Is spacetime curved in a near-earth-free-falling-object's frame of reference?
If I float motionless in outer space, I will measure spacetime to be approximately flat. By the equivalence principle, I will get flat spacetime if I am free falling near earth. Is this right? otherwise I can distinguish between the two scenarios.
Let say I declare myself (while free falling) to be at rest in an inertial frame of reference under GR. I see the earth accelerating towards me. What causes that acceleration? It's not the curvature of spacetime, right? I just measured it to be flat.
general-relativity gravity spacetime reference-frames equivalence-principle
whopwhop
$\begingroup$ See Why can't I do this to get infinite energy? and my answer to Curvature of Spacetime $\endgroup$ – mmesser314 Jun 28 '20 at 20:40
$\begingroup$ The first gives an elementary explanation of why time slows near the mass of the Earth. The second explains why this implies space is curved near Earth. $\endgroup$ – mmesser314 Jun 28 '20 at 20:47
If I float motionless in outer space, I will measure spacetime to be approximately flat. By equivalence principle, I will get flat spacetime if I am free falling near earth. Is this right? otherwise I can distinguish between the two scenarios.
Notice that the curvature of spacetime is a tensor, so it is a covariant quantity. In particular, if it is zero in one frame then it is zero in all frames and if it is nonzero in one frame then it is nonzero in all frames.
However, in order to measure spacetime curvature requires a sufficiently large region of spacetime. The equivalence principle only applies if you use a region of spacetime which is too small to measure any curvature. That caveat is very important for understanding the equivalence principle, it only applies locally where "locally" means over a small enough region of spacetime that the spacetime curvature cannot be detected.
Let say I declare myself an inertial frame of reference under GR. I see the earth accelerating towards me. What causes that acceleration? It's not the curvature of spacetime, right? I just measured it to be flat.
The floor accelerates because the ground pushes up on it. There is no gravitational force in this frame so there is no downward force to counteract the upward force. Therefore the floor accelerates.
Similarly, the ground immediately under the floor has two forces on it: a downward force from the floor and an upward force from the earth immediately under the ground. The upward force is larger than the downward force, so it accelerates upward.
This continues with the next layer of earth and the next, and eventually you get far enough away that the curvature becomes detectable over the region and the equivalence principle no longer applies.
DaleDale
The equivalence principle applies only locally. An inertial reference reference frame near the Earth does not extend as far as centre of the Earth (which determines the position of the surface of the Earth). Spacetime is not flat over regions where tidal forces (differences in gravitational motions) can be detected.
Compare this to observing that the surface of the Earth is close to flat in the region of your town. You cannot conclude the flat Earth theory from that observation.
Charles FrancisCharles Francis
$\begingroup$ Suppose I am currently 1m from the earth surface, are you saying if I measure 1 nano meter around me, then I get flat spacetime? but if I measure spacetime at 1 meter away, then I get a massive-earth-curvature (9.8g) in spacetime? $\endgroup$ – whop Jun 28 '20 at 19:28
$\begingroup$ No. The distance you can measure an inertial frame is limited by the detection of tidal forces which are extremely small. You would measure the surface of the Earth as accelerating towards you because it is supported by active forces, not by gravity. $\endgroup$ – Charles Francis Jun 28 '20 at 19:33
I found this question a very apt one to probe what it is we are saying about gravity in general relativity. My answer will agree with one already posted by Charles Francis, but perhaps I can help by spelling out the analysis in more detail.
I would begin by constructing or imagining a spacetime diagram. Draw on the diagram the worldline of yourself and of planet Earth. Both these worldlines are geodesics. Now measure the separation between the two worldlines, as a function of proper time along both of them. We find that the separation is changing quadratically with time. That quadratic change is the essential ingredient in one way of measuring spacetime curvature: the way called geodesic deviation. The notion of a distance between two geodesics a long way away from one another is rather hard to define, but for geodesics close to one another it can be done in a well-defined way. The essence of the observation is that if this distance is $\eta$ then for timelike geodesics and small $\eta$ it will obey the equation $$ \frac{d^2\eta}{d \tau^2} = K c^2 \eta $$ where $K$ is the local Gaussian curvature of spacetime (in a direction singled out by the chosen geodesics). What this means is that if I am sitting in my own little inertial frame of reference, in freefall, and near to me is another thing in free fall, and I notice that the distance between us is changing quadraticaly (not just linearly), then I have detected the presence of spacetime curvature.
Notice that spacetime curvature is well-defined even at a point, but its effects tend to zero in the limit of small times and distance at a rate that is at least quadratic in distance and time. The equivalence principle is based on that observation. The mathematical statement would be to say that the spacetime metric is not merely Minkowskian at any given event P (in suitable coordinates), but also at nearby events, in that its departure from Minkowskian form only appears at quadratic not linear order in the spacetime displacement from P: $$ g_{ab} = \eta_{ab} + O(x^a - x^a_P)^2 $$
After writing this much, I began to ask myself, "is it really that simple?" Is the Gaussian curvature of spacetime near the surface of the Earth equal to $-g / c^2 R_E$? The answer is yes, it is that simple, but a further remark is called-for. The quadratic effect that we considered here, involving timelike geodesics, would be correctly described by Newtonian gravity. So one might argue that we don't need to invoke the concept of curvature to describe them. That is correct: we could just say there is a gravitational force as in Newton's theory. The more interesting case would be to find a departure from Newtonian predictions for worldlines, or to detect the curvature of spacetime in the spatial direction, and that is harder to do. It turns out that it is of a similar size to the one we just calculated. The spacetime curvature in the vertical spatial direction near to the surface of the Earth is equal to $-g / c^2 R_E$, and this means that a circle oriented in the vertical direction, such as a bicycle wheel or the London Eye, would have a smaller area than you might expect on the basis of its circumference. It is a small effect though: just 230 nm$^2$ missing for the London Eye.
The distance $R_E$ in the above calculation is equal to the radius of the Earth, not the distance from the surface of the Earth. This is because the surface of the Earth is not in free fall, but the centre of the Earth is. So the two worldlines are that of the person in free fall near Earth and that of the centre of the Earth.
Andrew SteaneAndrew Steane
You are saying "If I float motionless in outer space, I will measure spacetime to be approximately flat.". Let's say this is correct in the voids of intergalactic space.
Now when you say "By the equivalence principle, I will get flat spacetime if I am free falling near earth.", this is not correct. Near earth, you are inside Earth's gravitational field, inside the Sun's gravitational field, and inside the Milky Way's gravitational field (etc.). You are in curved spacetime.
Now you are saying "I see the earth accelerating towards me. What causes that acceleration? It's not the curvature of spacetime, right? I just measured it to be flat." In reality, you do measure spacetime to be curved near Earth. This curvature is what causes the acceleration. The reason you are accelerating towards Earth is that curvature is dominated by Earth's stress energy near Earth, but you are still inside the Sun's gravitational field and the Milky Way's gravitational field too (etc.). Curvature in different directions does cancel out and the net effect is what you see that you are accelerating towards Earth.
Just like when you are at the center of the Earth, you do feel weightless. This is because curvature does cancel out from all directions. Does this mean there is no curvature at the center of the Earth? No! Contrary to popular belief, there is curvature at the center of Earth, but you do feel weightless. This is because the radial four acceleration is zero, but curvature is nonzero.
The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. We get the four acceleration using the geodesic equation: $$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} $$ where the $\Gamma^\mu_{\alpha\beta}$ are called [Christoffel symbols][5] (of the second kind) and they depend on the spacetime curvature. When $r = 0$ the Christoffel symbol $\Gamma_{tt}^r$ is zero and that means the radial four-acceleration is zero and that means you're weightless.
The answer to your question is that near Earth, in your case, when you are free falling towards Earth, there is still curvature, and that causes the acceleration.
This can be easily understood in another way, using the four velocity vector. If you accept that the Universe is set up so, that the magnitude of the four velocity vector is always constant. If you are initially stationary near Earth, you are still inside Earth's gravitational field. This causes time dilation, you will slow down in the temporal dimension. Since the four vector's magnitude is constant, your spatial component will need to compensate. You will start moving in space. You will start moving towards the center of gravity. Why will you accelerate? Because you are getting deeper and deeper inside the gravitational field of the Earth, curvature is getting stronger (still outside the surface), and the GR potential is getting stronger. You are moving towards Earth, and you are even more slowing down in the temporal dimension. Your spatial component needs to compensate even more, this you gain more speed, you are accelerating.
Now in your case, it is very important to understand the difference between curvature and potential.
Gravitational acceleration depends on the gravitational force, which is encoded (within a reference frame) in the components of $\Gamma_{ab}{}^{c}$ Time dilation effects depend on the gravitational potential, which is encoded, within a reference frame, in the components of $g_{ab}$.
Does spacetime curvature (for time dilation) cancel out at the point of center of mass (because curvature effects cancel out from all directions)?
So the answer to your question is, that you are accelerating towards Earth, because there is nonzero curvature, and the net effect causes a radial four acceleration vector to point towards the center of Earth. You are saying that while you are freefalling towards Earth, you are measuring spacetime to be flat, but that is not correct. If you measure spacetime curvature near Earth, inside Earth's gravitational field, you will see it is not flat, the curvature is nonzero, and the potential is nonzero too. If you do this outside Earth's surface, you will get a radial four acceleration that points towards the center of Earth, and that is why you accelerate.
Árpád SzendreiÁrpád Szendrei
The equivalence principle only applies if you use a region of spacetime which is too small to measure any curvature.
@Dale I am slightly doubtful about the application of this condition. This statement is misused many times. I think if the observer, as well as the space station within which he floats, freely falls towards the earth, he always measures the spacetime to be flat WRT him, and since the space station is freely falling along with him, all the experiments being performed by the observer produce the same results as those performed by an observer floating in an interstellar space away from any gravitational field. That is, whether the length of the space station is $1\space m$ or $1\space Km$, the spacetime is flat all inside it from the viewpoint of the observer inside the space station.
However, the observer can claim that the earth is mechanically accelerated towards him inside his flat spacetime.
The only thing that may make the observer doubtful about if the spacetime is entirely flat inside the space station is the difference of G-field obeying an inverse-square law, i.e., the gravitational field is not uniform in reality. This causes the nearer parts of the space station to the earth to accelerate with greater values compared to the farther parts. The rigidity of the space station's body makes the entire station accelerate with, say, an average value. This value slightly differs from the acceleration of the observer floating within the station, and thus the observer sees that the space station is accelerating WRT him though this acceleration is very small. On the other hand, the acceleration vectors are convergent towards the center of the plant, which causes tidal forces to be detected inside the space station as mentioned by some contributors here.
Nonetheless, if we can assume a uniform G-field like that near an infinitely large plate (instead of planet, assume that the space station and the observer floating within are both freely falling towards this huge plate, that is), the spacetime is entirely and exactly flat everywhere inside the space station regardless of how large the spatial dimensions of the space station are. Remember that, for such a plate, the G-field, similar to the E-field inside a parallel-plate capacitor, is exactly uniform.
Remember that whether you assume a planet or a massive plate, the observer floating in the freely-falling space station always claims that the spacetime is nearly flat inside the space station, which this flatness never justifies the considerable acceleration of the planet/plate towards the observer unless the observer assumes that the planet/plate is somehow mechanically accelerated towards him. In other words, we are not allowed to say that because there is a very small curvature inside the space station (in the example of the planet), this small curvature explains the acceleration of the planet towards the observer inside the space station. This phenomenon is more perceivable in the example of the huge plate: Can the observer inside the space station attribute the acceleration of the plate to the exactly zero-spacetime curvature inside the space station? The answer is clearly negative.
What causes that acceleration? It's not the curvature of spacetime, right? I just measured it to be flat.
Assume that, away from any gravitational field, an observer and an object are at rest WRT each other inside a flat spacetime. Suddenly, the object accelerates towards this observer. Can we say that the flat spacetime attached to the observer is abruptly bent just because an object is accelerating away from or towards him?
Therefore, I think that the spacetime is bent for the observer standing on the earth as well as for the observer attached to the accelerating object (in my later example), whereas it is flat for the falling observer in the falling space station as well as for the observer floating in an interstellar space just watching an object is accelerated towards him (in my later example).
edited Jul 8 '20 at 7:19
Mohammad JavanshiryMohammad Javanshiry
$\begingroup$ said "if the observer, as well as the space station within which he floats, freely falls towards the earth he always measures the spacetime to be flat WRT him". This is not at all correct, but more importantly that should be in a comment to my answer instead of in an answer to the question $\endgroup$ – Dale Jun 28 '20 at 22:21
$\begingroup$ See Sean Carroll's lecture notes: arxiv.org/abs/gr-qc/9712019 Particularly the discussion around p 98-100 or so. $\endgroup$ – Dale Jun 28 '20 at 22:30
$\begingroup$ @Dale The answer is edited. $\endgroup$ – Mohammad Javanshiry Jun 29 '20 at 7:04
If I float [...] in outer space, I will measure spacetime to be approximately flat. [...]
Strictly speaking, you floating all alone by yourself would not quite be capable of determining the curvature (values of curvature invariants) of the region that contains you, along your trajectory. Instead, you'd need the assistance of at least another four distinct and separated participants;
hence Synge's "Five-Point Curvature Meter" (Gen. Rel, p. 409).
And, in order to measure curvature accordingly, you (or any of your four assistants) wouldn't necessarily have to "float freely" (a.k.a. "in free fall"). The measurement of curvature (of a region, point by point) is somewhat separate from the measurement of anyone's "freeness", or acceleration;
cmp. How to express the magnitude of proper acceleration through spacetime intervals
The curvature of a region is a quantity to be measured unambiguously; regardless of any particular (and available) choice of frame of reference referred to for this purpose.
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