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  Does a moving object curve space-time as its velocity increases?

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We always hear how gravity bends space-time; why shouldn't velocity?

Consider a spaceship traveling through space at a reasonable fraction of the speed of light. If this spaceship, according to special relativity, gains mass as a factor of y as it approaches c, then its gravitational field should increase in strength as well. Hence, space-time should warp.

Note: Changes in space-time, gravity and mass should only be measureable by an outside observer with a different velocity. Those inside of the ship moving with it would not be able to measure the change in these properties.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Armend Veseli
asked Jan 24, 2014 in Theoretical Physics by Armend Veseli (35 points) [ no revision ]
Related: physics.stackexchange.com/q/3436/2451 and links therein.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Qmechanic
Re: "Changes...should be measurable only by an outside observer with a different velocity" ---- all observers must agree on the spacetime metric, and hence on the curvature tensor.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user WillO

6 Answers

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Start with the gravitational field of the Sun. We are effectively stationary with respect to the Sun, because our relative speed is much less than $c$, and the Sun is rotating at well below relativistic speeds so we expect its gravitational field to be well described by the Schwarzschild metric. And indeed this is true: Newton's law of gravity is the non-relativistic limit of the Schwarzschild metric.

The metric tensor is invariant with respect to coordinate transformations, so if we take some observer moving at near light speed they would also find the gravity round the Sun to be described by the Schwarzschild metric. It will not look the same in the observer's coordinates, that is the individual components $g_{ij}$ will be different, but it will be the same tensor. Since in the observer's frame they are stationary and the Sun is moving, the conclusion is that velocity does not change the spacetime curvature.

Incidentally, this is why a fast moving object does not turn into a black hole.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user John Rennie
answered Jan 24, 2014 by John Rennie (540 points) [ no revision ]
I would have to say that while a moving object close to the speed of light would effect space time, it would still require massive amount of energy to do so iirc. Just like how light moving at lightspeed is infinite energy despite being massless.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user C. Jordan
I disagree. Curvature is affected by all components of the energy-momentum tensor, including momentum – hence velocity. For an observer stationary with the Sun, the energy-momentum tensor has no momentum components; this observer interprets curvature to be affected by the energy & pressure components. For an observer in motion, the energy-momentum tensor has non-zero momentum components, and this observer interprets curvature to be affected by these components as well.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user pglpm
@pglpm Yes, but the changes in the different components of the SET balance each other out so the curvature remains unchanged. Consider quantities that describe the curvature like the Ricci scalar or Kretschmann scalar. These are scalar invariants so they cannot be affected by Lorentz transformations. The components of the Riemann curvature tensor will indeed change as we change coordinates, but the tensor itself will not.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user John Rennie
@JohnRennie I'm not disagreeing on that point: tensors are tensors. I'm disagreeing on "velocity does not change the spacetime curvature". Whether it does or not is an interpretation depending on the coordinate system. As you write, for the moving observer some $g_{\mu\nu}$ components will be different, and this observer will relate them to the different components in $T_{\mu\nu}$, where momentum – and velocity – appear. This observer interprets curvature to come partly from velocity-related gravitomagnetic-like effects.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user pglpm
@JohnRennie For instance, an observer using a coordinate system rotating w.r.t. the Sun will use non-zero $g_{0i}$ components, which will be related to the non-zero rotational velocity of the Sun (in these coordinates), roughly speaking proportionally to $\rho v_i$ (gravitational vector potential). For this observer, the non-zero velocity of the Sun "causes" specific metric effects.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user pglpm
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I assume you're asking whether a moving objects curves spacetime differently than a stationary one as its velocity increases.

Strictly speaking, no: it's the same spacetime geometry either way. The spaceship warps spacetime either way, and all we'd be is talking about it in a different frame. Because of this difference of frames, in a sense the gravitational field is different even though the geometry is the same.

If this spaceship, according to special relativity, gains mass as a factor of y as it approaches c, then its gravitational field should increase in strength as well.

This is not quite right. First, the spaceship does not gain mass. It's true that quantity $\gamma m$ is sometimes called relativistic mass, but this term is redundant with energy, bad at its intended purpose of preserving a superficial resemblance to Newtonian mechanics, and depreciated in physics. In special relativity, mass is invariant: $(mc^2)^2 = E^2 - (pc)^2$ is the same in all inertial frames.

Which is just as well, since the 'gravitational charge' isn't mass, but energy. But it is not a simple proportional increase when we view the spaceship's gravitational field in a frame in which it has a lot of energy.

This shouldn't be surprising if you know a bit of electromagnetism. A moving charge produces an electromagnetic field that has both electric and magnetic parts, since the motion of the charge, i.e. the current, matters. The electric field is enhanced in directions perpendicular to the direction of motion, which we can picture as the initially spherically symmetric field lines getting Lorentz-contracted, thus 'squishing' them close together in the perpendicular directions.

The gravitational analogue of electric current would be momentum, but because gravity is spin-2, stress in addition to energy density and momentum density is relevant to how spacetime is bent. You can see this described in the stress-energy tensor. So the gravitational field is more complicated, but it has an analogous behavior of being strengthened perpendicular to the direction of motion, although its quantitative behavior is different.

In the limit of lightspeed, the electromagnetic field of an electric charge turns into an impulsive plane wave, and the gravitational field of a point-mass behaves analogously, turning into a vacuum impulsive gravitational (pp-)wave.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Stan Liou
answered Jan 24, 2014 by Stan Liou (160 points) [ no revision ]
Shouldn't we be way beyond deprecation by now? Shouldn't we consider it both obsolete and harmful to learning? Because that is the overwhelming impression I get from this using site over many years.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user m4r35n357
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Looking at the (spatial) velocity of an object alone is like looking at a plane and considering only its north-south velocity and neglecting its east-west motion.

Timelike objects are like cars with only one gear, no brakes, and a gas pedal that is stuck in place. They continue to go forward in time no matter what; all they can change are their direction through spacetime--how much velocity is going forward in time vs. how much is used to travel distances in space. (The steering wheel still works, but that's all, so to speak.)

When you consider both of these velocities together as one quantity, you find that this "four-velocity" of a given object has constant magnitude: the speed of light itself.

Similarly, the "four-momentum" has constant magnitude, proportional to the rest mass of the object.

These are all results from special relativity. General relativity is a bit more nuanced (not in the sense that four-velocity and four-momentum no longer have constant magnitudes--they do--but in how those magnitudes are calculated using the metric).

While different observers will disagree on obviously frame-dependent things like spatial velocity, they will all agree on the magnitudes of vectors in this fashion. You've been told that gravity depends on mass. I'm telling you that it depends on rest mass--which all observers agree on in this way.

Because spacetime boosts are analogous to rotations, boosting a particle and looking at its gravity just results in all spacetime variables (like the metric) being boosted as well. These transformation laws can be tricky for some tensors, but the underlying physical system can still be understood as equivalent to that of a stationary particle. It's just being looked at through another observer's eyes.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Muphrid
answered Jan 24, 2014 by Muphrid (60 points) [ no revision ]
Without exactly disagreeing with you, I'm not sure just rest mass is all that matters. Consider something like a Z particle at rest decaying into an electron and positron. The overall energy is conserved and I'd thought that viewing the subsystem as a whole it would exert the same gravitational effect, which is not just the sum of the rest masses. Rather we have to look at the stress energy tensor as a whole?

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Keith
Muphrid uses a lot of technical jargon that I find hard to wrap my head around. I like physics but I'm just a freshman at community college. Are you saying that the question is not true?

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Armend Veseli
@Keith: sure, the stress-energy tensor is the ultimate source of all spacetime curvature. Everything I said should be considered valid only as much as it applies to the narrow case of an inertial point particle.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Muphrid
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When talking about gravity, mass, velocity, energy etc. it’s always good to give the context of the definition i.e. which theory we are discussing. Otherwise things get very confusing.

In General Relativity, gravity is usually said to be a consequence of the curvature (or bending) of space-time. Space-time curvature can be calculated from the total energy-momentum (the stress energy tensor) within a region of space.

In special relativity, the value of the mass/energy and velocity/momentum are relative quantities i.e. they depend on the observer and the measurement of another reference object. The total energy-momentum is invariant i.e. all observers can measure the same value without reference to other objects. The energy-momentum can be decomposed into values for the mass/energy and velocity/momentum but these can be different for different observers (as the observers can only measure them relatively).

For the spaceship in the question, one observer moving with the spaceship would say that the velocity/momentum of the spaceship is 0 and therefore not contributing to the total energy-momentum of the spaceship, another observer moving relative to the spaceship might say that the spaceship has a velocity/momentum which contributes to the total energy-momentum of the spaceship. Both observers will calculate the same value for the total energy-momentum for the spaceship however (as each will also calculate different values for the mass/energy of the spaceship). The same energy-momentum will result in the same space-time curvature. So usually, people don’t say that velocity causes space-time curvature because it is a relative quantity.

It is still an important point that moving objects do result in space-time curvature and therefore gravity – it is not possible for an observer to be at rest relative to all moving objects. This is true even without using General Relativity. A body of material (this could be anything, a lump of metal, a person, a star) is composed of huge numbers of atoms all oscillating in different directions – it is not possible for an observer to be at rest relative to every oscillation. The momentum of each moving component part contributes to the total energy-momentum of the object. In thermodynamics, the hotter the object the faster the component parts oscillate. So an observer at rest relative to the body of material as a whole will measure a higher mass for a hot body than a cold one. In Newton’s gravitational laws the greater the mass the greater the gravitational attraction it causes.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user user12345
answered Dec 29, 2017 by user12345 (30 points) [ no revision ]
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The answers given so far make many good points. One problem with the question is that it's implicitly simplifying the context. So I'd like to give a more detailed answer, which is a qualified "yes". Curvature is "affected" not only by energy-mass, but also by momentum – which is related to velocity – and also by energy & momentum fluxes, like heat and pressure (see the pedagogical article by Ehlers & al. below).

First let's be more precise on a couple of points:

  • "Curvature", in its most general form important for general relativity, is not something which can be quantified by one number. It is expressed by a set of numbers, called a tensor, and their values depend on the coordinate system which we adopt in a region of spacetime. Simplifying things a little, your "curvature in the $x$-direction" could be a combination of my "curvatures in the $x'$ and $y'$ directions", if we're using two different coordinate systems $(t,x,y,z)$ and $(t',x',y',z')$.

    One particular set of curvature numbers, called the Einstein tensor, can be represented by a 4-by-4 symmetric matrix of numbers $G^{\mu\nu}$. This matrix can also be represented by an ellipsoid (or hyperboloid) in spacetime. This is the curvature tensor that enters directly in the Einstein equations.

  • If at a spacetime point where there's a non-zero amount of matter, by which we mean some substance, or baryons or leptons, then we can associate to that matter at that point a four-velocity vector $U^{\mu}$. This is also a collection of four numbers, which again depend on the coordinate system. If we are using one temporal and three spatial coordinates $(t,x,y,z)$, then the spatial components of the four-velocity $(U^x, U^y, U^z)$ are proportional to the coordinate velocities. Clearly we can choose a coordinate system in which the coordinate velocities are zero (at least in a small region).

    We represent the energy and momentum (their densities and fluxes) of matter by an energy-momentum tensor, which can be represented by a 4-by-4 matrix $T^{\mu\nu}$ of numbers, again dependent on the coordinates. In the simple case of matter whose internal pressure is negligible, called "dust", its energy-momentum tensor is directly given in terms of its four-velocity: $$T^{\mu\nu} = \rho\,U^{\mu}\,U^{\nu}$$ where $\rho$ is the rest-energy density.


Now we can say this unambiguously: four-velocity "affects" curvature. In fact this comes directly from the Einstein equations: $$G^{\mu\nu} = \kappa T^{\mu\nu}$$ where $\kappa$ is a proportionality constant. In the case of "dust" matter we have $$ G^{\mu\nu} = \kappa \rho \,U^{\mu}\,U^{\nu} $$ which clearly says, for instance, that the curvature $G^{xy}$ is proportional to the $x$ and $y$ coordinate velocities. So yes, if the matter at a given point has a higher $x$-velocity, then it's also generating higher $G^{x\nu}$ curvatures there.

This will also be reflected in the metric tensor, which will contain off-diagonal terms proportional to $\rho U^x$, and called "gravitational vector potential". In a first approximation they are somewhat analogous to the magnetic field, but for gravitation. Indeed the curvature effects coming from velocity are called "gravitomagnetic effects". These effects are concretely considered by the International Astronomical Union in their modelling of the solar system; see references below.

[For the sake of correctness, the fact that the four-velocity (and hence the coordinate velocities) affect Einstein curvature is not a pure consequence of the Einstein equations, but a consequence of them together with the fact that the energy-momentum tensor of matter depends on the four-velocity. One can also create models of matter whose energy-momentum tensor does not depend on the four-velocity, for instance $T^{\mu\nu}=\lambda g^{\mu\nu}$, where $g^{\mu\nu}$ is the inverse metric tensor. Such matter would correspond to a cosmological constant; see the pedagogical article by Ehlers & al.]


But it's important to emphasize the coordinate-dependence of velocity components, curvature components, and their interpretation. A picture may help. Consider a point in spacetime where we have matter with four-velocity $U^{\mu}$ and Einstein curvature $G^{\mu\nu}$, related by the equation above. We can visualize them as follows:
Illustration of curvature and 4-velocity
Note that this is not a spacetime-diagram. It's just a (quite faithful) illustration of the four-velocity and of the Einstein tensor in two dimensions. Also, we are considering matter that does have internal pressure. There is a relation between four-velocity and Einstein curvature because if the four-velocity vector were longer, then the ellipse would have a longer major axis as well.

Now we can introduce a coordinate system $(t,x)$ as follows (the axes looks orthogonal, but that's unimportant):coordinates t,x
In these coordinates, the matter at this spacetime point has non-zero coordinate $x$-velocity, and the Einstein curvature has a particular $G^{xx}$ component, represented by the intersection between the ellipse and the $x$-axis. This value is proportional to the square of the $x$-velocity and to the pressure. Note how the ellipse is skew with respect to this coordinate system. We interpret this skewness as being an effect of the non-zero velocity with respect to this coordinate system.

But we can also introduce a different coordinate system $(t',x')$ as follows:coordinates t',x'
In these coordinates, the matter is at rest, with zero $x'$-velocity. The Einstein curvature has a particular $G^{x'x'}$ component, proportional to the pressure. Note how the ellipse is now not skewed with respect to this coordinate system. We interpret this absence of skewness as being an effect of the zero velocity with respect to this coordinate system.

Clearly the physical situation is the same.

References

For the representation of symmetric matrices as ellipses:

On gravitomagnetism there is introductory material in many places; for instance:

Gravitomagnetism in the International Astronomical Union specifications for coordinate systems for solar system and Earth:

On the cosmological constant, interpreted as a kind of matter:

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user pglpm
answered Jan 3 by pglpm (350 points) [ no revision ]
Really excellent answer! Somewhat tangentially - not that it changes the content of your answer - when you say ""Curvature" is not something which can be quantified by one number." you can with different notions of curvature (i.e. the Ricci scalar).

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user qwerty
@qwerty Cheers! That's an important point, so I added a short sentence about that.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user pglpm
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Equivalence requires acceleration to curve space the same way gravitating mass does. Hermann Weyl, Zur Gravitationstheorie, Annalen der Physik, 54, 117, (1917) argued that kinetic energy should curve space just as gravity and electromagnetic fields do by entering into the stress energy tensor.

The concept of inertial mass increasing at high speed has been used a lot in science, but is often replaced by non linear terms in the energy momentum equation which has better verification. An argument is made that particle accelerators running at high kinetic energy would not be able to levitate a beam of particles if they had relativistic gravitating mass unless the electric charge was also increasing relativisticly, which is not observed. To conserve the equivalence principle many scientists no longer use the concept of relativistic inertial mass.

In summary an object that is changing speed or direction is expected to generate a gravity wave, usually very small. It is proposed but not proven that an object of high kinetic energy could contribute to space curvature.

This post imported from StackExchange Physics at 2025-01-21 21:32 (UTC), posted by SE-user Jerry Decker
answered Sep 28, 2015 by Jerry Decker (20 points) [ no revision ]

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