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Author Topic: If I give an object some potential energy, does its mass increase?  (Read 95244 times)

Offline VernonNemitz

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Farsight, it occurs to me that another reason you are wrong about the plate acquiring significant potential-energy-as-mass relates to the photon I described at the end of my last message here.  The photon would come away from the black hole almost as blue as when it was created, if you were right about where the potential energy goes!  Therefore it is I that am correct; most of the energy that becomes potential, as a small body or photon escapes the gravitational field of a larger body, goes to increasing the mass-energy of the larger body, not the smaller body or photon.  Do remember that the most important thing about General Relativity is that it is about "mass-energy" regardless of form.  You can't have a double-standard, treating the plate one way and the photon another.
I don't treat them differently Vernon. The important  thing is that a photon doesn't actually change energy when it enters a gravitational field. There's no evidence of any energy transfer into the photon via unseen particles, the photon is the only particle there, and we must abide by conservation of energy. Yes, the photon appears to gain energy, for example it's measurably blue-shifted. But the frequency hasn't actually changed. Your measuring devices have, because they're subject to gravitational time dilation. Your clocks run slower, so you see the frequency as increased. It's a little like the way something feels warmer if you're cold.
You do indeed treat them differently, because for the plate you continue to say it gains mass when it is forced out of a gravitational field.  Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).  This means you are employing a double-standard, between your description of the plate and your description of the photon.  Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated plate, at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.  So, if you want to claim the mass of the plate has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also.  Since you don't, it means you are violating the basic principles of General Relativity; it means you don't know what you are talking about.
« Last Edit: 03/02/2010 14:36:36 by VernonNemitz »
 

Offline yor_on

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Okay Vernon, I'm still not sure how you see it, but you're at the start of it as I understood. Virtual particles are defined as being outside Planck time and GR says that there is no measurable 'energy' to space as I understands it. Let's put it like this, if space had an measurable energy then it would have to be a 'medium' too as I see it. Doesn't mean you can't have 'virtual particles' and 'vacuum energy', as long as they don't make any measurable 'dent' of their own at SpaceTime. That we see indirect evidence seems to be allowable by GR though?

I'm looking forward to see you 'flesh your ideas out' Vernon. I'm also wondering about 'times arrow' :)
 

Offline Geezer

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Hi!

We seem to be drifting off topic just a tad here.

Anyone is welcome to propose a new theory under that heading, but it might be best if we try to answer the topic question in terms of well accepted science.

Or should we simply lock the thread?

Geezer (Mod)
 

Offline yor_on

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Well Geezer, you're right :)

Thinking of my idea of if the plate would 'jiggle' less if lifted up on that table instead of being left at the ground made me reevaluate how I understand gravity. As I said I believe it to be a 'geometry'. And as such it's not a 'force', but it still doesn't explain how it can be in all points (matter and space) and also work from each one of those points as if that would be the definite 'center'. If you look at Earth we say that in the middle of it, gravity will be 'nulled' as the 'mass' around that specific point will take out all gravitational forces (as I understand it). So gravity act both as each point would be its center as well as being able to add those points together to create an additive 'force'. And yes, suddenly I'm back to call it a force. ain't I :)

The thing here is, how can a bare geometry be additive?
What allows it that?
 

Offline Geezer

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Thanks Yoron  ;D

There are several posts on this topic and I'm reasonably confident I'm not going to trawl through the lot, so could someone perhaps try to summarize for a poor old geezer where we are?

Something along the lines of:

According to classical mechanics, mass is not added because of  - insert formula here
According to GR, mass is added because of - insert formula here
According to so-and-so's theory, mass turns into a white rabbit because of - insert formula here

Something along those lines might help someone less "skilled-in-the-art" to understand what we're on about here.
 

Offline yor_on

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The white rabbit I might be able to help with? As for defining the math for mass in general relativity?
I'm afraid that that would be a thread all of its own :)

---Quote---

Generalizing this definition to general relativity, however, is problematic; in fact, it turns out to be impossible to find a general definition for a system's total mass (or energy). The main reason for this is that "gravitational field energy" is not a part of the energy-momentum tensor; instead, what might be identified as the contribution of the gravitational field to a total energy is part of the Einstein tensor on the other side of Einstein's equation (and, as such, a consequence of these equations' non-linearity). While in certain situation it is possible to rewrite the equations so that part of the "gravitational energy" now stands alongside the other source terms in the form of the Stress-energy-momentum pseudotensor, this separation is not true for all observers, and there is no general definition for obtaining it.

------end of quote---

Just as a by side if you look at this idea of using 'Stress-energy-momentum pseudotensors' Michael Weiss and John Baez have this to say about such.

"Mathematicians invented tensors precisely to meet this sort of demand -- if a tensor equation holds in one coordinate system, it holds in all.  Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles.  In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects.  (See the FAQ entry on black holes for some examples.) These pseudo-tensors have some rather strange properties.  If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime.  By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation.  For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy."

Now, what the heck did this mean "In fact, it turns out to be impossible to find a general definition for a system's total mass (or energy). The main reason for this is that "gravitational field energy" is not a part of the energy-momentum tensor;" ?

Well, 'gravitational field energy' I take to be gravity when seen/treated as a field.
And the 'energy-momentum tensor'? Hold on to your hats now.

--Quote(s)----

The stress-energy tensor (sometimes stress-energy-momentum tensor) is a tensor quantity (A generalization of the concept of a vector) in physics that describes the density and flux of energy and momentum in spacetime, generalizing the stress tensor of Newtonian physics. It is an attribute of matter, radiation, and non-gravitational force fields. The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass is the source of such a field in Newtonian gravity.

=
And
=

The Einstein field equations (EFE) or Einstein's equations are a set of ten equations in Einstein's theory of general relativity which describe the fundamental interaction of gravitation as a result of spacetime being curved by matter and energy.[1] First published by Albert Einstein in 1915[2] as a tensor equation, the EFE equate spacetime curvature (expressed by the Einstein tensor) with the energy and momentum within that spacetime (expressed by the stress-energy tensor).

---End of quote(s)---

Now;

"The math behind general relativity is called Einstein Field Equations. They are equations of the coupled hyperbolic-elliptic nonlinear partial differential type, which, in plain English, means that they are really, really hard. Einstein himself recognized the mathematical difficulties of general relativity as "very serious." He predicted it as being the primary hindrance of general relativity's development. The equation can be stated in a "symbolic form" that isn't very useful. Here it is:



It doesn't mean much to us, but you can see on the left of the equal sign the stuff that describes the curvature of space-time. On the right is the matter within space-time, and how it behaves. " Taken from Relativity-Math

Okay, the real problem with defining a mass in general relativity, as I understands it, have to do with that it's okay to labor with different definitions for 'mass' as long as we are doing it from a spatial infinity, that is, not trying to define it locally.

---Quote---

Since the 1970s, physicists and mathematicians have worked on the more ambitious endeavor of defining suitable quasi-local quantities, such as the mass of an isolated system defined using only quantities defined within a finite region of space containing that system. However, while there is a variety of proposed definitions such as the Hawking energy, the Geroch energy or Penrose's quasi-local energy-momentum based on twistor methods, the field is still in flux.

---End of quote-----

So we don't seem to have any stringent mathematical definition for mass locally in General Relativity as I understands it? And if I'm wrong :) Correct it please..

Mathematics_of_general_relativity
Einstein_field_equations
Mass_in_general_relativity

« Last Edit: 04/02/2010 22:50:15 by yor_on »
 

Offline Geezer

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Hey! That's a very helpful summary. Thank you sir.

As you suggest, it's certainly a great target to stick on the wall.

 
 

Offline VernonNemitz

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Just in case there was some misunderstanding about all the G.R. stuff a couple of posts ago, I'd merely like to point out that in G.R. the gravitational field is considered to be a form of energy, which contributes to the intensity of the gravitational field.  This contribution is used to explain some observed extra motion in Mercury's orbit, that classical gravitational physics could not explain.  But by inspection, it leads to an infinte series: if gravitation itself can enhance gravitation, then where do the enhancements end? Well, we know that some infinite series can be added up to a finite sum, and obviously G.R. has to do that (and Q.M. will need to do the same if it is ever to explain gravitation).  I think this is the cause of the mis-match in the equations that were being described in that earlier post.  You can't have mass without also having the gravitational field of that mass, and the associated self-enhancement.  So what magnitude of "mass" do you want to describe?  The initial mass without the gravity field, or the mass that includes the equivalence of the immediatedly-associated gravitational field, or the mass that includes both that immediate gravitational field and the first level of enhanced field, or...?

Meanwhile, every mass is located within the gravity well of some other mass!  In nuclear physics they talk about "negative binding energy" a lot; it is a trick that allows potential energy to be kept distinct from the mass of a particle.  Two particles closely interacting via the Strong Nuclear Force have a lot more negative binding energy than two widely-separated particles (but the separated particles have more potential energy).  Inside some other body's gravitational field, a mass can be associated with different amounts of potential energy (depending on location), and G.R. needs to take this into account, also, when trying to describe the mass of a particular object.

Personally, I think that some of the headache can be solved by simply saying that potential energy always takes the form of mass.  Farsight says the same thing, but we disagree on details (how some potential energy is divided between two interacting masses).
 

Offline yor_on

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Yes Vernon, gravitational waves seems to be some sort of energy?
But It seems to be so that both statements exist side by side?
Yep, my headache is getting worse again :)
 

Offline VernonNemitz

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Yes Vernon, gravitational waves seems to be some sort of energy?
But It seems to be so that both statements exist side by side?
Yep, my headache is getting worse again :)
Do not confuse a gravitational field with a gravitational wave; they are not identical things, like an electromagnetic field and an electromagnetic wave are not identical things (you don't confuse the field surrounding a magnet with a radio wave, do you?).  Both have energy, though; both are different forms of energy.  In Quantum Mechanics the two things are related (radio waves consist of real-energy photons and a magnetic field consists of virtual-energy photons), and should QM one day describe gravity adequately, a similar description could be expected regarding gravitational waves and fields.  General Relativity does not offer such a close relationship, though; the gravitational field is simply a fixed region of curved space, and a gravitational wave is a moving ripple of space.
 

Offline yor_on

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"General Relativity does not offer such a close relationship, though; the gravitational field is simply a fixed region of curved space, and a gravitational wave is a moving ripple of space."

Lovely expression Vernon, I quite agree. And as all ideas of 'energy' seems to build on a transformation can we say that gravity 'transforms' anything? Well, that gravitational wave might change (transform) your geometry but it still seems that from any chosen frame you will be the same, 'internally' in that frame, so to speak. So how about 'potential energy' then? What the heck is that 'energy' we're talking about. Isn't it just an abstraction describing a relation?
« Last Edit: 12/02/2010 21:02:02 by yor_on »
 

Offline VernonNemitz

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...as all ideas of 'energy' seems to build on a transformation can we say that gravity 'transforms' anything?
Try walking around on the surface of a neutron star, and your shape will be transformed significantly.  Something like a billion gees, there.  To properly answer your question we probably have to be sure we are talking about exactly the same thing.  What is "gravity"?  For this Message I will start by choosing to define that as, "It is a thing that can tend to keep all sorts of slow masses in close proximity to each other."  Thus an Answer to your question appears: a role that involves the word "keep" does not automatically include "change" or "transform".  We thus realize that the definition is incomplete; the more mass is involved, the more powerfully gravity can keep objects in the vicinity of each other.  Further, we know of an additional incompleteness in the definition, because we know that gravity can eventually cause initially distant things to collide.  THAT means that "change" does indeed need to be involved in the definition!  Newton's equation for gravity is a most excellent first-approximation definition.  And, despite more modern definitions, gravity can be called a "force" simply because it can cause masses to accelerate --the essence of "force" is the acceleration of a mass, after all.  Not even Einstein could change the definition of "force" in Physics.
So how about 'potential energy' then? What the heck is that 'energy' we're talking about. Isn't it just an abstraction describing a relation?
You can't get something for nothing, that we know of.  The acceleration of a mass is always associated with a change in the kinetic energy of the mass.  If it increases, we need to know where it comes from, and if it decreases, we need to know where it goes.  "Potential" energy is just a way to talk about "storage".  It gives us a place to say, "There is where the kinetic energy went!" when we hurl a rock upwards toward some lofty height, and see it slow down to a vertical velocity of zero.  And of course it gives us a place to say, "There is where the kinetic energy came from" when we see the rock acquiring an increasing downward velocity.

The whole point of this entire Message Thread is the attempt to answer a question regarding the FORM of "potential" energy.  What are the properties of that "storage system"?  Newton offered a nice explanation, which I shall phrase as: "Keeping the gravity gradient in mind, with distance from the center of mass of the main gravitating body, potential energy can be represented as a simple magnitude of its lofty height above the surface of that mass."  Since different masses can have different gravity gradients, two identical objects at equal heights above those different masses can have different amounts of potential energy.

My personal objection to that explantion has three parts.  First, we know by direct measurement that when potential energy is associated with the Strong Nuclear Force, the Weak Nuclear Force, and the ElectroMagnetic Force, the form of that potential energy is always "mass".  Second, Newton did not know that E=mc2.  And third, physicists desire to Unify/simplify the descriptions of the Natural Forces, including "Gravity".  This makes it very very easy for me to be willing to assume that with respect to Gravity, potential energy also takes the form of mass.

However, if the overall logic is to be completely consistent (and as I've pointed out to Farsight in various messages), The Answer To This Message Thread's Main Topic Question ("If I give an object some potential energy [presumably by lofting it], does its mass increase?") would then be: "Yes and no.  First, due to E=mc2, the initially-supplied kinetic energy, that becomes lofty potential energy, only becomes an exceedingly tiny quantity of mass in most environments influenced by Gravity (so if you can't measure an increase in mass, can you really say that its mass increased?).  Second, most of the mass that does appear needs to become proportionately more-added-to the main source of the gravitational field, than added-to the lofted object (making the amount of increase for the lofted object even less measurable)."  If the overall logic is to be completely consistent, I repeat!
« Last Edit: 16/02/2010 06:52:22 by VernonNemitz »
 

Offline Farsight

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I'm glad we seem to be agreed on that the potential energy goes into mass, as slight as it might be. But Vernon, you need to look at that distribution again. Think about ½mv² in the context of an explosion. Consider a 1kg cannonball fired by a 1000kg cannon. Conservation of momentum p=mv tells you the cannonball velocity is 1000 times the cannon recoil velocity. Now apply ½mv² and you see that more of the energy goes into the cannonball. Now embed the cannon firmly into the solid rock of the earth (mass 5.9736 × 1024 kg), and fire the cannonball upwards at 11.2 km/s. Most of the energy goes into the cannonball, and since 11.2 km/s gives it escape velocity, this energy is then lost from the earth system.   
« Last Edit: 16/02/2010 11:47:56 by Farsight »
 

Offline VernonNemitz

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Farsight, I see you have ignored my message of 03/02/2010 14:31:16 (slightly edited in copying below).  As I wrote earlier today, "If the overall logic is to be completely consistent", then I'm talking about consistency with other known gravitational effects, where General Relativity is better at describing things than Newton.
Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).  This means you are employing a double-standard, between your description of the plate and your description of [a] photon.  Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.  So, if you want to claim the mass of the [cannonball] has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also.  Since you don't, it means you are violating the basic principles of General Relativity; it means you don't know what you are talking about.
In other words, Farsight, you are not being consistent!  At the moment of acceleration, the mass-energy of the cannonball is increased significantly, due to kinetic energy being added to it.  Let us now imagine two photons, one (Photon A) equal in energy to the mass of the cannonball just prior to it being accelerated, and one (Photon B) equal to the total mass-energy of the cannonball just after its acceleration.  Upon being declared to have basically escaped the gravity field of the neutron star, we know that the two photons are seriously red-shifted (lost a lot of energy), and that the cannonball is moving lots slower (also lost a lot of energy).  But Photon B must still have the same total mass-energy as the cannonball.  You cannot say that all the original kinetic energy of the ball, that has now mostly become potential energy, has become extra mass for the ball, since you know you cannot say an equivalent thing for Photon B.  You can say that the energy of Photon B is a bit more than the original energy of Photon A, when it was at the surface of the neutron star (and was equal in energy to the mass of the not-yet-accelerated ball).  But only a bit more; the majority of energy-that-became-potential-energy-in-the-form-of-mass had to appear somewhere else:  It was added to the mass of neutron star, extracted from both the photons' original energy and from the accelerated cannonball's energy.  In simple-minded terms, as I explained in another message quite some time ago, we can say that gravitational interactions literally suck energy from escaping objects, and give energy to falling objects (by "pulling" on them, a thing typically associated with, yes, "force", and the-supplying-of-energy).  This is known as "balance", and is again consistent; it is even consistent with the idea of Gravitation being an aspect of a system (two interacting objects).  Your description, Farsight, basically breaks the system as soon as you apply sufficient kinetic energy to a mass, not after the mass has physically escaped.  (Why?  Because you want all the applied kinetic energy to be/stay part of the accelerated mass only!)  Sorry, but the system doesn't break that easily; Photon B and the cannonball will have the same mass-energy after escaping the star (--or the Earth, but the effects are much less obvious here).
 

Offline Farsight

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Farsight, I see you have ignored my message of 03/02/2010 14:31:16 (slightly edited in copying below).
Sorry, I simply missed it.

As I wrote earlier today, "If the overall logic is to be completely consistent", then I'm talking about consistency with other known gravitational effects, where General Relativity is better at describing things than Newton.
You know I agree with that. Let's have a careful look at what you're saying then.

Dude, mass-energy is mass-energy, and G.R. makes NO distinction about the FORM in which it appears; all forms interact gravitationally, equally-in-manner (differently depending only on magnitude).
I'd say energy is energy, and it can appear in a form that confers mass.

This means you are employing a double-standard, between your description of the plate and your description of [a] photon.
I'm not. But let's press on.

Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.
I think I see your problem. The total energy of the moving cannonball comprises the energy locked up as its mass, plus the kinetic energy. When it has escaped the neutron star, the total energy is unchanged. The same is true of the photon. 

So, if you want to claim the mass of the [cannonball] has increased, in acquiring potential energy, then you must also claim the photon has become equivalently more energetic, also. Since you don't, it means you are violating the basic principles of General Relativity...
I think the phrase mass-energy has caused confusion. We say "mass is invariant", and that the mass of a cannonball moving at 11.2 km/s is the same as that of a motionless cannonball at the same location. The total energy for the moving cannonball is however greater. 

In other words, Farsight, you are not being consistent! At the moment of acceleration, the mass-energy of the cannonball is increased significantly, due to kinetic energy being added to it.
The total energy or relativistic mass is increased significantly, but the mass is not. 

Let us now imagine two photons, one (Photon A) equal in energy to the mass of the cannonball just prior to it being accelerated, and one (Photon B) equal to the total mass-energy of the cannonball just after its acceleration.
OK.  

Upon being declared to have basically escaped the gravity field of the neutron star, we know that the two photons are seriously red-shifted (lost a lot of energy), and that the cannonball is moving lots slower (also lost a lot of energy).
Wrong. Conservation of energy applies. If they lost energy, where did it go? The cannonball kinetic energy has been transformed into cannonball potential energy. We know it's cannonball potential energy, because if that cannonball escapes the system, it takes it away.

But Photon B must still have the same total mass-energy as the cannonball.
Absolutely correct.

You cannot say that all the original kinetic energy of the ball, that has now mostly become potential energy, has become extra mass for the ball, since you know you cannot say an equivalent thing for Photon B.
I know I can. A photon does not lose energy when it climbs out of a gravitational field. It's appears red-shifted only because the environment is no longer subject to gravitational time dilation. Its frequency hasn't changed. 

You can say that the energy of Photon B is a bit more than the original energy of Photon A, when it was at the surface of the neutron star (and was equal in energy to the mass of the not-yet-accelerated ball).  But only a bit more; the majority of energy-that-became-potential-energy-in-the-form-of-mass had to appear somewhere else:  It was added to the mass of neutron star, extracted from both the photons' original energy and from the accelerated cannonball's energy.  In simple-minded terms, as I explained in another message quite some time ago, we can say that gravitational interactions literally suck energy from escaping objects, and give energy to falling objects (by "pulling" on them, a thing typically associated with, yes, "force", and the-supplying-of-energy).  This is known as "balance", and is again consistent; it is even consistent with the idea of Gravitation being an aspect of a system (two interacting objects).
That's wrong Vernon. That's the diference between Newtonian mechanics and relativity.

Now, take a look at that example I gave you which explains why most of the energy goes into the cannonball rather than the Earth.

 

Offline VernonNemitz

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Just imagine an extremely energetic photon having the same magnitude of mass-energy as the just-accelerated [cannonball], at the surface of a neutron star.  When both exit the star they must still have mass-energy equal to each other.
I think I see your problem. The total energy of the moving cannonball comprises the energy locked up as its mass, plus the kinetic energy. When it has escaped the neutron star, the total energy is unchanged. The same is true of the photon.
Wildly false, and the core of our disagreement.  In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy.  See this classic physics joke:
http://books.google.com/books?id=UGGhM2XKE_0C&pg=PA6&lpg=PA6&dq=%22soaked+in+a+gravitational+field%22&source=bl&ots=o6DWAZarZM&sig=nitPtT1Paa_loBYnLR3i2beP2PY&hl=en&ei=dc16S73oOYaN8AaOhoX0CQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAcQ6AEwAA#v=onepage&q=%22soaked%20in%20a%20gravitational%20field%22&f=false (scroll to see a bit more at the top of the page, after clicking the link)

Such "soaking" is exactly equivalent to a cannonball shot out of the gravity well; its observed total mass-energy after escaping is less than its total when just-accelerated.  You might want to claim that the total has remained the same, by having its initial kinetic energy become extra mass (potential-energy-stored-as-mass) as it arose/slowed, but there is no equivalent argument for the massless photon, and that is why your claim is invalid.

Now I'm aware you might want to argue that I'm talking about measurements in two different reference frames, at the surface of the neutron star and out in Space far away from it.  Such an argument is partly a "red herring", because it fails to take into account one key detail.  To see it, let's use the same measuring device in both frames.  Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) --but the device was stationary at the surface of the star!  You have to add a lot of kinetic energy to the measuring device, to get it away from the star!  When you do that, you are not adding energy to the photon (no need; since we know the photon can escape).  The synchronicity between the photon and the measuring device is now broken.  Thus, far from the star, the total mass-energy of the measuring device can (in my view "A") be considered to have diminished to approximately its original value, and therefore can detect some red-shift (loss of energy) of the photon, or (in your view "B") be considered to have not diminished, and therefore can detect an even larger red-shift for the photon!  (That's relativity in actuality!)  Note that either way, there is a red-shift measurable!  In short, your assumptions are faulty (especially the one about the photon's energy not changing), and therefore your argument is worthless.

Conservation of energy applies. If they lost energy, where did it go? The cannonball kinetic energy has been transformed into cannonball potential energy. We know it's cannonball potential energy, because if that cannonball escapes the system, it takes it away.
Conservation of energy most certainly applies, and you are once again ignoring something I wrote to you a while back.
Really bad logic. The [cannonball] is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/[cannonball] SYSTEM.  Therefore the Earth does not lose the mass of the [cannonball] when the [cannonball] is given an escape velocity; only the Earth/[cannonball] system loses it.  And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping [cannonball]; the [cannonball] most certainly does not have it while traversing interplanetary space.  All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the [cannonball].  That is, the amount of potential energy that the [cannonball] can acquire by falling down that well is defined by the planet at the bottom of the well.  If the planet has the same mass as the Earth, then the potential energy the [cannonball] can acquire will be the same as if it could acquire in falling to Earth.
This is why the cannonball and the photon can both lose energy, which becomes potential-energy-stored-as-extra-mass-of-the-main-gravitating-body.  And the next such body that either encounters will have the potential energy to give away, to anything falling toward that body!
« Last Edit: 16/02/2010 21:15:38 by VernonNemitz »
 

Offline yor_on

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Vernon I wasn't discussing a Black Hole there was I? I thought I was commenting on a gravitational wave :) Which we have a reasonable expectation to exist. As for "I'm glad we seem to be agreed on that the potential energy goes into mass, as slight as it might be." Farsight? When did I do that? I thought I called it a relation, just as I see gravity. Relations all of them :)

As for what Gravity is Vernon. To me it's the geodesics of SpaceTime.
Don't get why you read me different here?

Hope I cleared that up now.
 

Offline Farsight

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Wildly false, and the core of our disagreement.  In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy...
Because the instrument is in a different environment. It isn't subject to gravitational time dilation, so it of necessity measures the photon frequency as reduced. 

...Such "soaking" is exactly equivalent to a cannonball shot out of the gravity well; its observed total mass-energy after escaping is less than its total when just-accelerated. You might want to claim that the total has remained the same, by having its initial kinetic energy become extra mass (potential-energy-stored-as-mass) as it arose/slowed, but there is no equivalent argument for the massless photon, and that is why your claim is invalid.
It isn't invalid. Conservation of energy says it isn't. 

Now I'm aware you might want to argue that I'm talking about measurements in two different reference frames, at the surface of the neutron star and out in Space far away from it.
Yes, that's the crux of it.

Such an argument is partly a "red herring", because it fails to take into account one key detail. To see it, let's use the same measuring device in both frames. Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) - but the device was stationary at the surface of the star! You have to add a lot of kinetic energy to the measuring device, to get it away from the star!
If that device is at the surface, then to see the photon, the photon has to come back down. Sorry. 

When you do that, you are not adding energy to the photon (no need; since we know the photon can escape). The synchronicity between the photon and the measuring device is now broken. Thus, far from the star, the total mass-energy of the measuring device can (in my view "A") be considered to have diminished to approximately its original value, and therefore can detect some red-shift (loss of energy) of the photon, or (in your view "B") be considered to have not diminished, and therefore can detect an even larger red-shift for the photon! (That's relativity in actuality!) Note that either way, there is a red-shift measurable! In short, your assumptions are faulty (especially the one about the photon's energy not changing), and therefore your argument is worthless.
It isn't. Yes, we measure a redshift, but we know our instruments are affected by gravitational time dilation. You're not accounting for it.

Conservation of energy most certainly applies, and you are once again ignoring something I wrote to you a while back.
Really bad logic. The [cannonball] is considered to be separate from the Earth as soon as you start treating it separate from the Earth.  That means even when falling off a cliff on Earth, it is not part of the Earth; it is part of the Earth/[cannonball] SYSTEM.  Therefore the Earth does not lose the mass of the [cannonball] when the [cannonball] is given an escape velocity; only the Earth/[cannonball] system loses it. And the Earth still sucked about 11kps of velocity and associated kinetic energy from that escaping [cannonball]; the [cannonball] most certainly does not have it while traversing interplanetary space.  All it has is the potential to fall down a different gravity well, but that well is defined by the planet at its bottom, not by the presence of the [cannonball]. That is, the amount of potential energy that the [cannonball] can acquire by falling down that well is defined by the planet at the bottom of the well. If the planet has the same mass as the Earth, then the potential energy the [cannonball] can acquire will be the same as if it could acquire in falling to Earth.
This is why the cannonball and the photon can both lose energy, which becomes potential-energy-stored-as-extra-mass-of-the-main-gravitating-body. And the next such body that either encounters will have the potential energy to give away, to anything falling toward that body!
Vernon, no. When you give that cannonball its 11.2 km/s of kinetic energy, it has more total energy than it had when it was motionless. All of this total energy is lost from the earth/cannonball system when the cannonball departs that system. The earth does not retain this energy, otherwise where does the 11.2km/s of kinetic energy come from when the cannonball falls down to another Earth? This other earth gains the mass of the cannonball along with that kinetic energy. As a result the gravitational field of this new earth will be increased. Conservation of energy, Vernon. Don't forget it. 

 
 

Offline VernonNemitz

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Such an argument is partly a "red herring", because it fails to take into account one key detail. To see it, let's use the same measuring device in both frames. Sure, at first glance you might think the device that measured X photon-energy the surface of the star will also measure X photon-energy, for the same photon, far away from the star (per relativity; both the device and the photon are affected simultaneously/together/synchronously) - but the device was stationary at the surface of the star! You have to add a lot of kinetic energy to the measuring device, to get it away from the star!
If that device is at the surface, then to see the photon, the photon has to come back down. Sorry. 
You STILL are ignoring what I wrote!  Pay attention!
In more detail, consider a just-created photon at the surface of a neutron star; it contains some X amount of energy.  If the photon travels sideways and is absorbed by a measuring instrument, we detect X amount of energy.  But if the photon climbs out of the gravity well, it is red-shifted; an instrument will not detect all of that X amount of energy.
Care to try again, Farsight, with a less-obviously ridiculous remark?  (There's no reason for me to write more until after I know you fully understand the starting point of this thought-experiment --which, currently, and obviously, you don't, not in the slightest.)
« Last Edit: 17/02/2010 03:20:40 by VernonNemitz »
 

Offline VernonNemitz

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A photon does not lose energy when it climbs out of a gravitational field. It's appears red-shifted only because the environment is no longer subject to gravitational time dilation. Its frequency hasn't changed. 
Here's another thing about which you are wrong, if less obviously so.  Interpretations matter!  Consider that Shrodinger's Wave Equation relates the momentum of a particle of matter with a wavelength.  And I assume you know that the momentum of a particle can depend on its mass, not just its velocity.

SO:  If we are agreed that a cannonball or plate has less mass at the bottom of a gravity well, than at the top, then the Wave Equation automatically associates every particle of that mass (at the bottom of the well) with longer wavelengths, as if "time dilation" was in effect.  In other words, you don't need to assume that time dilation is an intrinsic part of a gravity well; it is a simple logical consequence --a mere interpretation!-- associated with the idea of potential-energy-as-mass being converted to other forms (and finally radiated away) when every particle of matter at the bottom of a gravity well got there and cooled down after colliding.

Well, if gravitational time dilation is just an interpretation, what of an incoming photon?  Obviously all it need do is also be affected, acquiring energy as it falls into a gravity well (just like happens to everything else that falls in; G.R. makes no distinction regarding the form of mass-energy, remember!), and losing it as it climbs out again.  A matter of interpretation, some might say --but yours is absolutely flawed because you are trying to create a distinction between photons and other forms of mass-energy, with respect to General Relativity).  Therefore the photon looks blue-shifted when arriving, and is seen by an instrument at the bottom of the well, because it is blue-shifted, and it looks red-shifted when leaving, and is seen by an instrument at the top of the well, because it is red-shifted.
 

Offline JP

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So to come back to the original topic, I see that only a couple of posts have mentioned the stress-energy tensor, which is the tool needed to describe how an object bends space-time.  The point is that you need the whole tensor to make the equations of GR invariant under a transformation of coordinates (i.e. reference frame), so it seems to me that you can't just pull out the "mass" (which should be contained in the energy part of the tensor) and say that's a physically meaningful quantity in all reference frames.  You need the whole tensor or else you're dealing with a quantity that is only valid in your particular frame. 

And if you want to also include gravitational energy in this discussion, you have to go further and define a psuedotensor, as yor_on pointed out (which isn't reference frame invariant).
 

Offline VernonNemitz

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So to come back to the original topic, I see that only a couple of posts have mentioned the stress-energy tensor, which is the tool needed to describe how an object bends space-time.  The point is that you need the whole tensor to make the equations of GR invariant under a transformation of coordinates (i.e. reference frame), so it seems to me that you can't just pull out the "mass" (which should be contained in the energy part of the tensor) and say that's a physically meaningful quantity in all reference frames.  You need the whole tensor or else you're dealing with a quantity that is only valid in your particular frame. 
We aren't dealing with just one mass here.  There is the mass of the main gravitating body, of course, and for it the appropriate tensors would be necessary.  But for a small body such as a plate or a cannonball, we've been discussing how it behaves in the field of the other mass, and haven't been paying attention to its own (obviously relatively trivial) gravitational field.  So, the phrase, "you can't just pull out the 'mass'" is a bit misleading, since we are not at all talking about pulling out any of the mass of the main gravitating body; we are discussing a system of two interacting bodies, one of which is so small as to have a basically ignorable gravity field, and it's mass (or more accurately, mass-energy) is the only one being "pulled out".  And a very-much-tinier piece of mass energy is actually the main thing being argued about by Farsight and myself --the mass-equivalent of the potential-energy-associated-with-height of a plate or cannonball.  Its gravitational contribution is even more ignorable.
 

Offline Farsight

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You STILL are ignoring what I wrote!  Pay attention! ...
Let's just agree to differ, Vernon.

Yes, mass in general relativity is rather slippery, JP.
 

Offline VernonNemitz

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Let's just agree to differ, Vernon.
That's just a feeble excuse to let yourself keep thinking you have a valid argument.  Today I'm going to address one of your more fundamental errors.  This is the notion that energy cannot be undetectably transferred between objects; that is, after all, your primary excuse for erroneously thinking that Conservation of Energy requires all the kinetic energy that might be given to a plate or a cannonball to be retained by those objects, albeit in some form other than actual motion, when climing out of a gravity well.

The example-of-the-moment is the "simple" collision of one billiard ball against another, on a pool table.  Turn on your "mind's eye ultramicroscope" and look at the event really closely.  We see the surface of each billiard ball as consisting of atoms, clouds of electrons surrounding atomic nuclei.  More, we "see" the electric fields of those electrons, repelling other electrons.  The whole surface of each billiard ball is a thin layer of electrostatic repulsion; at a distance each ball electrically neutral, of course, but up-close-and-personal the story is very different!  (You are aware, are you not, that the electric force is something like 1040 times stronger than the gravitational force?)

Obviously, when the first billiard ball gets close enough to the other, those repulsive surface fields start to interact.  The balls never actually touch!  They simply get close enough for those mutually repelling electric fields to cause each ball as-a-whole to do the things that we typically observe at the ordinary macroscopic level.  And as you know, energy and momentum can be transferred during that collision event --across a distance!  Sure, the distance here is very tiny, but that is simply due to the generic cancellation of opposite electric fields, as seen from a distance, of protons and electrons.  It is necessary that ordinary physical objects approach very closely for the fine details (non-cancelled fields!) to be able to play a role in events.

The point I'm attempting to make is simple: A tiny distance for energy transferrance still qualifies as a "distance".  You erroneously think that energy can be transferred only when the distance is zero, between interacting bodies.  You are wrong because all ordinary interactions take place across distances!  The fact that usually the relevant distances are very tiny is totally irrelevant --the only reason those distances are tiny is because of charge-cancellation, as described above.  We don't normally encounter any field-cancellation of gravitation; therefore masses can interact at significant distances, transferring energy and momentum between them.

Perhaps instead of the above I should have just mentioned the phrase "gravitational slingshot"; according to your erroneous thinking, regarding Energy Conservation, the phenomenon that NASA has used on a number of space probe missions, to increase probe velocity, should never ever be possible.
 

Offline Farsight

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I mentioned the gravitational slingshot when we were talking about gravity and work, Vernon. See http://www.thenakedscientists.com/forum/index.php?topic=27444.msg291021#msg291021. A better example for "never actually touch" would have been a magnet. The reason a nail is attracted to a magnet is because its in a magnetic field, which is local to the nail. It's similar for a gravitional field. A field has energy, and is a non-uniform disposition of the space/energy caused by one object that has some effect upon another. All I can do is reiterate that when you fire a cannonball up into the sky at 11.2 km/s it escapes the Earth's gravitational field, which is then diminished. You really should look at the example I gave which explains why the energy goes into the cannonball rather than the far more massive Earth which has no detectable recoil.     
 

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