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SergUpstart

The speed of propagation of gravity

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4 minutes ago, SergUpstart said:

Then the main question is, does the relativistic mass satisfy the classical definition of mass as a measure of resistance to changes in inertia?

Not sure why it matters, since you have implied that there is a medium with physical properties.  This just seems like a distraction.

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Yes it is a mass term. Any mass term must satisfy that definition under kinematics and the laws of inertia. 

The total mass (both types) is given by the energy momentum equation.

E=mc^2 only describes the invariant mass of a particle.

[math]E^2=\overbrace{(pc)^2}^{kinetic-energy}+\overbrace{(m_0c^2)^2)}^{potential-energy}[/math]

It is the momentum term in the above formula that allows photons to provide thrust to a solar sail.

Little trivia side note the reason to square the energy term is to denote positive norm. Ie no negative energy allowed. 

Edited by Mordred

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36 minutes ago, Mordred said:

Little trivia side note the reason to square the energy term is to denote positive norm. Ie no negative energy allowed. 

Is there a better way to phrase what this is supposed to mean? We notice that the identity only specifies the value \(E^2.\) Surely this quite definitely allows for two different final solutions for \(E\) itself in the nonzero case; one positive and one negative. 

Edited by taeto

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What happens when you square a negative number ? 

Then consider energy density is always positive. In essence all particles has positive energy. Secondly all particles will have positive mass. The positive norm also applies to particle states hence squaring the wavefunction

[math]|\psi^2\rangle [/math] for its probability amplitudes and density functions.

Edited by Mordred

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8 hours ago, taeto said:

Is there a better way to phrase what this is supposed to mean?

The identity given is the the norm (“length”, in some sense) of the energy-momentum 4-vector. The norm of a vector is always positive, since you cannot have vectors of negative length.

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14 hours ago, Mordred said:

What happens when you square a negative number ? 

Then consider energy density is always positive. In essence all particles has positive energy. Secondly all particles will have positive mass. The positive norm also applies to particle states hence squaring the wavefunction

|ψ2 for its probability amplitudes and density functions.

 

15 hours ago, Mordred said:

Yes it is a mass term. Any mass term must satisfy that definition under kinematics and the laws of inertia. 

The total mass (both types) is given by the energy momentum equation.

E=mc^2 only describes the invariant mass of a particle.

E2=(pc)2kineticenergy+(m0c2)2)potentialenergy

It is the momentum term in the above formula that allows photons to provide thrust to a solar sail.

Little trivia side note the reason to square the energy term is to denote positive norm. Ie no negative energy allowed. 

Another question. We found out that a relativistic mass is an inert mass, but is a relativistic mass a gravitational mass?

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19 minutes ago, SergUpstart said:

We found out that a relativistic mass is an inert mass, but is a relativistic mass a gravitational mass?

Yes, but not a straightforward way because it affects more components of the stress-energy tensor than rest mass. 

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Are you sure about that Strange ?
A rest ( invariant ) mass can never be reduced, while a relativistic ( variant ) mass,  can be reduced to zero in certain frames.
While I realize that the energy-momentum, the variant component of an invariant mass in motion, must have an effect, it can lead to troublesome conclusions. You could, arguably, have a large enough variant component that, a relativistically moving object, collapses gravitationally in one frame, while it doesn't in its rest frame ( credit to Swansont for pointing this out numerous times ). As a matter of fact, even the gravitational field itself must add another component to the stress energy-momentum tensor in the equation.
It seems 'mass' isn't that easy to define in GR.

Maybe Mordred or Markus can explain how variant and invariant masses are differently treated by GR.
( I have heard of ADM mass, and even Comar mass, related to Bondi energy, but have no understanding )

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25 minutes ago, MigL said:

Are you sure about that Strange ?

No. :)

My understanding is that an object moving (at relativistic speeds) will have increased gravitational effect (in the frame of reference of the "stationary" observer) but this is nowhere near as simple as substituting the relativistic mass into the Newtonian equation, for example. Factors such as momentum flow complicate the calculation. And I have no idea what the net result would be - maybe it all cancels out (which would address the gravitational collapse "paradox"; but I think that is simply explained by the fact that in its own frame of reference there is no relativistic mass increase).

29 minutes ago, MigL said:

Maybe Mordred or Markus can explain how variant and invariant masses are differently treated by GR.

Let us hope so!

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5 hours ago, SergUpstart said:

 

Another question. We found out that a relativistic mass is an inert mass, but is a relativistic mass a gravitational mass?

I believe you meant inertia mass not inert mass in the above.

You might want to kick yourself Strange as to gravitational mass. You are right GR does handle gravitational mass but there is a straightforward  explanation.

GR applies the principle of equivalence the gravitational mass is the same as the inertial mass.

[math] m_i=m_g [/math] this is oft described by the Einstein elevator. The four vectors of GR however will preserve the invariant (rest mass)

Edited by Mordred

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34 minutes ago, Mordred said:

You might want to kick yourself Strange as to gravitational mass.

I'm on it.

Ouch

Ouch

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14 hours ago, Strange said:

My understanding is that an object moving (at relativistic speeds) will have increased gravitational effect (in the frame of reference of the "stationary" observer)

Gravity is not an observer-dependent phenomenon; if you have some \(R^{\mu}{_{\nu \alpha \delta}} \neq 0 \) in one frame, then you will find the same in all other frames as well. When an observer and a gravitational source are in relative motion, what changes is only the form of the metric - however, the metric in the rest frame of the gravitational source and the metric in the rest frame of the observer will be related via a simple coordinate transformation, so we are actually dealing with the exact same spacetime. So very simply put, if a gravitational source is in relative motion, the metric might look “distorted” in some way to an observer, but it will produce the exact same physics. Essentially, relative motion just means we are using different coordinates to describe the same spacetime.

The relevant solution to the field equations for this scenario is called the Aichelburg-Sexl ultraboost.

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I'm a little confused about how GR handles variant mass also ( see my last post, 15 hr ago )
As per your explanation,  I can see how the metric is a simple coordinate transformation, as that is what general covariance implies.

But the variant mass would have the invariant component, a momentum component due to its motion, in common with SR, but GR would  possibly even have a gravitational component due to the fact that gravity gravitates..

Can you shed some light on this ( keep it simple ).

 

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2 hours ago, Markus Hanke said:

Gravity is not an observer-dependent phenomenon

So this seems to contradict the comment from Mordred, above (although that was brief, verging on cryptic, so I am have misunderstood what was intended!)

But what you are saying is that "relativistic mass" (which is observer dependent) does not contribute to the gravitational effect of an object?

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2 hours ago, Strange said:

So this seems to contradict the comment from Mordred, above (although that was brief, verging on cryptic, so I am have misunderstood what was intended!)

But what you are saying is that "relativistic mass" (which is observer dependent) does not contribute to the gravitational effect of an object?

The short form of this is "will an object become a black hole if it moves fast enough?" and the answer is "no"

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10 minutes ago, swansont said:

The short form of this is "will an object become a black hole if it moves fast enough?" and the answer is "no"

Indeed. And that is easily explained by the fact that there is no increase in relativistic mass (kinetic energy) in the object's own frame of reference.

But it isn't so obvious why the extra kinetic energy / relativistic-mass seen by a "stationary" observer as the object passes by does not increase the gravitational effect of the object.

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6 hours ago, swansont said:

The short form of this is "will an object become a black hole if it moves fast enough?" and the answer is "no"

Based on SR, a fast-moving object with a rest mass will become a black hole for an external observer, even if the relativistic increase in mass is not taken into account. It should only become a black hole by relativistically reducing its size when it becomes smaller than the Schwarzschild radius for its invariant mass.

5 hours ago, Strange said:

 

But it isn't so obvious why the extra kinetic energy / relativistic-mass seen by a "stationary" observer as the object passes by does not increase the gravitational effect of the object.

It can be explained by the action of a spin field (analogous to a magnetic field for gravity). If two moving charged particles interact, the Coulomb force decreases by a factor of K, where K=1/sqrt(1-v^2/c^2), due to the fact that the Coulomb force is partially compensated by the Lorentz force. In the gravitational interaction formula, this factor K will enter once in the numerator, due to the relativistic increase in mass, the second time in the denominator, due to the action of the spin field, and as a result, it will decrease

F=G*m1*m2*K/r^2K=G*m1*m2/r^2

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The above is completely wrong. The first paragraph is a common misconception of not recognizing the reference frame relations.

 The second part isnt inapplicable in that gravity doesn't behave in the same manner as the EM field ie the coulomb force isn't involved.

 

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5 hours ago, SergUpstart said:

Based on SR, a fast-moving object with a rest mass will become a black hole for an external observer, even if the relativistic increase in mass is not taken into account. It should only become a black hole by relativistically reducing its size when it becomes smaller than the Schwarzschild radius for its invariant mass.

You can’t have an event happen in one frame but not another 

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23 hours ago, MigL said:

But the variant mass would have the invariant component, a momentum component due to its motion, in common with SR, but GR would  possibly even have a gravitational component due to the fact that gravity gravitates..

GR does not make reference to any concept of mass within the gravitational field equations, it references only \(T^{00}\), which is energy density, as well as \(T^{\alpha 0}\) and \(T^{0 \beta}\), which is momentum density. Note that these are densities.

Gravitational self-influences do not explicitly appear in the field equations (they can’t, because the associated quantities are not covariant), but are encoded in the non-linear structure of the equations themselves. 

21 hours ago, Strange said:

But what you are saying is that "relativistic mass" (which is observer dependent) does not contribute to the gravitational effect of an object?

In GR, the source of gravity is neither invariant mass, nor relativistic mass, but the stress-energy-momentum tensor. This is a generally covariant object, so when you go into a different frame of reference, the components of the tensor may change, but they will change in such a way that the relationships between these components - and thus the overall tensor - remain the same. So, having relative motion may change how different observers measure individual quantities such as densities, momenta, stresses etc, but it will not change the source term in the gravitational field equations. This is also relevant in vacuum, because distant sources determine vacuum solutions in the form of boundary conditions.

So essentially what I am saying is that relative motion between test particle and source has no bearing on the geometry of spacetime due to that source, it only changes how the observer labels events in that same spacetime. This is of course provided that the test particle’s own gravitational influence is negligible (otherwise we have a GR 2-body problem, which is much more complex).

18 hours ago, Strange said:

But it isn't so obvious why the extra kinetic energy / relativistic-mass seen by a "stationary" observer as the object passes by does not increase the gravitational effect of the object.

Because in the energy-momentum tensor, a change in one component also implies potential changes in all other components. In this example, if momentum density becomes non-zero, then energy density and all other relevant components will also change in such a way as to “compensate” (so to speak) for that change. You are basically just shifting around things within the tensor, without changing the tensor itself. 

12 hours ago, SergUpstart said:

Based on SR, a fast-moving object with a rest mass will become a black hole for an external observer

This is wrong, SR says no such thing. In fact, SR is a model of flat Minkowski spacetime, it has nothing to say at all about the gravitational influence of anything.

 

Edited by Markus Hanke

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Just to add some details to the above. Let's look at how photons contribute to the stress tensor. Due to their momentum photons have a high momentum to pressure relation. Matter often described as dust exerts zero pressure.

Pressure in the stress energy tensor is described as flux in a given direction (think of the classical container walls).

Gravity is also defined by its flux influence in the same manner as pressure. The pressure terms are the diagonal terms the off diagonal terms being shear stresses.

[math]T=\begin{pmatrix}-\rho&0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\end{pmatrix}[/math]

The [math]T^{00}[/math] is the mass density in the rest frame. The other diagonal terms describe the flux {pressure}

[math]diag T^{xx},T^{yy},T^{zz}[/math] for simplicity. T^00 being the mass density entry.

(Trying to keep the above as simple as possible lol). There is quite a bit of math involved to fill in the entries which I won't go into.

Edited by Mordred

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2 hours ago, Markus Hanke said:

GR does not make reference to any concept of mass within the gravitational field equations, it references only T00 , which is energy density, as well as Tα0 and T0β , which is momentum density. Note that these are densities.

Gravitational self-influences do not explicitly appear in the field equations (they can’t, because the associated quantities are not covariant), but are encoded in the non-linear structure of the equations themselves. 

In GR, the source of gravity is neither invariant mass, nor relativistic mass, but the stress-energy-momentum tensor. This is a generally covariant object, so when you go into a different frame of reference, the components of the tensor may change, but they will change in such a way that the relationships between these components - and thus the overall tensor - remain the same. So, having relative motion may change how different observers measure individual quantities such as densities, momenta, stresses etc, but it will not change the source term in the gravitational field equations. This is also relevant in vacuum, because distant sources determine vacuum solutions in the form of boundary conditions.

So essentially what I am saying is that relative motion between test particle and source has no bearing on the geometry of spacetime due to that source, it only changes how the observer labels events in that same spacetime. This is of course provided that the test particle’s own gravitational influence is negligible (otherwise we have a GR 2-body problem, which is much more complex).

Because in the energy-momentum tensor, a change in one component also implies potential changes in all other components. In this example, if momentum density becomes non-zero, then energy density and all other relevant components will also change in such a way as to “compensate” (so to speak) for that change. You are basically just shifting around things within the tensor, without changing the tensor itself. 

Excellent explanation. Thanks.

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Maybe the question of whether gravity is a force can be solved like this. Is centrifugal force a force? Working in Cartesian coordinates, we work in a non-inertial frame of reference in which gravity is a force.

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1 hour ago, SergUpstart said:

Maybe the question of whether gravity is a force can be solved like this. Is centrifugal force a force? Working in Cartesian coordinates, we work in a non-inertial frame of reference in which gravity is a force.

We do  ??

The frame of reference has nothing to do with whether the analysis is Cartesian or not.

The same frame can be used to consider a problem in dynamics in two different ways.

1) Newton's Laws of motion (N1 to N3) can be applied to the real forces that are acting. The system is then not an equilibrium system.

2) An equal (but opposite to the resultant of the real forces acting) imaginary force can be introduced to reduce the system to an equilibrium one
    The system is then solved by means of the equations of equilibrium not N1 - N3.
     In your case this imaginary force is called the centrifugal force. The method is quite old and was introduced by D'Alembert.

Edited by studiot

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What is defined as the four forces involve whether or not that field has a gauge boson. Example the gauge boson for the EM field is the photon.

Gravity if it is a force would need the graviton as it's gauge boson. We haven't discounted the possibility yet.

Keep in mind were talking force fields.

Edited by Mordred

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