Is mass the source of spacetime?

[Angelika10]

Mass curves the spacetime. In electrodynamics, an electron curves an electric field (if there is one) - because it has an electric field of its own. The electron is the source of an electric field(maxwell equations) Is there an analogy to space-time, is mass the source of the space-time field as mass curves it?

 

[PeterDonis]

Mass curves the spacetime.

More precisely, stress-energy curves spacetime.

In electrodynamics, an electron curves an electric field

"Curves an electric field" makes no sense; an electric field is not the kind of thing that can be curved.

The electron is the source of an electric field

Of an electromagnetic field, In the sense that it has a charge-current density, yes.

Is there an analogy to space-time

Yes, but not the one you describe in the title of this thread. The analogy is:

Charge-current density is the source of the electromagnetic field;

Stress-energy density is the source of spacetime curvature.

Note that stress-energy is not "the source of spacetime"; it is the source of spacetime curvature. Not the same thing.

 

[Angelika10]

Thank you for your reply!
Why can't an electric field not be curved? If I imagine a constant electric field, with or without an electron, then the field is curved with the electron, isn't it?

In the solar system, mass is the source of the curvature, as you said. But the solar system experiences a background field in which the solar system is falling freely in direction to the Galaxy center.

 

[vanhees71]

Indeed the analogy goes further. The electromagnetic field is also a "curvature" in a very similar sense as the "curvature" of the pseudo-Riemannian spacetime of GR. It's also introduced in the same mathematical way, namely as the commutator of covariant derivatives, where in this case it's covariant in the gauge-theoretical sense.

 

[PeterDonis]

Why can't an electric field not be curved?

Because it's not the kind of thing that can be curved. The concept doesn't make sense.

If I imagine a constant electric field, with or without an electron, then the field is curved with the electron, isn't it?

No.

In the solar system, mass is the source of the curvature, as you said.

No, I said stress-energy is the source of curvature. But there is plenty of other stress-energy besides the solar system.

the solar system experiences a background field in which the solar system is falling freely in direction to the Galaxy center.

No. The galaxy is just more stress-energy, which contributes to the overall spacetime curvature in the solar system. There is no separation of "background field" and "solar system". We might choose to make a mathematical model in which we separate those things for convenience in calculation, but that's a feature of our mathematical model, not reality.

 

[PeterDonis]

The electromagnetic field is also a "curvature"

Yes, the EM field can be interpreted as a curvature in an internal, abstract "space". But that is not the same as saying that the EM field is curved. It's the internal abstract space that is curved (in this interpretation).

 

[PeterDonis]

the solar system experiences a background field

You appear to think that this "background field" is not spacetime curvature. But it is. A truly flat spacetime would have zero stress-energy everywhere--it would contain no gravitating masses whatever.

 

[Angelika10]

The Minkowski spacetime is flat everywhere, no masses which curve the spacetime (Minkowski Spacetime is the spacetime of special Relativity).

The solar system moves in direction to the center of the Galaxy in free fall. Therefore, within the solar system, there must be Minkowski spacetime as boundary condition, then the curvature of the sun and planets adds. Is it like that?

 

[PeterDonis]

The Minkowski spacetime is flat everywhere, no masses which curve the spacetime (Minkowski Spacetime is the spacetime of special Relativity).

Yes.

The solar system moves in direction to the center of the Galaxy in free fall. Therefore, within the solar system, there must be Minkowski spacetime as boundary condition, then the curvature of the sun and planets adds. Is it like that?

No.

 

[PeterDonis]

@Angelika10 I have moved your post in a new thread into this one, as it is part of the same topic. There is no need to start a separate thread if you are just following up on the same topic.

I also deleted your other new thread since it was asking the same question as the post I moved to this thread.

 

[Angelika10]

The galaxy is just more stress-energy, which contributes to the overall spacetime curvature in the solar system. There is no separation of "background field" and "solar system". We might choose to make a mathematical model in which we separate those things for convenience in calculation, but that's a feature of our mathematical model, not reality.

Ok, thanks!
But, if we calculate with GR (leading to Newton's equation) in the solar system, we look only at the gravitational field of the central mass, in a spacetime which has flat boundary conditions.

 

[Ibix]

Ok, thanks!
But, if we calculate with GR (leading to Newton's equation) in the solar system, we look only at the gravitational field of the central mass, in a spacetime which has flat boundary conditions.

That's an approximation. You would need extremely precise measurements, or extremely long term measurements to detect the effect of other star systems' gravity on our solar system dynamics. Doesn't mean it's not there - just that it's negligible for almost every purpose. The obvious exception being predicting where our star system will be in the galaxy in the future - then other systems' gravity is important.

 

[Angelika10]

Thanks for cleaning up! Sorry for causing such a mess!

 

[PeterDonis]

Ok, thanks!
But, if we calculate with GR (leading to Newton's equation) in the solar system, we look only at the gravitational field of the central mass, in a spacetime which has flat boundary conditions.

Yes, and if we're doing that, we're ignoring the galaxy altogether, so it makes no sense to then say the "background field" of the galaxy is flat--there is no such "background field" in the model at all.

If you want to say anything about the galaxy at all, you have to construct a larger model that includes the galaxy, not just the solar system in isolation. And in any such model, the spacetime of the galaxy is curved, not flat, and the free-fall motion of the solar system in the galaxy as a whole is free-fall motion in the curved spacetime of the galaxy.

 

[PeterDonis]

if we calculate with GR (leading to Newton's equation) in the solar system, we look only at the gravitational field of the central mass, in a spacetime which has flat boundary conditions.

Strictly speaking, the model you describe here is asymptotically flat, meaning that the flat "boundary" is at infinity. Which just emphasizes the point I made in my previous post just now, that this model does not include the galaxy, or indeed anything outside the solar system. The asymptotically flat boundary condition is certainly not any kind of claim that the "background field" of the galaxy is flat.

 

[Angelika10]

That's an approximation. You would need extremely precise measurements, or extremely long term measurements to detect the effect of other star systems' gravity on our solar system dynamics. Doesn't mean it's not there - just that it's negligible for almost every purpose. The obvious exception being predicting where our star system will be in the galaxy in the future - then other systems' gravity is important.

And the Galaxy center? We're moving with 220km/s around the galactic center, in comparison to 30 km/s Earth around the sun.
Therefore, the field of the galactic center is big at our place!
But, since we're moving in free fall (with the solar system), I would suppose we do not measure any derivation from the flat space in the solar system.

 

[PeterDonis]

We're moving with 220km/s around the galactic center, in comparison to 30 km/s Earth around the sun.
Therefore, the field of the galactic center is big at our place!

The "field" you describe here--basically the Newtonian "force" that determines the orbital velocity--is not the same thing as spacetime curvature. See below.

since we're moving in free fall (with the solar system), I would suppose we do not measure any derivation from the flat space in the solar system.

More precisely, we do not measure any effects of the spacetime curvature due to the galaxy as a whole, locally in our solar system. That is because spacetime curvature is tidal gravity. In simplest terms, it is the difference in the Newtonian "force" from one side of the system to the other. (The full description of spacetime curvature is somewhat more complicated, but the simple description will do for this discussion.) Even though the overall Newtonian "force" (as manifested in orbital velocity) of the galaxy in the solar system is larger than that of the Sun itself, the difference in the Newtonian force due to the galaxy from one side of the solar system to the other is far too small to measure. That is why we can model the solar system locally using asymptotically flat boundary conditions without worrying about any spacetime curvature due to the galaxy. But, as I've said, such a model only describes the solar system. It does not describe the galaxy, or indeed anything outside the solar system at all.

 

[Ibix]

And the Galaxy center? We're moving with 220km/s around the galactic center, in comparison to 30 km/s Earth around the sun.

The point, as Peter has said, is that it isn't the field you need to worry about. In GR terms that corresponds to first derivatives of the metric, which can be transformed away at a point. What matters is the tidal effects, the second derivatives of the metric, and how big they are across the solar system. In Newtonian terms these are smaller than the gravitational acceleration by a factor of approximately the radius of the solar system to the radius of our orbit in the galaxy - a few light hours to a few thousand light years.

 

[Angelika10]

"the difference in the Newtonian force due to the galaxy from one side of the solar system to the other is far too small to measure." ... "That is why we can model the solar system locally using asymptotically flat boundary conditions without worrying about any spacetime curvature due to the galaxy."

In my opinion/view of relativity, there are two possibilities: 1. Go inside the moving coordinate system (which is the solar system now). There, the solar system does not feel any curvature of spacetime because it is moving in free fall. Can I say, in the moving coordinate system is the minkowski spacetime?
2. looking from outside. There, the field is asymtotically flat as you say.

 

[Ibix]

1. Go inside the moving coordinate system (which is the solar system now). There, the solar system does not feel any curvature of spacetime because it is moving in free fall.

No. Curvature is not something you can transform away - it is always there. However, curvature due to sources other than the Sun is negligible on timescales below thousands of years at least - probably nearer millions - so you can nearly always model spacetime around the solar system as only curved due to masses in the system.

 

[PeterDonis]

In my opinion/view of relativity, there are two possibilities: 1. Go inside the moving coordinate system (which is the solar system now). There, the solar system does not feel any curvature of spacetime because it is moving in free fall. Can I say, in the moving coordinate system is the minkowski spacetime?
2. looking from outside. There, the field is asymtotically flat as you say.

All of this is wrong.

The asymptotically flat model of the solar system is not "looking from outside". As I have already said, repeatedly, that model does not even include anything outside the solar system at all. It is a model of the solar system only, "from the inside".

The "moving coordinate system" you describe would be a model of the solar system moving in the overall curved spacetime produced by the galaxy. It would model the solar system as moving along a geodesic of that curved spacetime. It would show that the tidal gravity effects of the galaxy as a whole were negligible on the scale of the solar system. It would certainly not be Minkowski spacetime; it would be Fermi normal coordinates centered on the solar system's worldline, with corrections for the solar system's own sources of gravity (mainly the Sun).

 

[Angelika10]

Hm. OK, curvature is still there, but as we're moving along it we do not "feel" it, inside the moving coordinate system?

If I look at the astronauts in the space shuttle, I suppose to see the minkowski space in which there are "in" at zero gravity: No gravity, no forces, no curved space time. They are in free fall.

 

[Angelika10]

Minkowski spacetime is a tangential spacetime at every point of the curved spacetime, isn't t?

 

[PeterDonis]

curvature is still there, but as we're moving along it we do not "feel" it

Assuming "we" are small enough that tidal gravity effects on our size scale are too small to feel, yes.

inside the moving coordinate system?

Whether or not we can "feel" any tidal effects is an invariant, independent of any choice of coordinate system.

If I look at the astronauts in the space shuttle, I suppose to see the minkowski space in which there are "in" at zero gravity: No gravity, no forces, no curved space time. They are in free fall.

Again, what you are describing is not Minkowski spacetime. It is not even "locally" Minkowski spacetime. It is Fermi normal coordinates centered on the space shuttle's worldline. (Here there are no corrections due to gravitating objects inside the shuttle.)

If you restrict yourself to a short period of time as well as a short distance in space, then you can pick one particular spacetime event on the space shuttle's worldline and construct Riemann normal coordinates centered on that event, which, over a small enough distance and time scale, will look like a small patch of Minkowski spacetime in standard inertial coordinates. But the period of time would be much, much shorter than the time it takes the shuttle to make one orbit around the Earth.

 

[Angelika10]

Why Minkowski spacetime is the spacetime which the curved space converges to in the infinity?
In my General relativity textbook, they derive the Schwarzschild metric by using Minkowski (A=B=1) in the infinity.

 

[Angelika10]

What I'm wondering about is - is there the possibility that the "dark matter" effect at the edges of the galaxy could come from using the boundary conditions of the solar system also for the whole galaxy?

 

[Angelika10]

I have nowhere read it (in papers or textbooks), just thinking... ? Because for deriving Newton from General relativity the limit in infinity minkowskian spacetime is used - and how do we know that this limit also is true for the galaxy as a whole?

 

[PeterDonis]

Why Minkowski spacetime is the spacetime which the curved space converges to in the infinity?

This is not true of all curved spacetimes. It's only true of asymptotically flat ones, by definition. Such models are used because they are useful approximations.

 

[PeterDonis]

is there the possibility that the "dark matter" effect at the edges of the galaxy could come from using the boundary conditions of the solar system also for the whole galaxy?

No.

for deriving Newton from General relativity the limit in infinity minkowskian spacetime is used

More precisely, for deriving Newtonian gravity as an approximation for isolated systems the asymptotically flat model of Schwarzschild spacetime is used.

how do we know that this limit also is true for the galaxy as a whole?

The galaxy as a whole is also an isolated system and can be modeled to a good approximation with an asymptotically flat model.

 

[Ibix]

It is worth noting that the idea that dark matter is actually the difference between a valid and an invalid simplification of general relativity is not new. A paper was recently posted here based on the idea that if one accounted for gravitomagnetic effects dark matter was not needed. I don't think many people at convinced at the moment.