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 possesses 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 possesses 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 referring 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 which 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: https://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 mass

That 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 which 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 which 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 referring to this Higgs thingy and some may respond by mentioning the relationship between the Coulomb force and the photon's proper mass. You were referring 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 that's somewhat secondary.. The most fundamental reason is that it is a gauge boson, and the U(1) gauge 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 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 existence 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 referring 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, G_k 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 referring 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 referring 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 existence 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 existence 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]

please follow the discussion

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
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 existence 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 which 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?