Why can't time be constant for everything in the universe?

[elfmotat]

There would be no gravity inside the shell, but the gravitational potential is the important term, and that's not zero.

 

It can be, by definition. You're free to set it to any value you like.

[swansont]

It can be, by definition. You're free to set it to any value you like.

 

Not after you've set the potential at infinity to zero, which was implied in the setup of the problem.

[Markus Hanke]

In the case of the cavity within the shell, we simply have the Minkowski tensor , which represents a completely flat region of space-time.

 

Having thought about it further, it would appear I made a serious mistake here, so I have to retract the above statement.

We will indeed find a region of completely flat space-time, but not the Minkowski metric; instead, we will have a metric tensor the elements of which are all constants, but not equal to +/- 1. These constants will be some function of the shell mass and the cavity radius. Physically this means there is no gravitational field in the cavity, but a clock located there is still dilated compared to a clock located at rest infinitely far away. This tallies nicely with the Newtonian shell theorem, and in terms of classical mechanics, can be thought of as a non-vanishing gravitational potential inside the cavity.

 

It should be noted that there are no potentials in GR, it is the metric tensor which is the source term for gravitational time dilation.

[robomont]

i like getting you guys going but now i dont have a clue.

for knowledge to be shared and learned .it some times has to be dumbed down for us rednecks.so what shape and size would a mass be to increase longevity of one person compared to another.like double lifespan.

[elfmotat]

Not after you've set the potential at infinity to zero, which was implied in the setup of the problem.

 

I don't see how setting the potential to zero at infinity is required in the proof that the potential in a spherical mass distribution is constant.

 

 

 

Having thought about it further, it would appear I made a serious mistake here, so I have to retract the above statement.

We will indeed find a region of completely flat space-time, but not the Minkowski metric; instead, we will have a metric tensor the elements of which are all constants, but not equal to +/- 1. These constants will be some function of the shell mass and the cavity radius. Physically this means there is no gravitational field in the cavity, but a clock located there is still dilated compared to a clock located at rest infinitely far away. This tallies nicely with the Newtonian shell theorem, and in terms of classical mechanics, can be thought of as a non-vanishing gravitational potential inside the cavity.

 

It should be noted that there are no potentials in GR, it is the metric tensor which is the source term for gravitational time dilation.

 

I don't believe you've made an error. After all, if you consider the weak field metric in the case of a constant potential then you can trivially change your coordinate system to be Minkowskian. I.e. if, for example:

 

[math]ds^2=-(1+2 \phi)dt^2 + (1-2 \phi)d \mathbf{r}^2[/math]

 

then you can simply define the new coordinates:

 

[math]T=t\sqrt{1+2\phi}[/math]

 

[math]\mathbf{R}=\mathbf{r} \sqrt{1-2\phi}[/math]

 

so that:

 

[math]ds^2=-dT^2 + d \mathbf{R}^2[/math]

Edited May 28, 2013 by elfmotat

[swansont]

I don't see how setting the potential to zero at infinity is required in the proof that the potential in a spherical mass distribution is constant.

 

It's not. That wasn't the claim.

 

If you set the potential at infinity to zero, the potential inside of a mass shell will be negative, i.e. it will not also be zero inside the shell.

[Markus Hanke]

I don't believe you've made an error. After all, if you consider the weak field metric in the case of a constant potential then you can trivially change your coordinate system to be Minkowskian. I.e. if, for example:

 

[math]ds^2=-(1+2 \phi)dt^2 + (1-2 \phi)d \mathbf{r}^2[/math]

 

then you can simply define the new coordinates:

 

[math]T=t\sqrt{1+2\phi}[/math]

 

[math]\mathbf{R}=\mathbf{r} \sqrt{1-2\phi}[/math]

 

so that:

 

[math]ds^2=-dT^2 + d \mathbf{R}^2[/math]

 

That's true, but you need to consider the boundary conditions of this problem as well. The metric inside the cavity needs to connect smoothly to the interior metric of the shell itself to ensure global differentiability, which then again needs to connect smoothly to the exterior Schwarzschild metric. It would be interesting to do the numbers here, but just by looking at it I don't think it will work if the metric in the cavity is Minkowskian. If that is the case we will get either a discontinuity at the boundary, or a hypersurface where the metric is smooth but not differentiable, both of which is unacceptable. I don't believe it is possible for the elements of the metric tensor inside the shell itself to be exactly +/-1 at any point including the boundary, since this would imply a Minkowskian vacuum with vanishing energy-momentum tensor; so, in order to maintain smoothness and differentiability at the boundary, the vacuum metric in the interior of the cavity cannot be +/-1 anywhere either, or else we have a boundary problem.

But then again, that's just my two cents' worth from the point of view of differentiable manifolds, I might well be wrong.

 

Has anyone got any references to a fully worked calculation for just such a case ? I couldn't find anything.

 

I think here's the solution : locally the metric inside the cavity is Minkowskian; however, if we want a global coordinate system which asymptotically approaches Minkowski at infinity, then the cavity will be flat, but not Minkowskian. This takes care of the time dilation issue, and tallies nicely with Newton.

Edited May 28, 2013 by Markus Hanke

[elfmotat]

It's not. That wasn't the claim.

 

If you set the potential at infinity to zero, the potential inside of a mass shell will be negative, i.e. it will not also be zero inside the shell.

 

I argued that you can set the potential to any value you like, via a redefinition, without changing any of the physics. Unless I'm interpreting you wrong (which might be what's going on), you said that the convention of setting the potential to zero at infinity is somehow implied when determining the potential inside a spherical mass shell. That's simply not true.

 

 

 

That's true, but you need to consider the boundary conditions of this problem as well. The metric inside the cavity needs to connect smoothly to the interior metric of the shell itself to ensure global differentiability, which then again needs to connect smoothly to the exterior Schwarzschild metric. It would be interesting to do the numbers here, but just by looking at it I don't think it will work if the metric in the cavity is Minkowskian. If that is the case we will get either a discontinuity at the boundary, or a hypersurface where the metric is smooth but not differentiable, both of which is unacceptable. I don't believe it is possible for the elements of the metric tensor inside the shell itself to be exactly +/-1 at any point including the boundary, since this would imply a Minkowskian vacuum with vanishing energy-momentum tensor; so, in order to maintain smoothness and differentiability at the boundary, the vacuum metric in the interior of the cavity cannot be +/-1 anywhere either, or else we have a boundary problem.

But then again, that's just my two cents' worth from the point of view of differentiable manifolds, I might well be wrong.

 

Has anyone got any references to a fully worked calculation for just such a case ? I couldn't find anything.

 

I think here's the solution : locally the metric inside the cavity is Minkowskian; however, if we want a global coordinate system which asymptotically approaches Minkowski at infinity, then the cavity will be flat, but not Minkowskian. This takes care of the time dilation issue, and tallies nicely with Newton.

 

I think you're taking coordinates themselves too seriously. The metric is not an observable, and there's a tremendous amount of gauge freedom that comes along with it. You can shift to a coordinate system that is Minkowskian inside the sphere without changing any of the physics.

 

I don't know whether or not an exact solution for this situation exists, but you can get a good feel for it by using the weak field approximation. We'll use the convention that [math]\phi (r= \infty)=0[/math]. If we have a shell of density [math]\rho = \frac{3M}{4 \pi (b^3-a^3)}[/math] with its inner wall at the coordinate r=a and outer wall at r=b, then:

 

[math]\phi ® = \begin{cases} k= -2 \pi G \rho (b^2-a^2) & \text{ if } r < a \\ -GM/r & \text{ if } r> b \end{cases}[/math]

 

(When a<r<b the potential will be some linear function of r which keeps the potential function smooth, but I can't be bothered to work it out right now.) Inside the shell the metric is given by:

 

[math]ds^2=-(1+2k)dt^2+(1-2k)dr^2+d \Omega^2[/math]

 

Outside the shell:

 

[math]ds^2=-\left (1-\frac{2GM}{r} \right )dt^2 + \left (1+\frac{2GM}{r} \right )dr^2+d \Omega^2[/math]

 

Now it's obvious from this that a person inside of the shell will age more slowly than a person outside of the shell. However, this does NOT mean that the spacetime inside of the shell isn't Minkowskian - it most certainly is. There are two ways of showing this: you can redefine the potential so that [math]\phi (r=0)=0[/math], which will make the potential outside of the shell positive everywhere (and asymptotically approaching [math]-k[/math] as you go to infinity). No physics has been changed - observers outside of the shell still age faster than those inside the shell, but now the metric is explicitly Minkowskian inside.

 

The second way of making the inner metric explicitly Minkowskian is what I alluded to in my previous post: we change to a new convenient coordinate system. We define the coordinates:

 

[math]T=t \sqrt{1+2k}[/math]

 

[math]R=r \sqrt{1-2k}[/math]

 

Inside the shell the metric is clearly explicitly Minkowkian, and outside of the shell we have:

 

[math]ds^2=-\left (\frac{1-2GM/r}{1+2k} \right )dT^2 +...[/math]

 

The 00-component of the metric still clearly has a greater magnitude outside of the shell, so again we find that observers inside the shell age more slowly. In general if the metric is proportional to [math]\eta_{\mu \nu}[/math] then you can trivially find a coordinate system where it is exactly the Minkowski metric. I.e. the physics of the two are equivalent. Coordinate transformations don't change any of the physics, but they may make certain things more or less apparent.

[swansont]

elfmotat, on 28 May 2013 - 17:16, said:

I argued that you can set the potential to any value you like, via a redefinition, without changing any of the physics. Unless I'm interpreting you wrong (which might be what's going on), you said that the convention of setting the potential to zero at infinity is somehow implied when determining the potential inside a spherical mass shell. That's simply not true.

No, that's not what I meant. To me, the statement of the problem here in this thread implied the commonly used convention of the potential being zero at infinity. So sure, you can set the potential inside the sphere to zero (or any arbitrary value), and then the potential at infinity will be some positive value (or larger value). Regardless of that choice, the potential inside the sphere is lower than it is outside, and a clock inside the sphere will run slow even though g=0.

[elfmotat]

No, that's not what I meant. To me, the statement of the problem here in this thread implied the commonly used convention of the potential being zero at infinity. So sure, you can set the potential inside the sphere to zero (or any arbitrary value), and then the potential at infinity will be some positive value (or larger value). Regardless of that choice, the potential inside the sphere is lower than it is outside, and a clock inside the sphere will run slow even though g=0.

 

Okay, I gotcha. I think we're in agreement.

[robomont]

and the redneck still has no clue.i thought this was a forum to learn stuff?

i think the formula is w+t+ f =¿ ^0 ,but my math may be off .

[elfmotat]

and the redneck still has no clue.i thought this was a forum to learn stuff?

i think the formula is w+t+ f =¿ ^0 ,but my math may be off .

 

As you suspected, surrounding yourself with a bunch of mass will make you age more slowly relative to someone far away from that mass. If you refer to my above post (#33), you'll see that the relevant number is:

 

[math]k=-\frac{3}{2} GM \left ( \frac{b^2-a^2}{b^3-a^3} \right )[/math]

 

Lets say, for example, that you surround yourself will a shell of mass that is 70,000,000 meters thick (roughly the radius of Jupiter), with a small hole in the middle for you to live in. The amount of mass in the shell required to roughly double your life span with respect to someone on Earth is approximately 16,000 times the mass of the Sun. This amount of mass compressed into a space the size of Jupiter (by far the largest planet in our solar system) is well over enough to ensure that it collapses into a black hole. So surrounding yourself with a bunch of matter is not a very likely or efficient way to travel to the future.

[robomont]

so if i have a space ship travelling at a certain speed then it has a certain kenetic energy or mass.if im travelling fast enough then my spacecraft turns into a blackhole?

[elfmotat]

so if i have a space ship travelling at a certain speed then it has a certain kenetic energy or mass.if im travelling fast enough then my spacecraft turns into a blackhole?

 

This is a different question. The short answer is no. The obvious reason is that, from your rest frame, you have zero kinetic energy and events themselves cannot differ between frames.

[robomont]

i may be at rest inside the craft but my craft isnt at rest.to all the other atoms outside the craft my craft is just another atom.please explain?

[Markus Hanke]

I'm in agreement with you, elfmotat. That is pretty much what I was saying anyway; it boils down to the fact that there is a difference in metric tensors between the observer at infinity and the observer inside the cavity, even though the actual values are not fixed and simply are a result of the boundary conditions imposed. Hence the time dilation.

 

So I think we are all good

[zapatos]

I've studied the above posts, as best I can. Yet they don't dispel this feeling: That right now - at this exact moment - an event is happening on Earth, and an event is happening in the M.31 Andromeda Galaxy, and the two events are simultaneous.

I think that this feeling is correct. There are two things happening right now, at the same time, here and in the Andromeda Galaxy.

It seems to me that the problem is determining which two events are happening simultaneously.

[elfmotat]

I think that this feeling is correct. There are two things happening right now, at the same time, here and in the Andromeda Galaxy.

It seems to me that the problem is determining which two events are happening simultaneously.

 

Global nonlocal simultaneity has literally no meaning at all. Though two events may feel simultaneous (or even be exactly in your rest frame), there's no true absolute simultaneity - though they are approximately simultaneous for anyone moving at everyday speeds relative to you, so much so that you don't notice. This is a clear example of when intuition completely breaks down at large distances and speeds.

[zapatos]

Would you mind explaining this to me a bit more, or point me to some references? I'm unsure what it means when you say that things are only 'approximately' simultaneous. It makes sense to me that you may not be able to point to two events and know that they are happening simultaneously, but it seems that given a fixed duration of time with multiple events happening during the entire duration, some things must be happening simultaneously.

[Delta1212]

 

Would you mind explaining this to me a bit more, or point me to some references? I'm unsure what it means when you say that things are only 'approximately' simultaneous. It makes sense to me that you may not be able to point to two events and know that they are happening simultaneously, but it seems that given a fixed duration of time with multiple events happening during the entire duration, some things must be happening simultaneously.

Simultaneous according to whom? Individuals with different relative velocities will disagree on which events are simultaneous, and since there is no preferred frame of reference, they are all equally correct.

[zapatos]

Simultaneous according to whom? Individuals with different relative velocities will disagree on which events are simultaneous, and since there is no preferred frame of reference, they are all equally correct.

As I conceded previously, I understand that we may not be able to know that two events are happening simultaneously. But that sounds to me like a problem stemming from our perception of the universe. Just because you and I cannot agree that two events are happening simultaneously does not mean that no two events ever happen simultaneously.

[elfmotat]

Would you mind explaining this to me a bit more, or point me to some references? I'm unsure what it means when you say that things are only 'approximately' simultaneous. It makes sense to me that you may not be able to point to two events and know that they are happening simultaneously, but it seems that given a fixed duration of time with multiple events happening during the entire duration, some things must be happening simultaneously.

 

"Approximately" meaning virtually undetectable given our locational/gravitational/velocity differences. Sure some things may be happening simultaneously in your approximate inertial frame (I say approximate because of course any frame at rest on Earth's surface is by definition not inertial). That doesn't mean that other observers will agree with you, and in general they won't. There's no "correct" frame, as is the essence of relativity.

 

As I conceded previously, I understand that we may not be able to know that two events are happening simultaneously. But that sounds to me like a problem stemming from our perception of the universe. Just because you and I cannot agree that two events are happening simultaneously does not mean that no two events ever happen simultaneously.

 

It's not our "perception" of the universe (whatever that means), so much as the way the universe works. As I said above, there's no "correct" frame in SR. In fact, in GR the notion of a correct frame becomes even more meaningless. Two events separated by vast distances cannot be meaningfully compared in terms of "when" they happened without choosing a particular coordinate system to work in, which usually entails considering the Earth to be at rest w.r.t. whatever we're comparing it to (which obviously can't be true in general), and then comparing proper times.

Edited May 30, 2013 by elfmotat

[swansont]

As I conceded previously, I understand that we may not be able to know that two events are happening simultaneously. But that sounds to me like a problem stemming from our perception of the universe. Just because you and I cannot agree that two events are happening simultaneously does not mean that no two events ever happen simultaneously.

 

With the exception of co-located events, yes, it does. Simultaneous is defined as two events happening at the same time, and we have a protocol for synchronizing clocks so that we know what "at the same time" means. But that can only occur if you are in the same reference frame. You can't synchronize clocks that are in different frames.

 

Put another way, two frames will not agree on simultaneity, and there is no preferred reference frame.

[Delta1212]

Let's flip it around. You can't state that two things took place at the same location if they happened at different times, except in a specific frame. Different frames will disagree on the location just as they disagree on the timing.

 

An asteroid passing by the Earth sees itself at rest and the Earth as moving past it. So you might go to bed one night and wake up the next morning in the same bed and say that you fell asleep and woke up in the same location, but at different times. The asteroid, meanwhile, would say that you woke up tens of thousands of kilometers away from where you fell asleep because that's how far the Earth moved in that time.

 

Events that do not take place at the same time in the same place cannot be said to take place at either the same time or the same place, because the distance and time that separate the events will vary with the velocity of the observer.

[zapatos]

So, events don't happen simultaneously because there is no way to measure two events as being simultaneous? And events cannot be measured as being simultaneous simply because different frames measure events differently?

Am I correct in stating that the concept of simultaneity is essentially meaningless given the universe we live in?

 

It seems everyone can agree to a duration that is defined by the Big Bang at one end and 'now' at the other. Since the universe is less than 14 billion years old, if there existed two stars that lasted a total of 7 billion years each, can't we say that for at least part of their lives they were 'burning' simultaneously? Were they burning simultaneously, even if we cannot agree on exactly when they were burning simultaneously? When I look at an example like this it seems like we simply cannot agree on simultaneity, not that it doesn't exist.

 

I suspect I am simply looking at what you mean by simultaneous in a non relativistic manner, and perhaps that is what is hindering me from fully understanding.