Special relativity effects in general relativity

[johnny_bohnny]

SR is a theory based on flat space-time, and all of its effects are there in a flat space-time framework.My question is, how is SR compatible with GR, since GR uses curved space-time?

Or to say it better how are time dilation, length contraction and relative simultaneity manifested in curved space-time, I hope somebody could give me a comparision.

Thanks in advance.

 

[A.T.]

SR is a theory based on flat space-time, and all of its effects are there in a flat space-time framework.My question is, how is SR compatible with GR, since GR uses curved space-time?

SR is an approximation for regions of space-time where the effects of curvature are negligible. In GR free falling frames locally approximate the inertial frames SR deals with.

 

[pervect]

To draw an analogy, the difference between SR and GR is the difference between living on a plane, and on the surface of a sphere. (There's nothing special about the sphere in particular, I am using it as a specific example of the more general notion of a curved 2 dimensional geometry).

Living on the surface of a sphere, you have to deal with the effects of curvature on the geometry, but they only become important if you make "long enough" journeys. For instance if the sphere is the surface of the Earth, and you are walking to the corner store, the effects of curvature are negligible and you can successfully navigate as if the curved surface of the Earth was a flat plane. If you are sailing across the ocean, you will encounter significant errors if you do not account for curvature.

The boundary of how large a distance you can describe without accounting for the effects of curvature depends on how much accuracy you need.

The notion can be made more rigorous, for instance by talking about the plane tangent to the surface of the sphere, and creating a mapping between points on the tangent plane and points on the sphere.

Euclidean plane geometry is analogous to the Minkowskii geometry of flat space-time, the geometry of the sphere is analogous to the geometry of General Relativity. At every point in the curved geometry of General Relativity, we have a tangent space that's perfectly flat. The geometry of this flat tangent space is described by special relativity.

To answer your specific question about time dilation, length contraction, and relative simultaneity, they all consequences of the Minkowskii geometry of space-time, and they manifest themselves in the locally flat tangent space as usual. GR adds additional effects on top of the effects due to special relativity due to the curvature of space-time which only become significant when you analyze a sufficiently large region of space-time.

One way of describing these "extra effects" is to introduce a metric. In a flat geometry, the metric coefficients are all unity. Thus in a spatial geometry, we have an invariant distance ds that is the same for all observers, which we can write as ds^2 = dx^2 + dy^2 + dz^2. In a flat space time geometry, we haven an invariant Lorentz interval ds which we can write as ds^2 = dx^2 + dy^2 + dz^2 - dt^2.

In a curved geoemtry, we need to introduce metric coefficients, which are in general a quadratic form. For instance on the surface of the earth, a change in lattiude of 1 degree always represents a constant distance, but a change in longitude of 1 degree does not, the distance corresponding to a longitude change of 1 degree varies with lattitude. This behavior can be mathematically described by a quadratic form involving lattitude and longitude. Without metric coefficients, there is no way to account for the fact that the distance represented by a degree of longiutde varies with lattitude.

 

[johnny_bohnny]

@Pervect, thanks for the great answer.

I was wondering how exactly the SR effects that I mentioned correspond to curved space time. Will a curve in space get length contracted relative to a moving frame, or not, how do effects from flat space-time transmit to curved space-time?

 

[WannabeNewton]

I was wondering how exactly the SR effects that I mentioned correspond to curved space time. Will a curve in space get length contracted relative to a moving frame, or not, how do effects from flat space-time transmit to curved space-time?

You can in GR do everything that you do in SR so long as the calculations are done at a single event. This means in particular that length contraction of a (Born-rigid) rod relative to some observer is only defined at the event at which the rod passes right by the observer. The same goes for SR time dilation, just replace "rod" with "ideal clock". The length contraction and time dilation so obtained will contain not only the usual SR kinematic effects but also the GR effects of gravity. In order to extend these concepts globally, to all of space-time, you need a field of observers whose world-lines fill all of space-time. Actually this is not even special to GR. The same goes for SR when dealing with fields of arbitrarily accelerating, non-intersecting observers.

I can provide some examples if you wish.

 

[johnny_bohnny]

You can in GR do everything that you do in SR so long as the calculations are done at a single event. This means in particular that length contraction of a (Born-rigid) rod relative to some observer is only defined at the event at which the rod passes right by the observer. The same goes for SR time dilation, just replace "rod" with "ideal clock". The length contraction and time dilation so obtained will contain not only the usual SR kinematic effects but also the GR effects of gravity. In order to extend these concepts globally, to all of space-time, you need a field of observers whose world-lines fill all of space-time. Actually this is not even special to GR. The same goes for SR when dealing with fields of arbitrarily accelerating, non-intersecting observers.

I can provide some examples if you wish.

It would be great if you could give me some examples, cause this is very fuzzy for me, I haven't had almost any experience with GR. In curved space time the shortest path between two points is a curve, in flat space-time it is a line. But can that curve contract in some way, like a line can in SR?

 

[johnny_bohnny]

I have many questions regarding this topic, so I'll ask some and I hope I will get some answers.
The first one is connected to the motion in GR. For instance, if somebody would travel from the Earth to Sun, he would do in a curved path, right? How is this different from the classical 'distance' in both Newtonian physics and in SR. Is the time dilation that he will undergo under that curved path simply a product of its velocity (and acceleration because of the curve) during the path or is he also affected by gravitational time dilation. I had an idea that maybe gravitational acceleration is what causes the curve of his path to the Sun, and I don't know is this correct, so maybe someone will correct me.

And when mentioning rods like you did before, WBN, if we compare SR and GR, it seems that a rod that is straight in SR becomes curved in GR, because the space itself is curved. Is this true, and is there a way to construct a rod, or any object that basically isn't curved in curved space-time.

 

[ChrisVer]

I don't understand what you are trying to ask in your last post...
The path won't be curved- the paths are always straight lines... the straight lines change according to your spacetime geometry... (am I wrong?)

I don't think that in general an extended object in GR is curved... I'd preferably say that it's subject to tidal forces and thus can be stressed, it can be curved but in fact it takes the form at which it can have the least action satisfied ...

 

[johnny_bohnny]

I don't understand what you are trying to ask in your last post...
The path won't be curved- the paths are always straight lines... the straight lines change according to your spacetime geometry... (am I wrong?)

I don't think that in general an extended object in GR is curved... I'd preferably say that it's subject to tidal forces and thus can be stressed, it can be curved but in fact it takes the form at which it can have the least action satisfied ...

How will the path not be curved since spacetime itself is curved?

 

[HallsofIvy]

I suspect Chris Ver should have said "geodesics", rather than "straight lines". Geodesics, locally, look like straight lines.

 

[johnny_bohnny]

I suspect Chris Ver should have said "geodesics", rather than "straight lines". Geodesics, locally, look like straight lines.

But traveling with geodesics still makes you accelerate, right, and therefore experience different time dilation at each point of the curve?

 

[A.T.]

But traveling with geodesics still makes you accelerate,

Travelling on geodesics worldliness means zero proper acceleration.
http://en.wikipedia.org/wiki/Proper_acceleration
Of course you can still have coordinate acceleration (dv/dt) in some reference frame..

and therefore experience different time dilation at each point of the curve?

Kinetic time dilation depends on speed, while gravitational time dilation depends on position.

 

[WannabeNewton]

It would be great if you could give me some examples, cause this is very fuzzy for me, I haven't had almost any experience with GR.

In that case I'm afraid my examples won't make any sense to you. I was going to give you examples from Schwarzschild space-time on calculating length contraction of rods, time dilation of clocks, and kinematic redshift of photons when Lorentz transforming from radially infalling observers to observers in circular free fall orbit but they all make use of the Lorentz frame formalism and if you haven't had any experience with GR it won't be of much use.

But traveling with geodesics still makes you accelerate, right, and therefore experience different time dilation at each point of the curve?

There is no gravitational time dilation in a freely falling frame. I can show you this explicitly if you wish but the comment from above will still hold.

My comment from the other thread still applies here. If you are truly determined to learn SR and GR then you need to start working through a textbook. It's how everybody else here learned it and it's the only way you can learn it. If you are under the impression that geodesics correspond to acceleration then it's clear that you need to start at the basics of GR before asking questions that are mired with much more subtelty such as whether spatial curves on space-like foliations have a notion of length contraction.

 

[PeterDonis]

How will the path not be curved since spacetime itself is curved?

Because path curvature and spacetime curvature are two different things, and they need to be carefully distinguished in order to properly understand GR.

Path curvature, physically, corresponds to proper acceleration, i.e., acceleration that you actually feel, i.e., feeling weight. So a freely falling object, which is weightless, feeling zero acceleration, is traveling on a path that is not curved--it is straight. These paths are called "geodesics" in order to make clear that it is only the straightness of the path, by the definition I just gave, that is being asserted. Saying that a given path is a geodesic says nothing about the curvature of the spacetime in which it is embedded, and vice versa--you can have straight paths (geodesics) in curved spacetime, and you can have curved paths (accelerated worldlines) in flat spacetime.

Spacetime curvature, physically, corresponds to tidal gravity. The way you test for tidal gravity is to compare nearby straight paths (geodesics) that start out parallel, where "parallel" here means "at rest relative to each other" if we are talking about the worldlines of freely falling objects. If nearby geodesics that start out parallel don't stay parallel, spacetime is curved; physically, if two nearby freely falling objects that start out at rest relative to each other don't stay at rest relative to each other, tidal gravity is present.

Bear in mind, again, that these definitions may not match up with your intuitions; but that's a problem with your intuitions, not with the definitions. These definitions are adopted in GR because they work: they give us a theory of spacetime and gravity that has passed very stringent experimental tests.

 

[johnny_bohnny]

Ok, thanks for the answers, even though I feel I need time and knowledge to really get to know them.

So for instance, if a rocket is traveling from Earth to the Sun, will it travel by geodesic or not? Since the only things that travel by geodesic, if I understand correctly, are free-falling objects?

 

[PeterDonis]

if a rocket is traveling from Earth to the Sun, will it travel by geodesic or not?

If it's firing its engines, so the rocket passengers feel weight, then no, it's not traveling on a geodesic. If it's not firing its engines, so the rocket passengers are weightless (which is how spacecraft we can make with today's technology travel most of the time), then it's traveling on a geodesic.

 

[analyst5]

If it's firing its engines, so the rocket passengers feel weight, then no, it's not traveling on a geodesic. If it's not firing its engines, so the rocket passengers are weightless (which is how spacecraft we can make with today's technology travel most of the time), then it's traveling on a geodesic.

I understand the basics, free fall acelleration and curved space-time upward acceleration cancel each other out so the free falling object is inertial. So basically speaking, the rocket will still travel on a curve, but it won't be a 'straight curve' or to say it better, a geodesic? If the rocket isn't free falling, then the only acceleration in this case that it gets is from curved spacetime (the upward one). So it will undergo kinematical time dilation because of velocity and gravitational time dilation because of the space time curvature and its acceleration that arises from this fact?

This is extremely hard to conceptualize, if I may add, but thanks for your effor, I guess everybody here had a similar problem in their first steps.

 

[A.T.]

So for instance, if a rocket is traveling from Earth to the Sun, will it travel by geodesic or not?

If its accelerometer measures zero proper acceleration, then the wordline (path in space-time) is a geodesic.

In other words, when no forces (other than gravity which doesn't count as a force in GR) are acting on the rocket (or if they all cancel), then its wordline (path in space-time) is a geodesic.

Since the only things that travel by geodesic, if I understand correctly, are free-falling objects?

Yes. With engines off in space you are usually free falling.

 

[analyst5]

In that case I'm afraid my examples won't make any sense to you. I was going to give you examples from Schwarzschild space-time on calculating length contraction of rods, time dilation of clocks, and kinematic redshift of photons when Lorentz transforming from radially infalling observers to observers in circular free fall orbit but they all make use of the Lorentz frame formalism and if you haven't had any experience with GR it won't be of much use.

You still may do so, I think it will help me. Especially the comparision between SR effects in flat space time vs curved space time

 

[PeterDonis]

I understand the basics, free fall acelleration and curved space-time upward acceleration cancel each other out so the free falling object is inertial.

That is not "the basics", it is incorrect. There is no such thing as "curved spacetime upward acceleration".

So basically speaking, the rocket will still travel on a curve, but it won't be a 'straight curve' or to say it better, a geodesic?

Do you mean a rocket with its engines firing, or not firing? That is the key physical difference between the two cases, so it gets frustrating when people keep describing scenarios without even specifying this key variable.

If the rocket's engines are firing, so its passengers feel weight, it is traveling on a curved path.

If the rocket's engines are not firing, so its passengers are weightless, it is traveling on a straight path.

That is all you need to know to determine whether or not the rocket's path is curved. You don't need to know *anything* about the spacetime curvature.

If the rocket isn't free falling, then the only acceleration in this case that it gets is from curved spacetime (the upward one).

No.

So it will undergo kinematical time dilation because of velocity and gravitational time dilation because of the space time curvature and its acceleration that arises from this fact?

No.

 

[A.T.]

This is extremely hard to conceptualize,

Check out the top diagram here:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

The two clocks are not free falling, so their paths are not geodesics, but curved. The free falling object has a straight path. Gravitational time dilation makes the upper clock get ahead, because all objects advance at the same rate in space-propertime.

 

[WannabeNewton]

I understand the basics, free fall acelleration and curved space-time upward acceleration cancel each other out so the free falling object is inertial.

This is the wrong way to look at GR. While it can be made rigorous in stationary space-times, it is certainly not what the conceptual core of GR says. The same goes for everything you say following the quoted statement.

I explained the pitfalls of your line of thinking in a different post (see the 2nd paragraph): https://www.physicsforums.com/showpost.php?p=4770244&postcount=13

 

[analyst5]

In other words, when no forces (other than gravity which doesn't count as a force in GR) are acting on the rocket (or if they all cancel), then its wordline (path in space-time) is a geodesic.

So what happens when a body is between two gravitational fields, for instance between Sun and Earth. Surely it's not free falling, so what would be the description of its motion

 

[A.T.]

So what happens when a body is between two gravitational fields, for instance between Sun and Earth. Surely it's not free falling,

Sure it is, if only gravity acts on it.

 

[PeterDonis]

So what happens when a body is between two gravitational fields, for instance between Sun and Earth. Surely it's not free falling

Why not? If it has no rocket engines, or the engines are not firing, and the passengers feel weightless, it's free-falling. Once again: whether or not the body and its passengers feel weight is *all* you need to know to determine whether the body's path is straight (i.e., free-falling) or not. You don't need to know *anything* about the spacetime curvature. Path curvature and spacetime curvature are two different things.

 

[analyst5]

This is the wrong way to look at GR. While it can be made rigorous in stationary space-times, it is certainly not what the conceptual core of GR says. The same goes for everything you say following the quoted statement.

I explained the pitfalls of your line of thinking in a different post (see the 2nd paragraph): https://www.physicsforums.com/showpost.php?p=4770244&postcount=13

Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up? When I said 'upward acceleration' I really meant upward force, like in this video: .

So when a rocket travels, with its engines firing so it could fight off what we used to call gravity, it travels to the Sun in some kind of path. It undergoes time dilation because of velocity and gravitational time dilation due to its position. If my line of reasoning is correct? Or you may explain to me, what is the cause, or condition, that determines why gravitational time dilation arises. An object that is stationary on the surface of Earth is said to accelerate in GR, right?

 

[WannabeNewton]

Or you may explain to me, what is the cause, or condition, that determines why gravitational time dilation arises.

It arises from acceleration. This is why gravitational time dilation vanishes for freely falling frames but exists in the frame of an observer stationary in a gravitational field, as per the equivalence principle. You don't need space-time curvature in order to have gravitational time dilation also for this reason but if you do have space-time curvature then it will indirectly affect the closed form expression for gravitational time dilation in those coordinate systems wherein it does show up.

An object that is stationary on the surface of Earth is said to accelerate in GR, right?

Yes.

 

[A.T.]

Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up?

They rather multiply.

When I said 'upward acceleration' I really meant upward force, like in this video: .

You talked about "canceling out". In that video there is no "canceling of forces" in Einstein's model, just in Newton's.

what is the cause, or condition, that determines why gravitational time dilation arises.

Gravitational time dilation arises between different positions in a gravitational field (along the potential gradient), or in an accelerating reference frame (along the frame's acceleration).

An object that is stationary on the surface of Earth is said to accelerate in GR, right?

It has proper acceleration, like the apple on a branch. Put a smart phone on a table, and it shows 1g upwards proper acceleration.

 

[WannabeNewton]

You still may do so, I think it will help me. Especially the comparision between SR effects in flat space time vs curved space time

Here are some examples from past threads of mine:

https://www.physicsforums.com/showthread.php?t=717211
https://www.physicsforums.com/showthread.php?t=718299
https://www.physicsforums.com/showthread.php?t=730009

Let me know if you want more examples. I spent a considerable amount of time last summer solving a lot of the exercises in "General Relativity: An Introduction for Physicists"-Hobson so I have a slew of examples I can provide you with if need be.

Here's another fun one: say we have an observer moving to the right across a linear platform at constant velocity $v$ relative to some inertial frame, with the platform accelerating upwards at the rate $g$. Then it is well known that a gyroscope comoving with the observer will be observed to precess in the observer's rest frame defined by the axes obtained by boosting the axes of a local Rindler frame fixed to the platform to the right at velocity $v$. There are numerous ways to explain why this happens using only SR, the most physically appealing of which is a standard relativity of simultaneity argument.

But say we wanted to use the equivalence principle instead. Then we can basically treat this as a GR problem (although it isn't a GR problem per say since there is no space-time curvature and this is also a computationally trivial problem). Imagine we are in Rindler coordinates with a field of Rindler observers $e_t = \frac{1}{z}\partial_t$ stationary in this gravitational field and a metric $ds^2 = -g^2 z^2 dt^2 + dx^2 + dy^2 + dz^2$ where $z = \frac{1}{g}$ represents in Rindler coordinates the stationary position of the observer standing on the platform. Now we introduce a field of observers obtained by boosting along $x$ with velocity $v$ ($\nabla v = 0$) from the global Rindler frame so that $e_{t'} = \Lambda_{t'}{}{}^{\mu}e_{\mu} = \gamma(e_{t} + ve_{x})$ and $e_{x'} =\Lambda_{x'}{}{}^{\mu}e_{\mu}= \gamma(e_x + ve_t)$; then $\xi^{\flat} = gze^t + ve^x$ where $\xi = \partial_t + v\partial_x$.

Note first that $[e_{t'},e_{i'}] = 0$ hence if we consider a gyroscope comoving with $e_{t'}$ at $z = \frac{1}{g}$ then the gyroscope precesses relative to the axes $\{e_{i'}\}$ with angular velocity $\omega^{\alpha}|_{z = g^{-1}}\partial_{\alpha} = -\frac{1}{2}(\xi_{\mu}\xi^{\mu})^{-1}\epsilon^{\alpha\beta\gamma\delta}\xi_{\beta}\partial_{ \gamma}\xi_{\delta}|_{z = g^{-1}}\partial_{\alpha} = \frac{1}{2}\gamma^2 vg \partial_y$. It's very easy to interpret this precession through the equivalence principle if we consider the metric $g_{\mu'\nu'}$ in the coordinates centered on the observer moving across the platform because $g_{t'x'} = \Lambda_{t'}{}{}^{\mu} \Lambda_{x'}{}{}^{\nu}g_{\mu\nu} = \gamma^2 v (1 - g^2 z^2)$ so there's a gravitomagnetic field $\vec{\nabla}\times\vec{g}$ proportional to $\gamma^2 v g^2 z$ hence the gyroscope at $z = \frac{1}{g}$ precesses at a rate proportional to $\gamma^2 vg$ due to this gravitomagnetic field.

 

[PeterDonis]

kinematical time dilation and gravitational time dilation do add up?

Under certain limited kinds of scenarios, yes, you can think of them this way. But rockets moving between the Earth and the Sun are not really suited to this kind of thinking.

The key point about gravitational time dilation is that it is between two objects which are both *stationary*, as WBN pointed out. For example, if you are standing on the ground and I am standing at the top of a tall tower, there is gravitational time dilation between us--your clocks "run slower" than mine in an invariant sense, as we can both verify by, for example, exchanging light signals and measuring their round-trip travel times.

What you are calling "kinematical time dilation" is due to relative motion, and obviously two objects which are in relative motion can't both be stationary. For example, if you're standing on the ground and I jump out of an airplane high above you, I'm not stationary (though you are, because you are at rest relative to the Earth, which is the source of the gravitational field). So there is no invariant way to specify our "time dilation" relative to each other (gravitational or otherwise).

But now consider this scenario: you are standing on the ground, and I am orbiting the Earth in such a way that I pass directly over you on each orbit. Now there *is* a way for us to compare our "rates of time flow" in an invariant sense: we each measure how much time elapses, by our own clocks, between my successive passages overhead. In other words, since my motion is periodic, it can be treated as "stationary" for this purpose. And in this scenario, my "rate of time flow" relative to you will be the sum of two effects: my altitude (which makes my clock run faster than yours), and my orbital velocity (which makes my clock run slower than yours). These effects work in opposite directions, so the net effect will depend on their relative magnitudes; if you work the numbers, my clock will run slower than yours for low Earth orbits, but for higher orbits, my clock will run faster (the break point is, IIRC, an orbit at 1.5 Earth radii).

When I said 'upward acceleration' I really meant upward force

Yes, but you said "curved spacetime upward acceleration". The upward acceleration isn't due to curved spacetime; it's due to the rocket engine, or whatever else is exerting the force (if you're standing on the surface of the Earth, it's the Earth exerting force on you and pushing you upward).

 

[WannabeNewton]

Ok, I guess I'm wrong, but kinematical time dilation and gravitational time dilation do add up?

Sort of, at least in simple circumstances. For example if we have an observer in circular free fall orbit in Schwarzschild space-time then the observer's "gamma factor" in Schwarzschild coordinates will be $\gamma = (1 - 2M/R - R^2 \omega^2)^{-1/2}$ where the $2M/R$ accounts for the gravitational part and the $R^2 \omega^2$ accounts for the kinematical part. Then $\gamma^{-1}$ represents the rate at which the circularly orbiting observer's clock ticks relative to a clock at spatial infinity.