Locally Inertial Frames: Freefall & Relative Velocities

[Ranku]

When an object is in freefall, it is in a locally inertial frame of reference. If two objects are in freefall, can their locally inertial frames of reference have different relative velocities?

 

[Ibix]

You can use or not use a local inertial frame of reference whether you are in free fall or not.

If the two objects are at the same event (or close enough) then it is meaningful to compare their chosen frames, and they may choose inertial frames that are in relative motion, yes.

 

[Dale]

When an object is in freefall, it is in a locally inertial frame of reference. If two objects are in freefall, can their locally inertial frames of reference have different relative velocities?

Yes, and in the presence of tidal gravity their relative velocity can change over time.

 

[Ranku]

Yes, and in the presence of tidal gravity their relative velocity can change over time.

So the object closer to the gravitational source will have greater relative velocity than the object further from it?

 

[Sagittarius A-Star]

So the object closer to the gravitational source will have greater relative velocity than the object further from it?

If the scenario is regarded as "local" (in all 4 dimensions), then tidal effects are neglected.

 

[jbriggs444]

So the object closer to the gravitational source will have greater relative velocity than the object further from it?

Wait, what? Exactly what two relative velocities are you comparing? And why are we using tangent inertial rest frames with different origins when our measurements are accurate enough to detect the effects of local tidal gravity?

You are comparing the speed of H (high object) in the tangent inertial frame of L with the speed of L in the tangent inertial frame of H?

Surely one would expect gravitational time dilation to mean that L's clocks to run slow so that H's trajectory runs fast.

 

[Ibix]

So the object closer to the gravitational source will have greater relative velocity than the object further from it?

Velocity relative to what? And there's no general relationship between altitude and speed, not in GR nor Newtonian gravity.

 

[Dale]

So the object closer to the gravitational source will have greater relative velocity than the object further from it?

I don’t understand this question.

 

[TonyStewart]

Two objects in freefall will see the same acceleration and can have different velocities if they do not fall at the same time or if they are in different gravitational fields.

There is also an equivalence principle, in general relativity, that states that in a small region of spacetime, the effects of gravity are indistinguishable from the effects of acceleration.

 

[Ranku]

Two objects in freefall will see the same acceleration and can have different velocities if they do not fall at the same time or if they are in different gravitational fields.

There is also an equivalence principle, in general relativity, that states that in a small region of spacetime, the effects of gravity are indistinguishable from the effects of acceleration.

That is what I am trying to get at. Just like relative velocities between two inertial frames of reference give rise to time dilation and length contraction, does relative velocities between two locally inertial frames of reference within freefall also give rise to the same phenomena? Does that explain why closer to a gravitational field, time runs slowly and length contracts?

 

[PeterDonis]

relative velocities between two inertial frames of reference give rise to time dilation and length contraction

No, relative velocities between objects give rise to apparent time dilation and length contraction.

does relative velocities between two locally inertial frames of reference within freefall also give rise to the same phenomena?

Relative velocities between objects that are moving on different inertial worldlines that, because of tidal gravity, separate or converge, can give rise to apparent time dilation and length contraction, yes. But typically such relative velocities are way too small for such phenomena to be observable.

That said:

Does that explain why closer to a gravitational field, time runs slowly

No. Gravitational time dilation is not symmetric and doesn't work like apparent time dilation due to relative velocity in SR.

and length contracts?

There is no such thing as "gravitational length contraction" so I don't know what you are talking about here.

 

[Dale]

relative velocities between objects give rise to apparent time dilation and length contraction

What do you mean by apparent time dilation?

 

[Ranku]

I don’t understand this question.

I am trying clarify the following situation: Two objects A and B are in freefall, and B is accelerating faster than the A, because it has been freefalling longer and is closer to the gravitational source. Will the local inertial frame of reference of B have a greater relative velocity than that of A? If so, does that explain relativistic effect like time running slowly for B, because it is closer to the gravitational source?

 

[TonyStewart]

I think ... When an observer is closer to a gravitational field, time runs more slowly compared to an observer farther away from the field. Similarly, lengths in the direction of the gravitational field appear contracted.

 

[PeterDonis]

What do you mean by apparent time dilation?

I mean that the terms "time dilation" and "length contraction" in SR refer to observer-dependent appearances: clocks moving relative to an observer appear to run slow to that observer, and rulers moving relative to an observer appear to be shorter to that observer. But those appearances are symmetric; a second observer moving along with the clock and ruler will not find the clock to run slow or the ruler to be shorter, but will find clocks and rulers moving along with the first observer to run slow/be shorter.

This is to contrast with, for example, gravitational time dilation, which is not symmetric; two observers at rest in a gravitational field will agree that the one at the lower altitude has his clock running slower.

 

[Ranku]

I think ... When an observer is closer to a gravitational field, time runs more slowly compared to an observer farther away from the field. Similarly, lengths in the direction of the gravitational field appear contracted.

So is that because the local inertial frame of reference of the observer has a higher velocity?

 

[PeterDonis]

So is that because the local inertial frame of reference of the observer has a higher velocity?

No. The whole concept of "velocity of the local inertial frame of reference" is not a good concept and you should not be trying to understand GR in terms of it.

 

[Sagittarius A-Star]

That is what I am trying to get at. Just like relative velocities between two inertial frames of reference give rise to time dilation and length contraction, does relative velocities between two locally inertial frames of reference within freefall also give rise to the same phenomena?

You could for example define locally an inertial reference-frame in which one of the objects is at rest and the other moving with constant $v$. Locally, SR is valid. That means, you have for the moving object time-dilation and length contraction. Theses effects depend on the reference-frame. In this inertial reference-frame, you have locally no gravitational time-dilation.

 

[TonyStewart]

I think... relative velocities can play a role in special relativity, where time dilation and length contraction are primarily caused by differences in relative motion, in GR, these effects arise from the curvature of spacetime due to gravity rather than differences in velocity between observers.

 

[PeterDonis]

I think... relative velocities can play a role in special relativity, where time dilation and length contraction are primarily caused by differences in relative motion

No, time dilation and length contraction are entirely due to relative motion in SR.

in GR, these effects arise from the curvature of spacetime due to gravity rather than differences in velocity between observers.

No. GR includes SR as a special case. Within a small enough patch of a curved spacetime, relative motion can give rise to time dilation and length contraction just as in the flat spacetime of SR.

You can also get "gravitational" time dilation in SR with accelerated observers; for example, if there are two observers in a rocket accelerating in flat spacetime, one at the bottom and one at the top, both will agree that the clock of the one at the bottom runs slower.

Spacetime curvature is tidal gravity, i.e., it breaks the usual assumption that two inertially moving objects, if they start out at rest relative to each other, will stay at rest relative to each other indefinitely.

Spacetime curvature also makes possible configurations that cannot exist in flat spacetime, for example, accelerated observers standing in a room on the surface of a planet, who locally see things the same as the two observers in the accelerating rocket, but who stay on the planet indefinitely instead of flying off into space.

 

[Sagittarius A-Star]

in GR, these effects arise from the curvature of spacetime due to gravity rather than differences in velocity between observers.

No, as @PeterDonis mentioned. Locally measured gravitational time-dilation is an SR effect in an accelerated reference-frame.

Consider as local scenario an elevator-shaft (height $h$, made of glass) in a tall building. When a lamp at the top of the elevator-shaft sends a light-pulse, the elevator-cabin starts free falling down from the top level. At the ground level, an observer $A$ stands near the elevator-shaft.

An observer $B$ in the falling elevator-cabin is at rest in an inertial reference frame.

• The lamp was at rest with reference to this free-falling frame, when it sent out the light pulse at $t=0$.
• The observer $A$ is accelerating upwards with reference to this free-falling frame and is moving into the light with $v\approx g t$, when the light pulse reaches his eyes at $t \approx \frac{h}{c}$.

From observer $B$'s viewpoint, the observer $A$ must see the light pulse Doppler-blue shifted by the factor approximately $1+\frac{v}{c} = 1 + \frac{gh}{c^2}$.

Observer $A$ would call the same blue-shift "gravitational blue shift" due to difference of gravitational potential $\phi=gh$ and related gravitational time-dilation by the factor $1 + \frac{\phi}{c^2}$.

 

[Ranku]

This is to contrast with, for example, gravitational time dilation, which is not symmetric; two observers at rest in a gravitational field will agree that the one at the lower altitude has his clock running slower.

Why is gravitational time dilation non-symmetrical between two observers, in contrast to the symmetry between two observers in SR?

 

[PeterDonis]

Why is gravitational time dilation non-symmetrical between two observers, in contrast to the symmetry between two observers in SR?

I'm not sure there's any useful way to even compare the two scenarios; they're just different. The fact that the term "time dilation" is used in connection with both of them does not mean there's any meaningful similarity. "Time dilation" is not a fundamental concept in relativity; it's just a name that, for historical reasons, gets used to refer to certain particular scenarios.

 

[Ranku]

No, as @PeterDonis mentioned. Locally measured gravitational time-dilation is an SR effect in an accelerated reference-frame.

Consider as local scenario an elevator-shaft (height $h$, made of glass) in a tall building. When a lamp at the top of the elevator-shaft sends a light-pulse, the elevator-cabin starts free falling down from the top level. At the ground level, an observer $A$ stands near the elevator-shaft.

An observer $B$ in the falling elevator-cabin is at rest in an inertial reference frame.

• The lamp was at rest with reference to this free-falling frame, when it sent out the light pulse at $t=0$.
• The observer $A$ is accelerating upwards with reference to this free-falling frame and is moving into the light with $v\approx g t$, when the light pulse reaches his eyes at $t \approx \frac{h}{c}$.

From observer $B$'s viewpoint, the observer $A$ must see the light pulse Doppler-blue shifted by the factor approximately $1+\frac{v}{c} = 1 + \frac{gh}{c^2}$.

Observer $A$ would call the same blue-shift "gravitational blue shift" due to difference of gravitational potential $\phi=gh$ and related gravitational time-dilation by the factor $1 + \frac{\phi}{c^2}$.

How would the above situation compare with an ascending elevator, with both observers in the elevator, with observer A at the top of the elevator and observer B on the floor of the elevator, in terms of doppler shift and time dilation?

 

[Sagittarius A-Star]

How would the above situation compare with an ascending elevator, with both observers in the elevator, with observer A at the top of the elevator and observer B on the floor of the elevator, in terms of doppler shift and time dilation?

The equivalence principle states, that measurements in an elevator cabin cannot help to distinguish between

• upward acceleration of the cabin in absence of a gravitational field,
• the cabin being locally at rest on the surface of the earth with it's gravitational field.

This statement includes measurements of a gravitational red/blue-shift.

In my above example in posting #21, observer $A$ and the lamp at the top of the elevator shaft are locally both at rest in an accelerated frame. The proper acceleration of each can be measured with accelerometers.

In your example, assuming you mean an accelerated ascending elevator, observer $B$ would see light, emitted by observer $A$, blue-shifted by the factor $1 + \frac{\phi}{c^2}$, with $\phi=gh$ and $g$ being the proper upward-acceleration of the bottom of the cabin and $h$ the height of the cabin.

 

[A.T.]

Why is gravitational time dilation non-symmetrical between two observers, in contrast to the symmetry between two observers in SR?

If the scenario is symmetrical then then both time dilations are symmetrical. But in GR this requires that both clocks are placed and move symmetrically relative to the gravitational sources.

 

[PeterDonis]

If the scenario is symmetrical then then both time dilations are symmetrical.

If both observers are placed symmetrically relative to the source of gravity (for example, at the same altitude above a spherically symmetric planet or star), the relative time dilation between them is zero. But for gravitational time dilation, that is the only symmetrical possibility. If the relative gravitational time dilation between the observers is not zero, it will not be symmetrical either; both observers will agree on which one's clock is running slower.