Layman's doubts about Gen Relativity

[Altabeh]

If there is no consensus over the term proper acceleration then try to answer the OP without using the term. For me, I would answer it as I did above: A free-falling object follows a geodesic which is defined by the covariant derivative and is therefore a tensor and agreed on by all coordinate systems. No need to introduce the "controversial" topic of proper acceleration.

Ignore the uses of this term in my previous post and everything is clear then. For me, the topic is so controversial as well but what is obvious is that the use of \frac{d^2x^a}{ds^2} as "proper acceleration" other than when it is used in a locally flat spacetime is completely wrong because as you're also proposing any free-falling object following a geodesic is defined by the covariant derivative.

but only the first answers the OP. From his comments about the elevator experiment the OP understands already that it is possible to construct local coordinates around the falling stone wherein the stone is inertial.

I don't see any comment made by the OP concerning the argument that he "understands" this; because he "feels" in his post #16 that my answer is valid which suggests he was doubtful on the issue before my reply.

What he is asking is if other coordinate systems also recognize the stone as being inertial. The answer to that is yes, because in all coordinate systems inertial objects have a 0 covariant derivative.

Yes he is asking this question and I answered the way I explained. This is your answer.

The ordinary derivative is not what defines an inertial object in other coordinate systems as the OP was asking about.

I attempted to show that the ordinary derivative cannot be made zero along the geodesic everywhere (if not specify the coordinate system being used) so that a free-falling stone cannot always be looked, from the perspective of any observer using an inertial coordinate system to measure the motion, as being "inertial" and this agrees to the equivalence principle and the flatness theorem. Let alone the other coordinate systems.

AB

 

[Dale]

the ordinary derivative cannot be made zero along the geodesic everywhere (if not specify the coordinate system being used) so that a free-falling stone cannot always be looked, from the perspective of any observer using an inertial coordinate system to measure the motion, as being "inertial"

This is simply wrong. The ordinary derivative is not relevant to whether or not an object is inertial in general coordinates. The fact that the ordinary derivative is not generally zero along a geodesic no more implies that the geodesic is non-inertial than does the fact that e is an irrational number. You are making a non-sequitur fallacy.

 

[Altabeh]

This is simply wrong. The ordinary derivative is not relevant to whether or not an object is inertial in general coordinates.

I'm not saying such nonsense. All I said was that to any inertial observer all particles following geodesics the locality of flatness only implies that the measurements done by that observer on the ordinary derivative are only valid in an infinitesimally small region.

The fact that the ordinary derivative is not generally zero along a geodesic no more implies that the geodesic is non-inertial than does the fact that e is an irrational number.

If not say wrong, this is completely illogical. The problem is that you don't pay a little attention to what I've trying to say in this thread. In an early post, I quoted from Papapetrou's book that

As Papapetrou says the Christoffel symbols describe, according to the principle of equivalence, the sum of the inertial and the gravitational accelerations

And now to make it more clear I want to complete this by another straight quote from the book:

The second term in this equation (geodesic equation) is the inertial acceleration of the particle, the existence of which is a consequence of the fact that we are now using a non-inertial frame. Thus we see that the inertial accelerations are described quite generally by the Christoffel symbols.

Thus tell me what is the relevance of statement "e is irrational" to the fact that ordinary derivative cannot be generally made zero everywhere on the geodesic from the perspective of all observers? This can happen only for those observers using Fermi normal coordinates and yet "we are specifying the coordinates" which means all observers measuring the motion of a free-falling particles will not find the particle to be in free-fall unless in a very small region in which their coordinates coincide so they get a consensus over "an inertial motion".

AB

 

[Dale]

Thus tell me what is the relevance of statement "e is irrational" to the fact that ordinary derivative cannot be generally made zero everywhere on the geodesic from the perspective of all observers?

It is absolutely and completely irrelevant, that is my point. It is as completely irrelevant as all of your comments regarding the ordinary derivative.

Answer this: what equation determines if a worldline is inertial?

 

[Altabeh]

It is absolutely and completely irrelevant, that is my point. It is as completely irrelevant as all of your comments regarding the ordinary derivative.

See no point here.

Answer this: what equation determines if a worldline is inertial?

Are you familiar with GR or just trying to wrap your head around basics by raising such questions? Wasn't I clear when I said

Since with these conditions we are now in the Minkowskian spacetime we have

\nabla \rightarrow \partial,

and thus

\frac{d^2x^a}{ds^2}

can now be pronounced as "proper acceleration" in GR that is obviously zero.

AB

 

[Al68]

If your frame of reference has a non-uniform, or accelerated motion, then the Law of Inertia will appear to be wrong, and you must be in a non-inertial frame of reference. Right now you're being pulled towards the surface of the Earth by gravity, but at rest relative to it's surface, so you feel no fictitious forces that would lead you claim you were not at rest.

It's objects in freefall, not objects at rest in an accelerated frame that experience fictional forces. And fictional forces aren't "felt", that's why they're fictional. Objects in freefall require fictional forces to account for their motion relative to an accelerated reference frame. Earth's surface is an accelerated reference frame in the same way the floor of an accelerating rocket would be. Being at rest in such a frame means feeling a real force, while fictional forces account for the motion of objects in freefall.

 

[Dale]

Wasn't I clear when I said ...

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".

 

[Austin0]

A free falling (or floating) point test particle falls inertially, which means it undergoes 0 proper acceleration. This view is valid for all observers.

A free falling spatially extended test body as a whole does not fall inertially. For instance in the Schwarzschild spacetime the front and back of such a body accelerate away from the center. This acceleration increases as the body gets closer to the center of gravity.

Doesn't this seem inconsistent??

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Wouldn't this be is dilated by exactly the same factor [due to G potential ] as the assumed relative local acceleration differential due to this same local G potential?

I.e. the measured coordinate velocity [and drived acceleration] would be greater at the front simply because the clocks are dilated wrt the rear.
Equivalent to photon redshift where ther is no implication of an actual change of the photon.

How does length contraction fit into this proposition??

If all these factors are considered in the theorem could you simply explain ,I.e. without math.

Thanks

 

[djy]

Doesn't this seem inconsistent??

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

First, suppose the collection of particles is "dust", i.e. no intermolecular forces. Then each particle will be on its own free fall trajectory, and geodesic deviation ("tidal forces" in Newtonian parlance) will distort the dust field with time.

If the collection of particles is being held together by internal electrostatic forces, then these forces will tend to resist tidal distortion. The particles do experience proper acceleration, but it is from the electrostatic force, not gravity.

 

[Passionflower]

Doesn't this seem inconsistent??

It really is not.

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

That is the basic premise of GR. Things can move in relation to each other, and even accelerate in relation to each other without any need for proper acceleration.

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

Well EM forces will try to tend to keep the body rigid. This process is an instance of proper acceleration.

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Well in general it would depend on the observer and coordinate system. Only proper acceleration is observer independent.

 

[Austin0]

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

Doesn't this also imply a spatial expansion which would neccessitate a Newtonian force in opposition to the internal nuclear and electrostatic tensile forces resisting expansion?

That is the basic premise of GR. Things can move in relation to each other, and even accelerate in relation to each other without any need for proper acceleration..

Understood. If you use apples for test particles then of course a sufficiently large cloud of apples would be expected to expand and have coordinate acceleration wrt each other.

But the hypothesis is based on a premise that this would apply equivalently to one large apple.Which is an entirely different thing. I may have confused things by using the term particle whenI should have said different sections of the system.

Well EM forces will try to tend to keep the body rigid. This process is an instance of proper acceleration..

I take it you mean upward proper acceleration?

Yet you seem to assume that the downward kinematic acceleration would overcome this resulting in overall expansion. is this correct?

What about the view that the EM proper acceleration would speed up the rear and slow down the front and it would just be a question of the propagation time of momentum , of equallizing the differential??

Wouldn't an apparent acceleration differential be assumed on the basis of local coodinate velocity and acceleration measurements (at the front and the back) being based on local time.

Wouldn't this be is dilated by exactly the same factor [due to G potential ] as the assumed relative local acceleration differential due to this same local G potential?

I.e. the measured coordinate velocity [and drived acceleration] would be greater at the front simply because the clocks are dilated wrt the rear.

Well in general it would depend on the observer and coordinate system. Only proper acceleration is observer independent.

I assumed that in this instance question it would be a radial line of Swarzschild observers extending upward through the course of free fall.

That the clocks would be dilated throughout.

That they would measure length increase and relative instantaneous velocities at the front and back as those points were colocated.

That the observers measuring the front would neccessarily measure greater velocities because their clocks were runnning slower than the clocks measuring the back.

That even if there was no actual length increase and no actual acceleration differential there would still be a measured coordinate velocity/acceleration differential due to the dilation.

Make sense?

 

[superkan619]

Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?

 

[Austin0]

Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?

IMO... It is safe to say it was an inertial object.

But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.

Relative to the Earth frame on the surface it would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

 

[Dale]

IMO... It is safe to say it was an inertial object.

But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.

Hmm, interesting take on his question. I assumed he meant "inertial object" and was just being sloppy. Maybe he actually meant inertial coordinate system.

superkan619, perhaps you can clarify your question better. A reference frame is a coordinate system. An object is obviously not a coordinate system, so an object cannot be a reference frame. Do you mean to ask about whether or not the object is inertial or do you mean to use the object to construct some coordinate system and then ask about the coordinate system?

 

[superkan619]

perhaps you can clarify your question better. A reference frame is a coordinate system. An object is obviously not a coordinate system, so an object cannot be a reference frame. Do you mean to ask about whether or not the object is inertial or do you mean to use the object to construct some coordinate system and then ask about the coordinate system?

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

 

[Altabeh]

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".

Again read the post carefully. I don't want to play the role of your primary school teacher and point directly at things that take your intelligence to be understood through reading a simple statement.

Since with these conditions we are now in the Minkowskian spacetime we have

\nabla \rightarrow \partial,

and thus

\frac{d^2x^a}{ds^2}

can now be pronounced as "proper acceleration" in GR that is obviously zero.

The equation is shouting that it is present here.

AB

 

[Dale]

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

It is a necessary designation. An object is a physical piece of matter. A reference frame is a mathematical construct, a coordinate system. One cannot be the other. However, since you are indeed asking (as I had understood) whether or not the object is inertial from the point of view of arbitrary reference frames then the answer is unambiguously "yes, it is inertial in all reference frames".

 

[Altabeh]

My stress is on the fact that the observing frame is on Earth; I think its not necessary for the observed one to be designated as either an object or a reference frame as long as we're outside. I actually thought of it as an object.

The point is that to any kind of observer (even when the frame is inertial) looking over a particle moving along a geodesic the "inertial proper acceleration" can only be defined in an infinitesimally small interval of the trajectory (or better say, in a sufficiently small region in which EP holds). But you can create a coordinate system around a timelike geodesic wherein such inertial proper acceleration exists along the entire of geodesic but now we're narrowing down the list of our observers to a "special observer" located somewhere in the neighbourhood of the trajectory.

AB

Edit: I gave a clear answer to your third question again, I guess. So the quote itself does not show my intention here.

 

[Dale]

The equation is shouting that it is present here.

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
\frac{d^2 x^a}{ds^2}\neq 0

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

 

[Altabeh]

AN extended body is just a collection of particles. How could a particle at one point be accelerated relative to another particle without itself having some degree of proper acceleration??

The statement itself is a little bit ambigious as you're proposing though it can be put in a better mold by taking a swipe at another explanation of the problem. Actually here you got it right that particles have to get a (very small) degree of proper acceleration to accelerate away along the geodesic but we don't really need to go far from the "sufficient" region wherein Equivalence Principle would generally hold in order to describe the chasm between the proper accelerations. Such discrepancies have only one way in GR to be understood unambiguously and that is the fact that by transforming into a locally inertial coordinate system we are not actually getting a Minkowskian spacetime along the geodesic in the small region but only at some point A this can be satisfied; around the point generally we must get

\frac{d^2x^a}{ds^2}\approx 0

which holds only in the "extended" points lying in the neighbourhood of A and the more this extension goes, the more we advance towards the outside of the "sufficient" region mentioned above that would no longer provide us with a good reason for how such differences in "proper" accelerations happen (we are no longer having such thing as "proper" acceleration outside of the "sufficient" region becasue there the frame cannot be asuumed as being inertial) So supposing that an extended body here is the one which fits within this region, we are able to say that your reasoning is completely logical.

AB

 

[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.