Gravity & Curvature: Explaining Geometrical Orbit of Particles

[blumfeld0]

I need help in trying to explain to someone how gravity creates curvature and paths of particles are curved around this curvauture?
like the sun on a trampoline and the Earth's path around it being bent as it orbits the sun. so why does the sun bend the space around it? and why does the Earth travel around this bent space? i am looking for a geometrical explanation. thank you

blumfeld0

 

[pervect]

"Why" questions don't really have answers. If you understand that the Earth orbital motion, which appears "curved", can also appear to be a straight line in curved space-time, you've got the basics.

Note that most of the "curvature" in space-time can be attributed to time, and not space. This is one of several areaa where the trampoline model isn't terribly good. The trampoline model invites one to think of space as curved, because it's easier to visulaze. The actual math of relativity is that space-time is curved rather than just space being curved. Furthermore the part of curvature that's most significant is due to time, not space.

We can actually measure this curvature by the facts that clocks tick more slowly the deeper they are in a gravity well.

It is possible, using the principle that clocks tend to move in such a manner as to xtremize proper time, to work out the orbits of slowly moving bodies.

Very fast moving bodies have to take into account spatial curvature too, but slowly moving bodies can be modeled very well by looking only at the time part of the curvature, namely gravitational time dilation.

Enough physics to appreciate "the principle of least action" helps a lot, unfortunately to get to this level rigorously requires a fair amount of math (calculus and calculus of variations).

 

[Chaos' lil bro Order]

I think when you consider that gravity, energy and matter are all equivalent, you can see that any of these, in a sufficient amount, will appreciably curve space. Density is the true key to curving space. A 1kg book will not appreciably curve space, but put that 1kg in a space confined to 1 nanometer squared and space will certainly curve appreciably.

 

[MeJennifer]

I need help in trying to explain to someone how gravity creates curvature and paths of particles are curved around this curvauture?

Here is the how and why:

Basically you have to start with the principle of relativity and the fact that the speed of light is constant in each frame of reference. One of the implications of this principle is that particles observe a contraction of the distance of approacing particles. Sometimes called Lorentz or length contraction.
As long as these particles approach with a constant rate we can use a linear calculation to determine the amount of contraction.
Basically this phenomenon of contraction can be modeled adequately in a flat 4-dimensional Minkowski space.
However once the rate of approach is no longer constant the calculation is no longer linear and we can no longer model this in a flat Minkowski space. In order for the model to work we have to allow for curvature in this 4-dimensional Minkowski space.
So we need a 4-dimensional Minkowski space with curvature to provide a framework that caters for the principle of relativity and acceleration.

Finally there is Einstein's principle of equivalence. What he is saying is that acceleration and gravity are equivalent.
Hence mass must curve the 4-dimensional Minkowski space just like acceleration does.

The last step is going from the model, the 4-dimensional curved Minkowski space to reality.
This space is an adequate representation for calculations of classical mechanics. But is the universe really a 4-dimensional Minkowski space with curvature? That of course we cannot prove, and surely the string theory advocates would prefer to add a couple of dimensions to that.

But that is basically where the curvature thing is derived from.

 

[pmb_phy]

Finally there is Einstein's principle of equivalence. What he is saying is that acceleration and gravity are equivalent.

That isn't quite right. The weak form of the equivalence principle states that a uniform gravitational field is equivalent to a uniformly accelerating frame of reference. The strong form of the equivalence principle states that locally a gravitational field can be transformed away at point in spacetime (i.e. at any event). The strong form has also been stated in other ways. But in general a gravitational field is not the same as acceleration. That was never how it was stated by Einstein.

Hence mass must curve the 4-dimensional Minkowski space just like acceleration does.

This does not follow from the equivalence principle. If there is matter at an event E in spacetime then the spacetime curvature at that event is non-zero. However the spacetime near that event can indeed vanish, depending on the exact field of course. Gravitational acceleration is not the same thing as spacetime curvature. Gravitational tidal forces are the same thing. Where there are tidal forces there is spacetime curvature and where there is spacetime curvature there are tidal forces. Curvature has an absolute existence and cannot be transformed away like the gravitational field can be.

Pete

 

[DaveC426913]

I think when you consider that gravity, energy and matter are all equivalent, you can see that any of these, in a sufficient amount, will appreciably curve space.

This is a bit misleading - it's like saying matches (or other fuel) and flames are equivalent, and that enough of either can start a forest fire. Yet you cannot have a flame without a combusting fuel.

You see, GR demonstrates that gravity is not a thing at all, it is nothing but an effect. Mass or energy density creates curvature in spacetime. Period. Gravity is nothing more than our perception and measurement of that curvature.

You can't have a bunch of gravity all by itself any more than you can have a bunch of flames without a fuel source.

 

[Garth]

This is a bit misleading - it's like saying matches (or other fuel) and flames are equivalent, and that enough of either can start a forest fire. Yet you cannot have a flame without a combusting fuel.

You see, GR demonstrates that gravity is not a thing at all, it is nothing but an effect. Mass or energy density creates curvature in spacetime. Period. Gravity is nothing more than our perception and measurement of that curvature.

You can't have a bunch of gravity all by itself any more than you can have a bunch of flames without a fuel source.

And yet gravitational waves carry off energy...

Garth

 

[Farsight]

Blumfeld: I'd like to back up Pervect and Dave. IMHO gravity isn't a thing that creates "curvature", the underlying mass affects spacetime like Pervect was saying. A simple analogy is that you've got two legs, but time runs slower in your left leg than your right leg, so you find yourself walking round in circles. You give this effect the name "gravity".

Garth: I don't know if it's relevant, but surely you can't have energy all by itself either.

 

[pmb_phy]

The relationship between gravitation and curvature is really quite simple. First of all GR does not explain gravity. It is a general theory of relativity. This means that it covers all kinds of coordinate systems rather than simply Lorentz coordinates (i.e. inertial frames). One can, as Einstein put it, "produce" a graavitational field by a change in spacetime coordinates. However this does not mean that one can create spacetime curvature by changing the coordinate system.

GR describes the exact same phenomena (gravitational acceleration and tidal acceleration) that Newton's theory describes. What is referred to in Newtonian physics as "tidal forces" is now called "spacetime curvature" in Einstein's theory. The relative acceleration of two particles in Newton's theory is describe in GR by two geodesics which deviate.

It is to be noted that according to Einstein, gravity and curvature are not identical phenomena. In Newtonian gravity one can have a gravitational field with no tidal forces (i.e. a uniform gravitational field). The spacetime curvature of such a field is zero. Recall Newton's law of gravitation in differential form. At a point r in space the Laplacian of the gravitational potential at r is proportional to the mass density at r. Howevever at positions outside the clump of mass which generates the field the mass density is zero. The tidal forces at such locations may or may not be zero. An example of such a field is that of a spherical cavity cut out of a sphere of uniform mass density. When the center of the cavity is displaced from the geometrical center of the sphere a uniform gravitational field will be found within the cavity. Likewise, Einstein's tensor, which kind of replaces the Laplacian, is proportional to the stress-energy-momentum tensor, i.e. that object which completely describes mass. Where there is matter there is spacetime curvature. The converse need not be true though.

Pete

 

[DaveC426913]

And yet gravitational waves carry off energy...

Garth

Let me point out a different analogy then.

It is like saying water and (water) waves are equivalent.

You cannot have (water) waves without the water in which they do their thing. The waves are simply an effect - a description - of the motion of the water.

Gravity is simply a description of the curvature of spacetime. Gravity, in and of itself, is not what causes the Earth to revolve about the Sun.

 

[pmb_phy]

Gravity is simply a description of the curvature of spacetime.

Not as the term "gravity" was used Einstein. See above for clarification.

Pete

 

[blumfeld0]

Hello. Thank you for your nice responses

""Gravity is simply a description of the curvature of spacetime. Gravity, in and of itself, is not what causes the Earth to revolve about the Sun.""

so why does the Earth go around the sun according to einstein?

thanks

 

[pervect]

Hello. Thank you for your nice responses

""Gravity is simply a description of the curvature of spacetime. Gravity, in and of itself, is not what causes the Earth to revolve about the Sun.""

so why does the Earth go around the sun according to einstein?

thanks

There are really a couple of different analogies that are popular, and not directly related, with respect to gravity.

The first is the famous "elevator" thought experiment.

The second is the idea that bodies, in the absence of any forces, follow the equivalent of straight lines, known as geodesic, in a curved space-time.

These ideas are not incompatible, but they are different descriptions of the same thing.

I get the impression from your original question that you are more interested in the second idea.

A very brief sketch of this is the famous "ants crawling around on an apple". The ants each follow as straight a path as they can. However, while they start out moving away from each other, they eventually start approaching each other again. One could attribute this motion to a "force of attraction between the ants", but as far as they are concerned, they are just following straight lines.

Google finds

http://homepage.mac.com/stevepur/physics/riding/Riding_session_1.pdf

as a reasonable example of this sort of explanation. There are some very good and detailed explanations of this sort in MTW's famous textbook "Gravitation" as well, however this book as a whole is not written on a popular level.

"Curvature", as formally defined, turns out not to play any significant role in the elevator gerdankenexperiment. Thus when you mix and match the different examples, you invite a certain amount of confusion if one adheres to all the technicalities. It's the physics equivalent of "mixed metaphors" in Engish text.

BTW, I wouldn't particulary hang onto Einstein as the best and only source of information on gravity. In some ways, Einstein's original views are a bit cumbersome. I think it is better to focus on the physics, and less on the personalities, myself.

 

[MeJennifer]

It is to be noted that according to Einstein, gravity and curvature are not identical phenomena. In Newtonian gravity one can have a gravitational field with no tidal forces (i.e. a uniform gravitational field). The spacetime curvature of such a field is zero. Recall Newton's law of gravitation in differential form. At a point r in space the Laplacian of the gravitational potential at r is proportional to the mass density at r. Howevever at positions outside the clump of mass which generates the field the mass density is zero. The tidal forces at such locations may or may not be zero. An example of such a field is that of a spherical cavity cut out of a sphere of uniform mass density. When the center of the cavity is displaced from the geometrical center of the sphere a uniform gravitational field will be found within the cavity. Likewise, Einstein's tensor, which kind of replaces the Laplacian, is proportional to the stress-energy-momentum tensor, i.e. that object which completely describes mass. Where there is matter there is spacetime curvature. The converse need not be true though.

Pete

Well, so are you saying that clocks don't run slower there?

 

[pmb_phy]

Well, so are you saying that clocks don't run slower there?

No. However the gravitational redshift of light can be detected in a uniform gravitational field. In fact this was the first field in which the phenomena was calculated.

Pete

 

[MeJennifer]

No. However the gravitational redshift of light can be detected in a uniform gravitational field. In fact this was the first field in which the phenomena was calculated.

Pete

So you are saying that gravitational reshift has nothing to do with a slower clock?

Let me get this right, clocks don't run slower AND there is gravitational reshift, and the redshift is perfectly explainable with a flat space-time.

Perhaps you could explain this, to me it does not make any sense whatsoever.

 

[pervect]

So you are saying that gravitational reshift has nothing to do with a slower clock?

Let me get this right, clocks don't run slower AND there is gravitational reshift, and the redshift is perfectly explainable with a flat space-time.

Perhaps you could explain this, to me it does not make any sense whatsoever.

If you have two clocks, one higher than the other, in the gravitational field of the Earth, you'll see a gravitational redshfit of the lower clock from the POV of the upper clock.

If you have two clocks, one higher than the other, in the "gravitational field" of an accelerating elevator, you'll see exactly the same effect.

But if you look at the pair of clocks in the accelerating elevator from the POV of an inertial frame (rather than from the POV of the clocks), you won't see any gravitational redshfit at all. Instead, you'll see a perfectly inertial coordinate system, along with some doppler shifts due strictly to velocity.

 

[RandallB]

But if you look at the pair of clocks in the accelerating elevator from the POV of an inertial frame (rather than from the POV of the clocks), you won't see any gravitational redshfit at all. Instead, you'll see a perfectly inertial coordinate system, along with some doppler shifts due strictly to velocity.

This does not ring true for me.
How can the POV of an inertial frame expect to see an inertial coordinate system for the source of light it observes from the elevator?
It must see an accelerating coordinate system with Doppler shifts affected by an increasing relative velocity, not a fixed velocity.

 

[MeJennifer]

For the good order it seems this is what is claimed:

In a uniform gravitational field the following is true:

- The space-time is completely flat
- The clocks do not run slower
- There is gravitational redshift

To me that sounds like a contradiction.

How could one possibly explain gravitational redshift, in a flat space and claiming there is no time dilation?

 

[pervect]

I think some of the statements complained about by MeJennifer are mutually contradictory too. Here's the situation as I would describe it.

A "uniform gravitational field" is described by coordinates which have an associated metric of

-(1+gz)^2 dt^2 + dx^2 + dy^2 + dz^2

1) The above space-time is completely flat, in the sense that the Riemann curvature tensor for the above metric is zero.

Other loose senses of what the term"flat" means are occasionally used, it's not really a very precise term. In the precise sense of having a zero Riemann curvature, the above metric is definitely flat, however.

This "flatness" can also be seen by the fact that there is a mapping via a change of variables to a flat-space metric.

-dtt^2 + dxx^2 + dyy^2 + dzz^2

This illustrates flatness because the Riemann curvature tensor for the above metric (the flat Minkowski metric) is widely known to be zero, and because a tensor that is zero in one coordinate system is zero in all coordinate systems.

The choice of coordinates that puts the metric in the above form represents an inertial coordinate system. (tt,xx,yy,zz).

Note that the choice of coordinates is completely arbitrary - it's a human choice.

It can easily (hopefully) seen, that for an object "at rest" in the inertial coordinates, dxx=dyy=dzz=0, and thus dtau = dtt, and there is no time dilation. Coordinate time is the same as proper time.

It can also be seen that for an object "at rest" in the accelerated coordinates, dx=dy=dz=0, and thus dtau = (1+gz)*dt, and there is gravitational time dilation. The relationship between coordinate time and proper time is a function of "height", the z-coordinate. This is time dilation, when coordinate time is not the same as proper time, and the amount of time dilation depends on the "height" of the object, i.e. it's z coordinate.

It should be understood that the object with xx=yy=zz=constant, which is "at rest" in inertial coordinates, is not following the same trajectory as the object with x=y=z=constant, which is "at rest" in the coordinate system used by the accelerated observer.

It should also be understood that a lot of the confusion arises from the use of concepts such as 'time dilation' that are coordinate dependent. When one chooses to use a coordinate dependent explanation (because of familiarity on the part of readers, usually), one has to live with the fact that it is coordinate dependent.

An explanation formulated purely in terms of coordinate independent quantites (like the Lorentz interval and abstract tensor notation) doesn't have to deal with these issues. But it is not generally as accessible.

[add]
Let me add a note on "curvature causes gravity". This is still a valuable notion, it comes directly from Einstein's equation

G_uv = 8 Pi T_uv

Here G_uv is a tensor, called the Einstein tensor, which is dervied from the Riemann tensor, which represents the curvature of space-time. If the Riemann tensor is zero, the Einstein tensor mus also be zero.

T_uv is another tensor, which represents the density of energy and momentum per unit volume.

Note what the equations says about the elevator experiment:

The equation says that there is no appreciable mass in the elevator experiment (just some small test masses, i.e. we are ignoring the gravity generated by the elevator car itself).

It also says that there is no Einstein curvature of the space-time in the elevator. We can go further than this - not only is there no Einstein curvature of the space-time in the elevator, there is no Riemann curvature either.

So Einstein's equation tells us that matter curves space-time, and it also tells us that in the elevator experiment, because there is no matter (or rather, because there is a negligible amount of matter, not worth worrying about), there is no curvature.

The difficulty is that people expect G_uv to represent "felt" gravity, but it doesn't. It's a measure of the curvature of space-time.

The common idea of "felt" gravity is basically the magnitude of the 4-acceleration of a particular observer. If the observer is stationary, this can be equated to the Christoffel symbols.

This differs also from what Pete says that Einstein likes to call gravity, which are the metric coefficients.

This may be a bit confusing but the bottom line is that metric coefficients, curvature tensors, and the magnitude of 4-accelerations are all precise things defined by physics, and all of them are relevant to the topic of gravity. It is the responsibility of the reader to make an informed decision about which of the above best matches their own idea of what "gravity" really is. I would suggest that the last quantity, the magnitude of a 4-acceleration, is what most people have in mind when they talk about "gravity". That's the force that one feels on the seat of one's pants when one sits down in a chair.

 

[MeJennifer]

This may be a bit confusing but the bottom line is that metric coefficients, curvature tensors, and the magnitude of 4-accelerations are all precise things defined by physics, and all of them are relevant to the topic of gravity. It is the responsibility of the reader to make an informed decision about which of the above best matches their own idea of what "gravity" really is. I would suggest that the last quantity, the magnitude of a 4-acceleration, is what most people have in mind when they talk about "gravity". That's the force that one feels on the seat of one's pants when one sits down in a chair.

Well, I guess I am totally wrong about it but my "I guess novice" understanding is that the force that we feel when we sit in our chair is NOT gravity, but actually the electro-magnetic force that stops us from traveling on our geodesic. So either I am completely wrong, or something is explained to people here completely wrong.
Also in my novice understanding when I hear that some fields shows a redshift it implies that space-time is curved there. Where else would it come from. And furthermore, and it seems that you but others disagree, in my limited understanding, I guess, is it the case that mass curves space-time.

I will wait for pmb_phy to provide an explanation since he is making all those statements that confuse the whole thing even more (e.g. space-time is flat, no time dilation, but nevertheless redshifted.).

So Einstein's equation tells us that matter curves space-time, and it also tells us that in the elevator experiment, because there is no matter (or rather, because there is a negligible amount of matter, not worth worrying about), there is no curvature.

Did he actually, or was he actually wrong. From this perspective it looks like gravity and accceration are not as equivalent as Einstein claimed it was.

Did you read pmb_phy's paper on Einstein vs the modern interpretation?
What I, presumably a misunderstanding on my side, get from that document is that: Einstein is right, the modern interpretation is right too, and when it contradicts it must be that we "misunderstood" Einstein.

 

[pervect]

Well, I guess I am totally wrong about it but my "I guess novice" understanding is that the force that we feel when we sit in our chair is NOT gravity, but actually the electro-magnetic force that stops us from traveling on our geodesic. So either I am completely wrong, or something is explained to people here completely wrong.

I wouldn't say that this is totally wrong.

The fact that you are not following a geodesic means that you must feel those electromagnetic forces.

If you didn't feel them, if you were acted on by no external forces, you would be following a geodesic.

Those forces that you feel are intimately realted to the fact that your 4-acceleration is non-zero.

There isn't any big conflict here. IMO.

Also in my novice understanding when I hear that some fields shows a redshift it implies that space-time is curved there. Where else would it come from. And furthermore, and it seems that you but others disagree, in my limited understanding, I guess, is it the case that mass curves space-time.

What do you mean by "curved"? Can you express it in terms of the behavior of the metric coefficients?

Some pepole say, loosely speaking, that having the metric coefficents vary at all is sufficient for there to be "curvature". This defintion of curvature is not very formal, and dependent on the coordinate system used. It is usually used to point out that in the presence of any gravitational field (including the pseudo-gravitational field inside a rocket) that the metric coefficients must vary.

Other people say that one must have a non-zero curvature tensor for space-time to be curved. This defintion of curvature is more formal, but does NOT depend on the coordinate system used.

This is a real conflict, but it is more in the nature of semantics - about what the loose, popularly defined term "curvature" really means. Does "curvature" really mean the Riemann curvature tensor, or is it some other, fuzzier, idea, like varying metric coefficients?

I will wait for pmb_phy to provide an explanation since he is making all those statements that confuse the whole thing even more (e.g. space-time is flat, no time dilation, but nevertheless redshifted.).

Feel free. This issue comes up a lot when Pete is around, and not so often when he is not.

I think that the resulting discussions can be useful to people who already know a lot about relativity, but are not so useful to those struggling to understand it, I suspect it totally blows them away :-(.

Unfortunately I don't know what to do about that - I really am trying to explain as simply as I can what all the fuss is about.

As far as what Einstein thought or did not think, I personally don't really care that much. A lot of people have an "Einstein" fetish, but to me, any physical insight is valuable, even if Einstein wasn't the one who happened to think of it. And in the years since Einstein first published GR, there have beena LOT of contributions to the field.

 

[MeJennifer]

IThose forces that you feel are intimately realted to the fact that your 4-acceleration is non-zero.

Yes, the way I understand it is that we are in constant acceleration when we stand on earth. In fact we are moving, only if we were to make a projection of the 4-d space-time manifold from our frame of reference with time being perpendicular we would find out we were not moving, right?

 

[pervect]

You can easily set up a coordiante system on the surface of the Earth where an object not moving with respect to the Earth is stationary, and where the space vectors of the coordinate system are orthogonal to the time vectors.

However, an object that is stationary in this coordinate system still has a non-zero 4-acceleration, even though it has a zero 4-velocity.

The trick is to take the covariant derivative of the 4-velocity with respect to proper time rather than the ordinary derivative. But unfortunately it takes a fair amount of knowledge of differential geometry to appreciate how the covariant derivative is defined.

The simplest thing I can say is that the metric coefficeints in this coordiante system (the one fixed to the surface of the Earth) can't be constant - they won't be inertial coordinates. One symptom of this is the existence of gravitational time dilation.

 

[Chaos' lil bro Order]

Matter = Energy

Matter & Energy curve space

Curved space appears as gravity

Light detoured from its geodesic by a clump of gravity, becomes stretched (red-shifted) as it follows a longer curved path.

Acceleration = gravity


Is this a fair description of all the correct points made above in this thread?

 

[pmb_phy]

So you are saying that gravitational reshift has nothing to do with a slower clock?

I don't know how you came to the conclusion that I said/implied this.

If two clocks are at different gravitational potentials then they will run at different rates as measured by a given observer. Redshift and slowing of clocks are identical phenomena, i.e. different names for exactly the same thing. I was stating that the spacetime need not be curved in order for this to occur. E.g. if you have a flat spacetime and then you choose a coordinate system corresponding to uniformly accelerating frame of reference then you will observe a gravitational redshift when the two clocks are located at two different gravitational potentials.

The relationship is given by dT = sqrt(g_00)dt where dT is the proper time on the clock and dt is the observer's time (i.e. time as measured by the observer who measures time with coordinate "t").

For a detailed calculation for the Schwarzschild gravitational field (e.g. that of earth) see

http://www.geocities.com/physics_world/gr/red_shift.htm

This issue comes up a lot when Pete is around, and not so often when he is not.

The reason being is that people never explain this fact and quite often walk away believing, incorrectly, that spacetime curvature is both a necessary and sufficient condition for the presence of gravitational redshift. This point was also made in the American Journal of Physics so it is not a trivial point. I even know an author of a well-known GR book who got this wrong at one time. I also make this point since Einstein also made it in at least two of his GR papers (1907 and 1911). In fact redshift in g-field was first redshift calculated in a flatspacetime (Einstein 1907).

Pete

 

[Jorrie]

Gravity = acceleration?

Matter = Energy

Matter & Energy curve space

I think you are right so far. My "engineering view" on your other statements are as follows:

Curved space appears as gravity

Wrong. Curved space appears as tidal gravity. As very ably explained by pervect before, space need not be curved for gravitational effects to appear.

Light detoured from its geodesic by a clump of gravity, becomes stretched (red-shifted) as it follows a longer curved path.

Wrong. When viewed at the same gravitational potential as the potential it was transmitted from, there is no redshift, only deflection and a Shapiro time delay of the light.

Acceleration = gravity

Only correct in a restricted sense - very localized, e.g. in a vanishingly small accelerating laboratory. That is unless the acceleration has a profile that precisely mimics curved space-time - maybe?

I would love to hear an expert's opinion

 

[pmb_phy]

Did you read pmb_phy's paper on Einstein vs the modern interpretation?
What I, presumably a misunderstanding on my side, get from that document is that: Einstein is right, the modern interpretation is right too, and when it contradicts it must be that we "misunderstood" Einstein.

Actually I wrote that paper because there is a disagreement between Einstein and modern day relativists on the definition of gravity. Einstein defined the presence of a gravitational field by the non-vanishing of the affine connection (gravitational acceleration). It was Max Von-Laue who seems to be the first to define presence of gravity by the non-vanishing of the curvature tensor (i.e. Riemman tensor).

A good example of Einstein's definition being important is found in cosmology. In cosmology there is a particular entity called a domain wall which is basically a sheet of matter which negative pressure. On the (infinitesimally thin) wall itself the Riemman tensor does not vanish. However at all locations outside the wall the Riemman tensor vanishes. However, objects which move inertially (zero 4-acceleration) outside the wall will accelerate away from the wall (i.e. there is a gravitational field outside the wall) as observed by an observer at rest with respect to the wall. Another object of interest is a straight cosmic string. Such an object is defined by its stress-energy-momentum tensor. In this tensor the mass-energy (per unit length) of the string has a value proportional to the negative of the pressure (pressure is <0). Thus on the string both the affine connection and the curvature tensor vanish. However off the wall both the affine connection and the curvature tensor vanish. The only thing different is the topology if the space around the string (The geometry of tghe space is cononical). So even though there is no gravitatonal field in either sense the overall geometry around the string has been altered by the string.

Pete

 

[pmb_phy]

I think you are right so far. My "engineering view" on your other statements are as follows:

Wrong. Curved space appears as tidal gravity. As very ably explained by pervect before, space need not be curved for gravitational effects to appear.
...
Only correct in a restricted sense - very localized, e.g. in a vanishingly small accelerating laboratory. That is unless the acceleration has a profile that precisely mimics curved space-time - maybe?

I'm unclear on what you mean by this. Please clarify. Using Einstein's definition the presence of a gravitational field at a point in space will depend on the spacetime coordinates of the observer. Not so for spacetime curvature. However the spacetime curvature, for a given experimental apparatus, may be neglected when the region of spacetime under observation is small enough so that the experiment is not sensitive enough to measure the curature. Then when the frame is a locally inertial one the spacetime coordinates are called "locally inertial."

As far as "matter = energy" goes - According to Einstein the term "matter" is defined such that at events where the stress-energy-momentum tensor does not vanish there matter is present, otherwise there is no matter present. This implies that any kind of electromagnetic field/radiation has matter.

However I prefer the term "mass" over "matter." The former has found universal acceptance in relativity where as the later finds little use.

Pete

 

[MeJennifer]

I don't know how you came to the conclusion that I said/implied this.

If two clocks are at different gravitational potentials then they will run at different rates as measured by a given observer. Redshift and slowing of clocks are identical phenomena, i.e. different names for exactly the same thing. I was stating that the spacetime need not be curved in order for this to occur. E.g. if you have a flat spacetime and then you choose a coordinate system corresponding to uniformly accelerating frame of reference then you will observe a gravitational redshift when the two clocks are located at two different gravitational potentials.

The relationship is given by dT = sqrt(g_00)dt where dT is the proper time on the clock and dt is the observer's time (i.e. time as measured by the observer who measures time with coordinate "t").

For a detailed calculation for the Schwarzschild gravitational field (e.g. that of earth) see

http://www.geocities.com/physics_world/gr/red_shift.htm

The reason being is that people never explain this fact and quite often walk away believing, incorrectly, that spacetime curvature is both a necessary and sufficient condition for the presence of gravitational redshift. This point was also made in the American Journal of Physics so it is not a trivial point. I even know an author of a well-known GR book who got this wrong at one time. I also make this point since Einstein also made it in at least two of his GR papers (1907 and 1911). In fact redshift in g-field was first redshift calculated in a flatspacetime (Einstein 1907).

Pete

I see what you are saying now!
Thanks!

 

[Jorrie]

I'm unclear on what you mean by this. Please clarify.

I'm sorry Pete, from my side I don't quite get your question. The part on "curvature not required for gravitational effects" (like redshift) is exactly equivalent to what you yourself stated before:

Redshift and slowing of clocks are identical phenomena, i.e. different names for exactly the same thing. I was stating that the spacetime need not be curved in order for this to occur.

As far as "gravity=acceleration" goes, I reckon an extended lab that is linearly accelerated with constant acceleration does not look (from the inside) exactly like a gravitational field. The acceleration will be the same everywhere inside the lab, which is not the case with gravitational acceleration, i.e. curved spacetime.

As far as my speculation on "profiling" the acceleration goes, I think I missed it! An extended accelerating lab can surely never replicate curved spacetime, no matter how you accelerate it – or is there a way?

I agree with you on using mass rather than matter.

Jorrie

 

[pmb_phy]

I'm sorry Pete, from my side I don't quite get your question.

In response to the comment "Curved space appears as gravity" you wrote "Wrong. Curved space appears as tidal gravity." In essence, what you've stated here is that "Curved spacetime = gravity." is a different statement than "curved spacetime = tidal gravity." The truth of this statement hinges on the definition of gravity as used in GR. According to Von-Laue "Gravity is identical to spacetime curvature" whereas according to Einstein "Gravity is identical to gravitational acceleration." (i.e. non-vanishing of affine connection).

As far as "gravity=acceleration" goes, I reckon an extended lab that is linearly accelerated with constant acceleration does not look (from the inside) exactly like a gravitational field. The acceleration will be the same everywhere inside the lab, which is not the case with gravitational acceleration, i.e. curved spacetime.

Gravitational acceleration is defined by the affine connection while while spacetime curvature is defined by the Riemann tensor. When you use the phrase "exactly like a gravitational field, i.e. curved spacetime" it implies that you're using the definition of gravity given by Von-Laue and not that given by Einstein. That's fine. However if you do so then the equality "Curved space appears as gravity" is valid.

Pete

 

[Jorrie]

Thanks for the explanation Pete. I rest my case.

 

[Chaos' lil bro Order]

I think you are right so far. My "engineering view" on your other statements are as follows:

Wrong. Curved space appears as tidal gravity. As very ably explained by pervect before, space need not be curved for gravitational effects to appear.

Wrong. When viewed at the same gravitational potential as the potential it was transmitted from, there is no redshift, only deflection and a Shapiro time delay of the light.

Only correct in a restricted sense - very localized, e.g. in a vanishingly small accelerating laboratory. That is unless the acceleration has a profile that precisely mimics curved space-time - maybe?

I would love to hear an expert's opinion

Sorry Jorie, I don't think you understand what you are talking about. You say space need not be curved for gravitational effects to appear. This seems very wrong to me, even in your oddly worded statement. Gravity IS the product of curved space, plain and simple. If you can point out a local region of space where a non-uniform gravitational field exists that is not somehow curved, please do so, otherwise you are wrong.

If you are arguing that gravity and acceleration cannot be thought of as equivalent, I suggest you read a book on Relativity, or if you have, reread it. A lab accelerating upwards will experience an effect that is equivalent to gravity, but your one correct point is that the graviational effect in the lab will be uniform and not dissipated by the inverse-square law like real gravity would. Cudos on 1/4.

 

[MeJennifer]

It can easily (hopefully) seen, that for an object "at rest" in the inertial coordinates, dxx=dyy=dzz=0, and thus dtau = dtt, and there is no time dilation. Coordinate time is the same as proper time.

Makes sense!

It can also be seen that for an object "at rest" in the accelerated coordinates, dx=dy=dz=0, and thus dtau = (1+gz)*dt, and there is gravitational time dilation. The relationship between coordinate time and proper time is a function of "height", the z-coordinate. This is time dilation, when coordinate time is not the same as proper time, and the amount of time dilation depends on the "height" of the object, i.e. it's z coordinate.

I suppose you mean the t-coordinate instead of the z-coordinate.

It should be understood that the object with xx=yy=zz=constant, which is "at rest" in inertial coordinates, is not following the same trajectory as the object with x=y=z=constant, which is "at rest" in the coordinate system used by the accelerated observer.

Right.

It should also be understood that a lot of the confusion arises from the use of concepts such as 'time dilation' that are coordinate dependent. When one chooses to use a coordinate dependent explanation (because of familiarity on the part of readers, usually), one has to live with the fact that it is coordinate dependent.

Well I agree.
Then why not stay with space-time coordinates? From here we can make any coordinate transformation.
Don't you think that people then will get a better understanding of the "magic" of contraction and "slower" clocks?

An explanation formulated purely in terms of coordinate independent quantites (like the Lorentz interval and abstract tensor notation) doesn't have to deal with these issues. But it is not generally as accessible.

Well what do you see as the problem with regards to using space-time coordinates?

 

[Jorrie]

Sorry Jorie, I don't think you understand what you are talking about. You say space need not be curved for gravitational effects to appear. This seems very wrong to me, even in your oddly worded statement. Gravity IS the product of curved space, plain and simple.

Maybe we are both just guilty of expressing ourselves badly - you probably meant "Gravity IS the product of curved spacetime…..", not just curved space
Now the interesting thing is that the weak equivalence principle alone is enough to create gravity – and it is without space curvature, just like in the small, uniformly accelerating lab. The weak equivalence principle manifests itself as gravitational redshift and also produces half of the gravitational bending of light. The other half is produced by space curvature. Tidal gravity, on the other hand, only appears in a curved spacetime environment.
It is from this point of view that I disagreed with the statements made in the OP.
Jorrie

 

[pmb_phy]

Sorry Jorie, I don't think you understand what you are talking about.

Jorrie seems well versed in GR in my opinion.

You say space need not be curved for gravitational effects to appear.

This is true depending on what one means by "gravitational effects." In Einstein's GR gravitational effects consist in part of gravitational acceleration, gravitational redshift. There is no requirement for spacetime to be curved for these effects to exist.

Gravity IS the product of curved space, plain and simple.

You've arrived at this conclusion by defining gravitational effects to be identical with tidal acceleration. So you haven't demonstrated anything other than you are using a definition which differs from Einstein.

If you can point out a local region of space where a non-uniform gravitational field exists that is not somehow curved, please do so, otherwise you are wrong.

This is nonsensical since Jorrie already clarified that tidal acceleration and spacetime curvature are identical, i.e. different names for the same phennomena. Thus there is no spacetime curvature in a uniform g-field because a uniform g-field is defined by requiring that the Riemann tensor vanishes in such a field.

I suggest you read a book on Relativity, or if you have, reread it.

Sorry but it appears quite clear that Jorrie knows what he's talking about. I recommend that you pick up the works of Einstein on GR and read them carefully and focus your attention on how Einstein defines gravity in GR.

A lab accelerating upwards will experience an effect that is equivalent to gravity, but your one correct point is that the graviational effect in the lab will be uniform and not dissipated by the inverse-square law like real gravity would. Cudos on 1/4.

The term "real gravity " is based on a biased view on gravity, i.e. that gravity is identified by spacetime curvature. In this view a uniform gravitational field is a contradiction in terms and it makes no sense to speak of gravity, never mind referring to a uniform field.

Pete

 

[pervect]

I suppose you mean the t-coordinate instead of the z-coordinate.

dtau = (1+gz)*dt

So dt/dtau = 1/1+gz

This is a function of z, i.e. a function of height. g here is a multiplicative constant for z, the acceleration of the spaceship, in case that isn't clear.

As far as space-time coordinates goes, there are many possible choices for coordinate systems. The fact that SR is Lorentz covariant says that we will get the same answer, in the end, regardless of which one of them we choose.

However, the path to the unique answer may not always appear to be the same on a popular level - i.e. we wind up talking about "gravitational time dilation" in some coordinates, and not in others. This is unavoidable if we use the coordinate approach - only a highly abstract approach, using geoemtric entities rather than coordinates, can avoid this.

 

[MeJennifer]

dtau = (1+gz)*dt

So dt/dtau = 1/1+gz

This is a function of z, i.e. a function of height. g here is a multiplicative constant for z, the acceleration of the spaceship, in case that isn't clear.

I see.

However, the path to the unique answer may not always appear to be the same on a popular level - i.e. we wind up talking about "gravitational time dilation" in some coordinates, and not in others. This is unavoidable if we use the coordinate approach - only a highly abstract approach, using geoemtric entities rather than coordinates, can avoid this.

Well it seems to me that transforming the local coordinates to space-time will give resolve these "abiguities". Comparing coordinate time with proper time will be an indication if there is any time "dilation". The terms "time dilation" and "length contraction" are prone to confusion IMHO, because they only have a meaning when you compare coordinate systems.