How does matter accelerate in a gravitational field

[peterf1]

I have been reading various forums etc but can't find a clear explanation of how the space-time warping causes an object to accelerate/gain energy.

I don't want pseudo explanations about potential energy etc.

Thanks

 

[WannabeNewton]

Are you sure you are phrasing your question right? A massive test particle in free fall (no non - gravitational interactions) has no 4 - acceleration i.e. $\triangledown _{U}U = 0$ where $U$ is the 4 - velocity tangent to the particle's world line.

 

[Dale]

Hi peterf1, welcome to PF!

The reason that you haven't found a clear explanation is because it is not correct. Suppose that you have an apple which falls and hits Newton in the head. In a local inertial frame the apple doesn't accelerate at all and its energy is constant, Newton accelerates so that his head hits the apple. In an earth-fixed non-inertial frame the apple accelerates and gains KE, Newton remains at rest.

In both cases the space-time warping is the same, what is different is merely the coordinate system. Energy is not an inherent property of an object, but rather a coordinate-dependent quantity.

 

[peterf1]

But if the apple hit newtown in the head just after it left the tree as opposed to when it is just about to hit the ground there is more energy in the collision isn't there?

 

[Drakkith]

But if the apple hit newtown in the head just after it left the tree as opposed to when it is just about to hit the ground there is more energy in the collision isn't there?

Absolutely.

 

[Dale]

But if the apple hit newtown in the head just after it left the tree as opposed to when it is just about to hit the ground there is more energy in the collision isn't there?

Yes. That is a coordinate independent fact that can be explained by spacetime warping. However, I will have to post that tomorrow.

 

[PeterDonis]

But if the apple hit newtown in the head just after it left the tree as opposed to when it is just about to hit the ground there is more energy in the collision isn't there?

Absolutely.

Yes.

Just after it left the tree? Meaning, say, Newton is standing on a ladder just under the apple, instead of sitting on the ground? That would mean *less* energy in the collision, because the relative velocity of Newton and the apple is smaller, therefore the kinetic energy is smaller.

 

[Drakkith]

Just after it left the tree? Meaning, say, Newton is standing on a ladder just under the apple, instead of sitting on the ground? That would mean *less* energy in the collision, because the relative velocity of Newton and the apple is smaller, therefore the kinetic energy is smaller.

Whoops, I must have read that backwards.

 

[SinghRP]

I read Peterf1's question at Post 2. Does anyone believe this question has been answered? Can anyone answer it without using matn?

 

[Dale]

But if the apple hit newtown in the head just after it left the tree as opposed to when it is just about to hit the ground there is more energy in the collision isn't there?

OK, to understand relativity, especially GR, the first thing that you need to understand is that it is fundamentally a geometrical theory. Physical things are mapped to geometrical things in the theory.

First, the position of an object as a function of time is mapped to what is called a "worldline". It is literally a geometrical line drawn in spacetime.

If the object is inertial (i.e. in free-fall so an attached accelerometer would read 0) then its worldline is straight in a coordinate independent sense. Another word for that is that the line is a "geodesic". Objects that are not inertial (i.e. not in free-fall so an attached accelerometer reads some non-0 value) have worldlines that are curved in that same coordinate independent sense. The direction of the accelerometer reading is the direction that it is curved and the greater the accelerometer reading the tighter the curve.

So, Newton's head is not in free fall and an accelerometer attached to Newton's head would indicate that the acceleration is up, therefore his head's worldline is curved upwards. The apple is in free fall so its worldline is straight. The apple hits Newton's head when their worldlines intersect. The angle that their worldlines form when they intersect gives their relative speed, the greater the angle the faster their relative speed, and therefore the more energetic the collision.

Newton's head and the apple are initially at rest wrt each other, geometrically their worldlines are initially parallel. The further apart they start the more "distance" (aka spacetime interval) that the head-worldline has to curve before it intersects the apple-worldline. So the further apart they start, the greater the curving, the greater the angle, and therefore the more energetic the collision.

I hope that helps.

EDIT: Please see the animated diagram that A.T. posted below.

 

[A.T.]

I have been reading various forums etc but can't find a clear explanation of how the space-time warping causes an object to accelerate/gain energy.

Space-time warping causes coordinate acceleration (dv/dt) in the rest frame of the surface. The proper acceleration (accelerometer) in free fall is zero in any frame. Here an animation that compares Newtons and GR gravity. Note that that in GR no force is acting in free fall, hence a straight world line and zero proper acceleration:

https://www.youtube.com/watch?v=DdC0QN6f3G4

 

[peterf1]

Ok I think I understand that but I don't see why the "greater the angle the more energetic the collision". Where does that energy come form?

 

[Nugatory]

Ok I think I understand that but I don't see why the "greater the angle the more energetic the collision". Where does that energy come form?

Intuitively: The greater the angle, the greater the closing speed, so the more forceful the collision. Think about driving a car into a brick wall at a shallow glancing angle versus a steep head-on angle.

 

[Chestermiller]

Space-time warping causes coordinate acceleration (dv/dt) in the rest frame of the surface. The proper acceleration (accelerometer) in free fall is zero in any frame. Here an animation that compares Newtons and GR gravity. Note that that in GR no force is acting in free fall, hence a straight world line and zero proper acceleration:

https://www.youtube.com/watch?v=DdC0QN6f3G4

Hi guys. I'm a little confused over this video, and maybe someone can help me out. First let me say that, as an conceptual depiction for how gravity works, it is a zillion times better than the "rubber sheet analogy". But, maybe I'm missing something. It doesn't look like any curvature of spacetime is involved. It just looks like there are curvilinear coordinates, but the coordinates are in a flat spacetime. Even when it is wrapped around a cone, the spacetime still seems flat to me (i.e., a cone surface is a flat space). I'm confused about how this example should be interpreted physically.

The gravitational analogy that works well for me is presented in MTW where two observers are moving (in free fall) north on the surface of a sphere along lines of constant longitude and are getting closer together. The latitudinal direction is the analog of time. At least in this case, the spacetime is curved, although the curvature is not caused by the bodies.

 

[PeterDonis]

Intuitively: The greater the angle, the greater the closing speed, so the more forceful the collision. Think about driving a car into a brick wall at a shallow glancing angle versus a steep head-on angle.

This isn't quite right, because all of the motion is radial; there is only one relevant space dimension, so there is no variation in spatial angle, and the brick wall analogy requires a variation in spatial angle. (Also, if the two collisions were both at the same relative velocity, the damage would be about the same in either case; it would depend more on the details of the crashworthiness design of the vehicle than on the geometry of the collision.)

The angle in the case of Newton and the apple is an angle in spacetime, and it represents relative velocity: the greater the angle, the greater the closing speed, as you say. But the spatial angle is the same.

 

[HallsofIvy]

We know that objects move in geodesics in curved "space-time". Why that is true, we don't know- any more than we knew why, in Newtonian physics, "gravity" caused all mass to attract one another.

 

[peterf1]

This is all good but I suppose it leads me to my next question -

How do collisions work (on an atomic level)?

 

[A.T.]

It doesn't look like any curvature of spacetime is involved.

Actual curvature is not relevant locally. It causes tidal forces over an extended area. But the local effect of "gravity pull" doesn't depend on curvature. On board of an rocket accelerating in flat spacetime you have gravity too, which is locally equivalent to the gravity on Earth's surface.

Even when it is wrapped around a cone, the spacetime still seems flat to me (i.e., a cone surface is a flat space). I'm confused about how this example should be interpreted physically.

The rolling of the diagram has no physical significance. It doesn't change the intrisic geometry of the space time patch. It is done solely for visualization purposes:
- the unrolled state seen initially shows the geodesic property of the falling worldline better
- the cone-like rolled state has better correspondence to the apple as we see it falling radially

The gravitational analogy that works well for me is presented in MTW where two observers are moving (in free fall) north on the surface of a sphere along lines of constant longitude and are getting closer together.

That is tidal gravity. It is connected to curvature and 2nd metric derivatives. But local "gravity pull" is connected to the 1st derivative of the temporal metric component (gradient of gravitational time dilatation).

 

[Dale]

Ok I think I understand that but I don't see why the "greater the angle the more energetic the collision".

The relative speed is the tanh of the angle between the worldlines in units where c=1. The greater the relative speed, the higher the energy of the collision.

 

[DrGreg]

Hi guys. I'm a little confused over this video, and maybe someone can help me out. First let me say that, as an conceptual depiction for how gravity works, it is a zillion times better than the "rubber sheet analogy". But, maybe I'm missing something. It doesn't look like any curvature of spacetime is involved. It just looks like there are curvilinear coordinates, but the coordinates are in a flat spacetime. Even when it is wrapped around a cone, the spacetime still seems flat to me (i.e., a cone surface is a flat space). I'm confused about how this example should be interpreted physically.

The gravitational analogy that works well for me is presented in MTW where two observers are moving (in free fall) north on the surface of a sphere along lines of constant longitude and are getting closer together. The latitudinal direction is the analog of time. At least in this case, the spacetime is curved, although the curvature is not caused by the bodies.

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[Chestermiller]

Thank you so much DrGreg. This is really wonderful, and captures the concepts (in my opinion) much more effectively. Marvelous illustrations.

 

[pervect]

I have been reading various forums etc but can't find a clear explanation of how the space-time warping causes an object to accelerate/gain energy.

I don't want pseudo explanations about potential energy etc.

Thanks

Energy in GR is unfortunately a rather advanced topic. So I don't think you'll really find an explanation that's both simple and correct for the energy part of your question.

"Geodesic deviation", which I think a few posters (such as AT and Dr. Greg) have explained via diagrams can explain the acceleration part of your question.

http://www1.kcn.ne.jp/~h-uchii/apple.html

has some information on this, you might find the posts here just as readable. A textbook reference (graduate level) is MTW's big black book "Gravitation". You can probably find a similar explanation in "Exploring BLack Holes" or in other undergraduate treamtments of GR, but I can't definitively point at one at the moment.

A correct but not particularly simple or intuitive explanation for the energy part of your question is that energy can be associated with a time translation symmetry by noether's theorem.

http://en.wikipedia.org/wiki/Noether's_theorem

Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law.

When the symmetry is a time translation symmetry, you get a conserved energy. When the symmetry is a space translation symmetry, you get a conserved momentum.

There are fundamental difficulties even defining energy in a general fashion in GR, so you're going to be running into lots of difficultes here, because you are asking "how is energy conserved" when you should reallly be asking "is energy conserved".

For a quick overview about the status of energy in GR try the sci.physics.faq http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

My general impression is that most laypeople who read it tend to "blow it off". There's not much I can do about that except try to say that It Really Is Like That, that we Really Don't Have a universally applicable notion of energy in GR, though we have several that work in important special cases.

To try and make this even simpler, if you ask "do we have ways of defining energy in the Scwhazschild metric of a black hole", the answer is yes. If you ask "can we define the energy of the universe" the answer is "not currently, at least not for a general universe or for our own". It's rather liikely that we'll never have a good defintition applicable to "the energy of the uiverse" but the arguments as to why it's difficult may not be totally conclusive, perhaps there is something that people who have been working on the issue for the last 100 years or so have missed.

 

[bahamagreen]

Be sure you aren't just mentally placing the objects in the field to see what happens...

The models may be confusing because inertial masses need to be thought as already traveling their geodesic paths (the masses have a "travel history") prior to the time segment of examination in the models.

It is possible conceptually to imagine a mass being placed at rest in curved space as an initial condition, but then that mass will not enjoy any impetus to establish geodesic motion nor any previous geodesic to maintain. Physically, this never happens.

 

[A.T.]

It is possible conceptually to imagine a mass being placed at rest in curved space as an initial condition,

That is not only conceptually possible to imagine. It is perfectly possible to do it practically. I do it all the time with stuff.

but then that mass will not enjoy any impetus to establish geodesic motion ...

A mass that you let go at rest in space, will advance on a geodesic through spacetime. The geodesic advance is in spacetime, not in space. See animation and picture above.

 

[Naty1]

pervect links to [post# 22] : http://www1.kcn.ne.jp/~h-uchii/apple.html

ok, pervect...this is just 'cruel and inhuman'...

when I search around in the index for
some interesting topics, like 'torsion'...lo and behold, it is in Japanese [I think]...

Not funny! [LOL]

Nice synopsis of 'energy in GR'...

 

[OmCheeto]

Whoops, I must have read that backwards.

At least you can read it. I don't understand any of this at all.

From Dalespam's inertial/non-inertial example, with the apple not accelerating, the earth-Newton system would have a kinetic energy of 72 billion megatons of TNT after one second, in relationship with the apple.

hmmm... The reminds me of the old philosophy forum, only with math.

Energy in GR is unfortunately a rather advanced topic. So I don't think you'll really find an explanation that's both simple and correct for the energy part of your question.
...

Ah ha! I interpret this as;

It's over your head Om. Go away.

Ok.

 

[Drakkith]

It's over your head, Om. Go away.

 

[49ers2013Champ]

AT:

"A mass that you let go at rest in space, will advance on a geodesic through spacetime. The geodesic advance is in spacetime, not in space. See animation and picture above. "

So if this particle is far enough away from other bodies so that it's not following any curvatures being created by their mass, energy, momentum (whatever can be plugged into Einstein's stress-energy tensor), the particle is still moving along a geodesic?

I thought--and I'm sure I'm wrong--that geodesics were only used to explain movement when relative motion--or perhaps the influence of gravity--was involved. Now, I learned yesterday that certain energy forms create spacetime curvature, but from what I read yesterday, those curvatures are pretty negligible--talking about if the particle was getting reached by light and the photons were curving spacetime.

Please help me out here. Wouldn't the particle be stationary in space--not in observable relative motion and not being influenced by gravity--and so wouldn't it be incorrect to say it was moving along a geodesic?

Yeah, if you put me out there--and let's assume I'm in an awesome spacesuit--I can see my skin aging and so forth, and so I can understand that I'm moving through time. But am I also moving along a geodesic?

Edit: I just re-thought about what is confusing me. Are you saying that one's individual curving of spacetime is enough to say that one if following a geodesic, or does one need to be following a curvature being created by another body? I think that's what's confusing me about the word geodesic. Sure, I understand that you're curving space (and I suppose time here) but is that particle or body or mass moving along a geodesic?

 

[WannabeNewton]

AT:
So if this particle is far enough away from other bodies so that it's not following any curvatures being created by their mass, energy, momentum (whatever can be plugged into Einstein's stress-energy tensor), the particle is still moving along a geodesic?

Yes it moves along the geodesics of flat space-time which are just straight lines.

I thought--and I'm sure I'm wrong--that geodesics were only used to explain movement when relative motion--or perhaps the influence of gravity--was involved.

There is no "relativeness" when it comes to which objects follow geodesics and which don't because proper acceleration is absolute. There will be no ambiguity: if an object has vanishing proper acceleration then it is following a geodesic and if not it won't be. This is again because acceleration can be measured absolutely (without reference).

There doesn't necessarily need to be gravity present to speak of geodesics. The concept of a geodesic makes sense in flat space-time as well, as noted above.

Please help me out here. Wouldn't the particle be stationary in space--not in observable relative motion and not being influenced by gravity--and so wouldn't it be incorrect to say it was moving along a geodesic?

A particle released from rest in space-time, under influence of no external forces, will subsequently follow a geodesic because it will be in free fall.

Edit: I just re-thought about what is confusing me. Are you saying that one's individual curving of spacetime is enough to say that one if following a geodesic, or does one need to be following a curvature being created by another body? I think that's what's confusing me about the word geodesic. Sure, I understand that you're curving space (and I suppose time here) but is that particle or body or mass moving along a geodesic?

When we refer to particles traveling along geodesics, we mean test particles which make no disruptive contribution to the space-time curvature. These geodesics are the geodesics of the space-time geometry generated by some other mass-energy distribution acting as the source.

 

[pervect]

pervect links to [post# 22] : http://www1.kcn.ne.jp/~h-uchii/apple.html

ok, pervect...this is just 'cruel and inhuman'...

when I search around in the index for
some interesting topics, like 'torsion'...lo and behold, it is in Japanese [I think]...

Not funny! [LOL]

Nice synopsis of 'energy in GR'...

I mostly liked that site for figure 4. The figure you'll find on the cover of "Gravitation".

 

[49ers2013Champ]

Wannabe:

One last question regarding your last sentence: "These geodesics are the geodesics of the space-time geometry generated by some other mass-energy distribution acting as the source."

The pronoun "These" is referring to geodesics in flat spacetime--the aforementioned scenario--or in spacetime where bodies are moving along lines being created by energy-matter distributions acting as the source?

 

[WannabeNewton]

The pronoun "These" is referring to geodesics in flat spacetime--the aforementioned scenario--or in spacetime where bodies are moving along lines being created by energy-matter distributions acting as the source?

In flat space-time "these" geodesics refers to the straight lines of Minkowski space-time. When there is mass-energy present then yes "these" geodesics refers to what you said in the latter part of your sentence.

 

[Naty1]

These two already posted statements together with DrGreg' charts summarizes all this nicely for me:

[No gravity:]

In flat space-time ... geodesics refers to the straight lines of Minkowski space-time.

[With gravity:]

... objects move in {curved} geodesics in curved "space-time". Why that is true, we don't know- any more than we knew why, in Newtonian physics, "gravity" caused all mass to attract one another.

To clarify these statements further: A geodesic in GR generalizes the notion of a "straight line" to include curved spacetime...so a geodesic IS a straight line path in flat spacetime and a curved path [or worldline] in curved spacetime [where gravity] is present. In other words, a freely falling particle [a particle not subject to other non gravitational forces like electromagnetic attraction or repulsion] always moves along a geodesic.

What 'causes' spacetime curvature: sources captured in the stress energy tensor [like pressure, energy, mass] If the graph paper of DrGreg's illustrations remains 'flat'...as in his first three illustrations...that is a 'no gravity' condition; gravity is illustrated via a 'curved' graph paper as his last illustration.

The curvature of the coordinate grid, which we seem to call in these forums 'apparent curvature' [for lack of a better term] it isn't spacetime curvature. This is not gravitational spacetime curvature, is not sourced from the stress energy tensor; GRAVITATIONAL spacetime curvature is frame invarient. [I mention 'apparent curvature' because I recently got people inadvertently confused in another thread by not carefully distinguishing between it and gravitational spacetime curvature.]

The introduction here is a decent overview:
http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

and discusses that massive particles follow time like geodesics while photons [massless] follow null geodesics...

 

[Nugatory]

Wouldn't the particle be stationary in space--not in observable relative motion and not being influenced by gravity--and so wouldn't it be incorrect to say it was moving along a geodesic?

No. Instead, you've just identified about the simplest example of a geodesic in four-dimensional space-time.

I'm floating motionless in an otherwise completely empty and curvature-free flat universe. My x, y, and z coordinates are not changing. But what's happening to my t coordinate? It's steadily increasing... So my path through space-time is a straight line oriented along the t axis.

It's sort of like how in a one-dimensional number line $x=3$ specifies a point, but in the two-dimensional Cartesian plane it specifies a line parallel to the y-axis. Likewise, a point (me at rest) in three-dimensional space is a line in four-dimensional spacetime.

 

[Naty1]

Wouldn't the particle be stationary in space...

In absolutely flat spacetime, a particle would theoretically 'sit still' in SPACE...but Nugatory points out it still moves through the time dimension of SPACETIME.