Why don't photons possess mass?

[misogynisticfeminist]

Yea, noob question here, if e=mc^2, and mass and energy is the same thing, why is a photon massless when it has energy?

[mathman]

Photons always travel at the speed of light and have 0 rest mass. For gravitational purposes, the energy acts like mass.

[jcsd]

Cos the full formula is E^2 = m^2c^4 + p^2c^2, whic gives you the energy of a particle in an inertial frame. Photons don't have mass but they do have momentum (p), so they have energy.

[selfAdjoint]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.

[Tom McCurdy]

how would you "measure" the mass of a photon? to test it w

[misogynisticfeminist]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.

so are you saying that photons do possess mass but it is just extremely miniscule?

[aekanshchumber]

E=mC^2 gives us the relation between the mass and energy. according to it mass and eenergy are interchangeble, not that mass always posses energy or anything having energy always have mass.

[selfAdjoint]

so are you saying that photons do possess mass but it is just extremely miniscule?

NO! That's why I put the stuff about an upper bound in caps. Theory says the photon has no mass, and experiments have found no mass, but being experiments they have a margin of error, and that margin is an extremely tiny number. Also whenever they figure out how to do a better measurment, the margin always goes down. All the experiments are consistent with the statement that the mass is zero.

[ahrkron]

E=mC^2 gives us the relation between the mass and energy.

As jcsd said, the complete relation has also a term for momentum.

[misogynisticfeminist]

Cos the full formula is E^2 = m^2c^4 + p^2c^2, whic gives you the energy of a particle in an inertial frame. Photons don't have mass but they do have momentum (p), so they have energy.

so, the energy comes from their momentum instead of their mass? but isn't p=mv, and if mass is 0, momentum is 0. Or is there another formula for it?

NO! That's why I put the stuff about an upper bound in caps. Theory says the photon has no mass, and experiments have found no mass, but being experiments they have a margin of error, and that margin is an extremely tiny number. Also whenever they figure out how to do a better measurment, the margin always goes down. All the experiments are consistent with the statement that the mass is zero.

ok.....gottit.

[aekanshchumber]

Photons of course have momentum and the momentum is due to the mass, but still the photons do not have any mass this is because,the energy that the photon had, changed in mass. But neither the change in the mass nor in the energy could be noticed, according to the hysenberg's uncertainty principle
dm*de=h/2*pi (transformation of the standerd equation)

[aekanshchumber]

Furthermore, photons carry EM radiation, and if they had mass EM radiation would have a longitudinal component. But no such component can be detected to ever increasing levels of accuracy. So the photon's mass has to be less than a number starting with a decimal and followed by dozens of zeros before a significant digit, and that's the UPPER LIMIT. So the theoretical requirement that it have zero mass is confirmed in experments up to the limit of experimental accuracy achievable.


If you are saying that the Photons contains EM radiations then i think you are seriously missconcepted about photon. In fact photon itself is a radiation as photons were introduced to explains the particle character of the EM waves.

[arivero]

This is a FAQ in relativity. The m in E=mc^2 was time ago the "relativistic mass". The m in the "complete" formula is the "rest mass". Modernly people prefers to use only the latter, to avoid this kind of misunderstandings.

[pmb_phy]

This is a FAQ in relativity. The m in E=mc^2 was time ago the "relativistic mass". The m in the "complete" formula is the "rest mass". Modernly people prefers to use only the latter, to avoid this kind of misunderstandings.For details on this subject/FAQ see the thread named "Mass of Light" in "General Physics". The same conversation is going on there.

As I've often reminded folks here, many people still refer to "relativistic mass" (aka inertial mass) simply as "mass".

E.g. the newest modern text that I have is Cosmological Physics, by John A. Peacock. When he speaks of mass in the first chapter he is not refering to rest mass. E.g. he refers to T00/c2 as mass density where, as you know, T00 is energy density.

The misunderstanding that arises on topics like this is due to a lack of knowledge on the subject. It is not due to "poor" or outdated knowledge. Ignoring it can still lead to misunderstandings. Only complete knowledge and understanding will lead to a minimum of misunderstandings.

Pete

[humanino]

The most beautiful story about the photon mass can be found in Feynman's lecture on gravitation. I am not going to try to reproduce Feynman's great style. Basically, he is telling the story of a game, where another famous physicist asked him to prove that the photon has no mass. Feynman answered he is willing to play, under the condition : "give me an upper bound, I'll answer. After that, you are not allowed to change the bound !" Of course, the other physicist cheats, and gives Feynman four or five small and smaller bounds.

Anyway, I am not home so I don't have the book with me, I could not even copy the two or three pages if I wanted to. Really, try to find the book. Feynman derives Einstein equations from the assumption "spin 2 massles", as if we did not know about gravitation, but understand QFT already.

[kurious]

Photons are massless according to Higgs theory because they do not interact with the Higgs field.Protons and electrons do.The Higgs field has not been proven to exist bu t even if it does not it would seem reasonable that a photon is massless because it does not interact with whatever field does allow protons and electrons to have mass.

[misogynisticfeminist]

Photons are massless according to Higgs theory because they do not interact with the Higgs field.Protons and electrons do.The Higgs field has not been proven to exist bu t even if it does not it would seem reasonable that a photon is massless because it does not interact with whatever field does allow protons and electrons to have mass.

interesting, so the answer to this actually lies in the higgs field? any reason as to why the photon don't react with the higgs field???

[kurious]

I don't know a physical mechanism but someone on here might have a mathematical reason!

[Sariaht]

It's a wave; do waves have mass? You could say they are mass, but they are just particles exchanging velocity (as in all waves?); as written by another scientist here at this forum, mass is caused by particles in movement, and there is really no other solution to the problem. Try to explain it yourself.

By the way: p=mv, so sure, photons are mass

[da_willem]

The energy and momentum of a photon are respectively hf and \frac{h}{\lambda}=\frac{hf}{c}. Putting this into the equation E^2 = m^2c^4 + p^2c^2 you will find m=0.

[Sariaht]

But after having reconsidered, hf/c2 = m

You dribbled me out there though, the photon is now as said a wave and the mass would be useless to talk about if it wasn't for that the speed of the particles in the wave (and that is the only thing that can cause mass) causes mass which must be true, no?

[da_willem]

Again. The equation E=mc^2 is only correct in the rest frame of a particle where p=0 so E^2 = m^2c^4 + p^2c^2 will yield the famous equation found by Einstein. A photon has no rest frame so you will need the full formula and the results from quantum mechanics wich give you the momentum and energy to yield zero 'invariant mass' m.

[Sariaht]

Massless things that carry energy and bends in gravityfields, pretty scary, no?

1. Waves are particles exchanging speed with one another
2. Particles moving causes mass

But this does not belong here, it belongs to the theory development forum, so you can forget me ever mentioning it.

[da_willem]

See also: http://www.physicsforums.com/showthread.php?t=40279

Massless things that carry energy and bends in gravityfields, pretty scary, no?

By the relation (again...) m^2c^4 = E^2 - p^2c^2 mass can be thought of as being part energy and momentum. So it is not weird that a particle can have energy and no mass, as long as it has a equal amount of momentum!

And the general theory of relativity states that not only mass but also energy and momentum generate gravitational fields. And that all particles are influenced by this 'gravitational field' (wich is nothing more than a distortion of the 'normal' flat geometry) by moving in geodesics.

So it's not that scary at all. You just have to get used to the new theories and that might take a while...

[pmb_phy]

By the way: p=mv, so sure, photons are massThat is incorrect. Photons aren't mass, they have mass.
Again. The equation E=mc^2 is only correct in the rest frame of a particle where p=0 ...
You and Sariaht are using "m" to mean different things. Sariaht is using "m" to refer to relativistic mass and you're using it to mean proper mass. Also E refers to energy, not proper energy E0. As you're using it E does not equal mc2. This is rest energy and therefore E0 = mc2.

And the general theory of relativity states that not only mass but also energy and momentum generate gravitational fields.
Its mass which is the source of gravity where mass and energy are proportional. It is for that reason that energy appears in Einstein's equations. As Einstein phrased it
The special theory has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symmetrical tensor of second rank, the energy tensor.

Mass in one frame is momentum and stress in another frame. Its the stress-energy-momentum tensor which describes the source of gravity. T00/c2 is (relativistic) mass density. Since mass is proportional to energy it really doesn't matter which one you say is the source since you can replace one with the other in Einstein's equation. In fact there can be a total absence of proper mass and there can still be a gravitational field.

And that all particles are influenced by this 'gravitational field' (wich is nothing more than a distortion of the 'normal' flat geometry) ..Gravity is not a distortion of spacetime. It can distort it in certain cases. But you can have a gravitational field in flat spacetime ... at least according to Einstein.

Pete

[da_willem]

But you can have a gravitational field in flat spacetime ... at least according to Einstein.


How can there be a graviational field in flat spacetime? Can you explain or give me a link to a internet document that explains it?

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity. In GR particles follow geodesics wich are just straight lines with a flat metric (this is g_{ab} = \eta_{ab}), and this would be interpreted in the classical theory as the absence of a gravitational field...

[pmb_phy]

How can there be a graviational field in flat spacetime?
Note that I said at least according to Einstein. By this I mean that what I said is consistent with the way Einstein interpreted his general theory of relativity, i.e. how he defined quantities in The Foundation of the General Theory of Relativity. According to Einstein the presence of a gravitational field is dictated by the non-vanishing of the affine connection, not the non-vanishing of the Riemann tensor. In fact he stated this explicitly in a letter he wrote to Max Von Laue.
Can you explain or give me a link to a internet document that explains it?
Sure. See Einstein's gravitational field at http://xxx.lanl.gov/abs/physics/0204044

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity.
I agree 100%.
In GR particles follow geodesics wich are just straight lines with a flat metric (this is g_{ab} = \eta_{ab}), and this would be interpreted in the classical theory as the absence of a gravitational field...The relation g_{ab} = \eta_{ab} means only that the frame of reference is inertial. If that holds at all points in spacetime in the given coordinate system then the spacetime is flat. In a non-inertial frame and there is a gravitational field, i.e. according to Einstein that is. And then the metric is not \eta_{ab}.

If there is a gravitational field present and the spacetime is flat then the spatial trajectories are not straight lines. The presence of a gravitational field is frame dependant as always. In a uniform gravitational field there are no tidal forces and therefore no spacetime curvature. But there is still a gravitational field unless you transform it away by transforming to a free-fall frame. Any of Einstein's texts is consistent with this.

Pete

[misogynisticfeminist]

The energy and momentum of a photon are respectively hf and \frac{h}{\lambda}=\frac{hf}{c}. Putting this into the equation E^2 = m^2c^4 + p^2c^2 you will find m=0.

the recent talk about relativity in this thread, got me lost somewhat. I don't know **** about relativity lol. Ok, yea, so, is this the eqn which actually explains everything, as in I just sub in vlues of E and P and actually find that the photon has no mass. is this the answer to my question?

[pmb_phy]

is this the answer to my question?Depends on the question. You were not precise about what you meant by "mass." Assuming you meant "proper mass" then there are different ways to answer it. Some may respond by refering to this Higgs thingy and some may respond by mentioning the relationship between the Coulomb force and the photon's proper mass. You were refering to mass and energy so that Higgs point is moot since you didn't ask why a photon has zero proper mass. You asked how it can have no mass when it has energy and E = mc2. Your answer lies in the definition of these terms and not having to do with Higgs stuff. That is a different question. After you understand the difference between mass and proper mass and you understand that a photon has zero proper masss and you understand the relationship between Coulomb's law and photon proper mass and you still want to know "why" a photon has zero proper mass then the Higgs thingy may be what you're looking for.

But "why" questions seek reason, not description.

Pete

[Haelfix]

Well yes, the photon does not acquire a Higgs mass, but thats somewhat secondary.. The most fundamental reason is that it is a guage boson, and the U(1) guage symmetry of electromagnetism is both a global symmetry and crucially a local one. Mathematically, that implies it has zero mass.

[da_willem]

is this the answer to my question?

I think it is the answer to your question, but maybe not the answer to why a photon has no mass! Your question came from a misunderstanding of the formula relating mass and energy. As said before E=mc^2 anly aplies to an object having zero momentum and otherwise you would need the full equation with all these squares and c's...

So I think it gives a satisfactory answer to your original question. But if you would want to know why a photon has no mass you would have to read some of the other posts more carefully.

[pmb_phy]

As said before E=mc^2 anly aplies to an object having zero momentum ..
That is incorrect. If one is using m to mean proper mass then E0 = mc2, not E = mc2.

Pete

[humanino]

pmb_phy : I don't see your point here. I think it has been posted several times that E^2 = \vec{p}\,^2 + m^2 and that the expressions you use are valid only in the rest frame where p=0. Unfortunately in the case of photon, there is no rest frame (I am aware that you know it...).

One can always formally define m=\sqrt{E^2 - \vec{p}\,^2} = \sqrt{p^2} but no physical interpretation in a rest frame can be obtained from it for a real photon (for which m=0). Then, for a virtual photon the mass defined in this way is perfectly legitimate.

I am sorry to be confused here :confused: As far as I undestand, Healfix's argument is sound (gauge symmetry). Now could you elaborate on several things :

you say :
Gravity is not a distortion of spacetime. It can distort it in certain cases. But you can have a gravitational field in flat spacetime ... at least according to Einstein.

In a flat spacetime there is no gravitational field. Usually one defines the graviton field as a perturbation from the flat metric (even though it might not be the best convention : see Rovelli which defines the graviton field as the vierbein instead).

you also said :

The relation g_{ab}=\eta_{ab} means only that the frame of reference is inertial. If that holds at all points in spacetime in the given coordinate system then the spacetime is flat. In a non-inertial frame and there is a gravitational field, i.e. according to Einstein that is. And then the metric is not \eta_{ab}.

The discussion might not be clear to me. This seems correct to me : consider a non-empty paracompact region A of spacetime : if the metric is globally flat in A then their can't be any gravitational field in it. I believe this is the meaning of your post. But, maybe due to foreign language, I feel I might be wrong. Is this the meaning of your posts ?

If you are saying : one can have a locally flat metric in a gravitational field, then OK, that's true, but that is the essence of GR.

[humanino]

Once again about the photon mass question : it is not the first thread about that. Really, Feynman's discussion in his "lecture on gravitation" is the best.

___________________________
EDIT : the discussion one can find there is entirely experimental. No theoretical argument there to motivate a vanishing mass of the photon.

[zefram_c]

Allow me to throw in my two cents:

SR allows for clear distinctions between massless and massive objects and describes the behavior of both. It doesn't, however, predict what has mass and what doesn't. Nor does it provide a fundamental explanation for mass; it is simply taken as an intrinsic parameter of the object under study. I'm not going into the various interpretations of E=mc2; the complete formula posted by jcsd leaves much less room for error - it's just not the one that's been popularized.

GR (someone correct me if I'm wrong) doesn't explain mass either. It establishes the relation between spacetime curvature and the stress-energy tensor, and the source's mass goes somewhere in the latter.

Maxwell's equations predict that the speed of light is always c. This, combined with SR, predicts photons are massless.

QFT doesn't explain mass either. Mass is a free parameter in the field equations for free fields as well as QED and QCD.

The weak interaction theory is somewhat different. Since the local gauge transformation rules are different for two components that make up the free particle wave function, the mass term breaks the invariance. Hence traditional mass terms cannot be allowed. To allow mass terms to appear in a gauge invariant way, another field - the Higgs field - must be introduced that is nonzero and asymmetric in the vacuum state. Now when the interaction with this field is accounted for, new terms appear in the equations that have the form of mass terms. In fact, they originate from the coupling of the original massless fields with the non-zero vacuum expectation of the Higgs field. The vacuum expectation is chosen in such a way as to be asymmetric with respect to the gauge mediators, but symmetric wrt a linear combination of them. This remaining symmetry of both the interaction and the vacuum is the U(1) symmetry of EM and accounts for a massless photon. (I note in passing that the local gauge symmetry is still present for the interaction). More importantly, the whole phenomenon of mass can be given a new interpretation as arising from the interaction of massless particles with a very strange vacuum.

All this theory doesn't offer any profound reason for masslessness of photons - the theory is built to be consistent with it. The current experimental upper bound on photon mass is on the order of 10-16eV for lab experiments.

[pmb_phy]

pmb_phy : I don't see your point here. I think it has been posted several times that E^2 = \vec{p}\,^2 + m^2 and that the expressions you use are valid only in the rest frame where p=0.

I don't believe that you were paying close attention to what I was saying. I stated quite cleary above to da_willem that he and Sariaht were using "m" to mean different things. Sariaht is using "m" to refer to relativistic mass and da_willem was using it to mean proper mass. Also E refers to energy, not proper energy E0. As da_willem is using "m" then it E does not equal mc2, its E0 = mc2 where E0 is E as measured in the rest frame of the particle.

The relation E^2 - (pc)^2 = m^2 c^2 is valid if an only if "m" is the proper mass of the particle. If, as Sariaht was using it (and how I use it) m is the relativistic mass of a particle. Defined as such the relation E^2 - (pc)^2 = m^2 c^2 is invalid. Also when m is the relativistic mass then the relation E = mc2 is valid.

One can always formally define m=\sqrt{E^2 - \vec{p}\,^2} = \sqrt{p^2} but no physical interpretation in a rest frame can be obtained from it for a real photon (for which m=0).I've never thought that defining "rest mass" in that way was a very good idea. Also one cannot legitimately define proper mass in that way since its a circular definition since the momentum of a tachyon is defined is defined in terms of proper mass. That is merely a relationship between energy, momentum and proper mass. It can't be taken as a definition of proper mass since that requires leaving momentum undefined - bad idea.

In a flat spacetime there is no gravitational field.Depends on how one defines "gravitational field" and as I said when I posted that at least according to Einstein and according to Einstein the existance of a gravitational field is frame dependant and one can "produce" a gravitational field by chaning coordinates. To be precise - the presence of a gravitational field is dictated by the non-vanishing of the affine connection (when the spatial coordinates are Cartesian) and not the non-vanishing of the Riemann tensor.

Usually one defines the graviton field as a perturbation from the flat metric (even though it might not be the best convention : see Rovelli which defines the graviton field as the vierbein instead) ...if the metric is globally flat in A then their can't be any gravitational field in it.That is a different notion than that given by Einstein. You're thinking of "gravitational field" as the non-vanishing of the Riemann tensor. I am not and Einstein did not. You can define "gravitational field" anyway you like but I have yet to see a good reason to define it in any other way than Einstein did, e.g. in his 1916 GR paper or any of his texts. In fact not everyone speaks of it that way in the GR literature.
Is this the meaning of your posts ?No.

Pete

[humanino]

I still don't get it. I am not doing it on purpose, i am truly sorry, but I want to understand.

Why do you introduce the dinstiction between Energy and Proper energy whereas it is irrelevant to the the Photon ?

What kind of gravitational field could have a vanishing Riemann tensor and a non-vanishing affine connection ? The action for the gravitational is determined by the integral of the curvature : if the curvature is zero, there is non action, no Gravitational field ! Where am I wrong ?!

Thank you for help !

[humanino]

Sure. See Einstein's gravitational field at http://xxx.lanl.gov/abs/physics/0204044

I read the paper you are reffering here :
1 It does not seem to be submitted to any journal
2 The email address given here is hotmail address. Does not this guy has a position in an institute ?
3 In the aknowledgement, the author states that the people whose name is mentionned are not responsible for the possible mistakes in the paper. How should the reader interpet this ?
4 The view proposed in this paper are supposed to be Einstein's one, with opposition to other physisicists such as : Hawking and Thorne. I need better references to challenge such physicists names, not just pretending you have Einstein on your side. I see this paper's view as a very old fashioned one. The formalism is old fashionned. It looks like a student work !

I believe I am paying attention to your posts. I demand more argumentation. I was not doubting you have a real point until I read your reference.

[humanino]

The most convincing argument I saw so far for what the gravitational field is that is : how it should be mathematically represented, comes from Rovelli's last book, where he motivates the vierbein as the gravitational field. The vierbein in turn can be looked as a "square root" of the metric. Fermions require the introduction of the vierbein.

[pmb_phy]

Why do you introduce the dinstiction between Energy and Proper energy whereas it is irrelevant to the the Photon ?

It was you who commented on the applicability of the relation E = mc2. I pointed out that relation is E0 = mc2 and, assuming m = proper mass, holds only for proper energy. I.e. you omitted the subscript denoting proper energy.

What kind of gravitational field could have a vanishing Riemann tensor and a non-vanishing affine connection ?

Any gravitational field with no tidal forces, E.g. a uniform gravitational field. The metric of a uniform gravitational field is

ds^2 = c^2(1+gz/c^2)^2dt^2 - dx^2 - dy^2 - dz^2

The Christoffel symbols are calculated at - http://www.geocities.com/physics_world/uniform_chris.htm. The only non-vanishing Christoffel symbols are

\Gamma ^0_{03} = \frac {g}{c^2}\frac{1}{1 + gz/c^2}
\Gamma ^3_{00} = -\frac {g}{c^2}\frac{1}{1 + gz/c^2}

where g = gravitational acceleration at z = 0. According to the equivalence principle a uniform gravitational field is equivalent to a uniformly accelerating frame of reference in flat spacetime. One can obtain the metric for the uniform g-field by transforming from an inertial frame in flat spacetime to a uniformly accelerating frame of reference. The Christoffel symbols will change from zero to the above values but the Riemann tensor will remain zero as it must (i.e. if a tensor vanishes in one coordinate system then it vanishes in all coordinate systems). The uniform gravitational field was the very first gravitational field that Einstein considered in his theory of general relativity. In fact one form of the equivalence principle is stated as follows

Einstein's Equivalence Principle: A uniform gravitational field is equivalent to a uniformly accelerating frame of reference.

A few examples of a gravitational field with zero spacetime curvature from the general relativity/cosmology literature are the gravitational field of a vacuum domain wall and a straight cosmic string. E.g. see

Gravitational Field of Vacuum Domain Walls, Alexander Vilenkin, Phys. Lett. 133B, page 177-179

Gravitational field of vacuum domain walls and strings, Alexander Vilenkin, Phys. Rev. D, Vol 23(4), (1981), page 852-857

Cosmic strings: Gravitation without local curvature, T. M. Helliwell, D. A. Konkowski, Am. J. Phys. 55(5), May 1987, page 401-407

(Note: The author of the Ma. J. Phys. article uses the term "local curvature" in the title of this paper but he is refering to Riemann = 0)

The action for the gravitational is determined by the integral of the curvature : if the curvature is zero, there is non action, no Gravitational field ! Where am I wrong ?!

What does the action have to do with gravitational acceleration?
The view proposed in this paper are supposed to be Einstein's one, with opposition to other physisicists such as : Hawking and Thorne.
It is a fact that Hawking and Thorne are in opposition to Einstein on this point. But then again they're in opposition to other GRist such as Tolman too.

In any case this is not about what Hawking and Thorne think. Its about what Einstein thought. You haven't supported your claim that Einstein associtated the non-vanishing of the gravitational field with the non-vanishing of the Riemann tensor. Would you care to support your claim?

Its funny that you should mention Thorne since it was Thorne who pointed out to me that the Riemman tensor for a uniform gravitational field is zero. In fact he was the one who sent me the references to the paper by Vilinken for the domain wall which also has zero spacetime curvature (in the region of the gravitational field outside the wall itself). When I asked him about this and his position on gravity = curavture he simply told me that it all depends on how you define gravity.

And is also not in opposition to the GR historian/GR expert Dr. John Stachel, Boston University, former head of the Einstein Papers project. His paper is referenced in that article. Did you even read the paper you're now complaining about? It was intented to explain all the details of why I said this since its too long to get into in detail in a forum.

Try reading The Foundation of the General Theory of Relativity, A. Einstein, Annalen der Physik, 1916 and/or The Meaning of Relativity, A. Einstein, Princeton University Press.

It's pretty clear what Einstein meant when he wrote in a letter to Max von Laue (Einstein to Max von Laue, September 1950)
... what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the components of the [affine connection], not the vanishing of the [components of the Riemann tensor]. If one does not think in such intuitive (anschaulich) ways, one cannot grasp why something like curvature should have anything at all to do with gravitation. In any case, no rational person would have hit upon anything otherwise. The key to the understanding of the equality of gravitational mass and inertial mass would have been missing.That comment cannot be taken in any other way.

But if you want more on this then all you need to do is go to the library and read How Einstein Discovered General Relativity: A Historical Tale With Some Contemporary Morals, J.J. Stachel, General Relativity and Gravitation, Proceedings of the 11th International Conference on General Relativity and Gravitation, (Stockholm,Cambridge University Press, Jul 6-12, 1986). As Stachel writes on page 202
Within a few year years, Lavi-Civita, Weyl and Cartan generalized the Christoffel symbols to the concept of affine connection. This concept served to make the relationship between the mathematical representations of various concepts much clearer. Just because it is not a tensor field, the connection field provides adequate representation of the gravitational-cum-inertial field required by Einstein's equivalence principle. There is no (unique) decomposition of the connection field into an inertial plus gravitational tesor.
Since Stachel is probably the world's leading authority on the history of general relativity, as well as being a noted GRist as well, I'm very comforatable with everything I've explained.

In any case this is all written in Einstein's work. All one has to do is pick it up and read it. Why would you think otherwise? On what basis do you hold that the Riemann tensor must vanish when there is no gravitational field?

A uniform gravitational field is defined as a gravitational field with no spacetime curvature. For this definition and a derivation of the metric based on this definition please also see Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173.

Pete

[pmb_phy]

Although this is getting way off topic I'm going to address one more point - that of da_willem

As I understand it the metric (as determinded by the energy content of space) is the 'replacement' of gravitational potential in general relatvity.

It can be shown (See - http://www.geocities.com/physics_world/gr/grav_force.htm) that the components, Gk of the gravitational force, G, on a particle moving in a gravitational field can be expressed in terms of the gravitational potentials g_{\alpha\beta} as

G_k = \frac{1}{2}mg_{\sigma\beta,k}v^{\sigma}v^{\beta}

where v^{\sigma} = (1, \bold v) and m = \gamma m_0. Its possible, as is the case for a uniform gravitational field, for G to be finite and non-zero and yet Riemann = 0. In the special case where the metric has constant spatial components then we can express the above in terms of \Phi = (g_{00} - 1)c^2/2 as

\bold G = -m \nabla \Phi

The metric for a uniform gravitational field is (Ref. - Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173)

ds^2 = c^2(1+gz/c^2)^2dt^2 - dx^2 - dy^2 - dz^2

for which Reimann = 0 but for which G is not zero.

Pete

[humanino]

Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime


a uniform gravitational field

Gravitational Field of Vacuum Domain Walls, Alexander Vilenkin, Phys. Lett. 133B, page 177-179

Gravitational field of vacuum domain walls and strings, Alexander Vilenkin, Phys. Rev. D, Vol 23(4), (1981), page 852-857

Cosmic strings: Gravitation without local curvature, T. M. Helliwell, D. A. Konkowski, Am. J. Phys. 55(5), May 1987, page 401-407

(Note: The author of the Ma. J. Phys. article uses the term "local curvature" in the title of this paper but he is refering to Riemann = 0)

seem to me to be specific or pathological examples. I certainly do not consider relevant the example of a constant gravitational field. Of course, by assuming a constant gravitational field, one can obtain many sorts of inconsistent results, but this is only due to the fact that a gravitational field is never constant. This is non-physical. There is a local approximation in which the gravitational field looks constant : this is only a local approximation.

The other examples you provide (as well as those provided in the reference physics/0204044) are either : non-physical, or at least the relevance of which is far from being proven : examples such as the Vacuum Domain Walls are nice mathematical constructions, but certainly not physical situations proven to occur in Nature. Or, the other examples such as Cosmic strings, are too specific situations : it is clear that GR is not perfectly suited to describe cosmic strings, but we are able to investigate them with GR, even though it requires re-constructing the geometrical shape of spacetime around the string "by hand", that is we have no general method to systematically handle those structures. At least, I consider the examples provided so far unable to motivate the assertion "the gravitational field is not equivalent to the curvature".



What does the action have to do with gravitational acceleration?

You were refering to the possibility that in an empty region of spacetime, a spurious gravitational-like field could artificially appear due to a pathological choice of coordinates. Otherly said : if gravitation is equivalent to acceleration, one could construct fake gravitational fields by choosing fuzzy coordinates. I was answering here that it seems to me, this pseudo-paradox has been understood now for a while. This assertion is wrong, and it is most easily seen with the lagrangian formulation :
(1) the integral of the curvature vanishes on any subdomain of A
(2) the gravitational field is zero on A
Assertion (1) is equivalent to assertion (2). This the way I understood GR so far, and I never faced any inconsistency.

[pmb_phy]

You're really taking this thread off topic. This thread is about the mass of photons. Not the definition of gravitational field. If you must keep this up then I suggest that you do this in PM so as not to take this thread more off topic than it already is.Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime ... seem to me to be specific or pathological examples.
That's quite meaningless. The term "pathological" does not apply in this context. The only meaning that can be given to your use of the term would be "abnormal" and all corodinate systems are equally normal in general relativity. It appears to me that you've simply chosen a term with negative connotations to describe something you don't like or have never considered before. But feel free to call it what you like. It does not detract from the fact that, as Einstein defined it, it is a gravitational field and this is about how Einstein defined it.
Of course, by assuming a constant gravitational field, one can obtain many sorts of inconsistent results,..Care to provide proof? If so then please do so in PM. I don't care to waste more space here on something as off-topic as this.

Even in Newtonian gravity an example of a uniform gravitational field which can be generated with a finite distribution of mass can readily be found. There is no basis for claiming that a uniform g-field is non-physical.

In any case, as Einstein said, you can "produce" a gravitational field merely by changing coordinates. To be exact, in his 1916 GR paper Einstein wrote
It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates.

Sorry but the rest of your comments are nothing more than comments based on your idea of what a gravitational field should be. This is simply off this side topic, i.e. as how Einstein defined it. It was already apparent that you don't like the way Einstein defined it.

In any case you've missed the point. The point was that the criteria for the presence of a gravitational fields is the non-vanishing of the the Christoffel symbols and not the non-vanishing of the Riemman tensor.

As far a "physical" you have not provided proof that domain walls have never existed and you have not provided proof that straight cosmic strings don't exist nor have your provided proof that a uniform gravitational field does not exit and you have not proved that a uniformly accelerating frame of reference is not equivalent to a uniform gravitational field.

The existance of a gravitational field depends on the choice of coordinates. That is the essence of the equivalence principle - i.e. the gravitational field can be transformed away at any point in spacetime, curvature cannot - one can only choose a region of spacetime small enough so that you can ignore the effects of tidal forces - and the interpretation of this point is pretty debated in the relativity community.

But as I've said, you can feel free to define it as you wish. But as I've also said that I was telling you how Einstein defined it. And the presence of a gravitational field does not require the presence of spacetime curvature.

Let me put this in the most simplest terms I can - As defined by Eintein (not Hawking etc) If there is a gravitational field at a point P in spacetime then it does not imply that the Riemann tensor vanishes at P. Also, the existance of the gravitational field is dependant on the choice of spacetime coordinates.

Do you care to prove that Einstein defined it as being different from what I've explained to you? If so then please do so in PM.

[humanino]

Dear Pete,

I wish to continue this discussion, at least I would like to go deeper in understanding your considerations. Since this is off-topic in this forum, I will post a new thread in the General physics forum.

[pmb_phy]

Dear Pete,

I wish to continue this discussion, at least I would like to go deeper in understanding your considerations. Since this is off-topic in this forum, I will post a new thread in the General physics forum.Please don't. The subject is general relativity so if you wish to start a new thread then I recommend that you place it in the relativity forum.

[humanino]

Sorry ! That's too late. A moderator could move the thread form General physics to S&G Relativity.

[pmb_phy]

Dear Pete : the examples you provide to illustrate/motivate the assertion that the gravitational field is not identical to curvature of spacetime ...seem to me to be specific or pathological examples. One last comment - It takes logic to prove an assertion is true. But it takes just one example to prove it's false.

I've provided two examples (vacuum domain walls and something strange involving strings) and that's twice as much as was needed to prove the gravity=curvature assertion false.

Pete

[pmb_phy]

Sorry ! That's too late. A moderator could move the thread form General physics to S&G Relativity.You understand, don't you, that you can delete a thread you started and place it in another forum by yourself, right?

Pete

[humanino]

Yes Pete ! I hear your argumentation, and I find it really intersting. I had elementary logic too in my math lectures :wink:
I am not trying to prove you wrong. I am trying to convince myself that indeed, "the gravity=curvature assertion false" as you said. Although, I really have not been convinced (yet?) by the examples you provided.

I hope you will participate the following of the discussion.

[da_willem]

That is incorrect. If one is using m to mean proper mass then E0 = mc2, not E = mc2.

Pete

But when p=0 there is no distiction between energy and proper energy, so you could equally well write E=mc^2 with m proper mass, right?!

If, as Sariaht was using it (and how I use it) m is the relativistic mass of a particle. Defined as such the relation is invalid. Also when m is the relativistic mass then the relation E = mc2 is valid.

My textbook on relativity sais the usage of relativistic mass is pretty outdated and hardly anybody uses it anymore. So when I wrote m, I thought everybody knew what I meant...

The existance of a gravitational field depends on the choice of coordinates. That is the essence of the equivalence principle - i.e. the gravitational field can be transformed away at any point in spacetime, curvature cannot

Although the equivalence principle is very useful for making the analogy between gravity and accelleration, the equivalence is not complete: wouldn't it be better to only speak of a gravitational field in the presence of mass/energy. I believe a field (at least an EM field) is an invariant entity wich is as real as a particle. Making it something you can transform away just by changing your coordinates seems pretty strange to me.

Even in Newtonian gravity an example of a uniform gravitational field which can be generated with a finite distribution of mass can readily be found

I can't think of any distrbution but a spherical shell, but there (in spite of the gravitational potential) the field is (uniform) zero... Could you please describe such a distribution to me?

[pmb_phy]

My textbook on relativity sais the usage of relativistic mass is pretty outdated and hardly anybody uses it anymore.

That is one person's opinion. There are others with the opposite opinion. E.g.

In defense of relativistic mass, T.R. Sandin, Am. J. Phys. 59 (11), November 1991

Concepts of Mass in Contemporary Physics and Philosophy, Max Jammer, Princeton Univ. Press, (2000)

Tell me something. How can something be called "outdated" when so many people use it in the modern relativity literature? E.g.

Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003).

Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995).

Especially when labs such as CERN have it on their websites? Any arguement that I've seen to the contrary, i.e. that rel-mass is outdated or a bad idea etc., are all flawed. I've outlined it all another paper I wrote. If you want to read it then go to http://www.geocities.com/physics_world/ and click on On the concept of mass in relativity.

Here's a good example. From Black Holes and Time Warps: Einstein's Outrageous Legacy, Kip Thorne, page 82

..the inertia of every object must increase so rapidly that, as its speed approaches the speed of light, that no matter how hard one pushes on the object, one can never make it reach or surpass the speed of light.

Teach every student that mass does not increase with speed - He reads this book. What is he to make of this comment?

So when I wrote m, I thought everybody knew what I meant...

Not everyone apparently. I know what you mean but when someone asks the question about the mass of light then you shouldn't assume it. I.e. if someone says that since light has energy and E = mc2 then why doesn't it have mass etc. then it should be apparent that they have relativistic mass in mind even though they may not know it.

Although the equivalence principle is very useful for making the analogy between gravity and accelleration, the equivalence is not complete: wouldn't it be better to only speak of a gravitational field in the presence of mass/energy. I believe a field (at least an EM field) is an invariant entity wich is as real as a particle. Making it something you can transform away just by changing your coordinates seems pretty strange to me.

In physics one should not speak of that which can't be measured. The detection of photons is frame dependant as is the detection of radiation. The detection of the electric field depends on the frame of reference. I.e. for some obervers it exists and for other observers it doesn't. Same with magnetic field.

If you have Thorne's book then take a look at Box 12.5. It states

This startling discovery revealed that the concept of a real particle is relative, not absolute; that is it depends on the frame of reference. Observers in a freely-falling frames who plunge through the hole's horizon see no real particles outside the horizon, only virtual ones. Observers in accelerated frames, who by their acceleration, remain always above the horizon see a plethora of real particles.


I can't think of any distrbution but a spherical shell, but there (in spite of the gravitational potential) the field is (uniform) zero... Could you please describe such a distribution to me?
Sure. Please see - http://www.geocities.com/physics_world/gr/grav_cavity.htm

Pete

[da_willem]

Thank you for the example of a uniform gravitational field, it's a very elegant example!

[I think I heard you say in some other forum topic you were open to corrections to your site: judging on your formula for g I guess G=1 in this text? I think you forgot a minus sign after the second equality sign in (3), and where does R come from in your definition of g, I guess z=||r_1-r_2||?]

..the inertia of every object must increase so rapidly that, as its speed approaches the speed of light, that no matter how hard one pushes on the object, one can never make it reach or surpass the speed of light.

The increase of inertia for moving objects can also be explained by another definition of momentum p= \gamma mv, and not assigning the gamma to m, right?


[Do you know Marc Lange or his book: "The philosophy of physics"? Many of my thoughts about mass and fields are affected by this book, and I don't know how correct his work is...

E.g. Could you say: "because proper mass is Lorentz invariant, it has a continuous identity (this piece here is the same as that piece then)"?

And he tries to prove that an electric field has invariant mass, and is thus real and with that he tries to save spatiotemporal locality...]

[pmb_phy]

Thank you for the example of a uniform gravitational field, it's a very elegant example!

[I think I heard you say in some other forum topic you were open to corrections to your site: judging on your formula for g I guess G=1 in this text? I think you forgot a minus sign after the second equality sign in (3), and where does R come from in your definition of g, I guess z=||r_1-r_2||?]
Thanks. I'm unable to sit any longer this afternoon (bad back). I'll get back to you later today.

The increase of inertia for moving objects can also be explained by another definition of momentum p= \gamma mv, and not assigning the gamma to m, right?

That is not a description of inertia. Inertia is defined as that which mass describes. You can say that the particle will not be able to accelerate to c because the momentum will become infinite. I think that is what you're refering to.

[Do you know Marc Lange or his book: "The philosophy of physics"? Many of my thoughts about mass and fields are affected by this book, and I don't know how correct his work is...

No.

E.g. Could you say: "because proper mass is Lorentz invariant, it has a continuous identity (this piece here is the same as that piece then)"?
I don't know what is meant by "continuous identity."

And he tries to prove that an electric field has invariant mass, and is thus real and with that he tries to save spatiotemporal locality...]What is the invariant mass of an electric field?

I'll get back later this afternoon as I mentioned. Until then keep this in mind; from The Evolution of Physics, Albert Einstein, Leopold Infeld, Simon & Schuster (1938) page 31
Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. In our endeavor to understand reality we are somewhat like a man trying to understand the mechanism of a closed watch. He sees the face and the moving hands, even hears its ticking, but has has no way of opening the case. If he is ingenious he may form some picture of a mechanism which could be responsible for all things he observes, but he may never be quite sure his picture is the only one which could explain his observations. He will never be able to compare his picture with the real mechanism and he cannot even imagine the possibility of the meaning of such a comparison. But he certainly believes that, as his knowledge increases, his picture of reality will explain a wider and wider range of his sensuous impressions. He may even believe in the existence of the ideal limit of knowledge and that it is approached by the human mind. He may call this ideal limit the objective truth.

Pete

[da_willem]

You can say that the particle will not be able to accelerate to c because the momentum will become infinite. I think that is what you're refering to.


That is what I was refering to.

What is the invariant mass of an electric field?

He says the (invariant) mass of an electric field is m_f = \sqrt{\frac{E_f}{c^4} - \frac{p_f^2}{c^2}} and in it's rest frame B=0 so the Poynting vecor is zero so it's momentum is zero, so the mass of the field is E_f / c^2. For the field energy he uses \int \frac{E^2}{8 \pi} dV

Although I read (parts of) your article in favor of relativistic mass, I also read a very good webpage confirming my own vision: http://www.wordiq.com/definition/Relativistic_mass

Among a lot of arguments I especially liked the following:If one truly wishes to retain the notion that mass measures the "resistance" to acceleration, then mass can no longer be treated as a scalar quantity. This is because it is easier to accelerate something perpendicular to the direction of motion than parallel to the direction of motion. In effect, an object would have more mass in one direction than other

And the quote from Einstein himself:
“It is not good to introduce the concept of the mass M = m/(1 - v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than ‘the rest mass’ m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion.” – Einstein, in a 1948 letter to Lincoln Barnett

[pmb_phy]

He says the (invariant) mass of an electric field is m_f = \sqrt{\frac{E_f}{c^4} - \frac{p_f^2}{c^2}} and in it's rest frame B=0 so the Poynting vecor is zero so it's momentum is zero, so the mass of the field is E_f / c^2. For the field energy he uses \int \frac{E^2}{8 \pi} dV

That is not an invariant. Consider the energy, E0, of the electric field between a parallel plate capacitor as measured in the rest frame of the capacitor. Let E be the energy as measured in a frame moving normal to the face of the plates. For your "mass" to be invariant the energy must transform as [itex]E = \gamma E_0[/tex]. However this is not the case. The energy transforms as [itex]E = E_0/\gamma[/tex]. For details please see
http://www.geocities.com/physics_world/sr/rd_paradox.htm

Although I read (parts of) your article in favor of relativistic mass, I also read a very good webpage confirming my own vision: http://www.wordiq.com/definition/Relativistic_mass

Sorry but that article is incorrect. It states

Gradually, as special relativity gave way to general relativity and found application in quantum field theory, it was realized that the invariant mass was the more useful quantity and people stopped referring to the relativistic mass altogether.

Today, when physicists talk about the mass of an object they always mean the rest mass.

That is simply incorrect as anyone can see from looking in many modern relativity texts and from online university lecture notes. Even in GR/cosmology texts/


Among a lot of arguments I especially liked the following:

And the quote from Einstein himself:
Einstein was inconsistent with his definition of mass. On the one hand he said that its not good to use a velocity dependant mass. Such as mass is defined as the m in p = mv. Yet in his GR text he actuallyt uses this mass. Although when he uses it he uses it for small speeds it is not the rest mass that you've been using. He uses the definition for which the mass changes as a result of the particle being in a gravitational field. This mass, m, is defined as

m = m_0 \frac{dt}{d\tau}

For a time orthogonal spacetime (i.e. g0i = 0)

m = \frac{m_0}{\sqrt{1 + 2\Phi/c^2 - v^2/c^2}}

When v << c m becomes

m \simeq \frac{m_0}{\sqrt{1 + 2\Phi/c^2}}

This is not an invariant quantity. It depends on the gravitational potential. Yet this is the mass Einstein speaks of in the GR section of his text The Meaning of Relativity. The last edition of which was written much later than he wrote that letter to Lincoln Barnett. Einstein also stated that light has mass. That is something that you wouldn't gather from his letter to Lincoln Barnett either.

By the way, there is a second part to that letter Einstein wrote to Barnett. It reads

I do not agree with the idea that general relativity is geometrizing Physics or the gravitational field. The concepts of Physics have always been geometrical concepts and I cannot see why tghe gik field should be called more geometrical than f.i. the electro-magnetic field or the distance between bodies in Newtonian mechanics. The notion comes probably from the fact that the mathematical origin of the gik field is the Gauss-Riemann theory of the metrical continuum which we are wont to look at as part of geometry. I am convinced, however, that the distinction between geometrical and other kinds of fields is not logically founded.Do you agree with this part of that letter to? :biggrin:

Pete

[pmb_phy]

That is not an invariant. Consider the energy, E0, of the electric field between a parallel plate capacitor as measured in the rest frame of the capacitor. Let E be the energy as measured in a frame moving normal to the face of the plates. For your "mass" to be invariant the energy must transform as [itex]E = \gamma E_0[/tex]. However this is not the case. The energy transforms as [itex]E = E_0/\gamma[/tex]. For details please see
http://www.geocities.com/physics_world/sr/rd_paradox.htm
When I wrote this I had a particular example in mind, the one in this link. I have not considered, as of yet anyway, how the energy between the plates transforms when the capacitor is moving in a direction parallel to the capacitor plates, only then the capacitor is moving in a direction which is normal to the capacitor plates.

Pete

[da_willem]

I don't have to book here, but I'll try to reproduce his reasoning. He said that when you compute the electric potential between two charged particles you bring together from infinity you get a negative number, but that is because in the initial situation the fields were not zero.

He computes the field energy between the two particles with a total electric field \vec{E}_1+\vec{E}_2:
E_f= \int \frac{ \vec{E}^2}{8 \pi} dV= \int \frac{ \vec{E}_1^2}{8 \pi} dV + \int \frac{ \vec{E}_2^2}{8 \pi} dV + 2 \int \frac{ \vec{E}_1 \cdot \vec{E}_2 }{8 \pi} dV

Where he assignes the first two terms to the respective particles, and the third term turns out to be exactly the binding energy. So now he has this positive energy assignable to the field, wich he can assign a mass to and show the existence of the EM field and save spatiotemporal locality...

[da_willem]

That is simply incorrect as anyone can see from looking in many modern relativity texts and from online university lecture notes. Even in GR/cosmology texts/

Here a quote by John Baez:
relativistic mass is of no use at all in general relativity

If a particle accellerates it doesn't excert a larger gravitational pull does it?


Do you agree with this part of that letter to? :biggrin:


Does he refer to his search for a mathematical theory that explains every field as of geometrical nature, just like he did with the gravitational field? He never really succeeded, right? It would be a very elegant theory though... :cry:

[pmb_phy]

I don't have to book here, but I'll try to reproduce his reasoning. He said that when you compute the electric potential between two charged particles you bring together from infinity you get a negative number, but that is because in the initial situation the fields were not zero.Whether the sign is positive or negative will depend on whether the signs are the same or whether they are opposite.
He computes the field energy between the two particles with a total electric field \vec{E}_1+\vec{E}_2:
E_f= \int \frac{ \vec{E}^2}{8 \pi} dV= \int \frac{ \vec{E}_1^2}{8 \pi} dV + \int \frac{ \vec{E}_2^2}{8 \pi} dV + 2 \int \frac{ \vec{E}_1 \cdot \vec{E}_2 }{8 \pi} dV

Where he assignes the first two terms to the respective particles, and the third term turns out to be exactly the binding energy.
It is only the binding energy which contributes to the mass of the system. Not the energy of the fields of the point charges, that part is infinite. I.e. the first two integrals diverge. Only changes in energy yield a change in mass in situations like this. You can read more on this in

Electrostatic potential energy leading to a gravitational mass change for a system of two point charges, Timothy H. Boyer Am. J. Phys. 47, 129 (1979)

Electrostatic potential energy leading to an inertial mass change for a system of two point charges, Timothy H. Boyer Am. J. Phys. 46, 383 (1978)

So now he has this positive energy assignable to the field, wich he can assign a mass to and show the existence of the EM field and save spatiotemporal locality...That was a very poor way of looking at it and has led him to an erroneous result. For instance - What is the mass associated with the field energy of a single point particle?
Here a quote by John Baez:..That's a load of bs. Whether this is useful or not is highly subjective. Perhaps he has never found it useful but I sure have. Different people study different aspects of GR.
If a particle accellerates it doesn't excert a larger gravitational pull does it?
What does acceleration have to do with this? We're talking about the dependance of mass on speed, not acceleration. If you're asking whether the gravitational pull is a function of the speed of the source then yes, absolutely

Measuring the active gravitational mass of a moving object, D. W. Olson and R. C. Guarino Am. J. Phys. 53, 661 (1985)

Abstract - If a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that M_rel = \gamma(1+2\beta^2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not M but is approximately [itex]2\gamma M.
I've worked out a few examples of this myself. Unfortunately I got sick before I was able to update my website to include them. People such as Baez willk inevidably say "Oh! That's not because of mass increase, that's because energy and momentum etc increase etc with speed". Rindler gets into this in his text. He places relativistic mass on the same footing as one would charge in EM. In fact there are equations in GR called the equations of gravitomagnetism which are almost identical to Maxwell's equations. Relativistic mass takes the place of charge and mass has been considered a "gravitational charge" even in Newtonian gravity.
Does he refer to his search for a mathematical theory that explains every field as of geometrical nature, just like he did with the gravitational field?I dunno. Each person has a different opinion on this. What one person thinks is "useful" another person thinks of as "useless".
He never really succeeded, right?I dunno. Try e-mailing him and asking. He's responded to me in e-mail often so he'll probably respond to you.

Pete

[da_willem]

Whether the sign is positive or negative will depend on whether the signs are the same or whether they are opposite

He uses two opposite charges so their fields cancel when brought together

What does acceleration have to do with this? We're talking about the dependance of mass on speed, not acceleration. If you're asking whether the gravitational pull is a function of the speed of the source then yes, absolutely

I meant a particle accellerating thus requiring a greater speed; I shold have said that. And now I think about it offcourse a moving object produces a larger gravitational pull. As you think about it this way; invariant mass would be the one having no place in GR because it is not the source of a gravitational field, energy (proportional to relativistic mass) is?

Thanks for the Am. J. Physics articles (do you know if it's possible to get them delivered in Europe?). I'm still reading your article on the concept of mass (It's quite large you know...).


Try e-mailing him and asking. He's responded to me in e-mail often so he'll probably respond to you.


Sure. Do you have his mailadress; A.Einstein@hotmail.com or something??! :smile:

[pmb_phy]

..invariant mass would be the one having no place in GR because it is not the source of a gravitational field, energy (proportional to relativistic mass) is?
Yup. I agree to a certain extent.

Thanks for the Am. J. Physics articles (do you know if it's possible to get them delivered in Europe?).
If you have access to a university library then they should have it.
I'm still reading your article on the concept of mass (It's quite large you know...).Unfortunately it was very large. I was hoping to keep it shorter when I wrote it but there is a lot to the subject and I chose thoroughness over shortness.

Sure. Do you have his mailadress; A.Einstein@hotmail.com or something??! :smile:baez@math.ucr.edu or john.baez@ucr.edu

This is a photo of John - http://andrej.com/mathematicians/B/Baez_John.html

Reminds of of Mr. Kotter. :laughing:

Saying that relativistic mass has no role/place in relativity is pretty silly since it has always been there and must always be there. Saying it doesn't is like saying that momentum has no role/place in relativity. E.g. consider the force on a particle with proper mass m0

\bold F = \frac{d}{dt}\left \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}}\right

Does v = dr/dt have a role? After all I never need to substitute v into any equation. But lets try it anyway.

\bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right

Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But lets try it anyway.

\bold F = \frac{d(\gamma m_0 \bold v)}{dt}

Does m = \gamma m_0 have a role? After all I never need to substitute m into any equation. But lets try it anyway.

\bold F = \frac{d(m\bold v)}{dt}

Does [itex]\bold p = m \bold v[/tex] have a role? After all I never need to substitute v into any equation. But lets try it anyway.

\bold F = \frac{d\bold p}{dt}

Seems that each time I defined a quantity the equations became simler. But I need not have defined serveral of those quantities, including velocity and momentum. Even force doesn't have to be defined. E.g. I can write the Lorentz "force" law as

\frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)

I can express the law of conservation of momentum by saying that mv is conserved. This holds in all cases, not just for tardyons, when m = relativistic mass.

I call the introduction of each term, i.e. v, m, p, gamma, f, etc. "Economy of thought."

Pete

[pmb_phy]

One can always formally define m=\sqrt{E^2 - \vec{p}\,^2} = \sqrt{p^2} but no physical interpretation in a rest frame can be obtained from it for a real photon (for which m=0). Then, for a virtual photon the mass defined in this way is perfectly legitimate.Just a Note: When discussing quantum mechanics one has to keep in mind that quantities which are meaningful in classical relativity become meaningless in relativistic quantum mechanics and quantum field theory. The above relation defines proper mass as mentioned above. Relativistic mass becomes ill-defined in quantum mechanics, as does \bold p = \gamma m \bold v and \bold F = d\bold p/dt since each is a function of velocity and velocity is not a well-defined quantity.

Relativity (SR and GR) is a classical theory and as such photons really have no place in it in principle. However that doesn't mean that its not highly useful.

Then again this is yet another discussion about terminology and semantics. But such conversations are not a waste of time since Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. - Einstein

Refering to proper mass as "mass" and labeling it "m" instead of m[sub0[/sub] is just a way to simplify things in a certain context. E.g. in particle physics one studies the inherent properties of particles. It'd be exhausing to keep using the qualifier "proper" evertime you spoke of the proper mass of a particle. State what you mean once and be done with it - whatever simplifies things for you and your reader per your taste. However this would mean refering to the proper lifetime of a particle simply as the "lifetime". In relativity the lifetime of a particle is not the same thing as the particle's proper lifetime, yet its rare to see the proper lifetime actually called the "proper lifetime."

Pete

[humanino]

photons really have no place in [a classical theory] in principle
I would really appreciate if you elaborate on this. Since light has such a central role in GR, I don't understand this issue.
Is this merely the fact that individual quanta of light appear only in quantum phenomena ?
Or do you mean to say that light is entirely described by a wave in classical theory ?
Is it something else ?

Thank you in advance.

[pmb_phy]

I would really appreciate if you elaborate on this. Since light has such a central role in GR, I don't understand this issue.Light, yes. Photons, no.

Is this merely the fact that individual quanta of light appear only in quantum phenomena ?

Yes.
Or do you mean to say that light is entirely described by a wave in classical theory ?No. In quantum theory you cannot assign a velocity vector to a photon whereas in classical mechanics/relativity you can. In relativity you should think of a photon as a classical particle which moves at the speed of light, not as a quantum particle, wave-partilce duality and all, until there is a full-fledged theory of relativsitic quantum theory.

Note that I didn't say that it wasn't useful or that I wouldn't speak of photons in SR/GR. On the contrary, I often do. But to be honest I think of them as classical particles which move at v = c.

Are you familiar with the tex Exploring Black Holes? If so then seen the acknowledgement section (http://www.eftaylor.com/pub/front_matter.pdf)

Philip Morrison made several suggestions and convinced us not to invoke that weird quantum particle, the photon, in a treatment of the classical theory of relativity (except in some exercises).

That was what I had in mind.

Pete

[humanino]

Thanks for clear and fast answer Pete.

[pmb_phy]

Thanks for clear and fast answer Pete.
Glad to help. But that was just an opinion and I'm not attached to it 100%.

Photons are funny animals. Even some of the most well know physicists are not even convinced that they exist. For instance, Willis E. Lamb, Jr., Nobel Laureate, wrote a paper entitled Anti-photon, Applied Physics B, 60, 77-84(1995) in which he argues that photons don't exist. And as you probably know, Lamb shared the Nobel prize in 1955 for "his discoveries concerning the fine structure of the hydrogen spectrum."

As the abstract states

It should be apparent from the title of this article that that author does not like the word "photon," which dates from 1926. In his view, there is no such thing as a photon. Only a comedy of errors and historical accidents led to its popularity among physicists and optical scientists. etc.

Good reading and well worth your time.

Pete

[pmb_phy]

I'm still reading your article on the concept of mass (It's quite large you know...).
At this point I'd say toss in the circular filing cabinet. I've grown tired over the years of this topic and I've chosen to (1) delete the paper from my web site and (2) say this only - Mass is what you define it to be and if you choose to defined it as I've explained then you can't go wrong.

Besides that I'd say don't bother with it. Its a waste of time worrying about a definition when you know what the definition is.

Pete

[pmb_phy]

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

Have a good one and it was a pleasure knowing you all.

Pete

[da_willem]

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

I'm sorry to hear that your situation is made worse by using the internet. Thanks though for your fast and accurate replies, and the discussions. I hope your back gets better and we will see you again on PF.

[nrqed]

Saying that relativistic mass has no role/place in relativity is pretty silly since it has always been there and must always be there. Saying it doesn't is like saying that momentum has no role/place in relativity. E.g. consider the force on a particle with proper mass m0

\bold F = \frac{d}{dt}\left \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}}\right

Does v = dr/dt have a role? After all I never need to substitute v into any equation. But lets try it anyway.

\bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right

Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But lets try it anyway.

\bold F = \frac{d(\gamma m_0 \bold v)}{dt}

Does m = \gamma m_0 have a role? After all I never need to substitute m into any equation. But lets try it anyway.

\bold F = \frac{d(m\bold v)}{dt}

Does [itex]\bold p = m \bold v[/tex] have a role? After all I never need to substitute v into any equation. But lets try it anyway.

\bold F = \frac{d\bold p}{dt}

Seems that each time I defined a quantity the equations became simler. But I need not have defined serveral of those quantities, including velocity and momentum. Even force doesn't have to be defined. E.g. I can write the Lorentz "force" law as

\frac{d}{dt} \frac{m_0 (d\bold r/dt)}{\sqrt{1-(d\bold r/dt)^2/c^2}} = q(\bold E + \bold v \times \bold B)

I can express the law of conservation of momentum by saying that mv is conserved. This holds in all cases, not just for tardyons, when m = relativistic mass.

I call the introduction of each term, i.e. v, m, p, gamma, f, etc. "Economy of thought."

Pete



The way you present it, the definitions you choose at every step are totally arbitrary. So, to someone who does not know physics, it might sound like physicists just defined things as they wish with no physical motivation. That's of course very misleading.

To make my point, consider the following step:



\bold F = \frac{d}{dt}\left \frac{m_0 \bold v}{\sqrt{1-v^2/c^2}}\right

Does = \gamma = 1/\sqrt{1-v^2/c^2} have a role? After all I never need to substitute [m = \gamma m_0 into any equation. But lets try it anyway.


So why not define scoobydoo = \frac{\bold v}{\sqrt{1-v^2/c^2}} instead of defining gamma and then putting the gamma with the mass? After all, someone who would not already know some SR would probably find it more natural to put all the v dependence together.

The reason of course is that the way things are grouped and the choice of new definitions are not at all arbitrary, as it might look from your post.
Things are defined according to how they transform under certain symmetries. That's what we always do in physics, from simple nonrelativistic mechanics to particle physics and GR. Symmetry is the key point. Unfortunately it's not pointed out often. For example, the quantity m_0 is a useful concept because it's invariant under Lorentz transformations. The quantity \bold p = \gamma m_0 \bold v is a useful concept because the sum of this quantity is conserved in a collision (no matter what frame we are observing the collision from). So we give names to these quantities. the quantity "scoobydoo" I defined above has no invariance property so it's not useful to give it a special name. The same things is true for the quantity \gamma m_0 . It has no interesting property so it should not really be given a special name. It's only if someone wants to keep using expressions familiar from nonrelativistic mechanics that there is some justification to treating this quantity in a special way (for example, if one really wants to keep using \bold p = m \bold v then one must define m = \gamma m_0 ). But there is no reason for wanting to keep using the nonrelativistic expressions. It's totally subjective. I think it makes more sense to accept the fact that expressions are different in SR than in classical mechanics and that the momentum is simply \gamma m_0 \bold v.

Pat

[da_willem]

It's totally subjective. I think...

I think this is the whole issue, it's just a discussion over semantics. And to a large extent in physics you are free to make your own definitions to make the equations 'more elegant'.

But anyway, Peter is not going to reply, see a few posts ago:

Note: I won't be able to discuss this or anythiing else in the near, or even the foreseeable, future. Due to the problem of my herniated disk I have to spend more time off my feet and out of my chair and spend more time on the couch on my back. The internet is too much of a temptation and what very little time I spend on it is turned out to be far too much and is causing to much damage. to the disk. I will therefore be offline, at least until my back is all better and probably for a while, if not permanently.

[pmb_phy]

I'm sorry to hear that your situation is made worse by using the internet. Thanks though for your fast and accurate replies, and the discussions. I hope your back gets better and we will see you again on PF.
Thanks. It seemed to get a tad better - then it got much worse. I have to see a neurosurgeon next week. I may need surgery. I have a doctors appointment today and there is a computer next door. This is the exception to the rule though.

The point of my post that Pat responded too was that "has no role" is a meaningless thing to say pertaining to the concept or rest mass. Regarding the expression p = M(v)v. As we all know when this is the momentun of a particle then M(v) = \gamma m_0 \bold v and m0 is an inherent property of the particle. This is often take to apply to everything and this may not be the case. Momentum may depend on the orientation of the body relative to its velocity.

There is an excellant example in

Mass renormalization in classical electrodynamics, David J. Griffith and Russell E. Owen, Am. J. Phys., 51(12), Dec. 1983.

The authors consider a dumbbell which consists of two identical charges held together by a rod. The momentum of the dumbbell has one value when the axis is parallel to the velocity and another value when perpendicular. In the arbitrary case where the axis makes an angle between those two values there is a range of values. And the momentum is not even parallel to the velocity in general! :surprised

[qote=Pat]The way you present it, the definitions you choose at every step are totally arbitrary.[/quote]Definitions are always "arbitrary" in that they are not set in stone by God. Perhaps Pat is unfamiliar with the saying
Physical concepts are free creations of the human mind, and are not, however it may seem, uniquely determined by the external world. - Albert EinsteinSee http://www.geocities.com/physics_world/physical_concepts.htm for full quote and reference.


So we give names to these quantities. the quantity "scoobydoo" I defined above has no invariance property so it's not useful to give it a special name. The same things is true for the quantity . It has no interesting property so it should not really be given a special name. That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful. Consider something else besides a point particle. Consider a continuos mass distrubution. Then the mass of the distribution is the inetral of dM = (1/c2)T00dV over the volume of the distribution. This integral is evaluated in the zero momentum frame. What is it that you think dM is? Consider the case of em radiation. Let all radiation in the region x>0 be moving in the +x direction and let all radiation in the region x<0 be moving in the -x direction. What do you think dM means in this case? I.e. when you're adding these quantities, what is it that you're adding?

Consider also that it is not m0 that is the source of gravity, \gamma m_0 is the source. Thus a beam of photons all moving in the same direction will generate a gravitational field. Consider a sheet of matter in the z = 0 plane. If you consider the mass of the particles that make up the sheet as \gamma m_0 then when you consider the fact that it behaves as if the gravitational mass increases like \gamma m_0 then it becomes a very interesting quantity.

Pat - Do you think that v is an "interesting" quantity?

It's only if someone wants to keep using expressions familiar from nonrelativistic mechanics that there is some justification to treating this quantity in a special way.. - This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition. Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.

Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point. They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

Take care all and see you the next time I have a doctors appointment.

Pete

ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm

There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation

[pmb_phy]

Thanks. It seemed to get a tad better - then it got much worse. I have to see a neurosurgeon next week. I may need surgery. I have a doctors appointment today and there is a computer next door. This is the exception to the rule though.

The point of my post that Pat responded too was that "has no role" is a meaningless thing to say pertaining to the concept or rest mass. Regarding the expression p = M(v)v. As we all know when this is the momentun of a particle then M(v) = \gamma m_0 \bold v and m0 is an inherent property of the particle. This is often take to apply to everything and this may not be the case. Momentum may depend on the orientation of the body relative to its velocity.

There is an excellant example in

Mass renormalization in classical electrodynamics, David J. Griffith and Russell E. Owen, Am. J. Phys., 51(12), Dec. 1983.

The authors consider a dumbbell which consists of two identical charges held together by a rod. The momentum of the dumbbell has one value when the axis is parallel to the velocity and another value when perpendicular. In the arbitrary case where the axis makes an angle between those two values there is a range of values. And the momentum is not even parallel to the velocity in general! :surprised

Note; - That is what the authors conclude in that paper. I am not convinced that they are correct. However they do speak of a paradox regarding this problem so I suspect a piece of this puzzle is missing.

[qote=Pat]The way you present it, the definitions you choose at every step are totally arbitrary.Definitions are always "arbitrary" in that they are not set in stone by God. Perhaps Pat is unfamiliar with the saying
See http://www.geocities.com/physics_world/physical_concepts.htm for full quote and reference.


That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful. Consider something else besides a point particle. Consider a continuos mass distrubution. Then the mass of the distribution is the inetral of dM = (1/c2)T00dV over the volume of the distribution. This integral is evaluated in the zero momentum frame. What is it that you think dM is? Consider the case of em radiation. Let all radiation in the region x>0 be moving in the +x direction and let all radiation in the region x<0 be moving in the -x direction. What do you think dM means in this case? I.e. when you're adding these quantities, what is it that you're adding?

Consider also that it is not m0 that is the source of gravity, \gamma m_0 is the source. Thus a beam of photons all moving in the same direction will generate a gravitational field. Consider a sheet of matter in the z = 0 plane. If you consider the mass of the particles that make up the sheet as \gamma m_0 then when you consider the fact that it behaves as if the gravitational mass increases like \gamma m_0 then it becomes a very interesting quantity.

Pat - Do you think that v is an "interesting" quantity?

- This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition. Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.

Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point. They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

Take care all and see you the next time I have a doctors appointment.

Pete

ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm

There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation[/QUOTE]

[nrqed]

That is quite untrue. You're considering this in a vaccum. Its not even clear what you mean by "no interesting property." Have you ever met two people who agreed on all points as to what is "interesting"? v is not invariant and yet I consider it useful.

Then let me be more specific: I think that when constructing a theory, the starting point is a question of symmetries and invariance. What are the symmetries that we think our system (and the equations) must possess? I think that's the starting point. Then one builds quantities that have specific transformation properties under those symmetries. Equations must relate things that have the same transformation properties, so these are the "interesting" things to work with. Of course, one can measure anything one wants! But the theory will involve quantities which have specific transformation properties. That's the whole point of using tensors to build relativistically covariant equations. If we did not pay attention to transformation properties, we would be trying a lot of crazy equations that would prove useless in the end. Why didn't Dirac try writing down equations involving only P_0 separately, or just P_x and P_y, etc?? Because he knew his equation would prove up useless if it was not Lorentz invariant.

*Of course*, when you change the applications of a theory, the symmetries change and the useful quantities become different. What were useful quantities in classical, nonrealtivistic mechanics are no longer useful in SR, and what is useful in SR may have to be modify when going to GR.

All I was saying is that I the quantity \gamma m_0 is not a useful concept in SR. The mass m_0 is a useful quantity, and the quantity \gamma m_0 \vec{v} is useful as the zeroth component of the four-momentum, but not \gamma m_0 .
If so, show me a relativistically invariant equation of SR which contains \gamma m_0 not as part of a four-vector!



Consider also that it is not m0 that is the source of gravity, \gamma m_0 is the source.


So we are talking about GR or SR? I thought we were discussing SR. In any case, show me one of the fundamental eqs of GR which contain \gamma m_0 by itself (not as a component of a tensor) and then I will agree with you that it is a useful quantity to define in itself.


Pat - Do you think that v is an "interesting" quantity?


In classical mechanics, it is, indeed. Because the symmetries that are interesting are the ones under Galilean transformations and one is dealing with inertial frames. In that contex, the quantity d \vec v/ dt is an invariant and that's why it appears in Newton's second law. (of course, we could do it the other way around and talk about the equation in order to define inertial frames, etc but I am following here the symmetry -> laws of physics approach).

Tell me, if you are saying that it's totally up in the air what physical quantities we single out, then why do we put the three components v_x, v_y, v_z in a vector in the first place?? It's because the concept of a vector is invariant under rotation of the coordinate system whereas the individual components are not! Of course, one can measure the indivudual components in an experiment, but from a conceptual point of view, it makes more sense to group the 3 components in a vector. Is your point of view that grouping the 3 components ina vector a totally arbitrary definition??

What about things like d^3 \vec v/ dt^3 ? Why don't we use that quantity in classical mechanics? Of course, it *can* be measured. And anyone could come along and claim it's a useful quantity and so on.


- This comment makes no sense to me. People don't define things so as to look like the non-relativsitic case. It is simply a definition.

I just don't see any reason to introduce gamma m_0 other than to have the momentum equation look like the NR expression. It does not transform a scalar, a vector or any type of tensor for that matter under Lorentz transfo. The rest mass is an invariant, and the four-momentum transforms as a 4-vector. But splitting the four-momentum in a way (as a product of this "relativistic mass" times something else) such that the two pieces have no particular property under Lorentz transfos is not a ueful step, IMHO.

Take as an example what happens in physics - Components of things like 3-force are actually what gets measured in the lab. These things are not invariant. The lifetime of a free neutron is not the same as the proper lifetime of 15 minutes. When the thing is moving in the lab then, on average, it lives longer when it is moving then when it is at rest. A moving rod is shorter than a stationary rod. That is physically measurable.


Of course, one can measure anything in a lab. Including the x component of the momemtum of an electron in any frame. By why didn't Dirac write an equation for the x component of the momemtum of an electron??? Because it is not a useful quantity to build the theory upon. And why? Because separating the x component hides the summetries underlying the theory!! Of course one could work out separate equations for the x, y,z and t components but then one would realize that they are linked in a nontrivial way . And that means that they are not useful quantities from a theoretical point of view because the symmetries are not made explicit. Same thing for \gamma m_0 .

Likewise for the length of the rod. Ofcourse, it is shorter in motion. But we don't have a separate equation for the length of the rod when it is moving at 0.1 c, at 0.2 c etc etc. The equations contain the proper length. If you want to fidn the length when it is moving at 0.1 c, you don;t look up an equation for that quantity, you use the Lorentz transfos to figure it out. So the useful quantity here is the proper length, not the length at 0.1 c. For the same reason, the useful information for a particle is its rest mass.



[
Its unwise to consider only the expression \bold p = \gamma m_0 \bold v as defining mass and to only consider it in a vaccum. There are extremely good reasons to define mass as m = \gamma m_0. Please don't be offended by this but its not like all these physicists for the last 100 years were so dense as not to think the way you do on this point.


I know it's an old debate. And I am not at my office so I can't give references but there are many people who do say that we should not teach the concept of relativistic mass anymore, that's it's an historical "faux-pas". It's a bit like using imaginary time in SR and GR. It used to be done but it is now realized that it's not conceptually a good approach (see for example MWT).

They've found by careful examination under various situations that the most reasonable thing to call "mass" as m = \gamma m_0. The way you define it doesn't always work. In fact it is invalid for a rod which is under stress and moving parallel to its length. That can be shown by looking at the stress-energy-momentum tensor.

The way "I define" it...you mean just m_0? Well, it should be clear if we are doing SR or GR. But in either case, all I am saying is that considering the quantity \gamma m_0 in isolation (and not as part of a four vector or a tensor) is incorrect. Of course, you can look at a certain mass distribution in a specific frame and say that it is something we can measure. All I am saying is that the meaningful quantity to discuss is the four-vector or the tensor, not a component in a specfic frame, for a particular mass distribution. That's what I mean by "useful" vs not useful quantities.

And I think that in teaching or presenting physics it's important to emphasize that anything can be measured and we can single out any combinations of of quantities we want, but the correct way to approach physics is through symmetries and that the useful quantities are the ones with specific transformation properties.


Take care all and see you the next time I have a doctors appointment.

Pete


Take care. I do very sincerely hope that you will be feeling better!


Pat


ps - Pat; to see more examples of where \bold p = \gamma m_0 \bold v fails, or where the definition of "invariant mass" of a system fails pleas see - http://www.geocities.com/physics_world/sr/invariant_mass.htm



There is also an imortant comment at the bottom of page 104 in Ohanian's text "gravitation and spacetime" regarding this. He explains the trouble with adding 4-momenta when the momenta are not constant and the particles have a spatial seperation
I'll look. However, my issue is not with the four-momentum, it's with singling out gamma m_0! But I'll look.

[pmb_phy]

Then let me be more specific: I think that when constructing a theory, the starting point is a question of symmetries and invariance. What are the symmetries that we think our system (and the equations) must possess?
To me the goal and purpose of a theory in the end is to describe nature and what we observe. What we observer are not invariant quantities.

But the theory will involve quantities which have specific transformation properties. That's the whole point of using tensors to build relativistically covariant equations.
The purpose of tensors in SR/GR is to meet the criteria of covariance in the laws of nature.


All I was saying is that I the quantity \gamma m_0 is not a useful concept in SR.

This is the point where opinion comes into it and we've left the area of calculation and prediction. In any case - Anything which is conserved is important. That is a pretty universal opinion. And \gamma m_0 is a quantity which is conser ved and that fact does not rely on the conservation of energy as many seem to think.
The mass m_0 is a useful quantity, ...I agree. Proper time is important as is coordinate time. Coordinate time is as important as proper time and mass is just as important as mass. But you seem to keep arguing in a vacuum. There is more than one reason why (relativistic) mass is so often spoken of in the relativity literature. It has all the properties that one associates with mass, i.e. the inertial properties, the active and passive properties of gravitational mass. Proper mass has none of those.

..the quantity \gamma m_0 \vec{v} is useful as the zeroth component of the four-momentum, but not \gamma m_0 .
If so, show me a relativistically invariant equation of SR which contains \gamma m_0 not as part of a four-vector!

That is incorrect. \gamma m_0 is the time component of 4-momentum. People replace it with inertial energy E but its more correct to use P0 = mc since it plays the role of time component perfectly since if dX is a spacetime displacement then dX0 = ct. So it t is the time component of dX then mc is the time component of P.

So we are talking about GR or SR? I thought we were discussing SR.
Why would you get that impression. Mass is mass. Relativity is relativity. There is no reason that I know of to define mass differently in SR than one does in GR. In fact it'd be a bad idea to do so in my opinion.

In any case, show me one of the fundamental eqs of GR which contain \gamma m_0 by itself (not as a component of a tensor) and then I will agree with you that it is a useful quantity to define in itself.

You're incorrectly demanding that mass must be a tensor in GR for it to be important. That's like saying that energy or pressure are not important in GR because they're components of tensors. That's your personal opinion. I don't expect you to agree with me and I don't expect you to think that its important. There's little use in wanting everyone to think in the exact same way. It'd be a pretty boring world and it would slow down the progress of physics.

However there are two equations which come to mind. One is in MTW. It reads

u*Tu = Mass

Where T is the energy-momentum tensor, u is the 4-velocity of the observer and "Mass" is what some people call "relativistic mass" or "mass-energy." MTW use the term "mass" in the section where they demonstrate the the energy-momentum tensor is symetric. This is in the beginning of the section in MTW where they speak of mass as the source of gravity

One can't really define these 4-tensors and 4-vectors without defining the components. I don't recall ever seeing someone define the energy-momentum tensor without ever speaking about the components. There are ways to speak of components as scalars. For example;

m = P*U/c2

Here m = mass (i.e. relativistic mass) of the particle whose 4-momentum is P as measured by an observer whose 4-velocity is U. Some people call this the energy as measured by the observer but it depends on what one calls the time component. As above u*Tu = Mass is the same thing. This is an observer-dependant invariant.

One can take something like the Faraday tensor and contract it with the 4-velocity of an observer and get the electric field 4-vector. Take the scalar product of the electric field 4-vector with a basis vector and get a component of the electric field. Notice that all of this is tensor manipulation, i.e. contractions of tensors etc.

For more on the electric field 4-vector see Wald or Thorne and Blanchard's text

http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf

You've been speaking of GR/SR completely from the geometric view and as if it were the only view that there is. There is also the 3+1 view that is used quite often as is the analytic treatment of tensors rather than the geometric treatment of tensors.

I've snipped the rest since its more of the same, i.e. more of "any number which is not a scalar is useless/not important." I see no use in going down that road since its all semantics.

I know it's an old debate. And I am not at my office so I can't give references but there are many people who do say that we should not teach the concept of relativistic mass anymore, that's it's an historical "faux-pas".
And there are many people who say that one should teach relativistic mass and that it is incorrect to claim that it is an historical "faux-paux". It is certainly no such thing. All arguements I've seen to back that claim up are flawed.

Some examples of authors whose texts use relativistic mass Wolfgang Rindler, Richard A. Mould, Ray D'Inverno, Bernard F. Schutz (see his new book), etc. (These are authors whose texts where published after 1991).


It's a bit like using imaginary time in SR and GR. It used to be done but it is now realized that it's not conceptually a good approach (see for example MWT).
[/qupte]
Why do you refer me to MTW? They themselves use relativistic mass although they call it "mass" or "mass-energy" rather than "relativistic mass."

[quote]
Take care. I do very sincerely hope that you will be feeling better!

Thanks. I go into surgery Monday. Wish me luck folks.

I'll look. However, my issue is not with the four-momentum, it's with singling out gamma m_0! But I'll look.Its all semantics anyway. To date nobody has stated anything that can be considered fact as to why one view is correct and the other incorrect. However I have seen people incorrectly think that E/c2 = p/v in all cases. That is an incorrect assumption. This is invalid in some instances when a body is under stress.

Up to this point its been philosophy more than relativity.

Pete

[da_willem]

Thanks. I go into surgery Monday. Wish me luck folks.

Good luck!! I hope this goes well, and you'll feel better afterwards...

[humanino]

Yes, good luck Pete ! We appreciate your discussions. I hope you feel better soon.

[marlon]

Good luck Pete, and get well soon.

We will be waiting for your return man...

regards
marlon :smile:

[Tom McCurdy]

GL Pete
it should go fine

[Sariaht]

If mass is energy, but energy is not mass cause it's a wave allthough particles are waves and have mass then I may very well be a wave moving through space in the speed of light with a hundred procent risk of turning into dust in the big nothing.

"Must be some cind of forcefield"

[Mwyn]

I love subatomic particles I find it a fascinating topic one of "the most' interesting in my opinion but there has been a couple questions that have made me ponder about the true nature of them............
1.are particles pieces of matter? or actual waves?
2.if they are made from actual particles or pieces of matter then are quarks composted of something smaller?
3.if they are not then how can it be possible for it to be fundamental?
4.& Finally if particles are actual waves then what are they waves of? and what prevents us from being see through if we are not made out of actual matter?

If any body can answer any of these questions you would of helped me out big time and I thank you very much. :confused:

[misogynisticfeminist]

to Mwyn:

1. Its like asking a hamephrodite (sp.) whether he/she is male or female.

2. Strings perhaps, but i don't think anyone in the physics community ever said that quarks are indeed the most fundamental, and absolutely ruled out anything more fundamental than that.

3. Waves basically carry energy from one place to another. Can't really answer the fourth question though....

: )

[marlon]

I love subatomic particles I find it a fascinating topic one of "the most' interesting in my opinion but there has been a couple questions that have made me ponder about the true nature of them............
1.are particles pieces of matter? or actual waves?
2.if they are made from actual particles or pieces of matter then are quarks composted of something smaller?
3.if they are not then how can it be possible for it to be fundamental?
4.& Finally if particles are actual waves then what are they waves of? and what prevents us from being see through if we are not made out of actual matter?

If any body can answer any of these questions you would of helped me out big time and I thank you very much. :confused:

Particles and waves are two different descriptions of the same thing. There are dual. This means that when trying to describe a particle you can look at them as "little bullets" that bounce on some detector giving a "click"-sound for example. A particle is caracterised by stuff like momentum and energy. Now de Broglie found out that momentum p is equal to h/l ; where l is a wavelength and h is the socalled Planck constant. Energy E is equal to hv ; where v is the frequence. Now l and v are quantities describing a wave. So particles are not waves of something. What you can do when you wanna describe a particle with energy E and momentum p is "construct" the equantion of a wave with v and wavelength l. You are describing the same thing but with another "language", if you will. So what this wave is, is of no relevance. The only thing that matters is that when you measure some wave (ie : you determin the v and the l) you are allowed to say that you measured a particle with certain mass, E and p...

This way of working is the fundament of quantummechanics.

Quarks are up till now the most fundamental particles that feel the strong force. This force keeps baryons and mesons together and it also binds an atomic nucleus together. Ofcourse it is possible that some more fundamental "thing" is found. Strings are a result of the attempt to unify all known interactions : gravitation, electromagnetic force, weak force (alpha beta and gamma decay) and the strong force. yet this is still a very young and unstable modell.

marlon

[marlon]

Well yes, the photon does not acquire a Higgs mass, but thats somewhat secondary.. The most fundamental reason is that it is a guage boson, and the U(1) guage symmetry of electromagnetism is both a global symmetry and crucially a local one. Mathematically, that implies it has zero mass.


INDEED HAELFIX, YOU WROTE THE ONLY GOOD ANSWER TO THE QUESTION OF THIS THREAD!!!


THIS IS THE REASON WHY PHOTONS DO NOT HAVE MAS!!!
The U(1)-symmetry is never broken. For this reason, there ain't no interaction with the Higgs-field...


marlon

[Mwyn]

So basically the question I want to know is that If I could shrink myself into a quantum universe were atoms are the size of basket balls. (highly unlikely yet metaphorically), what would I see? would I see orbs of energy? orbs that look and feel like marbles? or would they be orb like waves?

[selfAdjoint]

More like the waves. All the elementary particles in the Standard Model are treated as geometrical points, and no experients exist that give them firm non-zero dimensions, so you wouldn't see orbs. What you might see (assuming of course that you can "see" at this scale) is blobs with a clear gradient of blobness from the center out to trail off at the edges, and a max somewhere in the middle.

[Mwyn]

I have another question though? would energy itself or the stuff that bonds particles together be made from smaller particles too? like how electricity is composited of electrons, are the atoms blobs of matter or energy?