If something gives off light, is it losing mass?

jdweston · Dec 18, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>jdweston why are you splattering threads with that nonsense. Posted by Manwh0re</DIV> New to this board and was unsure of where to post it.  Please use the controls to remove if it is not welcomed.

BoJangles · Dec 18, 2008

nah its ok, all is well <div class="Discussion_UserSignature"> -------------- Let me start out with the standard disclaimer ... I am an idiot, I know almost nothing, I haven’t taken calculus, I don’t work for NASA, and I am one-quarter Bulgarian sheep dog. With that out of the way, I have several stupid questions... *** A few months blogging can save a few hours in research *** </div>

vogon13 · Dec 18, 2008

Even at absolute zero, do the constituent quarks and gluons of matter continue to move within the nuclei at relativistic speeds ?? How do we slow them down ?? And even at .000001K in a laboratory at Cal Tech, matter on earth continues to orbit the center of the galaxy at ~100km/sec.  Where does this energy (and it's mass equivalent) show up on the balance sheets ??      <div class="Discussion_UserSignature"> TPTB went to Dallas and all I got was Plucked !!So many people, so few recipes !!Let's clean up this stinkhole !! </div>

vogon13 · Dec 18, 2008

Let's say I have a good freezer, and a scale. I cool down a sample of matter on the scale.  Colder and colder we go. Eventually, I have a Bose-Einstein condensate. I keep cooling it, and it gets bigger. I really work at the cooling thingy, and eventually, my Bose Einstein condensate is 1 light year across. Does the reading on my scale change ??    <div class="Discussion_UserSignature"> TPTB went to Dallas and all I got was Plucked !!So many people, so few recipes !!Let's clean up this stinkhole !! </div>

DrRocket · Dec 18, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> Even at absolute zero, do the constituent quarks and gluons of matter continue to move within the nuclei at relativistic speeds ?? How do we slow them down ?? [/QUOTE}Speed bumps ? Replying to: <DIV CLASS='Discussion_PostQuote'>And even at .000001K in a laboratory at Cal Tech, matter on earth continues to orbit the center of the galaxy at ~100km/sec.  Where does this energy (and it's mass equivalent) show up on the balance sheets ??      Posted by vogon13</DIV>Speed is dependent on the reference chosen to measure it.  So is energy, and so is mass.  In the laboratory at Cal Tech the motion of the earth about the center of the galaxy is not a factor, since the quarks, gluons and the laboratory share that motion.  <div class="Discussion_UserSignature"> </div>

UncertainH · Dec 21, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>From a Wiki article on mass and energy in relativity:"This formula also gives the amount of mass lost from a body when energy is removed. In a chemical or nuclear reaction, when heat and light are removed, the mass is decreased. ...Posted by DrRocket</DIV> I don't mean to flog a dead horse but does this imply that if the higgs boson is responsible for mass that a photon contains a boson or a boson is destroyed when a photon is emitted ?

DrRocket · Dec 22, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> I don't mean to flog a dead horse but does this imply that if the higgs boson is responsible for mass that a photon contains a boson or a boson is destroyed when a photon is emitted ? Posted by UncertainH</DIV>No.BTW a photon is a boson. <div class="Discussion_UserSignature"> </div>

UncertainH · Dec 22, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>No.BTW a photon is a boson. Posted by DrRocket</DIV>Ok, so the photon is a guage boson. I'm still wondering where the mass comes from

DrRocket · Dec 22, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Ok, so the photon is a guage boson. I'm still wondering where the mass comes from Posted by UncertainH</DIV>E  = mc^2 = h*nuBut the rest mass is zero.  <div class="Discussion_UserSignature"> </div>

UncertainH · Dec 22, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>E  = mc^2 = h*nuBut the rest mass is zero.  Posted by DrRocket</DIV>I get the energy mass equivalence thing. I'm looking for a physical explanation. I thought that in the standard model that the higgs mechanism was responsible for mass. Presumably the electron when dropping to a lower orbit does not lose any mass but the rest mass of the atom has dropped by emitting a massless photon

DrRocket · Dec 22, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I get the energy mass equivalence thing. I'm looking for a physical explanation. I thought that in the standard model that the higgs mechanism was responsible for mass. Presumably the electron when dropping to a lower orbit does not lose any mass but the rest mass of the atom has dropped by emitting a massless photon Posted by UncertainH</DIV>The electron loses energy, hence loses energy.  The lost energy is the photon.If you want to try to follow this through the details of quantum field theory and the Higgs mechanism here is an outline:  http://en.wikipedia.org/wiki/Higgs_mechanism  <div class="Discussion_UserSignature"> </div>

derekmcd · Dec 23, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I get the energy mass equivalence thing. I'm looking for a physical explanation. I thought that in the standard model that the higgs mechanism was responsible for mass. Presumably the electron when dropping to a lower orbit does not lose any mass but the rest mass of the atom has dropped by emitting a massless photon Posted by UncertainH</DIV>The rest mass will remain the same for both the electron and the atom as a whole.  The actual equation is:E=(mc^2)^2 + (pc)^2where p is the momentum.  While it is fair to say that the photon is massless (or more precisely, has zero rest mass), it does still, in fact, have momentum.  When the electron emits a photon, the energy of the photon is defined by its momentum and via conservation laws pertaining to momentum, the electron loses an equal amount of momentum that the photon carries away.  Plug that number into the above equation and you will see that the electron loses energy and hence the electron and ultimately, the atom loses mass.Again, the rest mass never changes.  The "m" in the above equation is only considered rest mass when p=0.   When such is the case, the formula then, simply, becomes E=mc^2.  A photon, being massless, the equation becomes E= pc = h*nu.Now, you might ask yourself, where is the loss of mass?  Using the above formula, only momentum is lost.  As I understand it (there's two schools of thought on this of which I'm not very clear on), the concept of relativistic mass has fallen out of favor (though still useful as pedagogical tool).The above formula defines Mass as the total energy of the system... rest mass plus momentum.  By saying "relativistic" mass, one simply means the total energy.  Inertial mass is nothing more than resistance to a change in acceleration (m=f/a).  Acceleration is the rate of change in velocity.  Momentum is defined by p=mv.  If the electron loses momentum, it becomes easier to change its velocity.  If a change in velocity becomes easier,  one says it's mass has decreased.  Hence the loss of mass.It's interesting to note that the rest mass of the 2 up quarks and down quark that constitute a proton are individually less massive than when they are combined to make the proton.  This is due to the binding energy of the massless gluons.  The proton is 95% more massive than the particles that make it!!! http://www.physorg.com/news146415074.html A really obese individual weighing 1000kg could legally say their rest mass is merely 50kg and the reason they weigh so much is because they are so full of energy. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" /> I don't believe the Higgs mechanism has anything to do with the electron/atom losing mass when it emits a photon.  All the Higgs mechanism is attempting to describe is why some elementary particles have rest mass and others don't. <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

UncertainH · Dec 23, 2008

Thanks that is much clearer.

DrRocket · Dec 23, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>The rest mass will remain the same for both the electron and the atom as a whole.  The actual equation is:E=(mc^2)^2 + (pc)^2where p is the momentum.  While it is fair to say that the photon is massless (or more precisely, has zero rest mass), it does still, in fact, have momentum.  When the electron emits a photon, the energy of the photon is defined by its momentum and via conservation laws pertaining to momentum, the electron loses an equal amount of momentum that the photon carries away.  Plug that number into the above equation and you will see that the electron loses energy and hence the electron and ultimately, the atom loses mass.Again, the rest mass never changes.  The "m" in the above equation is only considered rest mass when p=0. Posted by derekmcd</DIV> Actually the equation that you gave either requires a square root or it applies to E^2 rathter than E.  Also, the mass term 'm" in that equation is the rest mass and not the relativistic mass.  It is the rest mass no matter what p is.  That equation is in fact equivalent to the usual E = mc^2 if m is the (relativistic) mass and not simply the rest mass.  E=mc^2 is perfectly valid, and always has been so long as one recognizes what m is.  http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4I am unaware that the concept of "relativistic mass" is unfashionable.  Unfashionable or not mass is a concept that is dependent on the observer.  <div class="Discussion_UserSignature"> </div>

derekmcd · Dec 23, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Actually the equation that you gave either requires a square root or it applies to E^2 rathter than E.  Also, the mass term 'm" in that equation is the rest mass and not the relativistic mass.  It is the rest mass no matter what p is.  That equation is in fact equivalent to the usual E = mc^2 if m is the (relativistic) mass and not simply the rest mass.  E=mc^2 is perfectly valid, and always has been so long as one recognizes what m is.  http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4I am unaware that the concept of "relativistic mass" is unfashionable.  Unfashionable or not mass is a concept that is dependent on the observer.  Posted by DrRocket</DIV>Good catch on the formula.  I actually meant to define rest mass vs relativistic mass in E=mc^2 (not the full formula).  I just flubbed that statement up a bit.As for the debate on relativistic mass:http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html Folks have even submitted papers on it.  Here's one in support of relativistic mass:http://arxiv.org/abs/0709.0687  <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

DrRocket · Dec 23, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Good catch on the formula.  I actually meant to define rest mass vs relativistic mass in E=mc^2 (not the full formula).  I just flubbed that statement up a bit.As for the debate on relativistic mass:http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.htmlFolks have even submitted papers on it.  Here's one in support of relativistic mass:http://arxiv.org/abs/0709.0687 Posted by derekmcd</DIV>Both of those articles explain quite clearly why relativistic mass is the most economical means of handling the concept of mass and inertia.It is interesting that many of the arguments for not using the concept of relativistic mass originate with the desire to use the equation F =ma based on the thoroughly mistaken idea that it is an expression of Newton's Second Law.  In fact Newton never said F = ma. but in fact stated the Second Law as F = dp/dt or F = d(mv)/dt which is precisely the equation that applies in special relativity.You can of course formulate relativistic mechanics without use of the notion of relativistic mass.  You can also formulate the equations of planetary motion in terms of a heliocentric model and epicycles.  But the purpose of physical laws is to explain the behavior of nature in as simple a manner as possible, and both formulations are just silly in most contexts.The equations of motion in relativistic models are simple if you simply use the proper formulation, F = dp/dt and the elegance that accompanies using the very same equation that applies is classical Newtonian mechanics is unmistakeable.  It is true that rest mass is invariant and mass is not, but it is the whole point of relativity that mass, distance, energy, and time are not invariant.  Yep it is not intuitive, and it sometimes confuses the newbies, but to attempt to cover up those issue does no one a service -- if you are going to understand relativity you simply have to get used to the fact that some things are indeed relative. The "debate" seems to be one involving only those concerned with teaching the subject to beginners, and seems to be based on the assumption that beginners are not very intelligent.  That assumption is invalid.  There is no point in teaching physics by dumbing it down and distorting it in the process.  <div class="Discussion_UserSignature"> </div>

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If something gives off light, is it losing mass?

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