Idea of increased mass at relativistic speeds

EnumaElish said:
That's a "wow." It's obvious when I think about it, yet... How does physics explain this? (I am sure the explanation is obvious to a physicist, too.)
The photon's energy is given by E = hv, where h is Planks constant and v is the frequency. It has momentum too, you just cannot use mass to calculate it. Its momentum is given by p = h/wavelength.

The usual equation for mass m^2 = E^2/c^4 - p^2/c^2 is equal to zero in the case of the photon because the wavelength times the frequency equals the speed of light c.
Or in other words E = hv = hc/wavelength.
So the equation for mass becomes,
m^2 = (hc/wavelength)^2/c^4 - (h/wavelength)^2/c^2 = 0.

Here is an interesting link http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html [Broken]
Which discusses the question of whether a photon has mass.
 
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Entropy said:
You know which is which because atleast one of the objects has been accelerated in its past, in other words it has "felt" a force. This information tells you which object's mass has changed.

What was the purpose of stating [tex]E = m c^2[/tex]??

Pete
 
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What was the purpose of stating [e=mc^2].
What? Stated it where?
 
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Entropy said:
You know which is which because atleast one of the objects has been accelerated in its past, in other words it has "felt" a force. This information tells you which object's mass has changed.



[tex]E = m c^2[/tex]



It doesn't matter what inertial frame you're in, the mass of an object appears the same to all observers.



Only when at rest. Photons carry energy and therefore mass except when they are at rest.
Even if a particle is neither moving nor part of a bound system, it has an associated energy, simply because of its mass. This is called the particle's rest energy, and it is related to the particle's rest mass as

rest energy = (rest mass)· c2

This in Einstein theory. E=Mc^2 has many fargilities as I already said in another threads.
 
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In post #17
Then why did you quote something different from that post?
 
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Does it help to consider relativistic mass equation as a tool with which one can explain the different perceptions regarding the momentum of a particle when observed from two different inertial frames?

By way of example consider two inertial frames one of which we label the "rest" frame. For convenience we can then just call the other frame the "inertial" frame which travels at a velocity of u, in relation to the the rest frame.

Observers in each of the frames are observing a particle which has a velocity of v according to the rest frame and v' according to the inertial frame. Therefore, the particle has momentum of p = mv according to the rest frame and p' = m'v' according to the inertial frame.

How does the observer in either frame explain how a single object can have a different momentum in another frame?

The terms I've used above give a clue - there are equations for translation of velocities between inertial frames so we can work out what v' is in terms of v and u (and c). We can then say that either there is "relativistic momentum" so that there is actually a real difference in the momentum (and hence energy) of a single particle when observed from two perspectives, or there is a difference in the mass as perceived from each of the frames and use the equations mentioned above to work out what that difference is - that is values of m and m' so that momentum is invariant.

The law of conservation of momentum is axiomatic, there is no reason to think that it is false and plenty to support it being true. It therefore also seems sensible to say that you can't change the momentum of a particle by just changing your perspective in respect to it - the implication is that something odd is happening with the mass, which is precisely what standard relativity tells us.

neopolitan
 

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neopolitan said:
Does it help to consider relativistic mass equation as a tool with which one can explain the different perceptions regarding the momentum of a particle when observed from two different inertial frames?

By way of example consider two inertial frames one of which we label the "rest" frame. For convenience we can then just call the other frame the "inertial" frame which travels at a velocity of u, in relation to the the rest frame.

Observers in each of the frames are observing a particle which has a velocity of v according to the rest frame and v' according to the inertial frame. Therefore, the particle has momentum of p = mv according to the rest frame and p' = m'v' according to the inertial frame.

How does the observer in either frame explain how a single object can have a different momentum in another frame?

The terms I've used above give a clue - there are equations for translation of velocities between inertial frames so we can work out what v' is in terms of v and u (and c). We can then say that either there is "relativistic momentum" so that there is actually a real difference in the momentum (and hence energy) of a single particle when observed from two perspectives, or there is a difference in the mass as perceived from each of the frames and use the equations mentioned above to work out what that difference is - that is values of m and m' so that momentum is invariant.

The law of conservation of momentum is axiomatic, there is no reason to think that it is false and plenty to support it being true. It therefore also seems sensible to say that you can't change the momentum of a particle by just changing your perspective in respect to it - the implication is that something odd is happening with the mass, which is precisely what standard relativity tells us.

neopolitan
I'm not sure what exactly you're saying. The momentum is different from a different perspective... This is true pre-relativity and post-relativity.
 
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Entropy said:
Then why did you quote something different from that post?
What does that have to do with my question? I was simply asking why you posted it in that post.

If you need to know why I didn't quote something else from there its because it was only that equation that I couldn't see why you made it where you did.

Pete
 
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If you need to know why I didn't quote something else from there its because it was only that equation that I couldn't see why you made it where you did.
Because mass can be converted to energy.
 
Mass and momentun are tools for calculations and lollypops for children. In classical mechanics we use inertial force to balance equations as a "pseudo force". Though we could consider inertial force as a "pseudo force" long ago, we cannot still consider mass as something "pseudo" (or if you like it - "relative"). Just as velocity is not an intrinsic property of an "object" and is irrelevant unless another object with a relative velocity is present as a reference; mass, momentum, energy all show up in calculations when some accident occurs, they are not intrinsic properties of "objects" . They are not realities in themselves but are measurable effects of some other realities.
 
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Entropy said:
Because mass can be converted to energy.
So what. Nobody seemed to care about the relationship between mass and energy. Why is E = mc^2 a good/meaningful response to "But it is not interchangeable and defining it to be so is not a useful thing to do."??????

Pete

ps - Its not quite correct to say that mass can be converted into energy. E.g. Neither mass nor energy changes in a nuclear reaction. What changes is the partioning of the energy from mass-energy to other forms of energy such as kinetic energy or potential energy.
 
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I don't know why you are getting all pissy about this. You're acting like I said something offensive.

So what. Nobody seemed to care about the relationship between mass and energy.
That is what this whole topic is about.

Why is E = mc^2 a good/meaningful response to "But it is not interchangeable and defining it to be so is not a useful thing to do."??????
Because matter and energy are inchangeable. That equation shows it.

Its not quite correct to say that mass can be converted into energy. E.g. Neither mass nor energy changes in a nuclear reaction. What changes is the partioning of the energy from mass-energy to other forms of energy such as kinetic energy or potential energy.
"Mass-energy?" Stop playing word games and just call it "mass." I'm not sure what point you're trying to make here. Calling something by a different word doesn't change what it is.
 
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Entropy said:
I don't know why you are getting all pissy about this. You're acting like I said something offensive.
You have read something into my posts which was definitely not there. Please try not to put words into my mouth or intentions into my tone.
"Mass-energy?" Stop playing word games and just call it "mass."
Nothing I post is ever a game unless I specifically state so. If I say something then I'm very serious about it. This is especially for this topic, i.e. "mass." Here you seem to be assuming that in all cases E = mc2 and this seems to have led you to this "correction" of yours and your comment regarding "word games." Quite untrue. That equation is valid only in certain special cases, e.g. for single particles and for systems of free particles. It is not generally valid for constrained systems. Thus the claim that mass is identical to mass-energy is invalid.

You may want to think of this as silly semantics but that is your choice and your opinion. I'm very very serious about terminology and nothing I post is any sort of "game."

Let me clarify - The way we think is deeply entrenched in the language we use. Different language -> different ways of thinking. As Einstein once wrote
The greated part of our knowledge and beliefs has been communicated to us by other people through the medium of a language which others have created. Without language our mental capacities would be poor indeed, comparable to those of higher animals; ...
I'm going to take a guess and assume that you think that this is all nonsense and that differences in language has little to do with differences in ideas. To that point please recall the language of the Hopi indian. In the language of the Hopi there is no word for time. The language contains no referance to "time," either explicit or implicit.

Oh how I wonder how a text on special relativity would be translated into the language of the Hopi! :smile:

Pete
 
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Quite untrue. That equation is valid only in certain special cases, e.g. for single particles and for systems of free particles. It is not generally valid for constrained systems. Thus the claim that mass is identical to mass-energy is invalid.
Alright then help me out here. What do you mean by mass-energy.
 

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mitchellmckain said:
Not only is it completely unnecessary but I think it creates more confusion than understanding.
How right you are!

The idea of mass increasing at relativistic velocities leads to the unavoidable conclusion that mass is relative just like velocity, which I find absurd. Reading some of the other post I see that it also leads to the conclusion that mass is different in different directions, which is even more absurd.
Yes and yes.
The total energy of a mass m at a relative velocity v and thus lorentz contraction factor gamma is given by E = gamma m c^2.
Yes.
Well what about kinetic energy? [...]
To handle the relativistic correction, we typically write
E = m c^2 + (gamma-1) m c^2
and we say that the first term is the mass energy (or rest energy) and the second term here is the relativistic kinetic energy, KE = (gamma-1) m c^2
Yes.
In this case we are obviously not thinking that the mass has increased by a factor of gamma at all, because the energy associated with mass has not changed. To say that the mass has increased by a factor of gamma would mean that all of the energy is a part of the mass and there is no energy of motion, and I don't think this helps in understanding special relativity at all.
Correct.
I guess the only way to make what I am saying clear is to look at an example. Suppose you accelerate a big ship to 1/2 the speed of light relative to the earth. If you have a medium ship inside the big ship then you can accelerate that medium ship to 1/2 the speed of light relative to the big ship. Then if you have a small ship inside the medium ship you can accelerate the small ship 1/2 the speed of light relative to the medium ship.

The energy requirements of all these acceleration depend on the rest masses of these ships (lets call them mbig, mmed, and msmall) in exactly the same way, using the KE shown above KE = (gamma-1) m c^2.
In each of the three cases gamma = 1/sqrt(1-.25) = 1.1547
First acceleration: energy required was KE = .1547 mbig c^2
Second acceleration: KE = .1547 mmed c^2
Third acceleration: KE = .1547 msmall c^2

If you want to talk about the resulting velocity with respect to the earth then you need the velocity addition formula v3 = (v1+v2)/(1+ v1 v2/c^2).
So the velocity of the medium ship with respect to the earth is (c/2+c/2)/(1+.25) = 0.8 c, and the velocity of the small ship with respect to the earth is (.8 c + .5 c)/(1+ .8x.5) = .92857 c. If you look carefully at the velocity addition formula you will see that if both v1 and v2 are less than c then v3 will be less than c, but if either v1 or v2 is equall to c then v3 will also be c.

The point is there is no increase of mass in this explanation nor should there be. The idea of mass increase promotes a misconception that something changes as you accelerate making an increase of speed more difficult. Absolutely nothing changes. The only limit is on relative velocity at which you see objects receding behind you. It does not even limit how fast you can travel to a destination.
Very good explanation. You obviously must be a qualified physicist unlike most on this forum.
The speed of light is unreachable because it is like an infinite speed in the sense that if you chase after a light beam your accelertion never reduces the relative velocity between you and the light beam you are chasing, the light continues to speed away from you at 3x10^8 m/s. You cannot catch the light no matter how fast you go, just as if the light were traveling infinitely fast. In fact, we know from the relativity of simultaneity that any travel faster than light would be equivalent to arriving at your destination before you left, leading to the same paradoxes as in time travel. Also if you think of the infinite speed as the limiting case where you get to your destination in no time at all, the speed of light is exactly such a limiting case.
You are the first person on this forum that I've seen correctly explain the issue of relativity and mass. Good work :smile:
 

Aer

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pmb_phy said:
There are at least two ways to define "mass" and each has its merits. Defined as m = p/v = mass, can be found in many places. Sayang one is the correct one and the other incorrect is nonsense. And I don't really care if paticle physicist is bothered by it or not.
m=p/v is incorrect for relativity. m=p/(γv) is correct for relativity. The former can only be used when v<<c and I hope is the only "many places" you are referring to, otherwise your sources are wrong.


pmb_phy said:
Please provide proof that was increasing mass was concocted as an explanatory tool
Actually, it was not concocted as an explanatory tool. In the old days, it was believed the relativistic mass was the real mass. It wasn't until quantum physics that physicists realized that invariant mass was the real mass. Since then, relativistic mass has all but been forgotten except in elementary physics classes - and even at a good university, it isn't included in the elementary physics classes either. Relativistic mass however is still very prevalent on the internet and with some older generation physicists.

pmb_phy said:
Physicists Einstein included, never intentionaly fool someone or "concoct" something for a teaching tool.
Einstein is included in the "early days" when relativity was new and the concept of relativistic mass was not fully understood to be the incorrect concept we know it to be today.


pmb_phy said:
I'll await proof for said claims.
How about you try to read an up-to-date text on relativity.

pmb_phy said:
This is your basic assumption? You don't like it so it must be wrong huh?
That would make you, as a person, wrong; so I wouldn't say that was his assumption. He hasn't assumed anything in fact, he is merely trying to teach you modern relativity.

pmb_phy said:
So. Did you read my entire paper? My website? Give it a whirl.
I've seen your website, the website would be better whirled down the toilet.

pmb_phy said:
In pre-Einstein physics rods don't contract, volumes don't change and there is non-similaneity is gone. Therefore do you wish to get rid of these thinhgs too?
Those are all direct results of the lorentz transformations. Relativistic mass was a definition that was incorrect.


pmb_phy said:
http://www.geocities.com/pmb_phy/mass.pdf

I've refered to it many times and guess what? Nobody has read the whole thing I bet.
Quite frankly, they'd be better off not clicking it, which consequently is my recommendation for everyone reading this thread.

The rest of your posts are so far off the mark it is just incredible :eek:
 

Tom Mattson

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Aer said:
Those (edit: length contraction, non-simultaneity, etc) are all direct results of the lorentz transformations.
Yes, that's true. It's also true that relativistic mass is not a consequence of the Lorentz transformations.

Relativistic mass was a definition that was incorrect.
Er...How can a definition be incorrect?

What predictions turn out wrong if you use the older convention?
 

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Tom Mattson said:
Er...How can a definition be incorrect?

What predictions turn out wrong if you use the older convention?
OK, I should have been more thorough in what I was saying. Relativistic mass was defined as γm since it was shown that E=γmc2 and from this definition, physicists thought relativistic mass was the true mass.

Now what is meant by true mass? Well that's very hard to say because of all of the different definitions people seem to attribute to mass. From quantum mechanics, it was found that invariant mass was the only useful quantity, but I would argue that this is just because QM doesn't deal with frames, which consequently is why QM and GR/SR are not unified in anyway. However, if we stick to just relelativity, we see also that the definition of mr=γm has less use than just the invariant mass as well. The only time γm really has any use is when dealing with the energy equation. The force equation for example would be F=*gamma;3ma, so "relativistic mass" doesn't really help out here any either.

Now let's get back to what was originally assumed when relativistic mass was defined:
Relativistic mass was defined as γm since it was shown that E=γmc2 and from this definition, physicists thought relativistic mass was the true mass. So we want to know what is true mass. I would put forth that an object, or series of objects (a la compound objects) true mass is proportional to the gravitational field it/they produce. Now when dealing with the curvature of spacetime, the kinetic energy of an object does not increase the curvature of spacetime around it, so relativistic mass in this way would not be considered true mass. Also, concerning thermal energies, thermal energy is just a term for the kinetic energy of atoms. In this way, thermal energy can be thought of as a compound object with constituent parts each having kinetic energies. So in this way, I don't think it makes much sense to say that thermal energy increases the "true mass" of an object either.

What we are left with (this is my personal opinion) is that "true mass" is a form of energy and not simply interchangable with energy. That is, mass and energy are only interchangable on the quantum level in which only quantum physics would be able to explain IMO.
 
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Tom Mattson

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Aer said:
Relativistic mass was defined as γm since it was shown that E=γmc2 and from this definition, physicists thought relativistic mass was the true mass.
Well, I'd ask them the flip side of my question: How can a definition be considered undeniably true? You can't show a definition to be either true or false.

Now what is meant by true mass? Well that's very hard to say because of all of the different definitions people seem to attribute to mass. From quantum mechanics, it was found that invariant mass was the only useful quantity,
That's very cute, the way you take your opinion and impute to it nothing less than the authority of quantum mechanics. But it is obivously not true that your opinion follows from QM, so let's not pretend like it does, OK?

but I would argue that this is just because QM doesn't deal with frames, which consequently is why QM and GR/SR are not unified in anyway.
Say what? Have you not heard of the Klein-Gordon equation? Or of the Dirac equation, or any of its higher-spin analogs? Or of QFT?

Quantum theory and special relativity have been completely unified. And even if they hadn't, your statement would still be wrong. I've "dealt with frames" even with the plain vanilla Schrodinger equation. There is nothing stopping you from performing a Galilean boost to find out how QM predictions in one inertial frame are related to those in another.

However, if we stick to just relelativity,
Yes, I think that would be for the best!

we see also that the definition of mr=γm has less use than just the invariant mass as well. The only time γm really has any use is when dealing with the energy equation. The force equation for example would be F=*gamma;3ma, so "relativistic mass" doesn't really help out here any either.
That's a bit uncharitable of you. You could define momentum as [itex]p=(\gamma m)v=Mv[/itex] and state that the force is [itex]F=\frac{dp}{d\tau}=\frac{d(Mv)}{d\tau}[/itex].

Now when dealing with the curvature of spacetime, the kinetic energy of an object does not increase the curvature of spacetime around it, so relativistic mass in this way would not be considered true mass.
Kinetic energy does, in fact, contribute to gravitation. All sources of energy and momentum are put into the energy-momentum tensor in GR.

You never did answer my questions on either how a definition could be considered wrong, or what predictions turn out wrong if the older convention of relativistic mass is used. But at least you admitted towards the end there that you are expressing your personal opinions about certain things.

Some things are simply a matter of opinion, and I think that the correctness or even the usefulness of relativistic mass is one of them.
 

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Tom Mattson said:
Well, I'd ask them the flip side of my question: How can a definition be considered undeniably true? You can't show a definition to be either true or false.
I didn't claim that it was a true definiton.

Tom Mattson said:
That's very cute, the way you take your opinion and impute to it nothing less than the authority of quantum mechanics. But it is obivously not true that your opinion follows from QM, so let's not pretend like it does, OK?
You read too much into the issue of bringing up QM. And it is not my opinion, it is the opinion of those that deal with QM that invariant mass is the only useful quantity - NOT MINE. Do not try to push that claim on me as if it is my opinion.


Tom Mattson said:
Quantum theory and special relativity have been completely unified.
No, you talk only of special cases, there is no complete unification. Gravity is unexplained in QM for example.

To see how QM and Relativity are not completely unified, we need not even consider the gravity issue, one only needs to consider a photon. In relativity, the energy of a photon is given by:

[tex]E = \gamma m c^2[/tex]

or

[tex]E^2 = (pc)^2 + (mc^2)^2[/tex]

which consequentyly is derived from [tex]E = \gamma m c^2[/tex] and [itex]p = \gamma m v[/itex].

Now pc in relativity for a photon would be: [tex]pc = \gamma m v c = \gamma m c^2[/tex], the familiar energy equation. However, [itex]\gamma m[/itex] is undefined for a photon because [itex]\gamma[/itex] is 1/0 where the denominator is exactly 0, which makes it undefined. If the denominator was only approximately 0, then it could be argued that γ is infinity for a photon, but that is not the case in current relativity theory. [itex]\gamma[/itex] is undefined for a photon. So the QM result that pc=hf cannot be explained in relativity for this very reason.

However, if one forgets the definition of p = γ m v, in which γ is undefined for a photon, and plugs pc = hf into the energy equation [tex]E^2 = (pc)^2 + (mc^2)^2[/tex], then one gets the correct result that E=hf. But this is an ad-hoc job because pc is undefined for a photon in relativity as I've repeated over and over so that you would understand.



Tom Mattson said:
Yes, I think that would be for the best!
No, let's consider more QM examples. Explain to me how QM explains gravity. I am aware of possible canidates that you will no doubt try to pass off as generally accepted, but I know better - in truth, they are just candidate theories and have problems when trying to unify the two theories.


Tom Mattson said:
That's a bit uncharitable of you. You could define momentum as [itex]p=(\gamma m)v=Mv[/itex] and state that the force is [itex]F=\frac{dp}{d\tau}=\frac{d(Mv)}{d\tau}[/itex].
You forgot to keep going... [tex]F=\frac{dp}{d\tau}=\frac{d(Mv)}{d\tau}=\frac{d(mv/ \sqrt{1-(\frac{v}{c})^2})}{d\tau}[/tex].
Remember, you need all terms that are going to differentiate with [itex]\tau[/itex] to be expanded out. I ask again, what was the use of M in the above? I could have just skipped to the last step if I never defined an M...

Tom Mattson said:
Kinetic energy does, in fact, contribute to gravitation. All sources of energy and momentum are put into the energy-momentum tensor in GR.
They are all put in, but some are refered to as "non-gravitational energy". Ho-hum, this disagrees with what you said about all energies contributing to gravitation:

The components T0b can therefore be interpreted as the local density of (non-gravitational) energy and momentum. The spatial components Tij correspond to components of local non-gravitational stresses, including pressure. This tensor is the Noether current associated with spacetime translations.Source: wikipedia

Tom Mattson said:
Some things are simply a matter of opinion, and I think that the correctness or even the usefulness of relativistic mass is one of them.
My opinion is that you are wrong, ok?
 
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DrGreg

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Can't the two sides of the mass debate agree to differ?

Like it or not, historically [at least] two different notions of mass have both been extensively used, and both are still in use today in different contexts.

"Relativistic mass" and "proper mass" (= "invariant mass" = "rest mass") are just two different tools in the relativist's toolbox. Each practitioner has their favourite tool and many will work exclusively with one tool and ignore the other. Each tool has its own merits and it doesn't matter which tool you use as long as it is fit for your particular task and you use it correctly. You will get into trouble if you try to use one tool as if it were the other.

You may well feel that your preferred tool is far superior to the other tool and that the other has no use. You are entitled to your opinion.

But surely we must all take a pragmatic view and accept that both tools have been used and are still in use? Anyone who wishes to read around the subject of Relativity needs to be aware that both tools exist and what is the difference between them. Otherwise you will have difficulty in understanding someone else who prefers a different tool to yours.

Unfortunately some people insist on referring to their preferred tool as "mass", rather than give the tool its full name of either "relativistic mass" or "proper mass" (or whatever). This can cause much confusion. Whenever you read about "mass", you need to know which of the two types is being referred to.

Neither side of the argument is going to admit defeat, so can't we just agree that there are multiple notions of mass and that we will always specify which we are talking about?
 

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DrGreg said:
Can't the two sides of the mass debate agree to differ?
I can agree to differ with the other side with the added context that the other side is just wrong. :biggrin:

DrGreg said:
"Relativistic mass" and "proper mass" (= "invariant mass" = "rest mass") are just two different tools in the relativist's toolbox. Each practitioner has their favourite tool and many will work exclusively with one tool and ignore the other.
You are correct to say that one may use proper mass and ignore relativistic mass forever and for every situation. However, the converse is not true. One using relativistic mass cannot ignore proper mass as it is in the definition of relativistic mass! Furthermore, when doing any detailed analysis, one is required to break relativitistic mass, M into &gamma; and m, as is even apparent in Tom's example of d(Mv)/dt above.


DrGreg said:
You may well feel that your preferred tool is far superior to the other tool and that the other has no use. You are entitled to your opinion.
One using relativistic mass is not free to not use proper mass in any context!

DrGreg said:
Neither side of the argument is going to admit defeat, so can't we just agree that there are multiple notions of mass and that we will always specify which we are talking about?
There are not multiple notions of mass, there is only one correct notion of mass and that is proper mass, rest mass, invariant mass (all the same thing).
 

Tom Mattson

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Aer said:
I didn't claim that it was a true definiton.
And I clearly said that I would ask "them" (meaning "those physicists who thought that relativistic mass was the true mass").

You read too much into the issue of bringing up QM. And it is not my opinion, it is the opinion of those that deal with QM that invariant mass is the only useful quantity - NOT MINE. Do not try to push that claim on me as if it is my opinion.
You stated the opinion, and you did not attribute it to anyone else. Any naitive English speaker would have concluded that the opinion is yours.

But fine: Let's not pretend that "their" opinion follows from QM, because it doesn't.

No, you talk only of special cases, there is no complete unification. Gravity is unexplained in QM for example.
I said that quantum theory and special relativity have been completely unified, and that is true. The status of quantum theories of gravity is irrelevant to that statement.

To see how QM and Relativity are not completely unified, we need not even consider the gravity issue, one only needs to consider a photon. In relativity, the energy of a photon is given by:

[tex]E = \gamma m c^2[/tex]
No, it isn't. If that were true then the energy of all photons would be undefined, which it isn't.

In relativity the energy of a photon is given by [itex]E=pc[/itex].

or

[tex]E^2 = (pc)^2 + (mc^2)^2[/tex]

which consequentyly is derived from [tex]E = \gamma m c^2[/tex] and [itex]p = \gamma m v[/itex].
The first equation is not derived from the second and third equations at all. The second and third equations only apply to massive particles, while the first applies to all particles, even those with zero mass.

Now pc in relativity for a photon would be: [tex]pc = \gamma m v c = \gamma m c^2[/tex], the familiar energy equation. However, [itex]\gamma m[/itex] is undefined for a photon because [itex]\gamma[/itex] is 1/0 where the denominator is exactly 0, which makes it undefined. If the denominator was only approximately 0, then it could be argued that γ is infinity for a photon, but that is not the case in current relativity theory.
This is complete BS. The process of equating [itex]pc[/itex] and [itex]\gamma mvc[/itex] for photons is incorrect on its face. There is nothing in the theory of relativity that would even suggest that it should be done.


[itex]\gamma[/itex] is undefined for a photon. So the QM result that pc=hf cannot be explained in relativity for this very reason.
Right, the explanation comes from the unification of special relativity and quantum theory, which has been done, despite the fact that you claim otherwise.

However, if one forgets the definition of p = γ m v, in which γ is undefined for a photon, and plugs pc = hf into the energy equation [tex]E^2 = (pc)^2 + (mc^2)^2[/tex], then one gets the correct result that E=hf. But this is an ad-hoc job because pc is undefined for a photon in relativity as I've repeated over and over so that you would understand.
You can repeat until you're blue in the face, but you would still be wrong. pc is not undefined for a photon.

No, let's consider more QM examples. Explain to me how QM explains gravity. I am aware of possible canidates that you will no doubt try to pass off as generally accepted, but I know better - in truth, they are just candidate theories and have problems when trying to unify the two theories.
Why would I need to explain quantum gravity, in order to defend my point?

You forgot to keep going... [itex]F=\frac{dp}{d\tau}=\frac{d(Mv)}{d\tau}=\frac{d(\frac{mv}{sqrt(1-{\frac{v}{c}}^{2})}{d\tau}[/itex]. Remember, you need all terms that are going to differentiate with τ to be expanded out.
I didn't forget to do anything. You can, in fact, write force as F=dM/dτ.

I ask again, what was the use of M in the above? I could have just skipped to the last step if I never defined an M...
Obviously, the use of M in that particular equation was to make the relationship more compact. Is that really too difficult to see?

They are all put in, but some are refered to as "non-gravitational energy".
If by a "non-gravitational energy" term you mean a term that describes "energy that does not contribute to gravitation", then kinetic energy is not one of them.

Ho-hum, this disagrees with what you said about all energies contributing to gravitation:
I said that all energies are put into the energy-momentum tensor, and that is a fact. Nothing you have referred to disagrees with anything that I said.

My opinion is that you are wrong, ok?
Unfortunately, you don't have a single good reason for holding that opinion.

Aer, you have been trying the patience of the staff with your insulting tone to others, and we have been watching. But in this post not only have you been patronizing, but you have descended into abject crackpottery. I must insist that you stop doing it.
 
Last edited:

Tom Mattson

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Aer said:
You are correct to say that one may use proper mass and ignore relativistic mass forever and for every situation.
That's true.

However, the converse is not true. One using relativistic mass cannot ignore proper mass as it is in the definition of relativistic mass! Furthermore, when doing any detailed analysis, one is required to break relativitistic mass, M into γ and m, as is even apparent in Tom's example of d(Mv)/dt above.
That's also true.

Now for the big question: So what?

This point of yours does nothing whatsoever to show that the definition of relativistic mass is wrong.

One using relativistic mass is not free to not use proper mass in any context!
Sure he is. A particle's proper mass would be its rest mass.

There are not multiple notions of mass, there is only one correct notion of mass and that is proper mass, rest mass, invariant mass (all the same thing).
There are multiple notions of mass that have equal claims to being "correct". Deal with it.
 

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