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

  • $\begingroup$ And mass is equivalent to energy by E = m*c**2. $\endgroup$ – jjack 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.

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    $\begingroup$ 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. $\endgroup$ – Stan Liou Aug 3 '15 at 21:52
  • $\begingroup$ @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. $\endgroup$ – Evan Rule Aug 3 '15 at 22:37
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    $\begingroup$ @EvanRule That is not gravitational curvature. That is just curvature of a worldline. $\endgroup$ – FenderLesPaul Aug 4 '15 at 0:02
  • $\begingroup$ @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. $\endgroup$ – Nimrod Sep 16 '18 at 21:55

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:

enter image description here

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.

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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".

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  • $\begingroup$ But gravity and acceleration are not equivalent globally... $\endgroup$ – honeste_vivere Sep 11 '16 at 18:08

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