# Does acceleration warp space?

I know that mass warps spacetime and gravity and acceleration are equivalent so does acceleration warp spacetime too?

• And mass is equivalent to energy by E = m*c**2. Aug 3 '15 at 20:45

Sort of. You are correct in saying (with some caveats) that gravity and acceleration are equivalent. According to general relativity, gravity is manifested as curvature of spacetime. As we know from special relativity and Einstein's famous equation $E = mc^2$, energy and mass are equivalent. As a result, any type of energy contributes to gravity (i.e. to the curvature of spacetime). This relationship can be seen directly from Einstein's Field Equations of General Relativity:

\begin{equation} G_{\mu\nu} = 8\pi T_{\mu\nu}, \end{equation}

where the left hand side of the equation (called the Einstein tensor) contains information about the curvature of spacetime and the right hand side (called the stress-energy tensor) contains information about the mass and energy contained in that spacetime.

Recall that Minkowski spacetime is the spacetime of special relativity. That is, it has no curvature (no gravity) and is the shape of spacetime when you are in an inertial (non-accelerating) reference frame. So, let's ask the question: what happens when you accelerate in Minkowski space?

The answer is that spacetime no longer looks flat to accelerated observers. This is precisely the equivalence principle; locally we cannot tell if we are in a gravitational field or accelerating. Thus, when we are in fact accelerating in a flat spacetime, everything will locally appear as though we are in a spacetime that is curved due to gravity.

There are other interesting similarities between accelerated observers in flat spacetime and observers in gravitational fields. For example, accelerated motion leads to horizons similar to the event horizon of a black hole because if you accelerate at a constant rate for long enough then there will be portions of the spacetime to which you can never send or recieve light signals. There is also an analog of Hawking radiation that occurs for accelerated observers in Minkowski space, called the Unruh effect.

• What do you mean by "[Mikowski] spacetime no longer looks flat to accelerated observers"? It surely does! The Riemann curvature vanishes in every coordinate chart, which of course includes the Rindler chart adapted to accelerated observers. Aug 3 '15 at 21:52
• @StanLiou Of course you are right! Sloppiness on my part. I should clarify that the effect is strictly local, and that, as you correctly point out, the Riemann tensor vanishes identically (since Rindler and Minkowski are related by a diffeomorphism). However, because the Christoffel symbols are not tensorial objects, one can find local frames which appear to have gravitational curvature viz. the geodesic equation. Obviously this is true of every spacetime, not just Minkowski. Aug 3 '15 at 22:37
• @EvanRule That is not gravitational curvature. That is just curvature of a worldline. Aug 4 '15 at 0:02
• @FenderLesPaul I think the question is whether an accelerating frame in flat spacetime can be described as observing curvature (obviously not because of any source of gravity). The answer is yes? Because there is an accelerating force but no motion w.r.t. the frame, there appears to be (fictitious) curvature. However, it is not super useful and there is no more explanatory power. Sep 16 '18 at 21:55
• @Evan Rule - Every material object has a gravitational field around it that is intrinsic and infinitely extensive. How can I place this object in a flat Minkowski space, if this object always curves all the space around it? Does this flat time space exist? Mar 15 at 2:27

Gravity and acceleration are not totally equivalent, only locally so. That's why you get tidal forces, so in some cases, you can detect if you are accelerating or in a gravitational field.

From Wikipedia Equivalence Principle

What is now called the "Einstein equivalence principle" states that the weak equivalence principle holds, and that:

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

Here "local" has a very special meaning: not only must the experiment not look outside the laboratory, but it must also be small compared to variations in the gravitational field, tidal forces, so that the entire laboratory is freely falling. It also implies the absence of interactions with "external" fields other than the gravitational field.

From Tidal Forces

Tidal forces, and a more precise definition

So far, so simple. Too simple, in fact, in several respects. Strictly speaking, all that was said about the equivalence of gravity and acceleration is true only for gravitational fields that are strictly homogeneous. Only in homogeneous gravitational fields are all bodies - per definition - accelerated in exactly the same way, namely in exactly the same direction and at exactly the same rate; as a result, it is indeed true that a researcher inside a cabin cannot distinguish acceleration from gravity. But real gravitational fields are always to a certain extent inhomogeneous.

Take, for example, the gravitational field of the earth. True, here on the surface, looking at experiments which take up only a very, very small fraction f the total surface area of the earth, the gravitational field is, to good approximation, homogeneous: all objects fall to the floor along parallel paths, in the same direction ("down") and with the same acceleration (at least as long as the effects of air friction can be neglected). But if we look closer, the situation is a bit more complex. Here is an example where the deviations from homogeneity are clearly visible - a truly gigantic elevator which contains two spheres, all falling towards the earth: This extreme example shows clearly: the elevator and the spheres do not fall in parallel. Instead, they fall towards one and the same point, the earth's centre of gravity. And while an observer inside the elevator does not see the common downward component of the fall, he or she will notice that the two spheres move slightly closer together.

This is what is called a tidal effect. Tidal effects are what tells a freely falling observer that he is in an inhomogeneous gravitational field, and thus definitely not in gravity-free space. Thus, a more precise formulation of the equivalence principle states that in any freely falling reference frame, the laws of physics are the same as in special relativity, as long as tidal effects can be neglected.

One can, in fact, be more specific as to how tidal effects can be kept small: first of all, by confining all observations to a small region of space: in the animation above, the effects are clearly visible because the distance between the two spheres is not that much smaller than their distance to the earth. For someone here on earth dropping two objects a mere few metres apart, the effect will be virtually undetectable. On the other hand, if you watch merely a brief excerpt from the above animation, you will hardly see the two spheres move towards each other.

Realizing that what matters are the size of the region, and the duration of our observations, we are led to a formulation in which the equivalence principle is not just a useful approximation, but exactly true: Within an infinitely small ("infinitesimal") spacetime region, one can always find a reference frame - an infinitely small elevator cabin, observed over an infinitely brief period of time - in which the laws of physics are the same as in special relativity. By choosing a suitably small elevator and a suitably brief period of observation, one can keep the difference between the laws of physics in that cabin and those of special relativity arbitrarily small.

Acceleration by itself won't warp spacetime, as far as I know, only mass energy will do that.

This question is really a duplicate of Warping Space Time, which you should read, as it explains your question much better than I ever could.

I am restating the question, for clarity:

I know that mass warps spacetime, and gravity and acceleration are equivalent, so does acceleration warp spacetime too?

This question is actually in the form of a logical argument:

IF mass warps spacetime, and this warpage is defined as "gravity" AND IF gravity and acceleration are equivalent THEN acceleration warps spacetime.

Alternatively you could phrase this as:

Mass warps spacetime; we call this effect, "gravity". Gravity and acceleration are equivalent. Therefore, acceleration warps spacetime.

It is important to note that this is a logically sound argument; if this argument's premisses are true, then it's conclusion must also be true... and if the conclusion is false, then one or more of its premisses must also be false!

I use the exclamation point because both of these premises are either essential logical assumptions of, or direct logical conclusions of, Einstein's general theory of relativity; and if they are wrong, then so was Einstein...

Fortunately, we are saved, because the conclusion that acceleration of matter must also warp spacetime is true, and its evidence is all around us to see. What is a rocket, if not a direct demonstration of this very principal? In fact, what is Newton's third law of motion (for every action there is an equal and opposite reaction) except a demonstration of this?

With any explosion, there is the dramatic acceleration of matter outward in all directions. In space, the ejecta expands spherically like a 3D ripple or wave. Within an atmosphere,the ejected material is slowed and stopped by air friction and impact with the earth, while the surrounding atmospheric pressure absorbs the rapid acceleration of air particles and sends all the air particles crashing (accelerating) back in a sonic boom.

That explosion of matter particles is in the shape of a bubble, a 3D space-time wave, the very same wave that can be focused by a rocket to travel in space. In effect, a rocket actually "surfs" on a 3D spacetime bubble, or "wave".

• But gravity and acceleration are not equivalent globally... Sep 11 '16 at 18:08

If the question is, does acceleration warp spacetime? then the answer is a strict and unequivocal no. Acceleration does not warp spacetime. What does warp spacetime is all the various components of the stress-energy tensor $$T^{ab}$$. Physically, these components are:

$$T^{00}$$ energy-density

$$T^{0i}$$ energy flux

$$T^{i0}$$ momentum density (v. closely related to energy flux)

$$T^{ij}$$ sheer stress (the off-diagonal components) and pressure (the diagonal components)

The amount of warping of spacetime can be expressed by a mathematical combination of partial derivatives called the Einstein tensor $$G^{ab}$$. The central equation of general relativity asserts that this is equal to the stress-energy tensor whose components are outlined above, multiplied by some constants.

To be precise about my assertion that acceleration does not warp spacetime, one could put it this way:

If two configurations of matter have the same stress-energy tensor but differing acceleration, they will produce precisely the same warping of spacetime.

But now one may raise the issue: can two matter distributions have the same stress-energy but differing acceleration? To answer this let's look at the equation of motion of the matter: $$\nabla_{\mu} T^{\mu b} = 0$$ where $$\nabla_{\mu}$$ is the covariant derivative, but it is not necessary to be familiar with this operation in order to follow the next bit. All you need to know is that this operation depends on the metric tensor, and the metric tensor in turn depends on the matter configuration, but it is not uniquely specified by the matter configuration. (For example, in the absence of any matter you can have either flat spacetime or gravitational waves). It follows that you can have the same $$T^{ab}$$ but differing acceleration in some region of spacetime, depending on whether there are any gravitational waves or similar.

But coming back to the start, the main point is that warping of spacetime is caused not by acceleration but by energy, momentum and force per unit area. Force per unit area can also cause acceleration, but it is not the same as acceleration and indeed you have have cases where force is present but balanced, so that there is no acceleration but there is an internal stress. It is such stress (along with energy density) which contributes to warping of spacetime.

Accelerating reference frame

But perhaps the question was intending to ask something else: maybe the question was trying to ask whether, if one is undergoing acceleration, does this result in an experience much as if space were warped (even if spacetime is not)? Here the answer is a qualified "sort-of".

To understand this, you can compare the situation to cutting a block of cheese. You can have an ordinary block of cheese, with no warping of the three-dimensional space, and then cut a curved shape out of it. The surface of that curved shape will not be flat. It might be a corkscrew shape or an egg shape or whatever. Similarly, when we talk about "space" as opposed to "spacetime" we are making a slice, only now we make a 3-dimensional "slice" out of a 4-dimensional spacetime. So we can if we like make a curved slice through a totally non-warped spacetime, and if we do then we shall have a warped three-dimensional space. This might seem an odd thing to do, but if one is living in an accelerating rocket then it can be a satisfactory way to organize one's measurements of distances and time intervals. So in this scenario, it is not quite that acceleration warps space, but rather: the choice made by an observer to define what part of spacetime is called "space" at any given moment might give a three-dimensional space which has non-zero curvature.

One can indeed get some nice insights into gravitational phenomena by this route. For example, on a rigidly rotating disk one can deduce that various contraction and precession effects occur, and one can infer that similar effects can be produced by gravitation.

The simplest answer is that energy (of any varying intensity) is affecting the flow of other energy around it from the fact that there is a dispersed field gradient and that gradient has the effect of varying the rate at which energy is flowing through it (most comprehensibly as a wave). Energy travels as a wave (even inside of particles), and since the wave finds a variance in its flow rate it changes its direction of flow from an outside observer. Thus the perception of the warping of space is perceived. Though at the minutest level, for each bit of the energy flowing through the field, it is always going in a straight line. As stated in the responses above. Locally (at the infinitesimal level) there is nothing affected. So since this effect is what causes matter to form from energy and persist as particles, the size that might define "local" and "unaffected" can change a lot. For those that would like the source of some evidence of this effect, you might note Maxwell equations on the propagation of radio waves. There is a small factor in the equations where the intensity of the energy involved affects the rate of propagation. Science News also had a small article stating this some years ago, it was relating to the rate of light's propagation through intense energy fields. Since acceleration is caused by gravity or some other applied energy that is playing, at the very least, with the energy in the mass generating particles it is acting upon that resist the change in speed that inherently must slightly rearrange their internal form to accept that change in relationships when being directly pushed or pulled, or with gravity the internal relationships are rearranged toward the source of greatest gravity. If you consider the amount of energy that is needed to make a particle and atom exist and how small it is, that will help to note the intensity of the flows involved in its existence.