Do Gravitating Bodies Warp the Fabric of Space?

[jaketodd]

Also, the graphic shows the object taking a path shorter than the curved, proper time dimension. This is incorrect because an object would take longer than the proper time since its motion causes time dilation. So its path should be longer than the proper time curve between the two end points of the object's path.

 

[jaketodd]

Where did I say "force" ? "Gravitation" refers to the general phenomena, not a specific model.

No. In GR you don't need a force to make things advance in spacetime. All objects advance in spacetime by default.

No. In GR you don't need a force to advance locally straight in spacetime. It is the default behavior of force free objects.

Yes, in GR within inertial frames, free falling objects are force free.

So objects fall according to what? The curvature of spacetime? What pulls them into a dimple? I guess you could say objects move by default to less dense areas of spacetime. But that brings up the question of what is the difference between tendency and force?

 

[A.T.]

Also, the graphic shows the object taking a path shorter than the curved, proper time dimension.

No idea what you compare here, A dimension doesn't "take a path of a finite length",

This is incorrect because an object would take longer than the proper time since its motion causes time dilation. So its path should be longer than the proper time curve between the two end points of the object's path.

Time dilation in this diagram means the object advances less along the proper time dimension. This might help you to understand the diagram better:
http://www.adamtoons.de/physics/relativity.swf
Set: intial speed : 0, gravity : 1.0 to simulate a free fall from rest towards a mass somewhere to the right.

So objects fall according to what? The curvature of spacetime?

Yes

What pulls them into a dimple?

You could just as well ask "What bends a cricle?". It is simply a geometrical consequence of the mathematical model (geodesics on curved manifolds).

I guess you could say objects move by default to less dense areas of spacetime.

You can say a lot of things. But the things GR says also fit the observation quite well.

 

[jaketodd]

I'm done debating this. We're going in circles or at least this thread as a whole is. I wish one of the people who have been recognized by the forum to be an authority would set the record straight.

 

[Dale]

I guess it works as a 2D analogy of a closed universe, but it doesn't help jaketodd, since inhabitants on the balloon surface cannot experimentally determine the direction of the curvature (positive if on the outside, negative if on the inside, but this is impossible for the 2-dimensional inhabitants to determine).

Actually, the curvature of the balloon surface is positive regardless of which "side" of the surface you are talking about. In fact, the concept of "inside" or "outside" the surface is only valid in the 3D embedding space and is meaningless within the 2D surface itself.

The curvature is an intrinsic property defined entirely within the surface itself. Experimenters within the surface can determine the positive curvature of a balloon by measuring the sum of the interior angles of a triangle and determining that it is greater than 180º. A 2D surface with negative curvature would be saddle-shaped in 3D, and again the "side" would have no bearing on the curvature. The sum of the interior angles of a triangle would be less than 180º.

 

[Buckethead]

If you think about space time as a balloon where the stretchiness of the balloon at a spot on its surface is determined by its mass/energy density, then the surface of the balloon will be dimpled. The rate of time and the spatial dimensions are all determined by the radius of the dimple. Motion across the surface of the balloon means that you will be moving through dimples in space time as well as causing a dimple to propagate over the surface.

This is a great analogy. I knew about the radius of the balloon represented time and the surface a 3D space, but when you consider that massive objects are actually living in the past since their clocks are slowed relative to empty space, they will of course dimple the surface partially into the past. Great visual!

 

[Passionflower]

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

I understand this is from a book by L Epstein. It clearly was never professionally reviewed as the concept of curved proper time is sheer nonsense.

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling. Curvature of spacetime simply causes objects to accelerate with respect to each other without any need for proper acceleration.

 

[A.T.]

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif

I understand this is from a book by L Epstein. It clearly was never professionally reviewed

Rickard Jonsson has derived the math of Epstein embeddings:
http://www.relativitet.se/Webtheses/lic.pdf (Chapter 6, page 53)

as the concept of curved proper time is sheer nonsense.

A single dimension cannot be intrinsically curved alone, that's why "curved time" is in quotes. But intrinsic curvature is not even necessary to have a "gravitational pull". The "gravitational pull" is related to the 1st derivates of the metric, while tidal forces are related to 2nd derivates (curvature). And the spacetime in the picture in fact doesn't have any intrinsic curvature. I agree that "curved time" is not a good title for the illustration.

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling.

So you think, greater proper acceleration means slower clock rate? Some counter-examples:

- A clock resting in the Earths center experiences less proper acceleration but runs slower than a clock resting on the surface.

- You can have a two clocks resting (relative to each other) in an uniform gravitational field experiencing the same proper acceleration, but running at different rates.

Curvature of spacetime simply causes objects to accelerate with respect to each other without any need for proper acceleration.

Yes.

 

[Passionflower]

- A clock resting in the Earths center experiences less proper acceleration but runs slower than a clock resting on the surface.

- You can have a two clocks resting (relative to each other) in an uniform gravitational field experiencing the same proper acceleration, but running at different rates.

A clock in the center of the Earth accelerates away from all directions because all the mass surrounding it tries to attract it. Proper acceleration does not necessarily result in relative movement.

A uniform gravitational field is a red herring, as the question often becomes how uniform is a uniform gravitational field really.

 

[A.T.]

A clock in the center of the Earth accelerates away from all directions because all the mass surrounding it tries to attract it.

That is just a very complicated way to say that the proper acceleration of a clock resting in the center is zero. This is less than the proper acceleration of a clock resting on the surface. Yet the center clock runs slower than the surface clock.

This is shows that your idea, that greater proper acceleration causes a slower clock rate, is false. Here your statement that I was objecting to:

The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling.

Do yon now understand, that it isn't the difference in proper acceleration that determines gravitational time dilation between two clocks?

 

[Passionflower]

An object can be both stationary and accelerating.

 

[Dale]

Not proper acceleration, that can only have one value, the one measured by an accelerometer. An accelerometer at the center of the Earth reads 0, and accelerometer at the surface of the Earth reads 9.8 m/s² upwards. A.T.'s counterexample is correct, gravitational time dilation is not due to differences in proper acceleration the way you suggest.

The uniform field is also a good counter example. Suppose you have an ideal gravitational field where the proper acceleration of a stationary particle is everywhere constant. In such a field a light pulse going "up" would be gravitationally red-shifted and therefore there would be gravitational time dilation despite the fact that the proper acceleration is constant.

 

[A.T.]

An object can be both stationary and accelerating.

I don't quite see how this addresses my counter example to your claim that greater proper acceleration causes a slower clock rate.

 

[starthaus]

That is just a very complicated way to say that the proper acceleration of a clock resting in the center is zero. This is less than the proper acceleration of a clock resting on the surface. Yet the center clock runs slower than the surface clock.

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_2=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field $$\Phi_1-\Phi_2$$.

Generalization:

At a distance $$r<R$$ from the center of the sphere, inside the sphere, the gravitational potential is:

$$\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})$$

The above gives:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1$$

For $$r=0$$ (clock2 at the center of the Earth) you recover the results from above.

For $$r=R$$ you get the expected:

$$\frac{d\tau_1}{d\tau_2}=1$$

 

[TCS]

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-2\frac{\Phi}{c^2})(cdt)^2+(1-2\frac{\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-2\frac{\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-2\frac{\Phi_1}{c^2}}{1-2\frac{\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_1=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-1/2\frac{GM}{rc^2}<1$$

So, $$f_1>f_2$$

In addition, the time dilation depends on the gradient of the gravitational field $$\Phi_1-\Phi_2$$, i.e., it depends on acceleration.

Shouldn't phi 1 be zero. I think that you are missing a little r in your calculation.

 

[starthaus]

Shouldn't phi 1 be zero. I think that you are missing a little r in your calculation.

No, the calculation is correct.

 

[TCS]

This link shows that gravitational time dilation is proportion to small g, which should be zero at the center of the earth.

https://www.physicsforums.com/library.php?do=view_item&itemid=166

 

[starthaus]

This link shows that gravitational time dilation is proportion to small g, which should be zero at the center of the earth.

https://www.physicsforums.com/library.php?do=view_item&itemid=166

The formulas on the link you cited are valid for outside the Earth. The potentials inside the Earth are different.You need to be careful with what expressions you plug in into your calculations. The calculations I showed are correct.

 

[TCS]

The formulas on the link you cited are valid for outside the Earth. The potentials inside the Earth are different.You need to be careful with what expressions you plug in into your calculations. The calculations I showed are correct.

That does make more sense to me because I had thought that time the contraction was determined by the energy density and those equations seemd to contradict my belief.

Is the difference related to change in potential energy of the clock?

 

[Dale]

Is the difference related to change in potential energy of the clock?

Yes. E.g. as a photon goes up it gains potential energy, loses kinetic energy, and therefore becomes redshifted. This indicates that time is slower lower in the potential well.

 

[Passionflower]

Not proper acceleration, that can only have one value, the one measured by an accelerometer. An accelerometer at the center of the Earth reads 0, and accelerometer at the surface of the Earth reads 9.8 m/s² upwards. A.T.'s counterexample is correct, gravitational time dilation is not due to differences in proper acceleration the way you suggest.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

An accelerometer at the center of the Earth reads 0

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

The uniform field is also a good counter example. Suppose you have an ideal gravitational field where the proper acceleration of a stationary particle is everywhere constant. In such a field a light pulse going "up" would be gravitationally red-shifted and therefore there would be gravitational time dilation despite the fact that the proper acceleration is constant.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Yes. E.g. as a photon goes up it gains potential energy, loses kinetic energy, and therefore becomes redshifted. This indicates that time is slower lower in the potential well.

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

 

[Dale]

Passionflower, you have a strong habit of making factually wrong statements and then when you are corrected by someone who actually knows what they are talking about making it worse by persisting in your error and trying to support it with completely irrelevant statements. You did it in the thread where you were confusing path length with distance and now in this thread.

Really? So when a ball explodes what single value is the proper acceleration of parts of the shell?
A clock is not a point mass. Are you suggesting there are no stresses on an object placed in the core of a massive object like the earth?

Irrelevant. Proper acceleration is defined along a worldline, stresses are not.

How do you know?
It is likely an accelerometer could not even operate properly at the center of the earth. By the way the fact that an accelerometer only measures an acceleration in one direction is not a limitation of nature but a limitation of the device.

Also irrelevant. As you yourself mention, measuring acceleration in one direction is a limitation of a specific type of accelerometer. Similarly, not operating properly at the center of the Earth is a limitation of a specific device, not a limitation of nature. In relativity when we talk about clocks, rods, or accelerometers we are always speaking of ideal devices.

You mean a gravitational field without tidal forces right?
Could you show me a physical configuration where we have such a "field"? Or is this good example in the same category as "when a chicken travels at the speed of light he cannot lay any eggs because time stands still"? Also please define "gravitational well" and "up" as well, not by using Newtonian gravity but by using GR.

Sure, no problem. A scalar gravitational potential can be defined in a static spacetime, such as the Schwarzschild solution. "Up" is the direction of the gradient of the potential, and the region around the minimum of a potential is described as a "well".

A emitted photon is absorbed redder or bluer not because something on its path affected it as you seem to suggest, it is simply because the clocks at the emission and reception run at a different rate.

And how would you make an experiment that would test this distinction? As far as I can tell it is two different ways of saying the same thing.

Again, all of this is in response to your factually incorrect statement that "The reason a clock attached to a ceiling runs faster than a clock on the floor is that the clock on the floor has a greater proper acceleration than the clock near the ceiling." A.T. provided two correct counter examples and none of your posts since then have addressed anything relevant, which is the exact same behavior as in the other thread.

 

[yuiop]

Start with the Schwarzschild solution in the weak field approximation:

$$(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...$$

For the case $$dr=d\theta=d\phi=0$$ you get the well known relationship:

$$d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt$$

Writing the above for two different gravitational potentials $$\Phi_1$$ and $$\Phi_2$$ you obtain the well-known time dilation relationship:

$$\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}$$

At the Earth surface :

$$\Phi_1=-\frac{GM}{R}$$

At the Earth center:

$$\Phi_2=-3/2\frac{GM}{R}$$

Now, due to the fact that $$\frac{\Phi}{c^2}<<1$$ you can obtain the approximation:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

In addition, the time dilation depends on the difference in the gravitational field $$\Phi_1-\Phi_2$$.

Generalization:

At a distance $$r<R$$ from the center of the sphere, inside the sphere, the gravitational potential is:

$$\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})$$

The above gives:

$$\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1$$

For $$r=0$$ (clock2 at the center of the Earth) you recover the results from above.

For $$r=R$$ you get the expected:

$$\frac{d\tau_1}{d\tau_2}=1$$

Here is a slightly better equation based on the interior Schwarzschild solution for $$dr=d\theta=d\phi=0$$ :

$$\frac{d\tau}{dt}=\frac{3}{2}\sqrt{1-\frac{2GM}{c^2R}}-\frac{1}{2}\sqrt{1-\frac{2GMr^2}{c^2R^3}}$$

where $$d\tau$$ is the proper clock rate of a clock located at r, $$dt$$ is the clock rate of a clock located very distant from the gravitational body and R is the the radius of the massive body with mass M, valid for $$r<=R$$. This equation has the slightly curious result that when $$R <(9/4)GM/c^2$$ the proper time of a clock at the centre runs backwards compared to the distant clock.

The equation assumes an even density distribution within the massive body and can be be expressed more generally for uneven density distributions (using units of G=c=1) as:

$$\frac{d\tau}{dt}=3/2\sqrt{1-2M/R}-1/2\sqrt{1-2 p (4/3)\pi r^2}$$

where p is the average density of the mass enclosed within a sphere of radius r where the clock is located. To compare the clock rates of two clocks located at a radius less than or equal to the radius of the gravitational body the following equation can be used:

$$\frac{d\tau_1}{d\tau_2}= \frac{3\sqrt{1-2M/R}-\sqrt{1-2 p_1 (4/3)\pi r_1^2}}{3\sqrt{1-2M/R}-\sqrt{1-2 p_2 (4/3)\pi r_2^2}}$$

For the case that one clock is located at the surface ($$dt_1 = dt_s$$) of the massive body ($$r_1 = R$$) and the other clock is located at the centre ($$dt_2= dt_c$$) of the massive body ($$r_2 = 0$$) the equation simplifies to:

$$\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}$$

which has a value of greater than unity for all $$2M<R<\infty$$

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

 

[starthaus]

$$\frac{d\tau_s}{d\tau_c}= \frac{2 \sqrt{1-2M/R}}{3\sqrt{1-2M/R}-1}$$

which has a value of greater than unity for all $$2M<R<\infty$$

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:$$\frac{d\tau_s}{d\tau_c}<1$$.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

 

[Passionflower]

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

That is certainly true.

However, in order to calculate such situations one has to start with some kind of perfect fluid solution as the Earth is not hollow. Geodesic paths, due to Ricci curvature, tend to get in in each others way in the center and create havoc to say the least.

 

[Dale]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

 

[yuiop]

You must have made an error somewhere. If you did your calculations correctly, you should have gotten:

$$\frac{d\tau_s}{d\tau_c}<1$$.

The clock at the center of the Earth (deeper in the gravitational well) should tick slower. You got the reverse.

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

$$\frac{d\tau_s}{d\tau_c}>1$$

So, $$f_1>f_2$$ where $$f_1$$ is the clock frequency on the Earth crust and $$f_2$$ is the frequency of the clock at the center of the Earth.

From this you should get:

$$\frac{f_1}{f_2} > 1$$

For example the clock rate ($$d\tau_s$$) of a clock on the surface of a body with a radius R=8M is 0.8666 seconds for every second that passes on a clock at infinity.

The clock rate ($$d\tau_c$$) of a clock at the centre of the same body is 0.7990 seconds for every second that passes on a clock at infinity (and slower than the surface clock).

The ratio is this example is:

$$\frac{d\tau_s}{d\tau_c} = \frac{0.8666}{0.7990} = 1.084 >1$$

Maybe there is a mistake in your calculations?

 

[Passionflower]

Dalespam, if the center of the Earth would not accelerate it would simply collapse into a black hole.

You just insist on making wrong statement after wrong statement. The interior Schwarzschild solution does not have a singularity nor an event horizon. Also the worldline defined by a constant r=0 is a geodesic in the interior Schwarzschild solution, meaning that it has no proper acceleration.

Really now, who is talking about particular solutions you or I?

Do you realize that the EM forces in all atoms inside the Earth's have to resist the tendency of gravity to collapse the Earth into a black hole. That resistance is called proper acceleration.

 

[starthaus]

The clock at the centre does tick slower than the clock at the surface and as a result the ratio should be:

$$\frac{d\tau_s}{d\tau_c}>1$$

The above signifies that the period of the clock at the surface is greater than the one at the ceneter of the Earth. This is incorrect.

From this you should get:

$$\frac{f_1}{f_2} > 1$$

Nope, you got this backwards:

$$\frac{d\tau_s}{d\tau_c}>1$$

implies:

$$\frac{f_1}{f_2} < 1$$

which is clearly wrong. Remember, clock frequency is the inverse of clock period.