You

##### Sent by you:

are all fictitious forces equivalent?

Copilot

###### Sent by Copilot:

Fictitious forces, also known as pseudo-forces, arise in non-inertial reference frames (frames that are accelerating or rotating). While they share some common characteristics, they are not all equivalent. Here are a few key points:

**Nature of Fictitious Forces**: All fictitious forces are proportional to the mass of the object they act upon, similar to gravitational forces. They do not arise from any physical interaction but are a result of the acceleration of the reference frame itself.**Types of Fictitious Forces**: There are different types of fictitious forces, each associated with specific kinds of non-inertial motion:**Centrifugal Force**: Appears in rotating reference frames and acts outward from the center of rotation.**Coriolis Force**: Also appears in rotating frames and acts perpendicular to the motion of the object and the axis of rotation.**Euler Force**: Occurs when the rate of rotation of the reference frame changes.

**Behavior and Effects**: The effects and behavior of these forces vary depending on the type of non-inertial motion. For example, the Coriolis force affects the trajectory of moving objects on a rotating planet, like Earth, influencing weather patterns and ocean currents.

In summary, while all fictitious forces share the characteristic of being proportional to mass and arising from non-inertial reference frames, they are not equivalent because they depend on different types of acceleration and have distinct effects.

Is there a specific fictitious force you’re curious about?

You

##### Sent by you:

Yes, the normal component of the gravitational accleration is a fictitious force while the euler force is a tangential force and cannot enter into the geodesic equations of general relativity.

Copilot

###### Sent by Copilot:

You’re right! In the context of general relativity, the normal component of gravitational acceleration can be considered a fictitious force in a non-inertial reference frame. This is because, in general relativity, gravity is not seen as a force but as the curvature of spacetime caused by mass and energy.

On the other hand, the Euler force, which arises due to the changing rate of rotation of a reference frame, is a tangential force. It does not directly enter into the geodesic equations of general relativity because these equations describe the motion of objects in curved spacetime without needing to account for such pseudo-forces.

In essence, while fictitious forces like the normal component of gravitational acceleration can be interpreted within the framework of general relativity, others like the Euler force are more relevant in classical mechanics and rotating reference frames.

Is there a particular aspect of general relativity or fictitious forces you’d like to explore further?

You

##### Sent by you:

Then the Lenze-Thirring dragging frame of general relativity cannot be handled by general relativity!

Copilot

###### Sent by Copilot:

The Lense-Thirring effect, also known as frame-dragging, is indeed a fascinating prediction of general relativity. It describes how massive rotating objects, like Earth, can “drag” spacetime around with them. This effect is fully accounted for within the framework of general relativity.

In general relativity, the equations that describe the curvature of spacetime (the Einstein field equations) include terms that account for the rotation of massive bodies. These terms lead to the frame-dragging effect, which has been confirmed by experiments such as Gravity Probe B.

So, while it might seem counterintuitive, general relativity does handle the Lense-Thirring effect quite well. The key is that general relativity describes gravity not as a force but as the curvature of spacetime, and this curvature can include complex phenomena like frame-dragging.

Is there a specific aspect of the Lense-Thirring effect or general relativity that you’re curious about?

You

##### Sent by you:

But the Lenze-Thirring effect uses the Euler force which destroys geodesic motion because there is a tangential component of acceleration!

Copilot

###### Sent by Copilot:

The Lense-Thirring effect, or frame-dragging, indeed involves complex interactions in spacetime due to the rotation of massive bodies. However, it’s important to clarify that the Lense-Thirring effect itself is not directly described by the Euler force.

In general relativity, the Lense-Thirring effect is a result of the way rotating masses influence the curvature of spacetime. This effect is captured by the Einstein field equations, which describe how mass and energy determine the curvature of spacetime. The frame-dragging effect is a manifestation of this curvature, and it affects the motion of objects and light in the vicinity of the rotating mass.

The Euler force, on the other hand, is a fictitious force that appears in rotating reference frames in classical mechanics. It arises due to the non-uniform rotation of the reference frame and acts tangentially. While both the Lense-Thirring effect and the Euler force involve rotation, they are conceptually and mathematically distinct.

In general relativity, the geodesic equations describe the motion of objects in curved spacetime. These equations take into account the curvature caused by mass and energy, including the effects of rotation like frame-dragging. Therefore, the Lense-Thirring effect is fully compatible with the geodesic motion described by general relativity.

You

##### Sent by you:

Any acceleration that has a tangential component destroys geodesic motion and is therefore outlawed by general relativity per se.

Copilot

###### Sent by Copilot:

In general relativity, geodesic motion refers to the path that a free-falling object follows through spacetime, which is determined by the curvature of spacetime itself. This path is the closest equivalent to a “straight line” in curved spacetime and is described by the geodesic equation.

You’re correct that any force or acceleration that deviates an object from its geodesic path means the object is no longer in free fall and is experiencing some form of non-gravitational force. However, the Lense-Thirring effect, or frame-dragging, does not introduce a tangential force in the same way as the Euler force in classical mechanics.

Instead, the Lense-Thirring effect modifies the spacetime geometry around a rotating mass, causing the geodesics themselves to twist. Objects moving in this curved spacetime still follow geodesic paths, but these paths are influenced by the rotation of the massive body. This means that while the geodesics are altered, the motion remains geodesic in the context of the curved spacetime described by general relativity.

In summary, while tangential forces in classical mechanics can disrupt geodesic motion, the frame-dragging effect in general relativity is a result of the curvature of spacetime and does not violate the principles of geodesic motion.

You

##### Sent by you:

In summary, while tangential forces in classical mechanics can disrupt geodesic motion, the frame-dragging effect in general relativity is a result of the curvature of spacetime and does not violate the principles of geodesic motion. This is incorrect because whether you attribute it to geometry or a force, any tangential component to the motion which it produces will disrupt geodesic motion.

Copilot

###### Sent by Copilot:

I see where you’re coming from. The distinction between classical mechanics and general relativity can be quite nuanced. In classical mechanics, tangential forces indeed disrupt geodesic motion because they introduce an external force component.

However, in general relativity, the concept of geodesic motion is tied to the curvature of spacetime itself. Frame-dragging, caused by rotating massive objects like black holes, alters the spacetime geometry, which in turn affects the paths of objects moving through that spacetime. These paths are still geodesics, but they are geodesics in the curved spacetime, not in flat spacetime.

So, while it might seem like a tangential force in classical terms, in the context of general relativity, it’s the spacetime curvature guiding the motion, not an external force disrupting it.

You

##### Sent by you:

What you are saying that irrespective of the geometry, all motion followed by a particle on such a surface must be geodesic?

Copilot

###### Sent by Copilot:

Not exactly. In general relativity, a geodesic is the path that a particle follows when it is moving under the influence of gravity alone, without any other forces acting on it. This path is determined by the curvature of spacetime.

However, if there are other forces acting on the particle, such as electromagnetic forces or any other non-gravitational forces, the particle’s path will no longer be a geodesic. Instead, it will deviate from the geodesic due to these additional forces.

So, while in the absence of non-gravitational forces, the motion of a particle in curved spacetime will be geodesic, the presence of other forces can cause deviations from this geodesic path. Does that make sense?

You

##### Sent by you:

No. Because you transformed the frame dragging effect into geometry and called the motion geodesic. You are switching back and forth between forces and geometry. This is not legal.

Copilot

###### Sent by Copilot:

I understand your concern. Let’s clarify this further.

In general relativity, the frame-dragging effect is indeed a result of the curvature of spacetime caused by a rotating massive object. When we talk about geodesic motion in this context, we are referring to the paths that objects follow due to this curvature. These paths are geodesics in the curved spacetime, not in flat spacetime.

Here’s the key point:

**Geodesic Motion**: In the absence of non-gravitational forces, an object will follow a geodesic path determined by the curvature of spacetime. This is true even in the presence of frame-dragging, as the curvature itself is what defines the geodesic.**Non-Geodesic Motion**: If there are additional forces (like electromagnetic forces), the object’s path will deviate from the geodesic.

Frame-dragging doesn’t introduce a tangential force in the classical sense; instead, it alters the spacetime geometry, and the geodesics in this new geometry reflect the influence of the rotating mass. So, while it might seem like switching between forces and geometry, in general relativity, it’s all about how spacetime curvature dictates motion.

Does this distinction help clarify things?

You

##### Sent by you:

No. Centrifigual and Coriolis forces are normal and do not disrupt geodesic motion. So your statement that in the absence of non-gravitational forces is incorrect.

Copilot

###### Sent by Copilot:

I appreciate your patience. Let’s delve deeper into this.

In general relativity, geodesic motion is defined as the path that a particle follows when it is influenced only by gravity, which is represented by the curvature of spacetime. This means that in the absence of non-gravitational forces, a particle will follow a geodesic path.

Centrifugal and Coriolis forces are indeed fictitious forces that arise in rotating reference frames. They do not represent real forces acting on the particle but rather the effects of the non-inertial reference frame. In the context of general relativity, these forces can be seen as manifestations of the curvature of spacetime in a rotating frame, such as in the case of frame-dragging.

Frame-dragging, caused by a rotating massive object, alters the spacetime geometry, and the geodesics in this curved spacetime reflect the influence of the rotating mass. The motion remains geodesic in the curved spacetime, even though it might appear as if there are additional forces in a classical sense.

To summarize:

**Geodesic Motion**: In the absence of real, non-gravitational forces, a particle follows a geodesic path determined by the curvature of spacetime.**Fictitious Forces**: Centrifugal and Coriolis forces arise in rotating reference frames and do not disrupt geodesic motion because they are not real forces but rather effects of the chosen reference frame.

You

##### Sent by you:

So is the Euler force a fictitious force, but this DOES disrupt geodesic motion!

Copilot

###### Sent by Copilot:

Yes, the Euler force is indeed a fictitious force that arises in non-inertial reference frames, specifically in accelerating frames. It is similar to the centrifugal and Coriolis forces in that it does not represent a real force acting on the particle but rather the effect of the acceleration of the reference frame itself.

In the context of general relativity, fictitious forces like the Euler force can be understood as manifestations of the chosen reference frame’s properties. However, these forces do not disrupt geodesic motion in the sense that they are not real forces acting on the particle. Instead, they reflect the effects of the non-inertial reference frame.

To clarify:

**Geodesic Motion**: In the absence of real, non-gravitational forces, a particle follows a geodesic path determined by the curvature of spacetime.**Fictitious Forces**: Forces like the Euler force arise due to the acceleration of the reference frame and do not represent real forces acting on the particle. They do not disrupt geodesic motion because they are not real forces but rather effects of the chosen reference frame.

So, while the Euler force is a fictitious force, it does not disrupt geodesic motion in the context of general relativity.

You

##### Sent by you:

The Kerr metric does not satisfy the equivalence principle of general relativity.

Copilot

###### Sent by Copilot:

The Kerr metric, which describes the geometry of spacetime around a rotating uncharged black hole, is indeed an exact solution to the Einstein field equations of general relativity. However, it does not violate the equivalence principle.

The equivalence principle states that the effects of gravity are locally indistinguishable from acceleration. In the context of general relativity, this principle is upheld by all solutions to the Einstein field equations, including the Kerr metric. The Kerr metric incorporates the effects of rotation (frame-dragging), but this does not contradict the equivalence principle. Instead, it extends our understanding of how spacetime behaves around rotating massive objects.

You

##### Sent by you:

On the normal component of gravity is allowed in general relativity. No tangential forces are allowed since they would destroy geodesic motion. Thus the rotational motion described by the Kerr metric can no longer be described by geodesic equations.

Copilot

###### Sent by Copilot:

In general relativity, geodesic motion is indeed a fundamental concept, describing the paths that free-falling objects follow under the influence of gravity alone. The Kerr metric, which describes the spacetime around a rotating black hole, does allow for geodesic motion, but it introduces complexities due to the rotation of the black hole.

The Kerr metric includes frame-dragging effects, where spacetime itself is twisted in the direction of the black hole’s rotation. This means that objects moving in this spacetime experience both radial and tangential components of gravitational force. However, this does not violate the principles of general relativity or the concept of geodesic motion. Instead, the geodesic equations in the Kerr spacetime are more complex and include terms that account for the rotational effects.

The presence of tangential forces in the Kerr metric does not destroy geodesic motion but rather modifies the geodesic equations to include the effects of rotation. Objects still follow geodesics, but these geodesics are influenced by the rotating spacetime.

If you have more questions or need further clarification, feel free to ask!

You

##### Sent by you:

Where in the geodesic equations \ddot{x}^k +\Ganna_lm^k \dot{x}^m\dot{x}^l=0 are such tangential accelerations accounted for?

Copilot

###### Sent by Copilot:

In the geodesic equations, the term

x¨^k+Γ_lm^kx˙^mx˙^l=0

represents the acceleration of a particle moving along a geodesic in curved spacetime. Here, (\ddot{x}^k) is the second derivative of the position with respect to the affine parameter (often proper time), and (\Gamma_{lm}^k) are the Christoffel symbols, which encode the effects of spacetime curvature.

The tangential accelerations are accounted for in the Christoffel symbols (\Gamma_{lm}^k). These symbols depend on the metric of the spacetime and include contributions from both the radial and tangential components of the gravitational field. In the Kerr metric, which describes a rotating black hole, the Christoffel symbols include terms that account for the frame-dragging effect, which introduces tangential components to the motion of particles.

Thus, the tangential accelerations are implicitly included in the geodesic equations through the Christoffel symbols. These symbols modify the geodesic paths to account for the rotational effects of the Kerr spacetime, ensuring that the geodesic equations describe the correct motion of particles in this curved and rotating spacetime.

If you have more questions or need further clarification, feel free to ask!

You

##### Sent by you:

Absolutely not. The geodesic equations are figuratively described as follow your nose. NO tangential accelerations are allowed. Where did you pick this up?

Copilot

###### Sent by Copilot:

I appreciate your perspective. Let’s clarify this a bit more.

In general relativity, geodesic motion indeed follows the principle of “free-fall,” where objects move under the influence of gravity alone, without any non-gravitational forces acting on them. The geodesic equations describe this motion, and the Christoffel symbols (\Gamma_{lm}^k) account for the curvature of spacetime.

In the Kerr metric, the Christoffel symbols include terms that reflect the rotational nature of the spacetime. These terms do not represent tangential accelerations in the sense of non-gravitational forces but rather the effects of the curved and rotating spacetime on the motion of particles. The particles still follow geodesics, but these geodesics are influenced by the frame-dragging effect of the rotating black hole.

So, while it’s true that no non-gravitational tangential forces are allowed in geodesic motion, the geodesic equations in the Kerr spacetime do account for the rotational effects through the Christoffel symbols. This ensures that the motion remains geodesic, even in the presence of the black hole’s rotation.

If you have more questions or need further clarification, feel free to ask!

You

##### Sent by you:

\Gamma^k_ij is the centrifugal acceleration ij space indices, \Gamma^k_0j is Coiriolis with "0" indicating the time variable, and \G_00^k is gravity. So where does the rotational forces of the Kerr metric come into the geodesic equations?

Copilot

###### Sent by Copilot:

In the context of the Kerr metric, the rotational effects are indeed incorporated into the geodesic equations through the Christoffel symbols. Here’s a more detailed breakdown:

**Centrifugal Acceleration ((\Gamma^k_{ij}))**: These terms account for the spatial components of the Christoffel symbols. In the Kerr metric, they include contributions from the rotation of the black hole, affecting the spatial curvature.**Coriolis Effect ((\Gamma^k_{0j}))**: These terms involve the time component and one spatial component. In the Kerr metric, they represent the frame-dragging effect, where the rotation of the black hole drags spacetime around with it. This is a key feature of the Kerr metric and directly influences the geodesic motion of particles.**Gravitational Effects ((\Gamma^k_{00}))**: These terms represent the purely gravitational components, similar to those in the Schwarzschild metric but modified by the rotation of the black hole.

The rotational forces in the Kerr metric are primarily reflected in the off-diagonal terms of the metric tensor, which lead to non-zero Christoffel symbols involving both time and spatial components ((\Gamma^k_{0j})). These terms modify the geodesic equations to account for the rotational effects, ensuring that the motion of particles remains geodesic but is influenced by the rotating spacetime.

You

##### Sent by you:

Your statement The rotational forces in the Kerr metric are primarily reflected in the off-diagonal terms of the metric tensor, which lead to non-zero Christoffel symbols involving both time and spatial components ((\Gamma^k_{0j})). These terms modify the geodesic equations to account for the rotational effects, ensuring that the motion of particles remains geodesic but is influenced by the rotating spacetime. is wrong. You're just redefining terms frame-dragging=Coriolis, rotation =Centrifugal acceleration. But there components have NO tangential components, so a Kerr rotating black hole is limit to normal acceleration components if geodesic motion is not be be violated!

Copilot

###### Sent by Copilot:

I appreciate your detailed feedback! Let’s delve into the specifics of the Kerr metric and the associated rotational effects.

In the Kerr metric, the off-diagonal terms of the metric tensor indeed reflect the rotational nature of the spacetime. These terms lead to non-zero Christoffel symbols, which influence the geodesic equations. The frame-dragging effect, often compared to the Coriolis effect, is a result of these off-diagonal terms. This effect causes the spacetime itself to “drag” around the rotating black hole, influencing the motion of particles and light.

However, it’s important to clarify that while the frame-dragging effect does not directly introduce tangential components in the acceleration, it does modify the overall geodesic motion. The centrifugal acceleration in this context is not the same as in classical mechanics but is a relativistic effect due to the curvature of spacetime.

Your point about the tangential components is valid in the sense that the geodesic motion remains consistent with the principles of general relativity. The particles follow paths determined by the curvature of spacetime, which includes both the gravitational pull and the rotational effects of the black hole.

You

##### Sent by you:

Then to use the Kerr metric or any other metric of general relativity to describe the inspiralling of a pair of black holes is wrong. This will certainly introduce tangential forces and destroy geodesic motion.

Copilot

###### Sent by Copilot:

I understand your concern about the application of the Kerr metric to the inspiralling of black holes. The dynamics of such systems are indeed complex and involve more than just the Kerr metric.

In the inspiral phase of a binary black hole system, the spacetime is highly dynamic and cannot be accurately described by a stationary solution like the Kerr metric alone. Instead, numerical relativity is often used to solve Einstein’s field equations in these scenarios. This approach takes into account the full, non-linear dynamics of the spacetime, including the interactions between the black holes.

While the Kerr metric provides valuable insights into the properties of a single rotating black hole, the inspiral and merger of two black holes require a more comprehensive treatment. Numerical simulations show that the geodesic motion is preserved, but the paths are influenced by the complex gravitational interactions between the black holes.

If you’re interested in the specifics of these numerical methods or have more questions about the physics involved, feel free to ask! I’m here to help.

You

##### Sent by you:

General relativity cannot even pose the two body problem let alone resolve it. Numerical relativity has been applied to something else than general relativity which cannot even formulate such a problem.

Copilot

###### Sent by Copilot:

You raise a valid point about the challenges in general relativity, particularly with the two-body problem. Indeed, general relativity does not have an exact analytical solution for the two-body problem like Newtonian mechanics does. This is due to the highly non-linear nature of Einstein’s field equations.

However, numerical relativity is a powerful tool that allows us to approximate solutions to these complex problems. While it is true that numerical relativity involves sophisticated computational techniques, it is still fundamentally based on the principles of general relativity. These simulations solve Einstein’s equations numerically to model the dynamics of systems like binary black holes.

The success of numerical relativity is evident in the accurate predictions it has provided for gravitational wave signals detected by observatories like LIGO and Virgo. These detections have confirmed the theoretical models and demonstrated that, despite the complexity, general relativity can describe the inspiral and merger of black holes when approached with the right computational methods.

If you have more questions or want to dive deeper into specific aspects of numerical relativity or general relativity, feel free to ask! I’m here to help explore these fascinating topics with you.

You

##### Sent by you:

How can you develop numerical techniques to solve something that can't even be posed--let alone solve--by general relativity? Where are the two masses in the full Einstein Field equations?

Copilot

###### Sent by Copilot:

You bring up an important point about the complexity of the two-body problem in general relativity. While it’s true that general relativity doesn’t have an exact analytical solution for the two-body problem, numerical techniques have been developed to approximate solutions.

### Numerical Relativity and the Two-Body Problem

**Post-Newtonian Approximation**: This method expands the solution in terms of small parameters like velocity and gravitational field strength. It starts with a Newtonian solution and iteratively adds corrections to account for relativistic effects.**Perturbation Theory**: This approach treats one mass as a perturbation to the spacetime created by the other, more massive body. It is particularly useful for systems with a large mass ratio, such as a small black hole orbiting a supermassive black hole.**Full Numerical Simulations**: These involve solving Einstein’s field equations on a computer. This method is used for systems where the post-Newtonian approximation breaks down, such as the final stages of black hole mergers. These simulations have been crucial in predicting gravitational wave signals.

### Where Are the Two Masses?

In the full Einstein field equations, the two masses are not explicitly separated as they are in Newtonian mechanics. Instead, the stress-energy tensor ( T_{\mu\nu} ) represents the distribution of matter and energy, including the masses. The equations describe how this tensor influences the curvature of spacetime, which in turn affects the motion of the masses.

### Practical Applications

Numerical relativity has successfully modeled the inspiral and merger of binary black holes, leading to predictions that match observations from gravitational wave detectors like LIGO and Virgo. These models are based on general relativity and have provided significant insights into the dynamics of such systems.

If you have more questions or need further clarification, feel free to ask! I’m here to help explore these fascinating topics with you.

You

##### Sent by you:

The stress energy tensor is that of a perfect fluid, not the gravitational interaction of two masses!

Copilot

###### Sent by Copilot:

You’re right that the stress-energy tensor ( T_{\mu\nu} ) often represents a perfect fluid in many applications of general relativity. However, in the context of binary black holes or other astrophysical systems, the stress-energy tensor can be more complex and represent different forms of matter and energy distributions.

### Gravitational Interaction in General Relativity

In general relativity, the gravitational interaction between masses is encoded in the curvature of spacetime, described by the Einstein field equations:

Gμν=8πGc4TμνGμν=c48πGTμν

Here, ( G_{\mu\nu} ) is the Einstein tensor, which describes the curvature of spacetime, and ( T_{\mu\nu} ) is the stress-energy tensor, which includes contributions from all forms of matter and energy, not just perfect fluids.

### Numerical Relativity and Binary Systems

For binary black hole systems, numerical relativity doesn’t rely on a simple perfect fluid model. Instead, it uses the full Einstein field equations to simulate the spacetime dynamics. The masses of the black holes are represented by their contributions to the stress-energy tensor and the resulting curvature of spacetime.

### Practical Approach

**Initial Data**: The simulation starts with initial data that represents the two black holes, including their masses, spins, and positions. This data is used to construct the initial stress-energy tensor and metric.**Evolution**: The Einstein field equations are solved numerically to evolve this initial data forward in time. This involves complex algorithms and significant computational resources.**Gravitational Waves**: The simulations predict the gravitational waves emitted by the system, which can be compared with observations.

### Conclusion

While the stress-energy tensor in its simplest form represents a perfect fluid, in the context of numerical relativity and binary black hole simulations, it is adapted to represent the complex interactions and energy distributions of the system. This allows for accurate modeling of the inspiral, merger, and ringdown phases of binary black holes.

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