User talk:Edgerck/archive1

Welcome and Policy

Welcome!

Hello, Edgerck, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

• The five pillars of Wikipedia
• How to edit a page
• Help pages
• Tutorial
• How to write a great article
• Manual of Style

I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question and then place {{helpme}} after the question on your talk page. Again, welcome! --Cronholm144 07:46, 22 May 2007 (UTC)

Help:Contents/Policies_and_guidelines WP:5P WP:MOS should help, and for math articles WP:FORMULA. Sorry no one welcomed you earlier, they probably assumed that you knew the relevant policy already based on your good edits.:)

Oh almost forgot... WP:WPM and WP:PHYS would probably like to have you as a member.

I just remembered a few more things... First, you can create subpages for your work and research by using your domain name http://en.wikipedia.org/wiki/User:Edgerck and then adding /Edgerck_special_relativity or something like that and then you can choose to edit the page, add your content, then oila!, you have a new subpage. Second, you can set your preferences to auto watch pages that you edit. I have 1000 pages on my watchlist as of now. This allows me to keep track vandalism and a variety of other things. Third, congratulations on surviving your first dispute thus far! They can get far more unpleasant than the one that you are having right now. Thankfully, math and physics articles are usually exempt from the worst of them. That's it and good luck!--Cronholm144 08:21, 23 May 2007 (UTC)

Thanks. Great hint on the sub-page. Will try it. I concentrated many great edits in 3 weeks and now I plan to be more in observation/maintenance mode, allowing time for the community to work on them. I believe this may prove to be a good method to reduce disputes, versus a less intense but longer period of editing work to do the same volume of changes. It is possible that shock/recovery works better to reduce resistance to change than a permanent edit irritant. Might be another lesson from the experiment.--Edgerck 08:52, 23 May 2007 (UTC)

Photolysis

I marked the photosynthesis section on the photolysis page as needing updating. Per request of tameeria, I updated it myself. Please see the article's talk page for discussion. Edgerck 15:02, 20 May 2007 (UTC)

Cross product

The result of a vector cross product between two vectors is a pseudovector, not a (true) vector. This is an important difference in mathematics and physics.

I cleaned up the previous definition, by explicitly introducing a right-handed coordinate system that allows the cross-product to be simply (but correctly) defined as a vector. Afterwards, I added a handedness discussion with a cross-product multiplication table for vectors and pseudovectors.

Complications arise here because the cross product is the three dimensional example of what is really a second order tensor. Gibbs is often made to be a villain in math literature for promoting such a cross-creature. However, the usefulness of the cross product -- when correctly used -- is that it makes life simpler for simple things. The flip side is that, up to a certain level, students are not motivated to grasp finer things that might be useful even if just for intellectual growth. Vector analysis (per Gibbs) is presented too often as the "end-all, be-all" of directed quantities.

Discussion at Talk:Cross_product #Pseudovectors

Anon rating question

I raised the issue of anon rating and its problems at Wikipedia talk:WikiProject Mathematics#Can people who don't edit under their real name rate articles?. Please read the comments there.

Mass-energy equivalence

I edited this page, which had a lot of incorrect and off-topic material. Well, editors are encouraged to be bold in WP. Please leave your comments in the talk-page for the article. Edgerck 14:39, 20 May 2007 (UTC)

I won't revert your deletions, but note that it's considered bad form to remove comments from your talk page, even the unsightly ones :-), except for archiving purposes. I agree that discussions about that article should take place on its talk page. Thanks and keep up the good work--just because you're being reverted doesn't mean you're wrong, it just means you need to overcome certain strong forces binding other editors to material in the article. You can do that with appropriate discussion and sourcing. Major changes to a highly evolved article just don't fly on Wikipedia, so you shouldn't be surprised if you're getting looks that say, "You must be new around here." :-) Cheers, Robert K S 14:53, 20 May 2007 (UTC)

Thanks! Comments here by other users were not deleted, they were just moved to the article's talk page. As you can see in other items above, I am in the (new and good) habit of streamlining my user talk page and moving discussions to the respective article talk page. Edgerck 14:57, 20 May 2007 (UTC)

3RR warning

You currently appear to be engaged in an edit war according to the reverts you have made on special relativity. Note that the three-revert rule prohibits making more than three reversions in a content dispute within a 24 hour period. Additionally, users who perform a large number of reversions in content disputes may be blocked for edit warring, even if they do not technically violate the three-revert rule. If you continue, you may be blocked from editing. Please do not repeatedly revert edits, but use the talk page to work towards wording and content which gains a consensus among editors.

Please note that you are now in violation of this policy, and the only thing stopping me from reporting it is that you have not been warned before. --EMS | Talk 02:39, 22 May 2007 (UTC)

EMS: Thank you for kindly calling my attention to this rule. A first interesting point is that the edits and reversals were always a bit different, so there was more of a "dialogue" than an edit war, the way I see it. A second interesting point is also that if the editor who reversed my edits 3x in a row (albeit with slightly different content each time) had been stopped at the third time (by the 3RR) and reversed the edit, I also would have stopped at the second time too -- even not knowing about the 3RR rule. A third point, that I find particularly interesting, is that the 3RR rule is not made known to every WP editor. Edgerck 07:09, 22 May 2007 (UTC)

more at Talk:Special_relativity#Mass-energy_equivalence_bullet

and more at Talk:Introduction_to_special_relativity#Common_misconceptions

You were given a link to it. See the Welcome and Policy entry above. That said, I do realize that it is a little much to ask you to review all of the rules and regulations of this forum at the outset. Now that you have encountered some trouble may be a better time to do so. In any case, I gave you a warning rather than contact the administrators since you needed to become aware of the rule.

As for the edits being different: It is quite common for a text to be refined over time. The 3RR rule therefore states that the effect of the change is more important than the text. --EMS | Talk 06:00, 23 May 2007 (UTC)

Dear EMS, at the time you gave him the warning there was no such link on Ed's page, I added the welcome after your warning to help Ed out, check the time stamp. thanks!--Cronholm144 06:18, 23 May 2007 (UTC)

EMS: as I wrote in your talk page, I think you can improve your approach. And, yes, I believe you acted incorrectly and in conflict of interest by giving me a 3RR (rather than who offended 3RR first) and then moving in to revert my (correct) text while you inserted your own incorrect text. Further, I am editing your note above because I consider it offensive, condescending, and a personal attack -- and this is my prerogative in my talk page. I also saw how you treated other people in WP. Please refrain from telling people how they should behave before they behave. It's just not good for the sense of community. Edgerck 06:38, 23 May 2007 (UTC)

{{helpme}} What do you need help with? -- John Reaves (talk) 07:02, 23 May 2007 (UTC)

Thanks, John. Am I correctly following WP policy in the above paragraph? I am new here and I just want to hear another opinion. Edgerck 07:25, 23 May 2007 (UTC)

Yes, you are allowed to remove personal attacks from your talk page. All users should remain civil at all times. -- John Reaves (talk) 07:40, 23 May 2007 (UTC)

Thanks. Edgerck 07:44, 23 May 2007 (UTC)

Wikipedia's meaning of "truth"

If you're asking for my opinion, I'll tell you: you'll get nowhere by pondering or debating the meaning of truth on Wikipedia. Wikipedia as an institution doesn't give the tip of a rat's tail about "truth", and, especially wherever they involve expositions long-winded and philosophical, such ruminations contribute nihil to achieving consensus on any matter. The only sheriff in these parts is attribution and citation (and they are distinct: see my rant here). Importantly, when someone tries to assert truth on the basis of logic/reasoning, personal experience/authority, interpretation, analysis/synthesis, or any other means that doesn't essentially border on copy & paste plagiarism from some source, with attribution to that source, then that someone will get smacked down with WP:NOR. So it's not worth it to argue about what's true. Just say what somebody else said is true (and that they said it), cite it, and if your POV is neutral enough (or at least represents a side of a valid controvery with a sufficiently large minority of supporters) to receive support from other editors, be cozily comforted that your edits will justly survive. That's the Wikipedia Way^TM. Robert K S 23:51, 23 May 2007 (UTC)

Context: Robert is kindly replying to my last question in his talk page, which you can see here.

I just submitted a comment to the talk-page of WP:V, which you can see here. --Edgerck 00:10, 24 May 2007 (UTC)

Ha! Like anyone would agree on what "objective verifiability" or "subjective truth" are. Good luck, anyway. :-) Robert K S 00:29, 24 May 2007 (UTC)

I exemplified in WP:V talk. Objective and subjective are quite clear terms. Others can submit their understanding too. I hope it is useful. Edgerck 00:37, 24 May 2007 (UTC)

Robert: I agree with you that "Just say what somebody else said is true (and that they said it), cite it, and if your POV is neutral enough (or at least represents a side of a valid controversy with a sufficiently large minority of supporters) to receive support from other editors, be cozily comforted that your edits will justly survive."

Those last three words "will justly survive" provided the main question (how long will it survive for?) behind the ongoing WP Reliance experiment, which BTW follows exactly the kosher recipe you gave -- the WP way. Edgerck 00:38, 24 May 2007 (UTC)

Except that all of your arguments have had as one of their premises the notion that you are capable of ascertaining truth. Whether you are or aren't is immaterial, and the sooner you accept this, the better you will understand the Zen of Wikipedia^TM. (I've never made up a faux ironic slogan before in my life, I swear, and now I'm two in one night. Heaven help me.) Objective and subjective may be quite clear terms, and humans may even be capable of thinking perfectly objectively. This, again, is immaterial, and we return to what I first said on this matter: the more one tries to pontificate about truth and/or assert what is true, the further one departs from consensus. Your first point above only indicates that you misunderstand completely what is meant by "...Wikipedia is verifiability". "Verifiable" in Wikipedia terms does not mean, as it means in science, that you can go out and perform an experiment to test a point's validity, or as it means in mathematics, that you can draw up an airtight proof. It means that you can visit the referenced source and check that what is being asserted to be said was actually said. Robert K S 00:50, 24 May 2007 (UTC)

Objective vs. subjective huh? the difficulty inherent in those terms is probably the exact reason that wikipedia's policy is established as verifiability. That said, as long as your edits are well-cited and the editors invested in the project understand the reason for your initial changes, I don't think there will be much problem ascertaining general decay/improvement rates in the articles you are studying.--Cronholm144 00:59, 24 May 2007 (UTC)

Wow! I understand this is a charged subject, from many discussions I (thankfully) did not participate. Thanks for summarizing it. But, basically, no, I don't think humans are capable of thinking perfectly objectively, in any shape or form. And, yes, I mean "verifiability" exactly as you say WP does, as almost any journalist does, and as some physicists do. And that is why it makes sense to talk about "objective verifiability" -- for example, a book that A. Crank did self-publish (perhaps with different texts for every copy) is not objectively verifiable. The position of the Sun in the sky is objectively verifiable. A paper in the Physics Review Letters is objectively verifiable (in addition to be peer-reviewed, which is a second level of verifiability albeit intersubjective).

More generally, in our interactions with the world, we can perceive the distinctive views:

• Objective Reality: money, market, companies, governments, law, books by reputable publishers, stars.
• Intersubjective Reality: medical diagnosis, bank-client relationship, majority of e-commerce, hacker organizations, criminal gangs.
• Subjective Reality: truth, thoughts, dreams.

which cannot be harmonized except to some extent. Truth is subjective. No wonder people have a hard time making truth be intersubjective (ie, acceptable by two people in a dialogue), or even attempting the impossible task of making truth objective. Further, objective does not mean immutable. The market is objective (as it would be very difficult for one person to single handedly control it), but changes every minute. Hope this is useful. Edgerck 01:27, 24 May 2007 (UTC)

You are still showing you don't get it when you say "The position of the Sun in the sky is objectively verifiable." I am afraid this is not the Wikipedia sense of verifiability. "According to the Astronomer's Solar Almanac, the position of the sun at X time in Y place is Z coordinates"--that is verifiable. Do you see the distinction? It is not the position that is verifiable (such requires an experiment). It is the fact that the cited source actually does present the contributed information that is verifiable; that is to say, you can "go look it up" without "going and looking up". Robert K S 01:42, 24 May 2007 (UTC)

The WP objective verifiability is in my list as "A paper in the Physics Review Letters is objectively verifiable". The counter example (which is NOT quite WP objectively verifiable) is in my list as "a book that A. Crank did self-publish".

The Sun example, as well as the market example, is there but are clearly a non-WP example. The concept of "objective verifiability" > WP.

Hope this is useful. Edgerck 02:06, 24 May 2007 (UTC)

Yes, now you're showing you understand. Robert K S 02:36, 24 May 2007 (UTC)

Intersubjectively OK, then. And you are exactly right: one thing is to understand, another is to show you understand. And you are also right when you say (above): the more one tries to pontificate about truth and/or assert what is true, the further one departs from consensus. The reason is that truth is subjective; as each person affirms their own subjectivity more, their positions move farther apart.

Now, just to make sure I understand it correctly WP-wise, let me present an example. Suppose I write (as I actually did, in the experiment dialogue above):

"...as predicted by his [Einstein] original equation (see Wheeler et. al., Gravitation, ISBN-13: 978-0716703440)".

What I am saying with my reference above is the fact that the cited source actually does present the contributed information in a manner that is objectively verifiable; that is to say, anyone (not just me or you) can "go and look it up".

In other words, exactly as it means in science, that anyone (not just me or you) can go out and perform an experiment that objectively tests the validity of my assertion, to "go and look it up". And, as the scientific method works in physics, anyone else can repeat the experiment ("go and look it up"), and will obtain the same results under the same circumstances.

The test of "objective verifiability" is thus quite different from "subjective truth".

Hope all this is useful. Edgerck 02:59, 24 May 2007 (UTC)

This is your talk page. See the + tab at the top?

Okay, there are new rules about talk pages now? May I suggest that talk pages are for talking, and you shouldn't be creating a labyrinth that has to be negotiated in order to leave you a message? Consider moving the things on your talk page to your main user page. Robert K S 00:03, 27 May 2007 (UTC)

Good point. I'll make the change. Thanks. Edgerck 00:09, 27 May 2007 (UTC)

Mass of composite systems

Point 11:

"...or write the energy-momentum relation m² = E² - p² and infer that, in a closed system, mass must be conserved because the energy momentum 4-vector is conserved."

Yes, but we are still allowed to consider the mass of the closed system, which is not the same as the sum of the masses that are in the system. If you let an atomic bomb go off inside an infinitely strong container, then the mass of the container is the same as it was before the explosion. Even if there is no container, one can still define the invariant mass of the system after the explosion, which is not the same as the sum of the invariant masses of the particles after the explosion. That sum will have become less, of course. Count Iblis 18:13, 28 May 2007 (UTC)

If you let an atomic bomb go off inside an infinitely strong container, then the mass of the container is NOT the same as it was before the explosion. See answer #6: Contrary to classical physics, an isolated (free) system can reduce or increase its mass by internal mass energy conversion. For example, mass is not conserved when an isolated body (in a system considered large enough to be closed) emits a photon, or undergoes nuclear fission or fusion (see references in article). Thanks. Edgerck 19:52, 28 May 2007 (UTC)

We are considering a closed system that does not interact with the outside environment. This means that the walls are insulated and cannot radiate any energy away to the outside environment. One can argue that such boundary conditions are not under all cirumstances physically realizable (neutrinos can easily escape and how does one pevent tiny amounts of gravitational waves from leaving the system), but that's a different issue.

If nothing can leave the system, then the total momentum and enrgy of the system remains the same. The mass of the system (considered as a big thing in itself) is the same. This mass is not the sum of the (rest) masses of the particles in the system. If I want to know how much force is needed to accelerate th system as a whole then I need the mass defined in this way.

Just like if you need to know the exact mass of the atom you need to take into account the internal energy of th atom and can't just add the rest masses of the electron to the rest mass of the proton. Although this internal energy is called "binding energy" it has a potential and a kinetic part. The Hamiltonian contains a kinetic term and a potential energy term and in the ground state both terms contribute to the total energy.

If we replace the atom by an "electron in a small box" filled with radiation, then we can talk about the mass of the box, which will remain the same, provided nothing (not even radiation) leaves the box.

The atomic bomb example is just a pedagogical example where we made this box absurdly large... Count Iblis 21:10, 28 May 2007 (UTC)

Even if nothing can leave or enter the system, the mass of the system may change. This is what answer #6, quoted above, says. This is mainstream knowledge today. Edgerck 21:18, 28 May 2007 (UTC)

That's not true, unless you are misinterpeting what the cited article says (in particular the "small print", e.g. nothing leaves but what about radiation?). Also, when considering some reaction involving particles, then the "default" meaning of total mass will usually be the sum of the rest masses of the individual particles and not the invariant mass of the system considered as a whole.

You do understand that a box filled with radiation has a larger mass than the same but empty box? Count Iblis 22:03, 28 May 2007 (UTC)

When physicists say "a system considered large enough to be closed" then nothing enters or leaves, radiation or not. What you call "default" meaning above does not exist -- the total mass of the helium nucleus is not the sum of the rest masses of it nucleons. Your last question is undefined. Nonetheless, in SR, a photon has no mass. So, if an atom emits a photon, the system atom+photon has less mass, even though it has the same energy. Hope this is useful. Edgerck 22:36, 28 May 2007 (UTC)

No, you are completely mistaken here. This is why you should read and understand the physics, and not just use quotes from articles and books as authoritative texts (I'm not saying that the articles you cite are incorrect, just your reading of them). A box filled with thermal black body radiation at some temperature T has a well defined internal energy and therefore it has a well defined mass. This mass has a well defined experimental interpretation as it gives you the inertia of the box.

"the system atom+photon has less mass". You are adding up the invariant masses of the atom in its ground state to the invariant mass of the photon. But this does not give you the invariant mass of the composite system. Now, I fully agree that in practice we are usually interested in the mass of the atom after it has emitted the photon, which is, of course, less than the atom in the excited state. The atom+photon" system's mass is perhaps of academic or pedagogical interest only. But in theory you can imagine capturing the atom plus photon in some system with perfectly reflecting mirrors and measuring the inertia of this system. If we place the excited atom in the box and wait until it has decayed then the the mass of the box will not change.

Have yo forgotten your own your text in which you wrote that the invariant mass of two photons, each with an energy of E can vary from 2E/c^2 (when the total momentum is zero) to 0 (when the total momentum is 2E/c)? Clearly, if you heat a box so that its interior is filled with radiation then, because the total momentum of the radiation plus the walls of the box remains zero, the total energy divided by c^2 will give you the invariant mass. This thus increases by E/c^2 were E is the total energy of the radiation. Count Iblis 23:13, 28 May 2007 (UTC)

I could probably cite hundreds or even thousands of books and articles of authoritative authors in Physics that agree with most of what you say above, including Einstein's 1906 paper on E=mc² when he included the mass of the emitted photon in the mass of the box. Today, however, we know that this is not correct because the photon is massless.

Regarding you conclusion above, it does not follow from my text on the mass of a two photon system (which I quoted from Wheeler, referenced). If you disagree, please try to publish it in a peer-reviewed physics journal and you will see what other physicists have to say about it. Thanks. Edgerck 23:47, 28 May 2007 (UTC)

Yes, quoted and referenced, but not understood correctly. I think this is perhaps an unexpected result of your wikipedia experiment. There is no substitute for discussing things from first principles. You can quote things as much as you like, but if you don't understand it, then the quotations aren't worth all that much.

Have you not noticed that if the mass of the two photon system is 2E/c^2 the total momentum of the two photon system is zero? Doesn't this ring any bells? If you start with para-positronium (bound state of electron and positron, total angular momentum is zero) then after it decays into two photons, the photons move in opposite directions (momentum is conserved). Count Iblis 00:04, 29 May 2007 (UTC)

Of course you know you're right. But unless you (one day) agree on what I wrote above:

Nonetheless, in SR, a photon has no mass. So, if an atom emits a photon, the system atom+photon has less mass, even though it has the same energy.

the best we can do right now is stop. This thread exemplifies what I observed before, that WP editing rules are nonsensical. People get angry over an impossible proposition. Truth is subjective. An editor cannot write what he does not agree on. But anyone can edit.

So, I find that WP's rules are putting us both at odds, and that is not what I personally want. It is not my truth. Thank you. Edgerck 00:17, 29 May 2007 (UTC)

Ok, but three final points. First of all I'm not angry (I'm used o discussing with people who I strongly disagree with). Anger has, i.m.o. no place in open and frank discussions. Second, I agree that the wiki rules all by themselves won't get you very far. A procedural discussion like X says P but Y says Q is not as insightful as discussing P and Q directly from first priciples while leaving persons X and Y outside the discussion (but they can be cited in the article).

Third, back to the physics, if the system is a hydrogen atom you agree that the mass is not the sum of the masses of the electron and the proton, but if the system is an atom plus photon you are using this rule, which, you yourself acknowledge, does not hold in general. The mass of the combined system is always the energy of the energy in the zero momentum frame. The value of that energy is, of course, due to the energies of the contituent parts, the binding energies etc. etc.

Perhaps your reasoning is that the difference in mass between proton plus electron and the sum of the two parts is due to binding energy bt the photon is not bound in the system atom plus photon.

However, you then forget that the binding energy of the electron in the hydrogen atom is part kinetic and part potential. The expectation value of the kinetic energy "<psi|p^2/2m|psi|>" is not zero. Without this, the binding energy would be even larger (i.e. ground state energy would be even less).

Anyway, i.m.o. you should take rules of physics more seriously than the wikipedia rules. Why one rule for one closed system (atom plus photon in a box, total momentum = 0) and another for another closed system (electron plus proton in a box)? Count Iblis 00:42, 29 May 2007 (UTC)

Thanks for the WP comment. I do not want to convince you about anything. I value free-thinkers. I think a starting point would be if you agree with those phrases. It's OK if you don't; I value you as a human being just the same. A lot of people don't even know the question and yet have perfectly nice and rewarding lives. Thanks. Edgerck 00:55, 29 May 2007 (UTC)

--

I wanted to point out the following from Taylor & Wheeler, "Spacetime Physics" in the section "Use and abuse of the concepts of mass", pg 247.

Q: Does the explosion in space of a 20 meaton hydrogen bomb convert .93 kilogram of mass into energy....

A Yes and no! The question needs to be stated more carefully. Mass of the system of expanding gasses, fragments and radiation has the same value immediately after explosion as before; mass M of the system has not changed.

This directly contradicts some of your points, which also seem to lack internal consistency.

i.e you write earlier in the thread

If you let an atomic bomb go off inside an infinitely strong container, then the mass of the container is NOT the same as it was before the explosion.

This is directly inconsistent with the above quote from Taylor & Wheeler (which you yourself quote from) - it is also inconsistent with your correct statements about the energy-momentum 4-vector of the system (in this case, the bomb) being both conserved and invariant.

If you write the energy-momentum 4-vector of the bomb before the explosion, and after the explosion, how can you possibly believe that energy is conserved (the same before and after the explosion), that momentum is conserved (the same before and after the explosion), but the invariant mass (E^2 - |p|^2 in geometric units) somehow changes?

Several other posters have pointed this out to you already, but I thought I would do so along with some references. Pervect 07:24, 31 May 2007 (UTC)

TALK page question on mass

...apologize if I misunderstood you, but I read you think that mass is conserved in SR. You seem to link it to the invariance and conservation of 4-momentum. Can you please clarify? Otherwise, we agree that mass of the system is defined to be the magnitude of the energy-momentum 4-vector and is thus an invariant. Thanks. Edgerck 23:10, 28 May 2007 (UTC) Retrieved from "http://en.wikipedia.org/wiki/User_talk:Sbharris"

Answer: Yes, invariant and also independently conserved. I link mass to invariance and conservation of 4-momentum, since the best definition of mass (particle and system mass) in SR is the norm of the summed 4-momenta of the system. Invariance and conservation are different things, of course, but 4-momenta has both properties. Energy and momenta are not frame-invariant due to depending on frame for definition, but both are conserved over time, if you stay in the same inertial frame. 4-momentum is BOTH conserved AND invariant, and so avoids this problem. What you weigh on a scale is actually the norm of 4-momentum of the system you weigh. Fly past at high V and E/c^2 changes, but the scale shows the same weight. You need weight (and therefore also mass) to be an invariant, and not just total energy/c^2. It is actually 4-momentum which is closely connected with what we normally call "mass" in GR. The stress-energy tensor is just 4-momentum plus 3 4-momentum fluxes at a point, and although the stress-energy is not Lorentz invariant, it is also only mass-energy concentration at a point, and needs integrating over a volume to give you a classical mass. When you do that, you find the result IS Lorentz invariant, so long as you integrate out to flat space so there's no question of having taken care of the warpage of the G field that results from shifting frames. Mass charge is rather like electric charge-- the field is compressed in different frames but a version of Gauss' theorem keeps the source constant. But the same cannot be said of total energy of a system (which just gets bigger and bigger the faster you fly by it) and so that's another reason that E/c^2 is a bad definition for mass. Obviously total energy has nothing to do with gravity, if you sum up gravity in all directions over a decent volume. 4-momentum is a better candidate for the conserved quantity which is connected to the G field, and which is invariant with proper integration techniques. SBHarris 00:09, 29 May 2007 (UTC)

Let's stay within SR and start with the 4-momentum, if that is OK for you. To me, if a quantity is invariant, then it will have the same measured value in any inertial reference frame; if a quantity is conserved, then its value, as measured in a particular inertial reference frame, does not change over time. Please just let me know if you agree, yes or no, before I continue. Thanks. Edgerck 00:44, 29 May 2007 (UTC)

Yep. Agree. 4-momentum for systems is both conserved and invariant. Energy and momentum for any closed system are conserved over time and over interaction, but (due to frame-dependence for their very definition) are NOT invariant. Moreover, since proper mass (also called invariant mass) of a system is the norm of its total 4-momentum, this quantity is also (like 4-momentum) both conserved and invariant. SBHarris 01:03, 29 May 2007 (UTC)

Mass is an invariant but is not conserved in SR. Mass can be converted to massless energy (photon), even 100% (mass annihilation). Thanks. Edgerck 05:28, 29 May 2007 (UTC)

Mass cannot be converted to single photons, which are the only ones that are massless. Sytems of photons, which is what mass converts to, continue to have a mass. You do know about the conservation of the energy momentum 4-vector for system reactions, do you not? It's a very basic conservation principle of SR. Well, the length of this system vector is the system invariant mass, which must also therefore be conserved through reactions. The end. What part of this don't you agree with? Be specific. SBHarris 15:43, 31 May 2007 (UTC)

The physics, even within SR. Take, for example, a laser medium as a closed system including some photons. These photons will be copied by the stimulated emission process, resulting in a field of photons all with the same momenta and collinear. According to "Spacetime Physics", these photons will have mass zero. However, they did take energy out of the laser medium, which energy resulted in a decrease in mass of each photon emitter. So, at the end of the process, in the closed system, mass decreased. Edgerck 17:20, 31 May 2007 (UTC)

Edgar - Here I think that you are wrong. Until the photons are emited, they are contributing to the mass of the laser. That they are going back and forth between the mirrors and are individually massless does not matter. What does matter is that the resistance to acceleration of the system is a function of the amount if energy localized within it. Simple example: If the system is accelerated, they energy and momentum need to be transfered to and from the photons to have them moving back-and-forth with the same energy in the new "moving" reference frame. Until that happens, they are "pushing" on the "back" end of the laser (away from the direction of acceleration) harder than they are pushing on the front end. Hence in a closed system even photons contribute to the total mass of the system. --EMS | Talk 18:25, 31 May 2007 (UTC)

I think it is simpler to see if you consider my example as a laser amplifier. The seed photons start at the one end, when we measure the system mass, and travel to the other end, where we measure the system mass when they arrive. The system remains closed. The photon field is built up by stimulated emission that reduces the laser medium's energy, resulting in exact copies (within Heisenberg's uncertainty) of the seed photons -- energy, polarization, phase, and direction of propagation. According to Spacetime Physics (simple application of the energy-momentum equation) the coherent photon subsystem at the end of the experiment has zero mass and a large momentum. Use that subsystem's 4-momentum in calculating the total system 4-momentum (laser medium + photon subsystem), and you will see that by conservation of energy the (laser medium + photons) must have less mass at the end. Edgerck 18:40, 31 May 2007 (UTC)

You are making the same mistake as in case of the atom-photon system, i.e. you are neglecting the change in kinetic energy of the laser that is producing the photons. Conservation of momentum implies that this kinetic energy cannot be zero both before and after the photons are emitted. This is how you should do the calculation. Count Iblis 19:50, 31 May 2007 (UTC)

Edgar - I read this slightly differently than Count Ibis: IMO you are not coinsidering the change in energy-momentum for the reflected photons. Photons which are going in the direction of acceleration get less energetic as they are reflected. Photons going against the acceleration get more energrtic as they are reflected. So the photons "push" the back more than the front and contribute to the whole system resisting acceleration more than the laser medium alone would. It brings us right back to E=mc². It is the energy enclosed in the system the matters, and not its form.

Note the the issue is not the masslessness of the photons IMO, but rather how the behave when reflected in an accelerating system. --EMS | Talk 20:03, 31 May 2007 (UTC)

BTW, the issue is even simpler. You can't willy-nilly use the "subsystem" of the laser photons or beam (or pulse, to make this easy) and add it up with the laser to get total mass. Invariant masses cannot be added legitimately in SR, and that includes the mass of photons (here zero) and the mass of laser. Of course they don't add up to the previous mass of the laser, if calculated separately, THEN added. But that does not mean that mass is not conserved for closed systems, as this can be true without mass addition being true.

So you can't select a subsystem, or two substystems, and add their mass up to get the mass of the bigger system. You can do that for energy, but not for mass. Why? Because momenta (which are part of mass but not total energy) may cancel. For mass, to get total system mass you need to add energy and momentum of substystems separately FIRST, THEN take the length of that vector, which is the invariant mass (this reminds me of certain system in quantum mechanics where you must add raw wavefunctions first, THEN square-- you can't get the answer by adding squared wavefunctions). Again, masses do not add in SR, but 4-momenta and 4-vectors do. SQRT(A) + SQRT(B) is not equal to SQRT (A+B). The total length of the sum of vectors, each of which may well be pointing in different directions, is not the sum total of their lengths. Even when the vectors are composed of aditive components; not least here where the 4-momentum is difference-vector. Edgerck, in assuming masses must add, is effectively assuming you can add vectors by simply adding their lengths. The photon E-M vector length is always zero, but photons have energy and momentum components which affect systems, even so.

Note: of course relativistic masses DO add, unlike invariant masses. But the laser beam pulse HAS a relativistic mass, so using that definition of mass is no help to Ed. Relativistic mass is conserved in closed systems also, because it's basically energy. But like energy, it's not invariant. SBHarris 20:58, 31 May 2007 (UTC)

Sbharris: I was, as I wrote, calculating using 4-momenta. The masses were not added, the resulting mass was calculated from the resulting 4-momentum. And, of course, mass means invariant mass.

BTW, here is an even simpler example that requires no calculation. Take two atoms that are separated by a light-like interval (ie, the space between the two atoms is exactly balanced by the time of flight of light between the two atoms). One atom emits a photon in the direction of the other atom. The photon is able to cross the light-like interval. When the photon is absorbed by the second atom, the second atom must increase its mass. So, mass was transfered from one atom to another even though the atoms were separated by a light-like interval (that matter cannot cross).

Now, to complete the example, let's ask the questions: Did the first atom reduce its mass, and when? When the photon was emitted or when the photon was absorbed? Since light is the limiting speed, the right answer is "the first atom reduced its mass when the photon was emitted". So (since matter cannot cross the light-like interval), while the photon was in transit to be absorbed, the closed system composed of the two atoms + photon must have had less mass before the photon was absorbed than before the photon was emitted.

Edgerck 23:01, 31 May 2007 (UTC)

Why so? I agree that the 1st atom loses mass (about) equal to the photon's energy when the photon is emitted. But while the photon is in flight, the photon adds precisely this amount of mass TO THE SYSTEM (which is drawn around both atoms, and at least must be drawn around the 1st atom and photon, otherwise it's not closed and you've let energy escape it, via the photon). Yes, the photon itself has no mass, but it adds mass to the system when in flight. That is because its momentum and energy must still be added to the system while it is in flight, and even though these mass and energy of the photon add to zero for it, individually as a subsystem (because they are equal and subtract, thus cancel), they still add mass to the system, because the photon's momentum cancels the first atom's momentum, allowing the energy of atom and photon to be added while it is in flight. So the photon's energy makes up precisely for the atom's loss of energy, and the mass of the system (which is the sum of these energies in the COM frame) remains constant. Do the math, for heavensake.

And by the way, Taylor and Wheeler do not contradict themselves (as you claim) about mass being invariant in a nuclear decay, or in a thermonuclear explosian of an H-bomb. First, what the authors say, is even of individual atoms emiting single photons (and can be amplfied to many atoms in a laser emitting a massless beam). In both cases the light is massless, but still adds mass to the system in flight, as discussed just above, because it allows for momentum to be cancelled so energies may be added to get a total system mass (system mass is system energy in COM frame). Second, even if you don't accept this argument, you can see that the photons in an exploding bomb are traveling in all directions in the bomb's initial frame (the COM frame), cancelling their momentum between them, and thus as a photon gas are likely to have invariant mass E/c^2 where E is the total photon cloud energy. So the bomb mass (and energy) doesn't decrease until these have been allowed to escape the system. Without a reference, you cannot remove this statement from the article. At worst (and this is dirty fighting) I have to note that your contention that Taylor and Wheeler contradict themselves would be news to them, and constitutes OR on YOUR part. Sauce for the goose is sause for the gander. If you like to use OR as a weapon, how do you like it when it's used on YOU? SBHarris 19:13, 1 June 2007 (UTC)

You did not calculate the four-momenta properly at all. In the atom+photon problem, you took the energy of the photon to be the diffrence in the rest masses of the atom and the excited atom. That's wrong because the kinetic energy of the atom changes. The velocity of the atom must change because momentum is conserved and the photon has a nonzero momentum.

And what happened to your insistence on using reliable sources etc. etc? Can you point out a single reliable source that writes that "invariant mass of a closed system can decrease"? No, you can't! Surely the fact that a closed isolated system can reduce it's rest mass would be a pretty surprising result in physics and you would expect that this topic would be covered in textbooks.

Instead you are now trying to prove the elusive effect by doing Original Reseach on your talk page. That's allowed, of course. But doesn't the fact that you can't find a direct quote from the literature plus the fact that you are trying to derive the result yourself ring any bells?

Now you have moved on to two atoms and a photon, lasers etc. What next? Anomalies in cold fusion experiments are caused by the (closed) system reducing it's rest mass????  :) Count Iblis 02:28, 1 June 2007 (UTC)

CI and all: My reading of WP:NPOV and WP:Verifiability is that original research is not allowed in WP. You can read my experimental conditions for reducing my influence on the WP experiment on reliance on information.

In summary, if any topic I selected for correct information is not entirely clear, not sufficiently referenced, or even proves to be wrong, then that topic will be eliminated from the evaluation. On purpose, I stayed away from "cold fusion" -- so that's not in question. Thanks. Edgerck 18:44, 1 June 2007 (UTC)

Here is my take on this:

• Stage 1: Atom A exited, Atom B ground state, no photon.
• Stage 2: Atom A ground state, Atom B ground state, photon going from A to B.
• Stage 3: Atom A ground state, Atom B excited, no photon.

Now in a given frame-of-reference, each component of the 4-momentum must be conserved. That means that for this system as a whole the invariant mass is conserved, including in the phase where the photon is in travel. I agree that the sum of the individual masses is lower in phase 2, but the issue for myself and the others is the system as a whole.

I do admit that it is wierd that a massless particle can contribute to the invariant mass of a system that includes it, but this is relatvity after all. --EMS | Talk 02:55, 1 June 2007 (UTC)

If you look at the example again, mass was transfered at the speed of light. This is not possible for anything that has mass. Further, for causality reasons (as indicated), there should be less mass in the system during the entire time of that transfer (suppose the atoms are separated by one light-year).

There is nothing in physics to state that mass cannot be transfered within systems at the speed of light. Indeed massless photons tranfer mass at the speed of light all the time. If THAT argument was correct, you could apply it to a shell of photons moving in all directions between nested spheres, and it would still be true. And yet I think even you have agreed that a photon gas with no net momentum must have a collective mass. That mass is being transported outward at the speed of light, if so. It may be news to you, but the sun loses mass all the time in just this way. The mass is retained by the "shell" of light it emits. Physics only says a massive particle cannot travel at the speed of light, but light itself has no mass, yet transports and adds mass to systems. Again, that's because mass doesn't add, so massless things can add mass to systems. If you'll think about it, this makes perfect sense when viewed from the invariant mass context. In a container of hot ideal gas, the kinetic energy of particles adds mass to the system, but for each individual particle, the invariant particle mass is its rest mass; its kinetic energy, which is busy adding mass to the system, DOES NOT APPEAR. So where is it? Where is this heat kinetic energy located physically, if it does not appear as any particle's invariant mass. I dunno. It's in the system, but no particle has it as invariant mass. Only the system has this extra mass, because the sum of masses for the particles which are in it, is the sum of their invariant (rest) masses, which is less than the total mass. For photons, the result is the same. Like kinetic energy, they don't appear as invariant mass when individual particles are examined, one by one. Yet their energy adds invariant mass to systems, while they exist. SBHarris 19:31, 1 June 2007 (UTC)

This example can also be quantitatively calculated showing that, indeed, the total system mass is less during the transit time. Thanks. Edgerck 19:04, 1 June 2007 (UTC)

No, you're completely wrong about that, and if you actually do a quantitative calculation to attempt to show that, you will either find out that invariant mass is conserved, or you will make an easily identifiable math or physics error. Try it here, if you don't believe me. Iblis or I will be glad to show you exactly where you go wrong. You can do it purely mathematically, or you can choose a real example like Tc99m decay.SBHarris 19:27, 1 June 2007 (UTC)

Sbharris and all: For WP purposes, I have spoken my peace for my experiment's purposes and I am happy to let the information be processed within the community. Now, if you could help me find verifiable sources that say that mass is conserved in closed systems at all times in SR, that would be useful as well. If any topic I selected for correct information is not entirely clear, not sufficiently referenced, or even proves to be wrong, then that topic will be eliminated from the evaluation. You can read my experimental conditions for reducing my influence on the WP experiment on reliance on information. Thank you for your kind help. Edgerck 19:47, 1 June 2007 (UTC)

• Just to be fair, you should note that I did find you such a reference: chapter 8 of Taylor and Wheeler. You THEN did your own OR and claimed Taylor and Wheeler were in error and contradicted themselves. So it did me no good to find you an authoritative reference, because you simply rejected the authorities as in error. Do you want a piece of it again? Fine. From Taylor and Wheeler, page 226:

  In summary, the mass of an isolated system has a value independent of the choice of reference in which it is figured. System mass remains unchanged by encounters between the consituents of the system. Any why? Because the system mass is the length (in the sense of the spacetime interval) of the arrow the total momentum-energy... System mass does make sense!

  And again, to a question of who draws boundards of the system (p 227):

  We do! We can draw the dashed line around any collection of objects whatever, subject to this one restriction: no object in our system may interact with any external object or experience a force from outside the system. Our system must be isolated. With that single limitation, the system we choose is arbitrary, has a conserved total energy, a conserved total momentum, and a system mass that is invariant-- a mass that is the same no matter in which free-float frame [inertial frame] it is reckoned.

  Now, what part of that isn't clear to you? SBHarris 20:44, 1 June 2007 (UTC)

Sbharris: conservation <> invariance. Conserved total energy, conserved total momentum, and system mass that is invariant are all fine. See the reaction 2H + 3H → 4He + n example by Frank Wilczek, in Physics Today (2004), given below for a counter-example to the text you attribute to Wheeler that "System mass remains unchanged by encounters between the consituents of the system." Edgerck 07:44, 2 June 2007 (UTC)

We are dealing with apples and oranges here. The sum of the masses of the components is not the mass of the overall system. In my example, we have (in geometrized units):

• For mass A: $m_{A}={\sqrt {(E_{A})^{2}-(p_{A})^{2}}}$
• For mass B: $m_{B}={\sqrt {(E_{B})^{2}-(p_{B})^{2}}}$
• For the photon: $m_{\gamma }={\sqrt {(E_{\gamma })^{2}-(p_{\gamma })^{2}}}=0$
• For the overall system: $m_{s}={\sqrt {(E_{A}+E_{B}+E_{\gamma })^{2}-(p_{A}+p_{B}+p_{\gamma })^{2}}}\neq m_{A}+m_{B}+m_{\gamma }$.

So you can argue all that you want that the total mass in the system has decreased when the photon is in flight. I won't say that this is wrong, but it does not affect the overall mass of the system, and the overall mass (instead of the total mass) is what we are talking about.

I agree that we need to fall back on authoritative sources here, but I hope that this will illuminate the nature of this discussion. --EMS | Talk 20:04, 1 June 2007 (UTC)

"There is no separate principle of mass conservation. Rather, the energies and momenta of such particles are given in terms of their masses and velocities, by well−known formulas, and we constrain the motion by imposing conservation of energy and momentum. In general, it is simply not true that the sum of the masses of what goes in is the same as the sum of the masses of what goes out. "

"To illustrate the problem concretely and numerically, consider the reaction 2H + 3H → 4He + n, which is central for attempts to achieve controlled fusion. The total mass of the deuterium plus tritium exceeds that of the alpha plus neutron by 17.6 MeV. Suppose that the deuterium and tritium are initially at rest. Then the alpha emerges at .04 c; the neutron at .17 c. In the (D,T) reaction, mass is not accurately conserved, and (nonrelativistic) kinetic energy has been produced from scratch, even though no particle is moving at a speed very close to the speed of light. Relativistic energy is conserved, of course, but there is no useful way to divide it up into two pieces that are separately conserved. In thought experiments, by adjusting the masses, we could make this problem appear in situations where the motion is arbitrarily slow. Another way to keep the motion slow is to allow the liberated mass−energy to be shared among many bodies."

by Frank Wilczek, in Physics Today, 57N12 10-11 (2004).Edgerck 07:35, 2 June 2007 (UTC)

Overall, that is another way of saying the same thing as what I am trying to tell you. In the 2H + 3H → 4He + n example, 17.6 Mev of energy either leaves the system or it redistributed amongst it's components. In the former case, the system is not closed and what poeple are arguing with you over is moot. In the later case, the system has an invariant mass that is conserved, but the (rest) masses of the components certainly has not been conserved. I really don't see what there is to argue over now. Instead, I think that it is time to put an end to this discussion and give all parties a chance to digest it for a bit. Then perhaps we can return to the SR page and reach a consensus of what we should be stating there regarding mass. --EMS | Talk 00:19, 3 June 2007 (UTC)