Gravity & Curvature: Explaining Geometrical Orbit of Particles

[Jorrie]

I'm unclear on what you mean by this. Please clarify.

I'm sorry Pete, from my side I don't quite get your question. The part on "curvature not required for gravitational effects" (like redshift) is exactly equivalent to what you yourself stated before:

Redshift and slowing of clocks are identical phenomena, i.e. different names for exactly the same thing. I was stating that the spacetime need not be curved in order for this to occur.

As far as "gravity=acceleration" goes, I reckon an extended lab that is linearly accelerated with constant acceleration does not look (from the inside) exactly like a gravitational field. The acceleration will be the same everywhere inside the lab, which is not the case with gravitational acceleration, i.e. curved spacetime.

As far as my speculation on "profiling" the acceleration goes, I think I missed it! An extended accelerating lab can surely never replicate curved spacetime, no matter how you accelerate it – or is there a way?

I agree with you on using mass rather than matter.

Jorrie

 

[pmb_phy]

I'm sorry Pete, from my side I don't quite get your question.

In response to the comment "Curved space appears as gravity" you wrote "Wrong. Curved space appears as tidal gravity." In essence, what you've stated here is that "Curved spacetime = gravity." is a different statement than "curved spacetime = tidal gravity." The truth of this statement hinges on the definition of gravity as used in GR. According to Von-Laue "Gravity is identical to spacetime curvature" whereas according to Einstein "Gravity is identical to gravitational acceleration." (i.e. non-vanishing of affine connection).

As far as "gravity=acceleration" goes, I reckon an extended lab that is linearly accelerated with constant acceleration does not look (from the inside) exactly like a gravitational field. The acceleration will be the same everywhere inside the lab, which is not the case with gravitational acceleration, i.e. curved spacetime.

Gravitational acceleration is defined by the affine connection while while spacetime curvature is defined by the Riemann tensor. When you use the phrase "exactly like a gravitational field, i.e. curved spacetime" it implies that you're using the definition of gravity given by Von-Laue and not that given by Einstein. That's fine. However if you do so then the equality "Curved space appears as gravity" is valid.

Pete

 

[Jorrie]

Thanks for the explanation Pete. I rest my case.

 

[Chaos' lil bro Order]

I think you are right so far. My "engineering view" on your other statements are as follows:

Wrong. Curved space appears as tidal gravity. As very ably explained by pervect before, space need not be curved for gravitational effects to appear.

Wrong. When viewed at the same gravitational potential as the potential it was transmitted from, there is no redshift, only deflection and a Shapiro time delay of the light.

Only correct in a restricted sense - very localized, e.g. in a vanishingly small accelerating laboratory. That is unless the acceleration has a profile that precisely mimics curved space-time - maybe?

I would love to hear an expert's opinion

Sorry Jorie, I don't think you understand what you are talking about. You say space need not be curved for gravitational effects to appear. This seems very wrong to me, even in your oddly worded statement. Gravity IS the product of curved space, plain and simple. If you can point out a local region of space where a non-uniform gravitational field exists that is not somehow curved, please do so, otherwise you are wrong.

If you are arguing that gravity and acceleration cannot be thought of as equivalent, I suggest you read a book on Relativity, or if you have, reread it. A lab accelerating upwards will experience an effect that is equivalent to gravity, but your one correct point is that the graviational effect in the lab will be uniform and not dissipated by the inverse-square law like real gravity would. Cudos on 1/4.

 

[MeJennifer]

It can easily (hopefully) seen, that for an object "at rest" in the inertial coordinates, dxx=dyy=dzz=0, and thus dtau = dtt, and there is no time dilation. Coordinate time is the same as proper time.

Makes sense!

It can also be seen that for an object "at rest" in the accelerated coordinates, dx=dy=dz=0, and thus dtau = (1+gz)*dt, and there is gravitational time dilation. The relationship between coordinate time and proper time is a function of "height", the z-coordinate. This is time dilation, when coordinate time is not the same as proper time, and the amount of time dilation depends on the "height" of the object, i.e. it's z coordinate.

I suppose you mean the t-coordinate instead of the z-coordinate.

It should be understood that the object with xx=yy=zz=constant, which is "at rest" in inertial coordinates, is not following the same trajectory as the object with x=y=z=constant, which is "at rest" in the coordinate system used by the accelerated observer.

Right.

It should also be understood that a lot of the confusion arises from the use of concepts such as 'time dilation' that are coordinate dependent. When one chooses to use a coordinate dependent explanation (because of familiarity on the part of readers, usually), one has to live with the fact that it is coordinate dependent.

Well I agree.
Then why not stay with space-time coordinates? From here we can make any coordinate transformation.
Don't you think that people then will get a better understanding of the "magic" of contraction and "slower" clocks?

An explanation formulated purely in terms of coordinate independent quantites (like the Lorentz interval and abstract tensor notation) doesn't have to deal with these issues. But it is not generally as accessible.

Well what do you see as the problem with regards to using space-time coordinates?

 

[Jorrie]

Sorry Jorie, I don't think you understand what you are talking about. You say space need not be curved for gravitational effects to appear. This seems very wrong to me, even in your oddly worded statement. Gravity IS the product of curved space, plain and simple.

Maybe we are both just guilty of expressing ourselves badly - you probably meant "Gravity IS the product of curved spacetime…..", not just curved space
Now the interesting thing is that the weak equivalence principle alone is enough to create gravity – and it is without space curvature, just like in the small, uniformly accelerating lab. The weak equivalence principle manifests itself as gravitational redshift and also produces half of the gravitational bending of light. The other half is produced by space curvature. Tidal gravity, on the other hand, only appears in a curved spacetime environment.
It is from this point of view that I disagreed with the statements made in the OP.
Jorrie

 

[pmb_phy]

Sorry Jorie, I don't think you understand what you are talking about.

Jorrie seems well versed in GR in my opinion.

You say space need not be curved for gravitational effects to appear.

This is true depending on what one means by "gravitational effects." In Einstein's GR gravitational effects consist in part of gravitational acceleration, gravitational redshift. There is no requirement for spacetime to be curved for these effects to exist.

Gravity IS the product of curved space, plain and simple.

You've arrived at this conclusion by defining gravitational effects to be identical with tidal acceleration. So you haven't demonstrated anything other than you are using a definition which differs from Einstein.

If you can point out a local region of space where a non-uniform gravitational field exists that is not somehow curved, please do so, otherwise you are wrong.

This is nonsensical since Jorrie already clarified that tidal acceleration and spacetime curvature are identical, i.e. different names for the same phennomena. Thus there is no spacetime curvature in a uniform g-field because a uniform g-field is defined by requiring that the Riemann tensor vanishes in such a field.

I suggest you read a book on Relativity, or if you have, reread it.

Sorry but it appears quite clear that Jorrie knows what he's talking about. I recommend that you pick up the works of Einstein on GR and read them carefully and focus your attention on how Einstein defines gravity in GR.

A lab accelerating upwards will experience an effect that is equivalent to gravity, but your one correct point is that the graviational effect in the lab will be uniform and not dissipated by the inverse-square law like real gravity would. Cudos on 1/4.

The term "real gravity " is based on a biased view on gravity, i.e. that gravity is identified by spacetime curvature. In this view a uniform gravitational field is a contradiction in terms and it makes no sense to speak of gravity, never mind referring to a uniform field.

Pete

 

[pervect]

I suppose you mean the t-coordinate instead of the z-coordinate.

dtau = (1+gz)*dt

So dt/dtau = 1/1+gz

This is a function of z, i.e. a function of height. g here is a multiplicative constant for z, the acceleration of the spaceship, in case that isn't clear.

As far as space-time coordinates goes, there are many possible choices for coordinate systems. The fact that SR is Lorentz covariant says that we will get the same answer, in the end, regardless of which one of them we choose.

However, the path to the unique answer may not always appear to be the same on a popular level - i.e. we wind up talking about "gravitational time dilation" in some coordinates, and not in others. This is unavoidable if we use the coordinate approach - only a highly abstract approach, using geoemtric entities rather than coordinates, can avoid this.

 

[MeJennifer]

dtau = (1+gz)*dt

So dt/dtau = 1/1+gz

This is a function of z, i.e. a function of height. g here is a multiplicative constant for z, the acceleration of the spaceship, in case that isn't clear.

I see.

However, the path to the unique answer may not always appear to be the same on a popular level - i.e. we wind up talking about "gravitational time dilation" in some coordinates, and not in others. This is unavoidable if we use the coordinate approach - only a highly abstract approach, using geoemtric entities rather than coordinates, can avoid this.

Well it seems to me that transforming the local coordinates to space-time will give resolve these "abiguities". Comparing coordinate time with proper time will be an indication if there is any time "dilation". The terms "time dilation" and "length contraction" are prone to confusion IMHO, because they only have a meaning when you compare coordinate systems.