Do change of energy of photons create force of acceleration of big mass when they fall to it?

[DimaMazin]

Photons are not mass therefore they don't create space/time. But when they are falling to big mass their energy is increasing. Does  the increse of their energy create space/time field?

[Markus Hanke]

2 hours ago, DimaMazin said:

Photons are not mass therefore they don't create space/time.

Spacetime is not created, so I am unsure what you mean by this. If you meant whether photons have themselves a gravitational effect, then the answer is yes they do, because they carry energy-momentum - which is the source of gravity (not just mass).

2 hours ago, DimaMazin said:

But when they are falling to big mass their energy is increasing.

That depends on the observer.

2 hours ago, DimaMazin said:

Does  the increse of their energy create space/time field?

No, because the blueshift is an observer-dependent phenomenon. Locally in a small enough spacetime neighbourhood of the photon itself, nothing changes.

[DimaMazin]

42 minutes ago, Markus Hanke said:

Spacetime is not created, so I am unsure what you mean by this. If you meant whether photons have themselves a gravitational effect, then the answer is yes they do, because they carry energy-momentum - which is the source of gravity (not just mass).

That depends on the observer.

No, because the blueshift is an observer-dependent phenomenon. Locally in a small enough spacetime neighbourhood of the photon itself, nothing changes.

Change of photon momentum in gravitational field is an action. What is counteraction? Does photon energy-momentum attract the big mass?

[swansont]

1 hour ago, DimaMazin said:

Change of photon momentum in gravitational field is an action. What is counteraction? Does photon energy-momentum attract the big mass?

Yes

[geordief]

3 hours ago, Markus Hanke said:

Spacetime is not created, so I am unsure what you mean by this. If you meant whether photons have themselves a gravitational effect, then the answer is yes they do, because they carry energy-momentum - which is the source of gravity (not just mass).

 

Could I ask you to say briefly what "energy-momentum" is,physically or mathematically in the example you have just used ?  (or just a link as there seemed to be one or two uses for the term when I googled it)

Does the dash (-) mean anything specifically? Not a ratio ,is it?  Or does it just mean "a combination of the two"?

[Markus Hanke]

4 hours ago, geordief said:

Could I ask you to say briefly what "energy-momentum" is,physically or mathematically in the example you have just used ?

What I was referring to is the energy-momentum tensor, which is the source term in the Einstein field equation. It physically represents the distribution of energy-momentum in spacetime. 

4 hours ago, geordief said:

Does the dash (-) mean anything specifically? Not a ratio ,is it?  Or does it just mean "a combination of the two"?

It’s called the “energy-momentum tensor” because it encapsulates both the distribution of energy and momentum in a system.

[DimaMazin]

On 11.07.2018 at 5:24 PM, Markus Hanke said:

What I was referring to is the energy-momentum tensor, which is the source term in the Einstein field equation. It physically represents the distribution of energy-momentum in spacetime. 

It’s called the “energy-momentum tensor” because it encapsulates both the distribution of energy and momentum in a system.

If big mass is rotating and has energy-momentum tensor, then the tensor creates spacetime? Does energy-momentum tensor of big mass attract photons or it can attract only objects with mass?

[swansont]

On 7/11/2018 at 10:24 AM, Markus Hanke said:

What I was referring to is the energy-momentum tensor, which is the source term in the Einstein field equation. It physically represents the distribution of energy-momentum in spacetime. 

It’s called the “energy-momentum tensor” because it encapsulates both the distribution of energy and momentum in a system.

That doesn't answer the question. You used the phrase "the distribution of energy-momentum in spacetime" The question is what is the "energy-momentum" part referring to?

[Markus Hanke]

2 hours ago, swansont said:

That doesn't answer the question. You used the phrase "the distribution of energy-momentum in spacetime" The question is what is the "energy-momentum" part referring to?

My understanding was that he asked what it was mathematically referring to, in the context of gravity. Did I misunderstand the question? If he meant what energy-momentum actually “is”, in an ontological sense, then that is not a very easy question to answer - the best I could do here is say that it is that property of a system that remains invariant under time translation symmetry, via Noether’s theorem. That’s quite abstract, but I don’t really know a better way to answer this.

2 hours ago, DimaMazin said:

If big mass is rotating and has energy-momentum tensor, then the tensor creates spacetime? 

It doesn’t “create” spacetime as such, but it does determine the geometry of spacetime in the interior of the rotating mass.

2 hours ago, DimaMazin said:

Does energy-momentum tensor of big mass attract photons or it can attract only objects with mass?

The wording here is a little odd, but I understand what you are asking. The answer is that everything is subject to the effects of gravity, including objects with no rest mass, such as photons. Note though that in a vacuum region the energy-momentum tensor identically vanishes, so the field equation for vacuum has no source term. The gravitational source here is considered “distant” (as opposed to local), and it enters the picture via the boundary conditions which you need to impose to find solutions to the system of differential equations.

Edited July 13, 2018 by Markus Hanke

[geordief]

1 hour ago, Markus Hanke said:

My understanding was that he asked what it was mathematically referring to, in the context of gravity. Did I misunderstand the question? If he meant what energy-momentum actually “is”, in an ontological sense, then that is not a very easy question to answer - the best I could do here is say that it is that property of a system that remains invariant under time translation symmetry, via Noether’s theorem. That’s quite abstract, but I don’t really know a better way to answer this.

I said "physically or mathematically". Does "physically"necessarily  have to mean "ontologically" ?

 

Although mathematically seems to be the key to the lock in this case ,I think I was also asking for any possible physical flesh on the bone in the way you were using the term in your example .if you were just using the term in the context of the energy -momentum tensor then that is fine .

 

But if you were using it in the sense of a physical  " source of gravity" (your quote)  then perhaps there might be a little extra  you could say  as to how it (presumably) causes space time to curve.

 

If one is calculating the curved coordinates in the vicinity of a massive object am I to expect that  we have also to take into account that massive object's velocity wrt the particular point on the coordinate map? Is that what this "energy-momentum tensor" provides for?

 

[swansont]

3 hours ago, Markus Hanke said:

My understanding was that he asked what it was mathematically referring to, in the context of gravity. Did I misunderstand the question? If he meant what energy-momentum actually “is”, in an ontological sense, then that is not a very easy question to answer - the best I could do here is say that it is that property of a system that remains invariant under time translation symmetry, via Noether’s theorem. That’s quite abstract, but I don’t really know a better way to answer this.

You could say what the elements of the tensor contain.

 

[DimaMazin]

3 hours ago, Markus Hanke said:

My understanding was that he asked what it was mathematically referring to, in the context of gravity. Did I misunderstand the question? If he meant what energy-momentum actually “is”, in an ontological sense, then that is not a very easy question to answer - the best I could do here is say that it is that property of a system that remains invariant under time translation symmetry, via Noether’s theorem. That’s quite abstract, but I don’t really know a better way to answer this.

It doesn’t “create” spacetime as such, but it does determine the geometry of spacetime in the interior of the rotating mass.

The wording here is a little odd, but I understand what you are asking. The answer is that everything is subject to the effects of gravity, including objects with no rest mass, such as photons. Note though that in a vacuum region the energy-momentum tensor identically vanishes, so the field equation for vacuum has no source term. The gravitational source here is considered “distant” (as opposed to local), and it enters the picture via the boundary conditions which you need to impose to find solutions to the system of differential equations.

Can energy-momentum tensor of rotating mass increase photon fraquency  without slowing of time in place where the photon is?

[geordief]

6 hours ago, geordief said:

 

 

But if you were using it in the sense of a physical  " source of gravity" (your quote)  then perhaps there might be a little extra  you could say  as to how it (presumably) causes space time to curve.

 

 

 

Oops,I need to rephrase that (too late to edit) .I think I  know that nobody actually knows that (the "how"") . I just meant to say "then perhaps there might be a little extra  you could say  as to what is "energy-momentum" as a physical phenomenon  in the context of  curving spacetime.  How would you go about measuring it (or perhaps its components ) at any specific region ?

Edited July 13, 2018 by geordief

[Markus Hanke]

5 hours ago, geordief said:

Oops,I need to rephrase that (too late to edit) .I think I  know that nobody actually knows that (the "how"") . I just meant to say "then perhaps there might be a little extra  you could say  as to what is "energy-momentum" as a physical phenomenon  in the context of  curving spacetime.  How would you go about measuring it (or perhaps its components ) at any specific region ?

The energy-momentum tensor captures a given system’s energy density, momentum density, momentum flux, pressure, and internal stress. You can measure these using a variety of instruments, but precisely what instruments they are really depends on what kind of a system you are dealing with - if the system is a fluid or a solid, you’d use equipment that measures flow, mass, density, and various forces in the interior of the fluid/solid. If it is an electromagnetic field, you’d use test charges etc. And so on. So it really depends on the specific situation.

I personally think of energy-momentum as being like a “river” that flows through spacetime (this is just an analogy though!) - when you insert a small surface in that river at a particular point, the surface will experience forces, motions and pressure, which depends on the aforementioned properties of the river. The energy-momentum tensor encapsulates this.

Energy-momentum is a “source” of gravity in the sense that it appears in the field equations that describe gravity. But do note that curvature is also self-interacting in a sense; the gravitational field itself contains energy-momentum too. This is why we can have gravity in vacuum, even though any sources of ordinary energy-momentum may be far away. This self-interaction doesn’t appear as a term in the equations (in particular, it is not part of the energy-momentum tensor), but is encapsulated in the non-linear structure of the field equations themselves. Non-linearity physically corresponds to self-interaction.

Does this help? I am still not entirely sure what your original question is aiming at, exactly.

12 hours ago, geordief said:

If one is calculating the curved coordinates in the vicinity of a massive object am I to expect that  we have also to take into account that massive object's velocity wrt the particular point on the coordinate map? Is that what this "energy-momentum tensor" provides for?

I don’t quite understand the question. Coordinate systems in GR are arbitrary, so they are not used as a basis for any physical effects. However, different observers may measure aspects of the same gravitational field in different ways, depending on their own state of relative motion. For example, a stationary and spherically symmetric mass will usually be described in terms of the Schwarzschild metric, for most observers. However, if the observer who encounters this object and its gravitational field moves at nearly the speed of light relative to it, then you’d get a different metric from the field equations (called the Aichelburg-Sexl ultraboost). It’s that difference in description that accounts for the motion effects. This doesn’t come from the energy-momentum tensor though (since we are in a vacuum), but rather from how you set up the metric Ansatz when you begin to solve the equations.

10 hours ago, DimaMazin said:

Can energy-momentum tensor of rotating mass increase photon fraquency  without slowing of time in place where the photon is?

Sorry, I don’t understand. Can you explain the question some more?

[studiot]

1 hour ago, Markus Hanke said:

The energy-momentum tensor captures a given system’s energy density, momentum density, momentum flux, pressure, and internal stress.

This tensor (like all of them) is a point function/property so isn't it stretching it a bit to call a point a system?

[geordief]

6 hours ago, Markus Hanke said:

The energy-momentum tensor captures a given system’s energy density, momentum density, momentum flux, pressure, and internal stress. You can measure these using a variety of instruments, but precisely what instruments they are really depends on what kind of a system you are dealing with - if the system is a fluid or a solid, you’d use equipment that measures flow, mass, density, and various forces in the interior of the fluid/solid. If it is an electromagnetic field, you’d use test charges etc. And so on. So it really depends on the specific situation.

I personally think of energy-momentum as being like a “river” that flows through spacetime (this is just an analogy though!) - when you insert a small surface in that river at a particular point, the surface will experience forces, motions and pressure, which depends on the aforementioned properties of the river. The energy-momentum tensor encapsulates this.

Energy-momentum is a “source” of gravity in the sense that it appears in the field equations that describe gravity. But do note that curvature is also self-interacting in a sense; the gravitational field itself contains energy-momentum too. This is why we can have gravity in vacuum, even though any sources of ordinary energy-momentum may be far away. This self-interaction doesn’t appear as a term in the equations (in particular, it is not part of the energy-momentum tensor), but is encapsulated in the non-linear structure of the field equations themselves. Non-linearity physically corresponds to self-interaction.

Does this help? I am still not entirely sure what your original question is aiming at, exactly.

 

Very much. It is helping me to familiarize myself with the model  although there can surely be no substitute for eventually familiarizing myself with some of the mathematics  (even unrealistic goals can be a help ) I am taking it now that "energy-momentum" as a phrase only needs to be understood as a mathematical definition in a mathematical framework...

6 hours ago, Markus Hanke said:

I don’t quite understand the question. Coordinate systems in GR are arbitrary, so they are not used as a basis for any physical effects. However, different observers may measure aspects of the same gravitational field in different ways, depending on their own state of relative motion. For example, a stationary and spherically symmetric mass will usually be described in terms of the Schwarzschild metric, for most observers. However, if the observer who encounters this object and its gravitational field moves at nearly the speed of light relative to it, then you’d get a different metric from the field equations (called the Aichelburg-Sexl ultraboost). It’s that difference in description that accounts for the motion effects. This doesn’t come from the energy-momentum tensor though (since we are in a vacuum), but rather from how you set up the metric Ansatz when you begin to solve the equations.

Sorry, I don’t understand. Can you explain the question some more?

Is it a fair comment that spacetime curvature is actually not an absolute "thing" but   has to be defined relative to an  observer? 

[Markus Hanke]

3 hours ago, geordief said:

Is it a fair comment that spacetime curvature is actually not an absolute "thing" but   has to be defined relative to an  observer? 

No, curvature (as given by the tensors that describe it) is a covariant quantity, so all observers agree on it. Where observers may differ is the metric they use to describe the same spacetime, i.e. they might determine different measurements of space and time, depending on their states of relative motion. The classic example here is the difference between a stationary observer far from a Schwarzschild black hole, and some other observer in free fall towards the black hole. They will determine different in-fall times, as measured in their own metric systems. At first glance the metrics they use look quite different, but actually they describe the same spacetime - just from different points of view. The aforementioned Aichelburg-Sexl metric is again the same spacetime, but from the perspective of an observer who uniformly moves relative to the black hole at nearly the speed of light. Using different metrics on the same spacetime basically boils down to labelling the same physical events with different coordinates, which is always allowed in GR, since it doesn’t make any difference to the actual physics. One just has to be careful to use the metrics in the right way, since they describe specific points of view onto the spacetime in question.

Whether two given (different) metrics describe the same spacetime can actually be quite a tricky question to answer, and takes more than just a cursory glance at the metrics.

8 hours ago, studiot said:

This tensor (like all of them) is a point function/property so isn't it stretching it a bit to call a point a system?

It is true that tensors are local objects - but they are objects defined at all events within some given region of spacetime, so in actual fact we are dealing with a tensor field. So for example, if you have a planet, then there will be an associated energy-momentum tensor field that spans that entire planetary volume at all relevant times, and smoothly varies from event to event. Generally speaking, the components of the tensor will thus be functions of the coordinates, rather than constants.

Edited July 14, 2018 by Markus Hanke

[studiot]

11 minutes ago, Markus Hanke said:

It is true that tensors are local objects - but they are objects defined at all events within some given region of spacetime, so in actual fact we are dealing with a tensor field. So for example, if you have a planet, then there will be an associated energy-momentum tensor field that spans that entire planetary volume at all relevant times, and smoothly varies from event to event. Generally speaking, the components of the tensor will thus be functions of the coordinates, rather than constants.

But as you pointed out elsewhere, we work with continuous manifolds which means there are no boundaries between the points.

So all your 'systems' have no boundaries and are therefore ill defined as systems.

[Markus Hanke]

22 minutes ago, studiot said:

But as you pointed out elsewhere, we work with continuous manifolds which means there are no boundaries between the points.

So all your 'systems' have no boundaries and are therefore ill defined as systems.

Yes, the manifold itself is always continuous (by definition!), but the distribution of energy-momentum on it may not be. For example, consider the case of a planet without any atmosphere. You have the interior of the planet itself, which is described by a non-vanishing energy-momentum tensor field, and then the exterior vacuum, where the energy-momentum tensor field vanishes. There is a sharp boundary between these regions at a precisely localisable region of space, in terms of the distribution of energy-momentum (i.e. the planet’s surface). When looking for a global metric that spans the entire spacetime, one then has to explicitly impose a continuity condition at the point of this boundary. The result is a piecewise-defined energy-momentum field, but a spacetime and metric that is still everywhere smooth and differentiable, including at the boundary.

An explicit example of this is the global Schwarzschild metric, which spans both the interior of the gravitating body as well as the exterior vacuum, with a sharp boundary, but a metric that is everywhere differentiable. In this situation, the interior metric smoothly transitions into the exterior one, without any discontinuities.

To make a long story short, my “system” was referring to the distribution of energy-momentum in spacetime, which can indeed have boundaries. However, sometimes the boundary may not be easily localisable - for example, when you have a planet with an atmosphere. It comes down then to your required degree of accuracy.

Edited July 14, 2018 by Markus Hanke

[geordief]

1 hour ago, Markus Hanke said:

No, curvature (as given by the tensors that describe it) is a covariant quantity, so all observers agree on it. Where observers may differ is the metric they use to describe the same spacetime, i.e. they might determine different measurements of space and time, depending on their states of relative motion. The classic example here is the difference between a stationary observer far from a Schwarzschild black hole, and some other observer in free fall towards the black hole. They will determine different in-fall times, as measured in their own metric systems. At first glance the metrics they use look quite different, but actually they describe the same spacetime - just from different points of view. The aforementioned Aichelburg-Sexl metric is again the same spacetime, but from the perspective of an observer who uniformly moves relative to the black hole at nearly the speed of light. Using different metrics on the same spacetime basically boils down to labelling the same physical events with different coordinates, which is always allowed in GR, since it doesn’t make any difference to the actual physics. One just has to be careful to use the metrics in the right way, since they describe specific points of view onto the spacetime in question.

Whether two given (different) metrics describe the same spacetime can actually be quite a tricky question to answer, and takes more than just a cursory glance at the metrics.

I see.

So, as per Swansont asked  earlier ,what do the elements of the momentum-energy tensor contain?

Equivalently (I hope) what actual physical measurements would be required of an observer approaching (or at a constant distance from) one of your atmosphere free planets in order to determine  the precise spacetime curvature at a point on, or at any distance from   the surface?

Speed of rotation,temperature,electric and other charges?Radiation?

Anything else?

Edited July 14, 2018 by geordief

[Markus Hanke]

12 hours ago, geordief said:

So, as per Swansont asked  earlier ,what do the elements of the momentum-energy tensor contain?

As listed a few posts ago, they contain energy density, momentum density, momentum flux, pressure, and internal stresses.

12 hours ago, geordief said:

Equivalently (I hope) what actual physical measurements would be required of an observer approaching (or at a constant distance from) one of your atmosphere free planets in order to determine  the precise spacetime curvature at a point on, or at any distance from   the surface?

One can measure gravitational time dilation across a small region around your point (using e.g. a Pound-Rebka style experiment), as well as tidal effects of gravity in the spatial directions. So long as we know that the spacetime is spherically symmetric and stationary, this will suffice to construct the Riemann tensor in that small region.

12 hours ago, geordief said:

Speed of rotation,temperature,electric and other charges?Radiation?

In Kerr-Newman type vacuum spacetime families, the metric is determined by just three free parameters - mass, angular momentum, and electric charge. The popular Schwarzschild solution is an example of this type of spacetime. 

[studiot]

2 hours ago, Markus Hanke said:

this will suffice to construct the Riemann tensor in that small region

This nicely embodies my comment, which is not about the general correctness, but about the detail or presentation.

I prefer the phrase "for every point in that small region."

I think it can be confusing for many to be emphasising the thermodynamic idea of a state function which is a single representation for a whole system, and then considering something like a tensor function which varies from point to point within a system and has no overall single representation. Indeed zero may feature as an overall average.

I know this and you know this but...

Sorry to be so pedantic.

 

[Markus Hanke]

57 minutes ago, studiot said:

This nicely embodies my comment, which is not about the general correctness, but about the detail or presentation.

I prefer the phrase "for every point in that small region."

I think it can be confusing for many to be emphasising the thermodynamic idea of a state function which is a single representation for a whole system, and then considering something like a tensor function which varies from point to point within a system and has no overall single representation. Indeed zero may feature as an overall average.

I know this and you know this but...

Sorry to be so pedantic.

 

A very valid point...I completely concur. Thanks for pointing it out, and for clarifying

[geordief]

3 hours ago, Markus Hanke said:

As listed a few posts ago, they contain energy density, momentum density, momentum flux, pressure, and internal stresses.

 

Yes, apologies I should have realized that .And thanks for your patience.

As an after thought I notice that you have included internal stresses as an element.

 

Can that be broadly understood as  reflective of physical non uniformity **of the  small region?

 

*perhaps non homogeneity is the correct term?

[Markus Hanke]

1 minute ago, geordief said:

Can that be broadly understood as  reflective of physical non uniformity **of the  small region?

Well, those are shear stresses, in the Newtonian sense:

https://en.m.wikipedia.org/wiki/Shear_stress

I can’t see how one could look at this in terms of non-uniformities.