied to keep the two particles at a constant distance.

The tidal forces in De-sitter space are conceptually the very same tidal forces as those due to a black hole, or due to the Earth.

In GR, we just calculate the magnitude of the components of the Riemann tensor (easy with software, not so easy to do by hand) in a frame-field, and we have the "tidal force" resulting from a metric, regardless of whether that metric is a black hole, De-sitter space, or whatever.

One thing that you may not appreciate is how tiny they are. The magnitude of the force in geometric units turns out to be H^2, where H is the Hubble constant.

The MKS units are (meters / sec^2 ) / meter = 1/sec^2. So there isn't any big unit conversion issue, you just need to express H in units of inverse seconds.

To get a number for the tidal force for two objects 1 km apart in m/s^2 I'd need an estimate of what H was during the inflation era, and I don't have a clue. (But I can tell you that it would be 1000/H^2, where H was expressed in seconds).

Sorry Pervect I do not follow you at all here.

The general theory of relativity postulates that rods get contracted due to space-time curvature and the special theory of relativity postulates that rods get contracted when observed in relative motion due to a rotation in space-time. So in other words, EM-forces follow space-time curvature. But now we a led to believe that it does not apply to the FRW metric or a De Sitter space.

What bothers me is the lack of (theoretical) criticism of those kind of cosmological models. Really we are just making it match our observations rather than presenting a sound theory.
Don't get me wrong there is nothing wrong about that but then treating it as something that magically rolled out of "the Einstein equations" is not really giving a right representation IMHO.

 

[pervect]

Sorry Pervect I do not follow you at all here.

The general theory of relativity postulates that rods get contracted due to space-time curvature and the special theory of relativity postulates that rods get contracted when observed in relative motion due to a rotation in space-time. So in other words, EM-forces follow space-time curvature. But now we a led to believe that it does not apply to the FRW metric or a De Sitter space.

I can see that you're not following me. I'm not sure what to do about it.

Modulo some issues regarding rotating frames, "tidal forces" can be regarded as measures of certain components of the Riemann curvature tensor, and vica versa.

This is one of the simplest ways to start appreciating the Riemann in "physical" terms, rather than as a mathematical abstraction. Parallel transporting vectors around closed curves is all well and good, and is one way to define the Riemann, but a tidal force is something that one can measure, directly, and it turns out that these tidal forces ARE (in non-rotating frames, anwyay) equal to specific comonents of the Riemann.

As far as refrences go, all I can suggest is to read up on the geodesic deviation equation.

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html
http://en.wikipedia.org/wiki/Geodesic_deviation

Armed with this information, we can find the tidal forces near a black hole, or the tidal forces in a De-sitter universe, simply by evaluating the Riemann tensor and finding the appropriate components.

The calculation itself is rather technical, so you'll just have to trust me when I say that the components of the Riemann are 1/H^2 in a "frame-field" in a De-sitter metric.

 

[oldman]

I've been having some second thoughts about my understanding of what you wrote, Pervect.

In the quote below I've shown the puzzles I have in bold -- they are mostly niggles about units and suchlike, but I'd like to get them straight so that I can combine what you wrote and my estimate of H during inflation (10 ^ 77 --- is this correct?) to estimate stretching forces in Newtons. Apologies for bothering you again!

The magnitude of the force in geometric units (what are geometric units ?) turns out to be H^2, where H is the Hubble constant. (H is the fractional change in proper separation between any two points, due to expansion, per second -- not so?)

The MKS units ((SI units of acceleration per unit length?) are (meters / sec^2 ) / meter = 1/sec^2. So there isn't any big unit conversion issue (in calculating accelerations and hence forces?), you just need to express H in units of inverse seconds (which are its SI units, instead of Km/s/Mpc?).

To get a number for the tidal force (the way you equate acceleration anf force means that you are considering the accn. of 1 Kg, not so?) for two objects 1 km apart in m/s^2 ... I can tell you that it would be 1000/H^2, where H was expressed in seconds (or do you mean in inverse seconds or times H ^ 2?).

 

[pervect]

Let's see if I can reply to this without it crashing before I put in a longer reply...

 

[pervect]

The magnitude of the force in geometric units (what are geometric units ?) turns out to be H^2, where H is the Hubble constant.

http://en.wikipedia.org/wiki/Geometrized_units

units where c=G=1/(4 pi epsilon_0) = 1

(H is the fractional change in proper separation between any two points, due to expansion, per second -- not so?)

In terms of the scale factor a(t), H = (da/dt) / a , which is equal to
(da/a) / dt, so I think this is basically correct. I'd replace "between any two points" with "between any two points with constant co-moving coordinates". Points with constant co-moving coordinates are points that are moving with the Hubble flow

The metric is (assuing spatial flatness, which I assumed BTW)

ds^2 = -dt^2 + a(t)^2 (dx^2 + dy^2 + dz^2)

where x,y, and z are "comoving" coordinates.

To get a number for the tidal force (the way you equate acceleration anf force means that you are considering the accn. of 1 Kg, not so?)

correct

for two objects 1 km apart in m/s^2 ... I can tell you that it would be 1000/H^2, where H was expressed in seconds (or do you mean in inverse seconds or times H ^ 2?).

Oops, 1000*H^2, 1000 for the 1000 meter (1 km) separation

As far as the magnitude of H goes, 10^77 * sec^-1 seems awfully high to me. This would imply that the universe doubled in size every ln(10)*10^-77 seconds.

i.e. a(t) = exp(Ht), da/dt = H exp(HT)

so when 1/H = 10^-77 sec, you have a doubling of the scale factor a about every 2*10^-77 seconds.

 

[oldman]

http://en.wikipedia.org/wiki/Geometrized_units

units where c=G=1/(4 pi epsilon_0) = 1

I should have looked at Wikipedia straight off. But thanks for this URL. That's fine.

As far as the magnitude of H goes, 10^77 * sec^-1 seems awfully high to me. This would imply that the universe doubled in size every ln(10)*10^-77 seconds.

i.e. a(t) = exp(Ht), da/dt = H exp(HT)

so when 1/H = 10^-77 sec, you have a doubling of the scale factor a about every 2*10^-77 seconds.

Yes, it does seem very high to me too. I wouldn't be surprised to find that I'm wrong by several tens of orders of magnitude. But whichever way you slice it, inflation is a very extreme process and I suppose one can expect extreme tidal accelerations as a matter of course here. I hadn't appreciated this feature of inflation until you pointed me in the right direction.

In another branch of physics, namely the elasticity of solids, the Hubble constant is exactly analagous to the dilation strain rate. In this context the dilation strain rate is usually small, say intermediate between the miniscule presently observed Hubble constant and its enormous and constant value during inflation.

One could view inflation in a Newtonian way, and very simply, as an (extreme) explosive dilation of the "cosmic fluid", driven by the mysterious tranformation of false to true vacuum. But it's not appropriate to talk of such matters as fluid explosives right now!

Thanks for your help in clarifying my understanding of such dilations.

 

[MeJennifer]

I can see that you're not following me. I'm not sure what to do about it.

Modulo some issues regarding rotating frames, "tidal forces" can be regarded as measures of certain components of the Riemann curvature tensor, and vica versa.

This is one of the simplest ways to start appreciating the Riemann in "physical" terms, rather than as a mathematical abstraction. Parallel transporting vectors around closed curves is all well and good, and is one way to define the Riemann, but a tidal force is something that one can measure, directly, and it turns out that these tidal forces ARE (in non-rotating frames, anwyay) equal to specific comonents of the Riemann.

As far as refrences go, all I can suggest is to read up on the geodesic deviation equation.

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html
http://en.wikipedia.org/wiki/Geodesic_deviation

Armed with this information, we can find the tidal forces near a black hole, or the tidal forces in a De-sitter universe, simply by evaluating the Riemann tensor and finding the appropriate components.

The calculation itself is rather technical, so you'll just have to trust me when I say that the components of the Riemann are 1/H^2 in a "frame-field" in a De-sitter metric.

And how exactly does this explain the claim that expansion of the universe does not apply to small objects like a meter stick?

So we are led to believe that while gravity reduces a volume of a sphere, expansion does not increase it and that that is all logical due to the Rieman curvature tensor?

The tidal forces in De-sitter space are conceptually the very same tidal forces as those due to a black hole, or due to the Earth.

Really, so do you claim that the inter-relationship between the time and space dimensions as defined in GR in a gravitational field are handled exactly the same as expansion in a De Sitter space or a FRW model?
It seems to me that time has simply a linear relationship in a De Sitter and FRW model and is fundamentally treated differently than time in GR.

 

[oldman]

And how exactly does this explain the claim that expansion of the universe does not apply to small objects like a meter stick?

So we are led to believe that while gravity reduces a volume of a sphere, expansion does not increase it and that that is all logical due to the Rieman curvature tensor?

I have always found expansion much more difficult to understand than the (it turns out unecessary) curvatures of space sections so beloved of those who write cosmology texts. You may be having similar troubles, MJ.

Nevertheless I think that you can rely on what Pervect says, and do as he suggested:

...so you'll just have to trust me...

.

But let me try and help --- I'm quite simple minded about such matters. Perhaps too simple minded. Does the following make sense to you?

The point is often made that the present expansion of the universe doesn't make objects like a meter stick expand. Let me construct a "straw man description" to explain why this is so.

Consider a planet exploding during a Star-Wars battle. Once the explosion has endowed the planet fragments with radial velocity it provides no further driving force -- it's all over and done with in an instant. But as the fragments (meter sticks and suchlike) fly apart they lose speed as the gravity of "interior" debris pulls them back. (By interior I mean debris closer to the centre if the explosion than the piece of debris being considered).

The fragments also experience tidal forces from the interior debris. They will be distorted by these forces to an extent that depends on how "stiff" they are. Some meter sticks may be stretched by these forces. (Others will be compressed --- it depends on their alignment with respect to the centre of the explosion).

But there is no uniform expansion of fragments. Why should there be?. Here my "straw man description" and the universe's expansion are well matched.

Nor can the universe's uniform expansion produce any tidal forces. Because such expansion has high symmetry, in that it is isotropic and the same everywhere, it is impossible to pinpoint a "centre of the explosion", or to define "interior debris" that can produce tidal forces. So there are none!

Cosmologists and general relativity folk have a horror of the kind of "straw man analogy" that I have simple-mindedly given. But as Hermann Bondi in Cosmology explained long ago, more simple-minded (such as Newtonian) treatments of the universe's expansion give very similar answers to the vastly more sophisticated machinery of general relativity.

And this is true of the "inflationary scenario" as well.

Fred Hoyle in a disparaging mood coined the now-universally-adopted name "Big Bang" for the universe's origin, a denigration initially resisted by cosmologists. But later they invented inflation, a violent process near "the beginning" that involves a sudden huge release of energy (in 10 ^ -34 sec) and a vast expansion (by a factor of 10 ^ 43).

If this isn't appropriately described as an explosive "Big Bang, what is? And during this process, while the release of energy was proceeding, there must have been huge tidal forces, originating as Pervect described. Unlike the "coasting" present expansion of the universe, influenced only by gravity (if you ignore the recently discovered acceleration of expansion), inflation is continually driven by energy release which maintains the exponential expansion and produces tidal forces.

Before this thread I hadn't appreciated this aspect of inflation, and suspect that many cosmologists haven't either.

 

[pervect]

And how exactly does this explain the claim that expansion of the universe does not apply to small objects like a meter stick?

So we are led to believe that while gravity reduces a volume of a sphere, expansion does not increase it and that that is all logical due to the Rieman curvature tensor?

Using Baez's & Bunn's "The meaning of Einstein's equation" approach, the volume of a sphere of coffee ground decreases if the sphere contains mass. I assume that's what you are referring to.If you have a universe containing matter and no cosmological constant and a sphere of "test particles", the test particles are attracted to the center of the sphere by the enclosed mass.

This shows that matter never causes expansion to speed up, it always causes expansion to slow down.

Furthermore, the tidal forces will always be compressive in a universe composed of matter with no cosmological constant. If you imagine the ends of the bar moving on geodesics, they'd "naturally" move towards the center. It's only the pressure in the bar that keeps them apart. But the magnitude of the forces involved are very small.

So, looking at the normal universe case, we have an expanding universe, but tidal forces that work to oppose the expansion. The expansion slows down as a result of these forces. And we have compression in a bar, due to the gravitational attraction of the ends of the bar to the center. This assumes the bar is in a region that has the "normal" amount of matter, that the matter is smoothly and continously distrbuted throughout the universe. We know that this is true on the average by the so-called cosmological principle (which is basically isotropy).

It's a bit backwards to think of the tidal forces as being due to the expansion of the universe. It's not a "force due to expansion". Rather, the tidal forces are due to the matter in the universe, and they slow down the expansion of the universe, because test particles are attracted to the center of the sphere.

We've ignored the region external to the sphere , but this isn't an issue, as it has no net effect by Birkhoff's theorem, similar to the way that Newtonian gravity has no effect in the inside of a spherical shell.

You might expect the same of an expanding gas cloud with Newtonian physics - gravity acts to oppose the expansion.

Oldman has made very similar points, I see.

This is quite consistent with the Riemann tensor approach, too. It's just a different way of viewing the same result.

Things change when we add a cosmological constant. Things start to get weird. The vacuum contains a positive energy, but a negative pressure.

Assuming that the universe is unchanging with time (as it will be if there is no matter in it), the effect of this negative pressure and positive energy denisty is a sort of anti-gravity. The enclosed mass (Komar mass) in the sphere is equal to volume * (density + 3*pressure) (and because the pressure is negative and equal to the density, the enclosed mass in a sphere is negative.

This is how a cosmological constant can fuel an accelerating expansion. Particles in a sphere start to accelerate away from the sphere, because empty space has a negative Komar mass.

The Komar mass formula isn't particularly intuitive, but it's one of the simpler formulas for mass in GR, and applies as long as the system is static. You can read the Wikipedia article if you like, but it's a bit advanced.

http://en.wikipedia.org/wiki/Komar_mass

BTW, this is a fairly recent addition that I recently wrote :-).

The tidal forces due to the cosmological constant cause tension, rather than compression, because the bar ends want to move away from each other.

This may seem a bit weird, but that's how it works.

Really, so do you claim that the inter-relationship between the time and space dimensions as defined in GR in a gravitational field are handled exactly the same as expansion in a De Sitter space or a FRW model?
It seems to me that time has simply a linear relationship in a De Sitter and FRW model and is fundamentally treated differently than time in GR.

I don't follow you. De Sitter and FRW models are both GR models.

 

[MeJennifer]

Things change when we add a cosmological constant. Things start to get weird. The vacuum contains a positive energy, but a negative pressure.

Well at least something we can agree on.

To me it looks like a plain old patching to make the theory match experimental results.

This whole notion of a cosmological constant is preposterous to me, but I realize that for others it is like an obvious piece of Gospel that is both perfectly explanable and logical.

Assuming that the universe is unchanging with time (as it will be if there is no matter in it), the effect of this negative pressure and positive energy density is a sort of anti-gravity. The enclosed mass (Komar mass) in the sphere is equal to volume * (density + 3*pressure) (and because the pressure is negative and equal to the density, the enclosed mass in a sphere is negative.

Let's for the sake of argument assume what you say is correct, then explain to me how it is consistent to have a linear dependency on time in such a anti-gravitational field (for instance the scale factor in a FRW metric) but a non linear dependency on time in regular gravitational field?

If you think I am wrong prove me wrong by expressing -a(t) as a scaling factor in a Schwarzschild geometry similarly as done in the FRW metric.

 

[oldman]

I've taken some time off to think about what I learned from your kind replies in this thread, Pervect, and I've come across one point that still bothers me quite badly. Please forgive me for being so slow and obtuse.

I gather that in many (but not all) respects general relativity gives the same answers as Newtonian gravity. Indeed, remarks you've made in the post below (I've picked them out in bold) seem to support this view.

Using Baez's & Bunn's "The meaning of Einstein's equation" approach, the volume of a sphere of coffee ground decreases if the sphere contains mass...

If you have a universe containing matter and no cosmological constant and a sphere of "test particles", the test particles are attracted to the center of the sphere by the enclosed mass.

This shows that matter never causes expansion to speed up, it always causes expansion to slow down.

Furthermore, the tidal forces will always be compressive in a universe composed of matter with no cosmological constant. If you imagine the ends of the bar moving on geodesics, they'd "naturally" move towards the center. It's only the pressure in the bar that keeps them apart. But the magnitude of the forces involved are very small.

So, looking at the normal universe case, we have an expanding universe, but tidal forces that work to oppose the expansion. The expansion slows down as a result of these forces. And we have compression in a bar, due to the gravitational attraction of the ends of the bar to the center. This assumes the bar is in a region that has the "normal" amount of matter, that the matter is smoothly and continously distrbuted throughout the universe. We know that this is true on the average by the so-called cosmological principle (which is basically isotropy).

It's a bit backwards to think of the tidal forces as being due to the expansion of the universe. It's not a "force due to expansion". Rather, the tidal forces are due to the matter in the universe, and they slow down the expansion of the universe, because test particles are attracted to the center of the sphere.

We've ignored the region external to the sphere , but this isn't an issue, as it has no net effect by Birkhoff's theorem, similar to the way that Newtonian gravity has no effect in the inside of a spherical shell.

You might expect the same of an expanding gas cloud with Newtonian physics - gravity acts to oppose the expansion...

This is quite consistent with the Riemann tensor approach, too. It's just a different way of viewing the same result.

I have difficulty reconciling Newtonian gravity with your remarks about tidal forces.

Newtonian tidal forces occur only in the presence of gravitational gradients. They do not occur when a uniform gravitational field accelerates (or decelerates) an extended object. Specifically, when in a uniform gravitational field (approximated by the field we live in), an arrow is shot vertically upwards it decelerates, but experiences no tidal forces.

The same is true of a linear array of "coffee grounds" projected simultaneously upwards with identical initial velocities (in a laboratory vacuum). The initial separations of grains remain constant throughout flight. (This situation mustn't be confused with sequential stroboscopic photos of a projected ball, common in elementary physics texts, where the sequential separations decrease).

In the case you were considering -- of a matter-only universe -- while gravity of course acts to slow expansion, it would seem to me that the "tidal forces" you mention in this homogeneous and uniform situation have no analogue in Newtonian gravitation. I am perturbed by this and can't help wondering if in this case you have the wrong end of the meter stick (as it were!) when you state above that:

"... the tidal forces will always be compressive in a universe composed of matter with no cosmological constant. If you imagine the ends of the bar moving on geodesics, they'd "naturally" move towards the center. It's only the pressure in the bar that keeps them apart. But the magnitude of the forces involved are very small."

In fact I doubt if there are such tidal forces. But I've been wrong so often. Could you clarify this apparent dichotomy, please?

 

[George Jones]

Specifically, when in a uniform gravitational field (approximated by the field we live in), an arrow is shot vertically upwards it decelerates, but experiences no tidal forces.

I don't know if the following has any relevance.

Using Newtonian gravity, consider small spherical subset of a large, uniform, spherically symmetric distribution of matter. The large and the small spheres don't have to be concentric. Consider test particles moving in this small sphere that don't interact with the matter in the sphere - in your case an arrow shot out in any driection (along a negligibly small tunnel) from the centre of the small sphere.

The matter outside the small sphere has, to a good approximation, no tidal gravitational influence on the arrow. The tidal force of gravity (due to stuff inside the small sphere) inside the tunnel is, I think, proportional to the distance from the small sphere's centre, so the business end of the arrow experiences more tidal force towards the centre of the small sphere than does than the feathered end.

Newtonian tidal forces compress the arrow.

 

[oldman]

I don't know if the following has any relevance.

Using Newtonian gravity, consider small spherical subset of a large, uniform, spherically symmetric distribution of matter. The large and the small spheres don't have to be concentric. Consider test particles moving in this small sphere that don't interact with the matter in the sphere - in your case an arrow shot out in any driection (along a negligibly small tunnel) from the centre of the small sphere.

The matter outside the small sphere has, to a good approximation, no tidal gravitational influence on the arrow. The tidal force of gravity (due to stuff inside the small sphere) inside the tunnel is, I think, proportional to the distance from the small sphere's centre, so the business end of the arrow experiences more tidal force towards the centre of the small sphere than does than the feathered end.

Newtonian tidal forces compress the arrow.

Thanks for this help, George. But I find your argument difficult to apply to a uniform and homogeneous universe.

In the case you consider all "small spheres" on whose surfaces the arrow lies produce gravitational forces (with gradients) directed towards their centres. These forces cancel out because of spherical symmetry, and there is no net gravitational and tidal force!

Maybe I'm missing your point?

 

[pervect]

Imagine a very large (but finite) dust cloud, in a Newtonian universe.

The dust cloud will want to collapse under its own gravity - it won't just "float" there, serenely.

I think you'll agree that a finite dust cloud wants to collapse, gravitationally. I'm just saying that an infinite dust cloud would want to do the same thing, due to itself gravity.

If you think about such a large, but finite, dust cloud, I think you'll also realize that particles in it experience tidal forces.

It's the issue of infnity that's obscuring the physics here, I think.

If you like, imagine that the universe is not quite infinite, but that some original finite universe was tremendously blown up by an inflationary process into something very much larger than we can ever hope to observe. But it's still finite, though inconceivably large. Call this the "greater universe".

Call the part of the greater universe we can observe and interact with the "observable universe".

Then think about the physics of that situation. There will be two sorts of effects - "edge effects" due to the sub-position of the observable universe in the greater universe, and the effects due to the matter in the observable universe itself.

The argument is that if the edge of the greater universe is far enough away, it doesn't cause any tidal forces in the observable universe. Tidal forces drop off as 1/r^3, remember. The area goes up as r^2. So the tidal effects drop off as 1/r.

You might worry about "edge effects" causing a uniform gravitational field in the observable universe, because gravity drops off as 1/r^2, and the area increases as r^2. But think about how we could observer a uniform gravitational field permeating the universe. It wouldn't have any effects we could measure - you can think of it as just a coordinate change, a uniform gravitational field is the same thing as changing your point of view.

 

[MeJennifer]

I think you'll agree that a finite dust cloud wants to collapse, gravitationally. I'm just saying that an infinite dust cloud would want to do the same thing, due to itself gravity.

A finite dust cloud must have a center of gravity but does an infinite one have?
To me that does not seem to be the case.
It seems to be more logical in the case of an infinite cloud to assume cluster formation due to random density fluctuations in the infinite cloud.

 

[CarlB]

A finite dust cloud must have a center of gravity but does an infinite one have? To me that does not seem to be the case.

Another way of putting the same argument, I think. If I am subject to tidal forces, then one part of my body is getting pulled more by gravity than another part. This seems to define a direction, and a direction is incompatible with the lack of the existence of a center to the dust cloud.

So the tidal force must be non directional.

Carl

 

[oldman]

Thanks for taking yet more time with my difficulties, Pervect. But I now suspect that we are talking about different aspects of the question in different posts, which is leading deeper into a morass of misunderstanding.

Your last post (#26) addresses the question of whether there are tidal forces in an infinite, uniform and homogeneous gravitating dust cloud, like that in the standard model of cosmology. From a Newtonian perspective I am still convinced that there are no tidal forces, just as it has been accepted since Newton's time (and by that Man himself, I believe) that there is no tendency for such a cloud to collapse or expand (although its equilibrium is unstable).

I think your argument in favour of tidal effects fails because of spherical symmetry. The argument runs as follows:

Consider the tidal effects generated at any point, by any spherical shell of matter centred on that point. (Remember that the "tidal force" is in fact "the tidal stress"; a second rank tensor.)

Look at (for instance) two components of the tidal stress tensor arising from an element of the shell; one, the radial pull-apart tensile tidal stress and two, the circumferential squashing compressive tidal stress. These are exactly canceled by the stresses from other element(s) in the shell 90degrees away from the first element. By pairing elements in this kind of way all components of the stress tensor arising from the shell (and indeed any other such shell) can be shown to cancel. Spherical symmetry strikes again.

The question of whether tidal forces can arise through the dynamics or kinematics of such a cloud, if it expands, (exemplified by its "Hubble constant", H) that I was trying to dismiss in a Newtonian way in post # 23is quite another matter.

With regard to this question, in your earlier posts written from a General Relativistic (GR) perspective, you have explained that the tidal effects may be calculated from the components of the Riemann tensor, and that:

...we can find the tidal forces near a black hole, or the tidal forces in a De-sitter universe, simply by evaluating the Riemann tensor and finding the appropriate components.

The calculation itself is rather technical, so you'll just have to trust me when I say that the components of the Riemann are H^2 in a "frame-field" in a De-sitter metric.

.

I accept that (GR) provides a deeper perspective on the behaviour of the gravitating fluid of the standard model than Newton had. For instance GR shows that such a fluid must expand (perhaps a divergence from the Newtonian view?). Hence the expanding universe.

If H is involved we are dealing with the dynamics and kinematics of the cloud, and here I must trust your expertise with GR.

But I'm still perturbed by the apparent dichotomy between the perspective of GR and the Newtonian view, which I'm convinced excludes tidal stresses; even in a cosmic fluid whose expansion is slowed down by the "body force" of gravity .

Perhaps its just the way things are in GR that in the extreme de Sitter case of exponentially rapid inflation the inflaton field produces tidal stress components that are all? proportional to H^2. But if the inflaton field acts like a negative gravity or body force, maybe these stresses have no effect on matter.

 

[oldman]

...So the tidal force must be non directional.

Yes indeed. And the only "force" I know of that is "non-directional" is pressure. I think that gravitation, in the General Relativity-ruled standard model of cosmology and the inflaton-driven field of the inflationary scenario, acts like a pressure (positive or negative) that retards or drives expansion equally everywhere.

If this is so, then in this situation gravity does not produce (uniaxial) forces. These are associated only with pressure gradients. Gravitation in the standard model fluid acts as a non-tidal stress that has only three equal and diagonal components. Gravity is an interaction that, uniquely among those in nature, acts on all kinds of mass/energy. Therefore solids objects like meter sticks, liquids, molecules, atoms and nuclei whose dimensions are controlled by other interactions will suffer dilation or compressive strains, and develop appropriate opposing stresses, in a changing cosmic environment. These stresses are imperceptibly small now, but were stupendously large in the inflationary instant --- if that ever happened!

I think the crux of my difficulties has been semantic. Thanks for your input, CarlB.

 

[pervect]

Rather than argue about Newtonian theory, let's try a different aproach.

Consider the GR case, as outlined in http://math.ucr.edu/home/baez/einstein/

more specifically
http://math.ucr.edu/home/baez/einstein/node3.html

In any event, we may summarize Einstein's equation as follows:

{\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z).<br />

This equation says that positive energy density and positive pressure curve spacetime in a way that makes a freely falling ball of point particles tend to shrink. Since E = mc^2 and we are working in units where c = 1, ordinary mass density counts as a form of energy density. Thus a massive object will make a swarm of freely falling particles at rest around it start to shrink. In short: gravity attracts.

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

So, the second derivative of the rate of change of the volume of a sphere of coffee grounds is proportional to the volume of the coffee grounds multiplied by the density of matter - plus, a pressure term.

Note that, as Baez & Bunn explain, this depends only on the amount of matter contained within the ball. External gravity via external matter does not cause the volume to shrink. If we put the coffee grounds near the Earth, there will be the tension and compression terms you noted. The result will be that the ball will grow longer in one direction, shorter in the other two, and the second derivative of the volume will be zero.

Now, skip on a bit to:

http://math.ucr.edu/home/baez/einstein/node7.html

They apply the above result to a sphere of coffee grounds, placed at rest relative to each other.

The result is that the ball shrinks, due to gravity, and that the accleration required is just the volume of the ball * (rho+3P).

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I thought that this was easy to explain from a Newtonian viewpoint, but we got tied up in a lot of difficulties. So perhaps the GR argument as presented by Baez & Bunn will actually make the explanation simpler.

I do agree that the force is independent of direction. I'm replacing the ball by a very-lightweight and open "rigid bar". And the goal is to find the tension (or compression) in the bar. The endpoints of the bar will correspond with two of the particles in Baez's ball of coffee ground. The fact that the ball of coffee grounds shrinks means that the bar would also shrink, unless a force opposes it.

Thus the B&B argument shows that a bar, in free space, would experience compressive forces. This is not due to the expansion of space, but is due to the matter contained within space, so that one has to be careful that the bar experiment doesn't change the density of the local matter distribution.

 

[MeJennifer]

Rather than argue about Newtonian theory, let's try a different aproach.

Consider the GR case, as outlined in http://math.ucr.edu/home/baez/einstein/

more specifically
http://math.ucr.edu/home/baez/einstein/node3.html



So, the second derivative of the rate of change of the volume of a sphere of coffee grounds is proportional to the volume of the coffee grounds multiplied by the density of matter - plus, a pressure term.

Note that, as Baez & Bunn explain, this depends only on the amount of matter contained within the ball. External gravity via external matter does not cause the volume to shrink. If we put the coffee grounds near the Earth, there will be the tension and compression terms you noted. The result will be that the ball will grow longer in one direction, shorter in the other two, and the second derivative of the volume will be zero.

Now, skip on a bit to:

http://math.ucr.edu/home/baez/einstein/node7.html

They apply the above result to a sphere of coffee grounds, placed at rest relative to each other.

The result is that the ball shrinks, due to gravity, and that the accleration required is just the volume of the ball * (rho+3P).

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I thought that this was easy to explain from a Newtonian viewpoint, but we got tied up in a lot of difficulties. So perhaps the GR argument as presented by Baez & Bunn will actually make the explanation simpler.

I do agree that the force is independent of direction. I'm replacing the ball by a very-lightweight and open "rigid bar". And the goal is to find the tension (or compression) in the bar. The endpoints of the bar will correspond with two of the particles in Baez's ball of coffee ground. The fact that the ball of coffee grounds shrinks means that the bar would also shrink, unless a force opposes it.

Thus the B&B argument shows that a bar, in free space, would experience compressive forces. This is not due to the expansion of space, but is due to the matter contained within space, so that one has to be careful that the bar experiment doesn't change the density of the local matter distribution.

While I don't disagree with anything here I fail to see how this explains anything. I am sure the problem is mine and not yours.

Clearly the effect represented by the cosmological [ initially and incorrectly written "gravitational". meJennifer] constant is treated differently from {\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z)..
I cannot see any explanation other than , "we'll just make some constant to even it out". The same goes for the numerous "expansion factors", to me they just seem ad hoc additions to match experimental data.
By the way there is *nothing* wrong with that, but what would be wrong is to assert that this all logically follows from GR.

Perhaps I have a fundamental misunderstanding here, I love to be educated here. I simply fail to see why a simple constant for "dark matter" would suffice while for regular matter we would need a more complicated method.

 

[pervect]

While I don't disagree with anything here I fail to see how this explains anything. I am sure the problem is mine and not yours.

Clearly the effect represented by the gravitational constant is treated differently from {\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z)..
I cannot see any explanation other than , "we'll just make some constant to even it out". The same goes for the numerous "expansion factors", to me they just seem ad hoc additions to match experimental data.
By the way there is *nothing* wrong with that, but what would be wrong is to assert that this all logically follows from GR.

Perhaps I have a fundamental misunderstanding here, I love to be educated here. I simply fail to see why a simple constant for "dark matter" would suffice while for regular matter we would need a more complicated method.

What makes you say that?

The only real difference with the above expression is that in GR, pressure causes gravity. In Newtonian physics, this isn't the case. This doesn't matter much until we get to "dark energy" and its negative pressure. It doesn't affect "dark matter" at all.

Note that we can apply Gauss's law to show that something is defintely fishy about the previous argument that was being presented about the gravitational field in an infinite and/or very large universe.

If we draw a sphere somewhere in this universe, and integrate the Newtonian gravitational field perpendicular to the surface of the sphere, we must get the enclosed mass, by Gauss's law.

The argument that the field is everywhere zero can't be corrrect - for then the surface intergal would be everywhere zero, and the enclosed mass would be zero. But we just got through saying that we wanted a universe that did have matter in it.

So let's go back to the finite case, of a very large (but not infinte) spherical planet, and forget about infinites for a bit.

By applying Gauss's law, we can conclude that the field always points towards the center of the sphere. In the constant density case, we find that the field is proportional to the distance away from center.

This is because F = GM/r^2 and M = rho * r^3, therefore F = g M rho r

Note that this sort of Hooke's law force is what we'd get for a constant density Earth.

Note also that this sort of Hooke's law force implies a tidal force in the radial direction, because the force increases with distance.

 

[oldman]

Rather than argue about Newtonian theory, let's try a different approach.

I quite agree. Thanks for the URL's.

From a Newtonian perspective all I could conclude, from the symmetry of the situation, is that the only possible manifestation of gravity in the "cosmic fluid" would be pressure, rather than the mistaken idea of tidal forces I was hung up on. But I can't calculate what that pressure would be. For this, it seems, GR is needed.

The conclusion you come to:

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I agree, even though I can't see it from a Newtonian perspective. GR rules.

When it comes to estimating these forces, or rather pressure (which I failed to calculate in a Newtonian way), I have one question about your earlier comments. It probably arises because I had muddied the waters by talking of tidal forces instead of pressure.

The pressure isn't given by the H^2 you mentioned earlier (as found from the components of the Riemann tensor), is it?

If it were, I would worry that the dynamics have something to do with determining this pressure, which seems ridiculous. Of course right now, when H is about 10 ^ - 19 per sec, nobody would give two hoots about such a tiny quantity as H^2 --- it would be a mere scientific curiosity. But during inflation, when I estimate that H is huge, H^2 has a startlingly large value which might have physical consequences if the pressure you mention were to depend on H.

I expect that this quibble arose from crossed wires, though.

 

[pervect]

I will agree with one thing about the Newtonian analysis, though. The finite spherical case I was talking about has a preferred direction, towards the center of the sphere.

I suspect that the Newtonian limit for an infinite universe may not really be well defined, that it may depend on how exatly one approaches infinity (infinte sphere, infinite plane, etc). I still don't believe the "no force" solution, as I explained earlier, because it conflicts with Gauss law.

The infinite GR case won't have such a preferred direction and doesn't run into this issue though. In the neighborhood of a point, everything nearby the point follows a geodesic will accelerate towards that point - given that there is no a positive matter density and no "dark energy" or "qunitessence".

 

[oldman]

I suspect that the Newtonian limit for an infinite universe may not really be well defined, that it may depend on how exactly one approaches infinity (infinite sphere, infinite plane, etc).

This may well be so.

I still don't believe the "no force" solution, as I explained earlier, because it conflicts with Gauss law.

Here we disagree. Your explanation was, I assume:

Note that we can apply Gauss's law to show that something is definitely fishy about the previous argument that was being presented about the gravitational field in an infinite and/or very large universe.

If we draw a sphere somewhere in this universe, and integrate the Newtonian gravitational field perpendicular to the surface of the sphere, we must get the enclosed mass, by Gauss's law.

The argument that the field is everywhere zero can't be correct - for then the surface intergal would be everywhere zero, and the enclosed mass would be zero. But we just got through saying that we wanted a universe that did have matter in it.

So let's go back to the finite case, of a very large (but not infinite) spherical planet, and forget about infinities for a bit.

By applying Gauss's law, we can conclude that the field always points towards the center of the sphere. In the constant density case, we find that the field is proportional to the distance away from center.

This is because F = GM/r^2 and M = rho * r^3, therefore F = g M rho r

Note that this sort of Hooke's law force is what we'd get for a constant density Earth.

Note also that this sort of Hooke's law force implies a tidal force in the radial direction, because the force increases with distance.

There is conflict embedded here that I don't understand.

You can show that there is no field at any point in the "cosmic fluid" by integrating the field inside spherical shells centred on the point from zero radius out to infinity. And inside each shell Gauss's law shows that the field due to that shell is zero. So the integration must also give zero. But, as you point out, then using Gauss's law to find the mass inside any sphere will then give zero, which is wrong. I don't understand this.

A simpler and in my view more believable argument is just to say that the gravitational field is zero throughout the cosmic fluid because the fluid is everywhere always isotropic, i.e highly symmetic.

I suspect that this conflict is somehow connected with the applicability of Gauss's law to "compact" domains. The cosmic fluid isn't compact; it may be infinite. I then agree that:

The infinite GR case won't have such a preferred direction and doesn't run into this issue though. In the neighborhood of a point, everything nearby the point follows a geodesic will accelerate towards that point - given that there is no a positive matter density and no "dark energy" or "qunitessence".

Whatever the resolution of the Gauss's law argument, my main question remains: how do you quantify the above acceleration in an inflating universe? as H^2?

 

[oldman]

I claimed in my previous post (#36) that:

You can show that there is no field at any point in the "cosmic fluid" by integrating the field inside spherical shells centred on the point from zero radius out to infinity. And inside each shell Gauss's law shows that the field due to that shell is zero. So the integration must also give zero. But, as [Pervect, post#33] points out, then using Gauss's law to find the mass inside any sphere will then give zero, which is wrong. I don't understand this.

A simpler and in my view more believable argument is just to say that the gravitational field is zero throughout the cosmic fluid because the fluid is everywhere always isotropic, i.e highly symmetic.

This post is a note to show that the view I've taken in this last paragraph is, I think, consistent with Einstein's law of gravity. The argument (which may be quite wrong!) runs like this:

Accept for a moment that the Newtonian gravitational field is indeed zero everywhere in an infinite cosmic fluid, as I've argued above. This means that there is no spatial gradient of field anywhere, no net force on any element of the fluid and no tidal stresses of the tear-apart or squash-together kind caused by force-gradients.

But a non-directional compressive or dilational stress, namely pressure is permitted. It must be the same everywhere --- it's not allowed to have gradients, because of high symmetry and the forces gradients would create, which are taboo.

Now pressure is the flow of momentum in a fluid, due to the motion of its elements. And momentum implies kinetic energy, which is equivalent to mass. The pressure that is allowed in the cosmic fluid is therefore the contribution to the mass/energy density of the fluid of the motion of its elements, in this case provided that such motion preserves the high symmetry of the cosmic fluid.

Expansion (or contraction) is just this kind of motion. It seems to me that a uniform pressure in the cosmic fluid will arise because of its expansion (or contraction), in exactly the way laid down by Einstein's law of gravity, which is, from

http://math.ucr.edu/home/baez/einstein/node3.html:

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

Of course this doesn't reconcile the paradox raised by Pervect (see above) with the no-field view. I still don't know how this is to be settled.

 

[pervect]

A few technical notes:

The method of measurement of tidal force is that basically one has a rigid bar, of fixed length. The ends of the bar are thus NOT following geodesics. One measures the stress, or strain, in the bar required to keep the ends of the bar at a constant distance.

For the flat FRW metric

ds^2 = -dt^2 + a^2(t) (dx^2 + dy^2 + dz^2)

the components of the tidal acceleration in the x direction can be computed via the geodesic deviation equation

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html

Note that this computes the tidal force for an observer following a geodesic, hence the name. It won't compute the correct tidal force for an accelerating or rotating observer. For instance, a non-accelerated observer in flat Minkowskian space-time won't experience any tidal forces, but an accelerated observer in flat Minkowskian space time will experience tidal forces according to GR. See for instance Bell's spaceship paradox. This effect is not one that one would expect from Newtonian physics.

The tensor used to calculate the tidal force doesn't have an input for the acceleration of the observer, it only computes the tidal forces for a geodesic observer.

When the center of the bar is following a geodesic, and the bar is short, the equation can also be used to find the stress in the bar as below.

The equation is:

a_tidal = R^\hat{x}{}_{\hat{t}\hat{x}\hat{t}}

Note that the four-velocities in Baez's equation are unity, and serve to pick out a particular component of the Riemann. I've not included them above, because they are unity. The "hats" indicate that I'm using a frame-field basis. This can be thought of as a local coordinate system that "wears the hats" that is locally Minkowskian.

Note that the y and z components are identical by symmetry.

The value for a general expansion is \frac{-\frac{d^2 a}{d t^2}}{a} where a is the scale factor.

While I computed this via the Riemann tensor, it is possible to take a simpler more physical approach.

We are trying to find the separation between geodesics. We know that comoving objects in the "cosmic flow" follow geodesics. We know that the distance as a function of time for comving objects goes as (coordinate-difference) * a(t). Thus we can write

accel = K d^2 a / dt^2

where K is a constant equal to the difference in comoving coordinates.

Tidal acceleration is just acceleration per unit length. Since length = K * a, the tidal acceleration is just

(d^2 a / dt^2 ) / a

This is our previous result, except for the sign, which I've been sloppy about keeping tract of.

For the particular case of a De-sitter universe, a(t) = exp(Ht) where H is Hubble's constant. We can then see that for this case, the above expression reduces to H^2.

In the general case, the tidal force can be derived from the decleration parameter q and the Hubble constant

http://scienceworld.wolfram.com/physics/DecelerationParameter.htmlAs far as your remarks about pressure in the fluid go, I'll have to study them at more length to make sure we aren't "crossing wires".

 

[oldman]

A few technical notes...

It's good to see actual machinery in motion. I guess that most of my puzzles have been resolved, thanks to you, and that we're now on the same page. It's been quite a long road, and I'm grateful that your persistence has prevailed.

I have one remaining question about the interpretation of the word "pressure" in general relativity (GR). It is often made clear that in GR pressure contributes to gravitating mass. You have shown that in the "cosmic fluid" the component of the Riemann? tensor is, in the x direction:

a_tidal = R^\hat{x}{}_{\hat{t}\hat{x}\hat{t}} {and that} ...the y and z components are identical by symmetry

By taking into account these three equal diagonal components you can calculate "the pressure", which in the case of an inflating cosmic fluid turns out to be H^2, and you seem to regard this as a symmetrical "tidal force" . I hope I've got this right.

In ordinary physics, by "pressure" one usually means phenomena like hydrostatic pressure, gas pressure, degeneracy pressure, or radiation pressure, which one understands intuitively to be a driver of compression (or dilation).

In GR, where pressure gravitates, I suspect that the interpretation of pressure is a bit wider than this. In GR, pressure seems to be a convenient (and quite appropriate) label for the flux of momentum, and through this for the mass equivalent of the energy of motion of constituent masses. Formally, of course, this label has the same derivation as "ordinary" pressure. In a gas, for example, GR pressure can be regarded as the mass equivalent of the kinetic energy of the random motion of gas molecules, via m = this energy/c^2.

Is this interpretation correct?

If it is, then is not the pressure you calculate for an inflating cosmic fluid nothing more than the mass equivalent of the energy of the motion of expansion of the cosmic fluid elements? In which case I can't quite make out why it would induce strains in a "rigid" bar that could be measured with strain gauges, as if the bar were at the bottom of a sea.

But perhaps I have got entirely the wrong idea about "pressure" as it is used and calculated in GR.

 

[pervect]

The pressure that is the source term in GR is not related to the tidal forces that we just calculated.

For instance, if we have a "dust" universe, the pressure, as used in Einstein's equations is zero - the tidal forces, however, we just got through calculating and won't be zero in a dust universe.

While your remark that what we've calculated is essentially a pressure is true, you shouldn't confuse it with the sort of pressure that's the source term in Einstein's equations, the pressure that's part of the stress-energy tensor.

To compute the pressure as it is usually definied, we look at the total force on the wall of a box of fixed volume, when the box is of infinitesimal size.

Now, you are probably thinking "well, the universe is expanding, so that there is a force on the box walls due to the geodesic motion of the particles, therefore there is a pressure on the walls".

But if we work out the pressure due to cosmological expansion, it depends on the size of the box, and goes to zero when the box has an infinitesimal volume. At small volumes, all particles essentially have the same velocity, so there is no force on the walls of a small box. In other words, the force on our box walls is proportional to the length of the side of a box, and when this length goes to zero, the force is zero. Therfore the pressure, which is the limit of the force / area as the size goes to zero, is also zero.

You can think of it this way.

If you have take the particle swarm model of a fluid, and all the particles at any point have the same velocity, the pressure in the fluid is zero.

I.e. if all particles at one particular spot have the same velocity in magnitude and direction, the fluid is pressureless.

However, if you have random, thermal motion, like that in an ideal gas, so that at anyone spot the "swarm of particles" are all moving in different directions, that represents a fluid with pressure.

The cosmological "pressure" for a dust universe can thus be seen to be zero, because all the individual particles are moving in one direction, the direction of the Hubble flow.

A universe with radiation represents a universe with pressure, a universe with just comoving dust has no pressure.

 

[oldman]

The pressure that is the source term in GR is not related to the tidal forces that we just calculated.

The argument you have given to justify this clear statement doesn't convince me, I'm afraid. We seem to be talking past each other, as here and in your post #38 you keep mentioning a "tidal force" that is being calculated, whereas I keep on banging on and on about "pressure".

I think that what you are calculating is neither a "force" nor "tidal", but the accelerated motion of fluid elements.

From the geodesic deviation equation you can calculate accelerations, which can be associated with proportional "forces" onlywhen the fluid elements have mass . The dust particles you mention in your post #40 don't have mass, so unsurprisingly there is indeed no pressure in a dusty fluid. You can't use a dusty fluid as an example to justify the statement quoted above. When fluid elements have mass, their accelerated motion must contribute to momentum flux and therefore act as a source term.

Then the adjective "tidal" is inappropriate in the cosmic fluid, as I realized some time back in this thread. As Peacock puts it in Cosmological Physics (p.43): in a tidal field "any tendency for test particles to approach one another along one direction means that they will be separated along the other direction". The pressure we are discussing is therefore not a tidal phenomenon.

Lastly, your argument that in the cosmic fluid:

At small volumes, all particles essentially have the same velocity, so there is no force on the walls of a small box... the pressure due to cosmological expansion, it depends on the size of the box, and goes to zero when the box has an infinitesimal volume. ... If you have take the particle swarm model of a fluid, and all the particles at any point have the same velocity, the pressure in the fluid is zero.

I.e. if all particles at one particular spot have the same velocity in magnitude and direction, the fluid is pressureless.

I can't accept this. All volumes of the cosmic fluid, no matter how small, are expanding and partake of the Hubble flow, and neighbouring particles never have the same velocity, no matter how close they are. "Essentially" is a bit of a weasel word!

I believe that I am not the person who is guilty of confusing ordinary pressure --- the kind that in a gas produces forces on the walls of a box with "the sort of pressure that's the source term in Einstein's equations, the pressure that's part of the stress-energy tensor." But I surely don't want to turn an amicable disagreement into an acrimonious argument. I hope I'm not doing this!

 

[pervect]

Let me provide a reference for what I'm trying to say:

http://en.wikipedia.org/wiki/Fluid_solution

In general relativity, a fluid solution is an exact solution of the Einstein field equation in which the gravitational field is produced entirely by the mass, momentum, and stress density of a fluid.

...

Several special cases of fluid solutions are noteworthy:

* A perfect fluid has vanishing viscous shear and vanishing heat flux:
* A dust is a pressureless perfect fluid:
* A radiation fluid is a perfect fluid with μ = 3p:

Dust solutions by defintion do not have a pressure by the standard defintion of the term, as per the Wikipedia defintion.

If you want to call what we have been talking about a "pressure" rather than a "tidal force" or "tidal acceleration", that's OK with me as long as I know what you're talking about.

But what you must not do is to confuse the concept we have been discussing with the standard defintion of the pressure of a fluid, the one that is used in the GR defintion, the P in rho + 3P.

And if you call the concept that we have been discussing just "pressure" without any other modifier, I don't see any way to avoid confusion, having two different ideas with the same name.

To try and be really specific:

A solution to the FRW equations where rho > 0 and P = 0 is called a 'dust solution' (per the wikipedia article).

[add] It's a red herring to talk about whether any partricular grain of dust has mass. The important thing is that in the specific solution I'm talking about, the average density of matter per unit volume is non-zero.

Because rho > 0, the volume of a sphere of comoving coffee grounds will contract due to gravity in the above solution. This leads to the effects we've been discussing.

Because P is defined to be equal to zero by the standard defintion, the rate of contraction of the volume of the sphere of comvoing coffee grounds cannot be considered to be a measure of pressure of the fluid, since the former is non-zero and the later is zero in the example case of a FRW dust solution.

 

[oldman]

If you want to call what we have been talking about a "pressure" rather than a "tidal force" or "tidal acceleration", that's OK with me as long as I know what you're talking about.

I agree that this is sensible. For the purposes of this discussion, then, let's call it "coffee grounds motion", say CGM for short, rather than pressure of any kind.

But what you must not do is to confuse the concept we have been discussing with the standard defintion of the pressure of a fluid, the one that is used in the GR defintion, the P in rho + 3P.

And if you call the concept that we have been discussing just "pressure" without any other modifier, I don't see any way to avoid confusion, having two different ideas with the same name.

Yes, I may well have been rather stupidly sowing confusion here. My apologies for this.

A solution to the FRW equations where rho > 0 and P = 0 is called a 'dust solution' (per the wikipedia article).

Thanks for this URL about various solutions. With the Wikipedia available, one hardly needs to attend grad school these days!

It's a red herring to talk about whether any partricular grain of dust has mass. The important thing is that in the specific solution I'm talking about, the average density of matter per unit volume is non-zero...Because P is defined to be equal to zero by the standard defintion, the rate of contraction of the volume of the sphere of comoving coffee grounds cannot be considered to be a measure of pressure of the fluid, since the former is non-zero and the later is zero in the example case of a FRW dust solution.

What you say is mostly correct. But, as I discuss below, is the specific dust solution the one that applies to CGM? Or is this so only by definition?

Because rho > 0, the volume of a sphere of comoving coffee grounds will contract due to gravity in the above solution

I agree; as you said earlier, this is why the expansion of the universe slows down, and why Baez and Bunn at

http://math.ucr.edu/home/baez/einstein/node7.html say:

...this is the same as the equation of motion for a particle in an attractive force field. In other words, the equation governing this simplified cosmology is the same as the Newtonian equation for what happens when you throw a ball vertically upwards from the earth!

But in real situations coffee grounds do have mass, even if this can often be neglected in applying an idealised solution. In the universe, for example, the coffee grounds are galaxies. CGM then does have momentum and energy, and hence mass. Surely this must contribute to the gravitating mass?

I believe (firmly at the moment) that what you describe as "the pressure term in the standard definition of the pressure of a fluid, the one that is used in the GR definition, the P in rho + 3P" represents (quite appropriately) nothing more than the mass-equivalent of the kinetic energy in this "standard" situation. If I'm wrong here then I've misunderstood the fundamentals of general relativity quite badly.

The heart of my difficulty seems to be this: you are looking at the situation using a specific established solution in the literature, which is appropriate on the largest scales because the universe is mainly empty, whereas I'm trying to understand things on a scale where one talks of bars being compressed and strains being measured with stick-on strain gauges.

As a result I still can't decide whether the effects we've been discussing actually exist, and if they would be so huge during inflation.

 

[pervect]

I believe (firmly at the moment) that what you describe as "the pressure term in the standard definition of the pressure of a fluid, the one that is used in the GR definition, the P in rho + 3P" represents (quite appropriately) nothing more than the mass-equivalent of the kinetic energy in this "standard" situation. If I'm wrong here then I've misunderstood the fundamentals of general relativity quite badly.

The heart of my difficulty seems to be this: you are looking at the situation using a specific established solution in the literature, which is appropriate on the largest scales because the universe is mainly empty, whereas I'm trying to understand things on a scale where one talks of bars being compressed and strains being measured with stick-on strain gauges.

As a result I still can't decide whether the effects we've been discussing actually exist, and if they would be so huge during inflation.

The pressure term is not the mass-equivalent of the kinetic energy, unfortunately. I think at one time I had this same (incorrect) idea. However, in any given frame, the intergal of T⁰⁰ gives the total energy content, including kinetic energy. More specifically, adopt a locally Minkowskian frame. Then in a unit volume dV in this locally Minkowskain (i.e. flat) frame, the local energy contained in a volume element dV is just.

\int_V T^{00} dV

where T^{00} = \rho does not include any pressure terms.

If you consider any complete closed system, however, the pressure terms will integrate to zero, so they don't contribute to the system mass. Systems with finite volume that aren't closed are very tricky to deal with, and I'd suggest avoiding them as much as possible.

You can "close" an otherwise unclosed system just by adding in neglected components of the system. When you do so, you have a complete, closed system and the pressure terms then will integrate out to zero.

I talk a little bit about some of the fine points regarding mass in

http://en.wikipedia.org/wiki/Mass_i...simple_examples_of_mass_in_general_relativity

the question and answer section is the most relevant.
(note that I'm the author of this article), and that it also discusses some unrealted stuff as well as related stuff. This has a reference to the Carlip paper, for instance, which discusses the relativistic virial theorem which is what makes the pressure terms integrate out to zero. Unfortunately I haven't seen this (the relativistic virial theorem) disucssed much in any of the textbooks I happen to own, so the Carlip reference is the best I have on this somewhat obscure but interesting topic.

As far as the rest goes, I'm personally quite convinced that there is no special difference between calculating tidal forces due to the Schwarzschild metric and tidal forces due to the FRW or De-Sitter metrics.

The only thing to beware of in my opinion is that when one calculates the tidal forces using the geodesic equation, one calculates the tidal forces for a non-rotating observer following a geodesic, that observers following non-geodesics may experince different forces (as may the "centrifugal forces" for rotating observers).

 

[oldman]

The pressure term is not the mass-equivalent of the kinetic energy, unfortunately.

You have seriously undermined my firm belief that this was so! It is now clear to me from your Wikipedia article and the Carlip article that gravitating mass/energy is an altogether more difficult concept in general relativity than I'd ever realized. Indeed my confusion about pressure gravitating (I'd thought that pressure exists only because elements of mass move, and that because the kinetic energy of their motion gravitated, pressure itself gravitated) is not entirely dispelled by statements like those below, which I of course accept:

We can thus tell our students with confidence that kinetic energy has weight, not just as a theoretical expectation, but as an experimental fact.

and

If two objects have the same mass, and we heat one of them up from an external source, does the heated object gain mass? If we put both objects on a sensitive enough balance, would the heated object weigh more than the unheated object? Would the heated object have a stronger gravitational field than the unheated object?

The answer to all of the above questions is yes...

I guess I'll have to accept that the answer to the question in my original post:

Consider an observer... in an inflating flat FRW universe that begins to expand exponentially rapidly after he has set up his local inertial frame. Suppose he ... releases two test particles (joined by string) from rest in this frame... What happens as the scale factor, and its derivatives with respect to time, change exponentially with time? ...

is that the string breaks because expansion in this case speeds up -- very rapidly indeed, since H^2 is so huge!

Thanks for your patience in bringing me to this conclusion.