Layman's doubts about Gen Relativity

[Altabeh]

Since you are unwilling to clarify your position then I will make my best guess as to what you mean. Feel free to correct my guess of your intention since I am a notoriously poor mind-reader and have already asked you to do so 3 times now. Considering the abysmal quality of your writing in the future you should be more helpful in providing clarification when asked.

You believe that in general an object is inertial iff
\frac{d^2 x^a}{ds^2}=0

This is can be shown to be false simply by considering an inertial object in inertial spherical coordinates where even though the object is inertial

There is no specification of any coordinate system here and of course having countlessly been said in this thread before, this is only correct in a very small interval of spacetime along the geodesic. But okay though you are now trying to battle with thousands of Physicists, which later I'll enucleate the precise names with their own approaches to the problem here, I'm enthusiasticly waiting for a proof of why such equation is false "by considering an inertial object in inertial spherical coordinates".

AB

 

[Altabeh]

The correct equation would be that in general an object is inertial iff
\nabla_{\tau}U^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}=0

It is so interesting that the fundamental approach to deriving this formula uses the famous "locally inertial" coordinates. Besides, any particle following geodesics in GR would only be seen as moving inertially in an infinitesimaly small region (yet I don't want to specify the coordinates here by considering all observers except a non-Fermi observer).

AB

 

[Dale]

this is only correct in a very small interval of spacetime along the geodesic

Which is why it is not relevant to the OP's question.

 

[Altabeh]

Which is why it is not relevant to the OP's question.

Why are you insisting on such nonsense? You can't even prove my assertation wrong even at one point by considering an inertial object in inertial spherical coordinates because you made a fallacy in your previous post.

AB

 

[yuiop]

That is not an equation. So again, I ask: in general what equation determines if a worldline is inertial? The answer should be something like "in general a worldline is inertial if and only if A=B".

We have shown in other threads that we can generally (in the Schwarzschild metric anyway) calculate the predicted gravitational acceleration (a_p) of any particle according to any observer. For example the predicted radial acceleration of a particle according to a stationary (non inertial) local observer was given as:

a_p = - \frac{M}{r^2}\left(\frac{1-(dr '/dt')^2}{\sqrt{1-2M/r}}\right)

(See https://www.physicsforums.com/showpost.php?p=2747788&postcount=345)

If we consider proper acceleration as accounting for the difference between the predicted acceleleration (a_p) and the actual acceleration measured using the rulers and clocks of a local observer (a_m), then we can say that a particle has inertial motion if and only if a_p - a_m = 0 according to any given observer.

The above linked post only considers purely radial motion, but I am sure the general idea can be extended. I am also sure that this is not a complete answer, but it might suggest a way forward in the discussion.

 

[superkan619]

unambiguously "yes, it is inertial in all reference frames".

Any objections to this?

 

[yuiop]

Any objections to this?

I would go along with it. A particle is inertial if it proper acceleration is zero (as measured by an accelerometer attached to it). As far as I can tell it is not possible to transform away this physical fact.

 

[Dale]

I'm enthusiasticly waiting for a proof of why such equation is false "by considering an inertial object in inertial spherical coordinates".

You can't even prove my assertation wrong even at one point by considering an inertial object in inertial spherical coordinates because you made a fallacy in your previous post.

Without loss of generality consider the inertial worldline in spherical coordinates:

x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)

Using your equation:

\frac{d^2 x^a}{ds^2}=\left(0, \frac{\gamma^2 r_0^2 v^2}{(t^2 v^2+r_0^2)^{3/2}},0, \frac{2 \gamma^2 r_0 t v^3}{(t^2 v^2+r_0^2)^2} \right) \neq 0

Further consider the non-inertial worldline
x=(t,r,\theta,\phi)=(t,r_0,\frac{\pi}{2},\omega t)

Using your equation:
\frac{d^2 x^a}{ds^2}=0

So your equation does not hold even for a local observer using spherical coordinates. It is not even valid locally in flat spacetime, and certainly not in general.

 

[yuiop]

A gravitational field does NOT change the frequency of a photon. It remains the same as it was when it was emitted, as seen by any fixed observer.

However, time runs at slightly different rates at different potentials. The fractional difference in the time rate for an object at two different positions is effectively the same as the difference in potential energy divided by the total energy.

This means that if you compare observations of the same photon from two locations at different potentials, its frequency appears to differ because of the difference in clock rates.

You have to be careful about how the above is understood.

Let us say that some physical process such as the transition of an electron gives off a photon emmision of a characteristic frequency (f). To an observer that is local and at rest with the emmision event, the characteristic transition frequency will always be f. If the emmision occurs lower down than the observer, then the observer measures the frequency of the photon to be lower than f, at the reception event. Now if we interpret this physically as the frequency of the photon "really" remaining constant, but only being measured to be lower because of the difference between clock rates at the emission location and the reception location, then we have to understand that the same logic means the speed of light is "really" slower lower down. It then comes down to how we define "real". If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

 

[yuiop]

...
3)When looked upon from Earth does a falling stone makes non-inertial frame of reference?

It depends upon how your question is interpreted. If by "When looked upon from Earth.." you meant the pointof view of an observer standing on the surface of the Earth watching the falling stone, then the answer is no, because the observer standing on the surface of the Earth has non zero proper acceleration as is therefore not an inertial observer. If you meant the point of view of an observer co-free-falling with the stone, then the co-free-falling observer is "locally" in an inertial reference frame. As Altabeh suggested, the local vicinity of the free falling observer is Minkowskian, but further away from the free falling observer, it becomes increasingly obvious that the observer is not in flat space. As for the stone itself, a very small free falling stone is on average, an inertially moving object (rather than an inertial frame of reference), but as Passionflower has already pointed out, parts of a very large stone furthest away from the centre of mass of the stone, are not inertially moving.

 

[Jonathan Scott]

If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

No, I don't buy that at all.

If you think that it could ever make sense to say that a wave is slowing in frequency as it rises, where do the extra cycles go? For example, replace the photon's path with a rod rotating about its axis. If the whole rod did not rotate at the same frequency, it would gradually get more and more twisted. Observers at different potentials may consider that the whole rod is rotating at slightly different speeds, depending on their potential, but it doesn't make sense to suggest that the lower end is rotating at a different frequency from the upper end.

 

[yuiop]

No, I don't buy that at all.

If you think that it could ever make sense to say that a wave is slowing in frequency as it rises, where do the extra cycles go? For example, replace the photon's path with a rod rotating about its axis. If the whole rod did not rotate at the same frequency, it would gradually get more and more twisted. Observers at different potentials may consider that the whole rod is rotating at slightly different speeds, depending on their potential, but it doesn't make sense to suggest that the lower end is rotating at a different frequency from the upper end.

Modern relativity is not about what makes sense. If you continue down that road, you end up in the LET camp It is about what you measure. That is the reality. If you can't measure it, it is not real. Logic, rationality and common sense does not come into it.

Now let us embelish your example. You did not specify which axis of the rod to rotate about, but I assume you mean one of the short axes. Let us say we have a huge ferris wheel and your rod is represented by two opposite spokes. At the top and bottom of the wheel are observers with clocks and rulers calibrated in the normal way. The obersers at the top of the wheel measure the instantaneous tangential velocity at the top of the wheel to be 0.2c and the observers at the base of the wheel measure the instantaneous tangential velocity of the bottom of the same wheel to be 0.4c. The observers at the bottom are wondering how this can be? How can the bottom of the wheel be rotating slower than the top of the wheel without the wheel breaking apart? The observers at the top realize what is happening and shout down "Hey you idiots down there! Of course the bottom appears to be moving faster according to your measurements, because your clocks are running slower than our clocks." So the top and bottom observers decise to use a different method to synchronise the clocks and use a clock at the top as a master clock and synchronise all clocks so that they run at the same rate. Signals sent from the top at a frequency of once per second are now measured as arriving with a frequency of once per second by the lower observers with their alternatively synchronised clocks. The same is true the other way. Signals sent once per second from the bottom now arrive once per second at the top. WIth this new method of synchronisation the tangential velocity at the top is now the same as the tangential velocity at the bottom. This seems to explain why the wheel does not fall apart. Is this a better description of reality? Some would say no. If the observers at the bottom now try to measure the speed of light with the newly synchronised clocks, they will find the speed of light is locally less than c. Others would say that the speed of light "really" is slower, lower down in a gravitational field, clocks "really" run slower and all physical processes "really" happen slower, lower down. Indeed the slower speed of light can even be measured by using the clock synchronisation method I have just described.

So we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.

 

[Jonathan Scott]

...
So we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.

I was actually talking about something like a thin rod or rigid cable connecting a point of low potential and a higher point and rotating about a line along its core, although in many ways this has similar results, in that the frequency at the top and bottom cannot actually differ as observed by anyone observer.

In a gravitational field, the speed of light as observed from another location does indeed vary with potential. The way in which it varies depends on the coordinate system (and may be different in different directions), because there is no way to define a flat space coordinate system which both matches local time and distant space. This is like comparing distances on maps. If I have a map of the Earth centered on my own location, it indicates distances to scale and angles correctly locally, but as the Earth is curved, it cannot do so accurately at points further away.

 

[superkan619]

I would go along with it. A particle is inertial if it proper acceleration is zero (as measured by an accelerometer attached to it). As far as I can tell it is not possible to transform away this physical fact.

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

 

[Austin0]

Modern relativity is not about what makes sense. If you continue down that road, you end up in the LET camp It is about what you measure. That is the reality. If you can't measure it, it is not real. Logic, rationality and common sense does not come into it. .

Hi kev

Couldn't it be said that SR came into being as a rational attempt to make sense of the illogical invariance of light as indicated by Maxwell and MM ??

In doing this the system itself is rational and logically consistent even if some of its implications were counter-intuitive and may actually be inconsistent.

GR was derived through logic and conceptualization, not through mathematical induction from measured reality. That came later.

To take the position that only what is measured is real, and only what can be mathematically described from those measurements is reality , I think you either deny an objective reality [a metaphysical view] or you accept a false view of reality because we can't really believe that description has any real correspondance to that reality.

How can it be a positive position to think as long as the math works out, no further logical thought is required. Is in fact discouraged. ?

[QUOTESo we have two choices of how we describe reality:

1) Buy that the frequency of a photon "really" is constant as it rises upwards, at the cost of acknowledging that the speed of light is not "really" constant in a gravitational field.

2) Buy that the frequency of a photon changes as it rises, at the cost of claiming clocks do not "really" run slower lower down and acknowledging that certain things do not "make sense."

You seem to have bought into the first choice.[/QUOTE]

1) WHy should we think that light is "really" constant in a gravitational field?

To be consistent with the 2nd P??

2) If you accept clocks "don't really run slower" at a lower potential this has other implications .eg in the "real" results of the twins journey.

Making sense does not only apply to our physical conceptions and assumptions but also to consistency and logic between those assumptions, doesn't it?

BTW I wouldn't say I have bought into the first choice but so far it seems more consistent with the other applications of time dilation.

 

[Austin0]

You have to be careful about how the above is understood.

Let us say that some physical process such as the transition of an electron gives off a photon emmision of a characteristic frequency (f). To an observer that is local and at rest with the emmision event, the characteristic transition frequency will always be f. If the emmision occurs lower down than the observer, then the observer measures the frequency of the photon to be lower than f, at the reception event. Now if we interpret this physically as the frequency of the photon "really" remaining constant, but only being measured to be lower because of the difference between clock rates at the emission location and the reception location, then we have to understand that the same logic means the speed of light is "really" slower lower down. It then comes down to how we define "real". If we define "real" as being defined by what we actually measure (which the way real is usually defined in SR and GR) then the frequency of a photon "really" gets slower as it climbs out of a gravitational well. If you define real as what an intelligent being would conclude is happening "behind the scenes" to explain what is measured, then the frequency of the photon is "really" constant as it rises. The modern interpretation of Relativity does not consider what is happening "behind the scenes", only what is measured.

Are you saying that the relative time dilation at different potentials is not measured?
Is not real?

What kind of measurements are possible while in transit that would make this interpretation "real ".?

Isn't the measured dilation differential by your definition "real" and the interpretation of slowing photons "behind the scenes" ?

 

[Altabeh]

It depends upon how your question is interpreted. If by "When looked upon from Earth.." you meant the pointof view of an observer standing on the surface of the Earth watching the falling stone, then the answer is no, because the observer standing on the surface of the Earth has non zero proper acceleration as is therefore not an inertial observer. If you meant the point of view of an observer co-free-falling with the stone, then the co-free-falling observer is "locally" in an inertial reference frame. As Altabeh suggested, the local vicinity of the free falling observer is Minkowskian, but further away from the free falling observer, it becomes increasingly obvious that the observer is not in flat space. As for the stone itself, a very small free falling stone is on average, an inertially moving object (rather than an inertial frame of reference), but as Passionflower has already pointed out, parts of a very large stone furthest away from the centre of mass of the stone, are not inertially moving.

Exactly.

AB

 

[Altabeh]

Without loss of generality consider the inertial worldline in spherical coordinates:

x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)

Using your equation:

\frac{d^2 x^a}{ds^2}=\left(0, \frac{\gamma^2 r_0^2 v^2}{(t^2 v^2+r_0^2)^{3/2}},0, \frac{2 \gamma^2 r_0 t v^3}{(t^2 v^2+r_0^2)^2} \right) \neq 0

Further consider the non-inertial worldline
x=(t,r,\theta,\phi)=(t,r_0,\frac{\pi}{2},\omega t)

Using your equation:
\frac{d^2 x^a}{ds^2}=0

So your equation does not hold even for a local observer using spherical coordinates. It is not even valid locally in flat spacetime, and certainly not in general.

Completely nonsense and irrelevant. What is your reason that x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right) is inertial and what does your "inertial" mean here? Physics is not about numerology but rather precise derivations and equations. Here you have not even tried to show that how these so-called "inertial" or "non-inertial" observers manage to observe a particle following a geodesic in a very small region along the trajectory which requires you to find a metric transformation of the form g_{ab}\rightarrow \eta_{ab} (locally inertial condition) around the trajectory so automatically the Christoffel symbols vanish and thus \frac{d^2 x^a}{ds^2}=0.

No such things as your nonsense "calculations" are cared about in case there are great proofs and derivations. If you want to learn what is going on and to know where your "fallacy" arises from, see the following sources in order:

Papapetrou "A. Lectures on GR", 1974, pp. 56-58.
Weinberg S. Gravitation and Cosmology: principles and applications of the GR, 1972, 70-73.
Schutz B. F. "A first course in GR", 1985, pp. 154-156, 158-160.

AB

 

[yuiop]

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

Yes, that is proper acceleration and the definition of inertial motion, in a nutshell.

 

[superkan619]

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

I've concluded that Altabeh's argument was right. The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

 

[Dale]

Completely nonsense and irrelevant. What is your reason that x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right) is inertial and what does your "inertial" mean here?

That is the parametric equation of a straight line in spherical coordinates where r0 is the radius of closest approach, v is the velocity, and the line lies in the equatorial plane. Any arbitrary straight line may be brought into this form by a suitable rotation, so it is the simplest form without loss of generality.

In flat spacetime this equation, being the equation of a straight line, is inertial by definition. Your equation is wrong because it falsely identifies this worldline as being non-inertial, even locally. Furthermore, as I showed above it falsely identifies uniform circular motion in spherical coordinates as being inertial. Your equation is not even generally valid in flat spacetime, let alone in curved spacetimes.

 

[Dale]

The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

On the contrary, it does. Whether or not something is inertial is not a relative quantity, it is absolute and frame invariant.

In GR the worldline of an inertial object is a geodesic in spacetime, and all coordinate systems agree on whether or not a given worldline is a geodesic since it is determined by a covariant equation. Therefore, if an accelerometer reads 0 we know that it is a geodesic (inertial) in all coordinate systems, and vice versa. In our coordinate system (for us) it is an inertial geodesic, just as certainly as it is an inertial geodesic in local coordinates (for it).

 

[atyy]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

 

[Altabeh]

That is the parametric equation of a straight line in spherical coordinates where r0 is the radius of closest approach, v is the velocity, and the line lies in the equatorial plane. Any arbitrary straight line may be brought into this form by a suitable rotation, so it is the simplest form without loss of generality.

Your nonsense here doesn't prove anything. The line you're defining this way in a spherical coordinates is only "locally straight" and itself as a special example can be used to prove my assertation is completely correct by a suitable metric transformation around the trajectory that such "inertial" (it is not so ever) observer follows. Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore. Basic stuff.

In flat spacetime this equation, being the equation of a straight line, is inertial by definition. Your equation is wrong because it falsely identifies this worldline as being non-inertial even locally.

Nonsense. See the page 56 of Papapetrou to get where this another non-sequiter fallacy comes from. Your problem is that you try to jump into discussions that you have no prior study on.

Furthermore, as I showed above it falsely identifies uniform circular motion in spherical coordinates as being inertial. Your equation is not even generally valid in flat spacetime, let alone in curved spacetimes.

The problem has been identified above. Your nonsense doesn't seem to be in agreement even with SR, let alone GR. You have not even identified what gives you this motivation to use "non-inertial" or "inertial" for those equations in a non-flat spacetime associated with a spherical coordinates. Read the sources I provided you with.

AB

 

[djy]

Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore.

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

 

[Dale]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

I think this is a correct assessment of the disagreement between Altabeh and myself. However, the OP clarified in post 45 that he is interested in whether or not the object is inertial and not so much interested in the coordinates.

 

[Dale]

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

Precisely and concisely stated.

 

[atyy]

I think this is a correct assessment of the disagreement between Altabeh and myself. However, the OP clarified in post 45 that he is interested in whether or not the object is inertial and not the coordinates.

I see. Yes, a free falling object undoubtedly can be defined to be inertial in a sense which is frame invariant by defining it via the magnitude of its 4-acceleration. However, I also think Altabeh's definition is equivalent, since in post 26 in he specifies dx_a/ds, in Fermi normal coordinates on a geodesic - his emphasis being that that definition makes sense as it reduces to the proper acceleration in flat spacetime. (Of course it wouldn't make any sense if x were taken as a non-inertial coordinate, as you pointed out.)

 

[Dale]

The line you're defining this way in a spherical coordinates is only "locally straight"

It is a geodesic in a flat spacetime. It can't get more globally straight than that.

Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore. Basic stuff.

No, the spacetime is flat, changing coordinates does not turn a flat spacetime into a curved one. If you disagree then please tell me which component of the curvature tensor is non-zero.

 

[atyy]

Yes he did a great job but I still don't believe that physics community has an agreement on such definition as Rindler is (let's say) the only one using this. I have wormed partially through Schutz, Letctures on GR by Papapetrou, D'inverno, Weinberg, the first part of Wald,

Wald p152 gives the proper acceleration of an observer who is static in the Schwarzschild spacetime as a=(1-2M/r)^-1/2M/r².

 

[Altabeh]

It is a geodesic in a flat spacetime. It can't get more globally straight than that.

I thought your background metric belongs to the spacetime around some gravitating body (related to the OP's question) which now is only described in spherical coordinates instead of Cartesian coordinates. Yet again this doesn't change anything ever because though the dynamics of spacetime does now seem to have changed, but we don't need any more of metric transformation around the trajectory in some small region to bring the metric into a locally flat phase: there is a globally valid coordinate transformation to do so:

\bar{t}=t,
r=\sqrt{x^2+y^2+z^2},
\phi=\arccos(z/ \sqrt{x^2+y^2+z^2}),
\theta=\arcsin(y/ \sqrt{x^2+y^2}). if x>=0 (if x<0, use \theta=\pi-\arcsin(y/ \sqrt{x^2+y^2}))

This transformation is valid everywhere so along the trajectory of an "inertial" observer you can see that generally the claim given in the page 56 of Papapetrou is true. Another example looking alike yours is the Rindler coordinates for which the same approach happens to be applicable.

No, the spacetime is flat, changing coordinates does not turn a flat spacetime into a curved one. If you disagree then please tell me which component of the curvature tensor is non-zero.

I do agree, of course!

AB

 

[George Jones]

we don't need any more of metric transformation around the trajectory in some small region to bring the metric into a locally flat phaseAB

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

 

[atyy]

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

 

[Altabeh]

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

Yes he does call it so several times.

AB

 

[Altabeh]

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

I believe that Poisson himself agrees to what Schutz says about local flatness and his treatment on the problem as you can see this from Poisson mentioning the name of Schutz's book in his references.

AB

 

[atyy]

I guess there could be several objections to "local flatness":

1) the metric is not flat anywhere, not even at a point, since flatness is judged by second derivatives of the metric, not first derivatives (objection to "flat")

2) it's ok to use local flatness for just making the first derivatives zero, but only at the origin of the normal coordinates, so we should not use the term if we talk about a small region (objection to "local")

Any more?

 

[George Jones]

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

I have my hard-copy of Poisson home with me, and, now that you point this out, I remember that Poisson uses the term this way.

Yes he does call it so several times.

AB

I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

 

[yuiop]

I was actually talking about something like a thin rod or rigid cable connecting a point of low potential and a higher point and rotating about a line along its core, although in many ways this has similar results, in that the frequency at the top and bottom cannot actually differ as observed by anyone observer.

I like your example of a long vertical cylidrical rod spinning about its long axis better than my example. Your example, is a clear and simple demonstration that clocks really run slower at lower gravitational potentials. Unfortunately this implies that the speed of light must also really be slower at lower gravitational potentials, to explain why local observers lower down measure the local velocity of light to be c.

...
Couldn't it be said that SR came into being as a rational attempt to make sense of the illogical invariance of light as indicated by Maxwell and MM ??

On a historical sidenote, the MM played no part in Einstein's formulation of SR and is not mentioned in his 1905 paper. We have also shown in other threads that the MM is inconclusive as far as SR is concerned, because it can be explained by a ballistic theory of light. Maxwell's equations are far more significant, in that they make it clear that a ballistic theory of light is ruled out.

BTW I wouldn't say I have bought into the first choice but so far it seems more consistent with the other applications of time dilation.

The problem is that local observer's always measure the speed of light to be c, while a coordinate measurement indicates the speed of light varies with potential. Jonathan Scott's thought experiment indicates that the coordinate speed of light has some objective reality. However, the coordinate speed of light tends to zero at the event horizon, while the local speed of light is always c and this brings about a clear contradiction between the two views at the event horizon. Which is the real measurement? Does light "really" stop at the event horizon?

Are you saying that the relative time dilation at different potentials is not measured?
Is not real?

Well, there is a fine distinction between measured and calculated. I think in general, local observations are considered to be "measured", while comparisons of clock rates at different altitudes (a coordinate measurement) is considered to be "calculated". In other words, the calculated rate of a clock according to a distant observer is considered to be less valid or less real than the measured rate of the clock by a local observer. Above the event horizon, the distinctions are largely philosophical, but if we wish to extend our observations to predicting what happens exactly at the event horizon and below, then we probably need to decide whether coordinate or local measurements are a better representation of physical objectivity.

Isn't the measured dilation differential by your definition "real" and the interpretation of slowing photons "behind the scenes" ?

Yes, the time dilation differential can be measured. Slowing photons "is behind the scenes" if we ignore the fact that we measured the time dilation differential to be real. If we acknowledge that clocks really do run slower at lower potentials, we can synchronise clocks to take this into account and then we would directly measure the local speed of light to vary with potential. The slowing of photons would no longer be behind the scenes using such a clock synchronisation method. In effect, what we do when we measure the vertical speed of light is this. We take a vertical rod and a clock and calibrate the apparatus using the assumption that the speed of light is constant everywhere. We then use said apparatus to measure the speed of light and then declare "lo and behold, the speed of light is constant everywhere!"

 

[Altabeh]

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

I think I have answered this in post #81. But to shed more light on why bringing a curved metric into a flat form cannot be done through intoducing an explicit coordinate transformation, I want to say that I myself have proved that if one defines a coordinate system with n free numbers in any n dimensional spacetime the only metric that can be transformed into Minkowski locally is a diagonal one because then we have a system of linear equations in the new coordinates with 0 degrees of freedom which requires you to adjust the free numbers around some arbitrary point according to the coordinates of this point; leading to a perturbed metric and that local flatness is again guaranteed. In general, for a symmetric metric there exists n(n+1)/2 components and thus equations involving new coordinates which means the number of equations exceeds that of unknowns; leading to an overdetermined system (where the method of least squares can be used to find an approximate solution which is the best one.) So in general local flatness through making use of this way can be weakly evaluated to be true.

AB

 

[Altabeh]

I have my hard-copy of Poisson home with me, and, now that you point this out, I remember that Poisson uses the term this way.


I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

More mathematical . Most of Physicists do prefer to use "locally flat", "locally inertial" or "flat in a sufficiently small region" but they barely utilize terms like "in the neighbourhood of xxxx" or even "in an infinitesiammly small region". For instance, when it comes to EP the use the term "flat in a sufficiently small region" to emphasize that flatness is not bound to a neighbourhood but it may exist within a larger distance or region which is translated as "sufficiently small".

AB

 

[Altabeh]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

Yes. But remember that the Fermi coordinates calls for its own observer; not every observer. So the general answer to the OP's question is no.

AB

 

[yuiop]

The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

I've concluded that Altabeh's argument was right. The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

If the object has inertial motion according to an accelerometer "actually functioning in the frame of reference of the object itself", then it has inertial motion according to all observers, whether they are inertial or non inertial, at rest or with relative motion and whether accelerating or not. Inertial motion is a universal observation, as Dalespam has pointed out several times.

Consider a space station orbiting the Earth in a highly elliptical orbit. Its angular and radial coordinates are continually changing and its angular and radial acceleration is continually changing, but to an observer within the space station itself (no windows), the space time is essentially flat to a high degree of accuracy. Even using modern technology, I think it would be extremely difficult for the observer inside the space station to prove that the space station is not stationary in flat space, using only measurements made inside the confines of the space station.

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

An Earth bound observer is a non inertial observer. He knows he is accelerating (as he can measure this using an accelerometer) and he can conclude that the apparent changing motion of the inertially falling stone can be explained entirely by his own acceleration.

 

[Altabeh]

The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

Exactly.

 

[atyy]

I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

So your objection is that a point is not a neighbourhood?

How about "ultralocal pseudoflatness"?

 

[Altabeh]

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

AB

 

[Altabeh]

How about "ultralocal pseudoflatness"?

A nice one if not so much of a pain to read!

AB

 

[yuiop]

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

 

[Altabeh]

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

AB

 

[yuiop]

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

 

[atyy]

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

Yes, the EP which gave birth to GR is only heuristic. In GR, it holds only at points, and only for first derivatives (all Christoffel symbols, which are first derivatives of the metric, can be made zero at a point in normal coordinates), but not second derivatives (components of the curvature tensor cannot all be made zero). In a strict mathematical sense, all derivatives are local, since they are limits defined at a point. But in the heuristic sense, higher order derivatives are more nonlocal, as you point out. So yes, the EP only applies locally in two senses 1) at a point 2) and only up to first derivatives at the point. But that's more a problem with trying to derive GR from an EP that is not sharply defined until after we already have the theory. The other motivating principle of GR "general covariance" is also problematic for deriving GR. However, GR itself is fine for the moment, having passed many experimental tests.