Is Relativistic Mass Plausible?

[planck42]

I've read contrasting sources concerning the concept of an object increasing in mass at relativistic velocities. Some of my older calculus texts mention this as being accepted by physicists, while a newer(by comparison) textbook called Principles of Physics: A Calculus-Based Text by Serway and Jewett claims that relativistic mass is outdated. Normally, the newer book would be correct, but I don't think there is any other reasonable explanation of the momentum of photons. Can the good people at PF please shed some light on this matter?

[keniwas]

I've read contrasting sources concerning the concept of an object increasing in mass at relativistic velocities. Some of my older calculus texts mention this as being accepted by physicists, while a newer(by comparison) textbook called Principles of Physics: A Calculus-Based Text by Serway and Jewett claims that relativistic mass is outdated. Normally, the newer book would be correct, but I don't think there is any other reasonable explanation of the momentum of photons. Can the good people at PF please shed some light on this matter?

Its really becomming a very outdated concept as you say.

As for the momentum of photons, its actually not a relativistic result at all. Classical electromagnetism predicts the momentum of electromagnetic waves. This relation is carried over into the quantum world (which is where photons live) when you quantize the radiation field (or more simply, when you postulate as planck did that the energy in light had to exist in discrete quantities). In the end you get the photon momentum without any mention of relativity.

[planck42]

Thank you for that explanation

[Wallace]

Note that what is changing/has changed is just the terminology that people see as being most helpful. The underlying physical theory hasn't changed. Just making that clear.

[JANm]

Thank you for that explanation
Hello planck42
So you are easily satisfied. I think that to say that mass-velocity relation is outdated is not a scientific statement. Science should be timeless. Theories are prooved true or false. So do you think that Serway and Jewitt mean in their book that there is no such thing as relativistic mass depending on velocity?, or is it depending on velocity in another way?
greetings Janm

[Wallace]

JANm, as stated in my post above, the scientific theory hasn't been changed, just the prevailing view on the clearest way to think and talk about the concepts. The answer to any question that relates to what an observer would actually measure (which relates to what can be experimentally verified) hasn't changed.

[Naty1]

As noted above it's all relative....rather than "right" or "wrong"

Wikipedia actually has a good discussion regarding your question....and these quotes:

It is not good to introduce the concept of the (relativistic) mass of a moving body for which no clear definition can be given. It is better to introduce no other mass concept 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.

– Albert Einstein in letter to L Barnett (quote from L. B. Okun, “The Concept of Mass,” Phys. Today 42, 31, June 1989.)




Contemporary authors like Taylor and Wheeler avoid using the concept of relativistic mass altogether:

"The concept of "relativistic mass" is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself."[17]


For more try here: http://en.wikipedia.org/wiki/Relativistic_mass#The_relativistic_mass_concept

[Naty1]

In the prior Taylor and Wheeler quote, am I supposed to find it easier to relate an increase in momentum with the geometries of spacetime rather than an increase in mass? Why?

Or is it that an increase in mass relates somehow to the time component of the four vector and that is supposedly a less plausible relationship?

I think the above might imply that because we think we have a fundamental concept of mass, a structure we think we understand that we lack with momentum, for example , then it's more difficult to relate to a purported change in that structure.

I wonder if maybe it's plausible our understanding of mass needs some updating...that, for example, maybe we only have a low energy understanding of mass and are not even aware of that limitation....just as Newtonian physicsts were unaware their understanding was a low velocity picture. To say it another way, if time and space and energy change at high velocity why not mass??

[Klockan3]

To say it another way, if time and space and energy change at high velocity why not mass??
Because it is much cleaner to say that mass is a kind of energy and that all kinds of energy have properties we normally associate with mass than to add mass to every object containing energy. Relativistic mass might be easier to understand but it usually gives people extremely bad understanding of what actually happens.

[DaleSpam]

I think the above might imply that because we think we have a fundamental concept of mass, a structure we think we understand that we lack with momentum, for example , then it's more difficult to relate to a purported change in that structure.I think this is actually reasonably accurate. The point, IMO, is that we think of mass as being an inherent property of an object itself, rather than a relationship between it and some observer. That means that the intuitive idea (or as you put it "fundamental concept") of mass must be relativistically invariant.

[Fredrik]

In the prior Taylor and Wheeler quote, am I supposed to find it easier to relate an increase in momentum with the geometries of spacetime rather than an increase in mass? Why?

I think you're missing their point, which is that we intuitively connect the word "mass" to internal structure. It doesn't sound strange to say that the particle's energy is different in different frames (because energy depends on velocity), but it sounds very strange to say that the particle's internal structre is different in different frames.

[JANm]

Relativistic mass might be easier to understand but it usually gives people extremely bad understanding of what actually happens.
With this "extremely bad understanding" some physici mean: getting of the track of believing in relativity. It could also mean it is pedagogically better not to speak of it because it can get you of the track of Klassical Mechanics, which used to be an exact science next to mathematics itself.
Janm

[Fredrik]

With this "extremely bad understanding" some physici mean: getting of the track of believing in relativity. It could also mean it is pedagogically better not to speak of it because it can get you of the track of Klassical Mechanics, which used to be an exact science next to mathematics itself.
Janm
:confused: What are you talking about? Special relativity is a more exact science than non-relativistic classical mechanics. (The theory is just as well-defined, and it makes much more accurate predictions about the results of a much wider range of experiments).

[JANm]

:confused: What are you talking about? Special relativity is a more exact science than non-relativistic classical mechanics. (The theory is just as well-defined, and it makes much more accurate predictions about the results of a much wider range of experiments).
Hello Fredrik
PLease don't think I want to go back to the days of classical mechanics. But there were two laws of conservation then: the one of mass and the one of Energy. It seems to me that if relativity theory wants to be right there would be one combining these two laws in one conservation-law. So it is not the prediction part of relativistics I am judging, but rather the theoretical part in which i miss something...
greetings Janm

[DaleSpam]

But there were two laws of conservation then: the one of mass and the one of Energy. It seems to me that if relativity theory wants to be right there would be one combining these two laws in one conservation-law. There is, it is called the conservation of four-momentum. It combines the classical conservation laws of energy, momentum, and mass into one unified law.

[JANm]

There is, it is called the conservation of four-momentum. It combines the classical conservation laws of energy, momentum, and mass into one unified law.
Hello DaleSpam
Wow that is nice. Could you elaborate a little more about this. I know mass and energy are somewhat bounded by E and mcpower2 and that they are in someway scalar functions. I suppose the momentum comes in as a vector function or what.
Ýou got me interested
Greetings Janm

[Klockan3]

Hello DaleSpam
Wow that is nice. Could you elaborate a little more about this. I know mass and energy are somewhat bounded by E and mcpower2 and that they are in someway scalar functions. I suppose the momentum comes in as a vector function or what.
Ýou got me interested
Greetings Janm
In relativistic theory you let time and length have the same units, so velocity is unit-less meaning that mass, momentum and energy can all be put on the same diagram without getting anything strange.

Anyway, to get the momentum 4 tensor you take the velocity 4-tensor and tacks on mass on it. This gives the normal relativistic momentums in the spatial dimensions while in the time dimension you get energy. Also since all of these needs to be rotational symmetric you need the length of this to be invariant you get that this is the invariant mass. (This is assuming c=1, otherwise you need to tack on a few c's, like E/c and mc)

Though, note that Minkowski space have a strange norm where time is seen as negative distance so you wont get the result I mentioned above if you do not use that.

[DaleSpam]

Could you elaborate a little more about this. I know mass and energy are somewhat bounded by E and mcpower2 and that they are in someway scalar functions. I suppose the momentum comes in as a vector function or what.In relativity I am sure that you have heard that time is fourth dimension. This is represented mathematically using four-vectors: (ct,x,y,z) = (ct,\mathbf{x}). See the hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c1) page and the Wikipedia (http://en.wikipedia.org/wiki/Four-vector) page (on both pages be sure to follow the links to further details).

Taking this approach you can do a couple of things that are mathematically very elegant and can give some physical insight. First, you can represent the Lorentz transform as a matrix. When you do so you notice that it has the form of a peculiar sort of a rotation matrix. Second, you can look for a norm which does not change under this rotation. This is the spacetime interval (http://en.wikipedia.org/wiki/Spacetime_interval#Spacetime_intervals) s^2 = -c^2 t^2 + x^2 + y^2 + z^2 = -c^2 t^2 + \mathbf{x}^2 it can be considered the "length" of a four-vector and all reference frames will agree on it.

Now, if you work through it you will find that the four-vector that contains momentum in the spacelike part has the form (E/c,p_x,p_y,p_z) = (E/c,\mathbf{p}) which is called the four-momentum. This means that energy and momentum have the same relationship to each other as time and space do. Now, if you take the "length" of this four-vector you find that it is the rest mass, aka the "invariant mass" or just "mass". All observers agree on this quantity, and it simplifies to the famous E=mc² equation for an object at rest.

Now, you can get to the part that you were interested in, the conservation law. Analogously to Newtonian mechanics, the four-momentum of a system is the sum of the four-momenta of its constituent particles, and the four-momentum of the system is conserved across any interaction, including particle anhilation and creation interactions. This means that a system's energy (timelike component of four-momentum), momentum (spacelike component of four-momentum), and mass ("length" of four-momentum) are also conserved and you get one conservation law which unifies three separate conservation laws from classical mechanics. To me it is one of the most elegant and compelling facets of relativity.

[Naty1]

PLanck: a general feature of physics is that as you learn more and more, different insights begin to emerge...for example Dalespam said above :

This means that energy and momentum have the same relationship to each other as time and space do.

That's enough to rock me back on my heels still and I've know about it for a while ...and while not smart enough to extrapolate that relationship further, I know enough to realize it's hinting at other things we likely don't understand fully yet. Just like E = Mc2...two pieces of a puzzle with many other pieces still missing....

I can't help but wonder if time and mass and energy and space all popped out, apparently togther in some sort of a bang, that we'll ultimately understand how they are all related at a fundamental level....maybe a fundamental constitutent from which these emerge, perhaps like quantum vacuum fluctuations...

[matheinste]

In relativity I am sure that you have heard that time is fourth dimension. This is represented mathematically using four-vectors: (ct,x,y,z) = (ct,\mathbf{x}). See the hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c1) page and the Wikipedia (http://en.wikipedia.org/wiki/Four-vector) page (on both pages be sure to follow the links to further details).

Taking this approach you can do a couple of things that are mathematically very elegant and can give some physical insight. First, you can represent the Lorentz transform as a matrix. When you do so you notice that it has the form of a peculiar sort of a rotation matrix. Second, you can look for a norm which does not change under this rotation. This is the spacetime interval (http://en.wikipedia.org/wiki/Spacetime_interval#Spacetime_intervals) s^2 = -c^2 t^2 + x^2 + y^2 + z^2 = -c^2 t^2 + \mathbf{x}^2 it can be considered the "length" of a four-vector and all reference frames will agree on it.

Now, if you work through it you will find that the four-vector that contains momentum in the spacelike part has the form (E/c,p_x,p_y,p_z) = (E/c,\mathbf{p}) which is called the four-momentum. This means that energy and momentum have the same relationship to each other as time and space do. Now, if you take the "length" of this four-vector you find that it is the rest mass, aka the "invariant mass" or just "mass". All observers agree on this quantity, and it simplifies to the famous E=mc² equation for an object at rest.

Now, you can get to the part that you were interested in, the conservation law. Analogously to Newtonian mechanics, the four-momentum of a system is the sum of the four-momenta of its constituent particles, and the four-momentum of the system is conserved across any interaction, including particle anhilation and creation interactions. This means that a system's energy (timelike component of four-momentum), momentum (spacelike component of four-momentum), and mass ("length" of four-momentum) are also conserved and you get one conservation law which unifies three separate conservation laws from classical mechanics. To me it is one of the most elegant and compelling facets of relativity.

That is a very enlightening presentation. Seeing how all these things knit together is, to me, far more interesting and worth the effort of learning before worrying about some of the more often discussed esoteric topics such as time travel, wormholes, black holes etc.

Matheinste.

[JANm]

See the hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c1) page
Hello DaleSpam
Clearly time*c is related to spacevector as energy is related to impulse. I understand that, but there is a mistake in the fourimpulse P_b. There is a ct standing there instead of energy...
greetings Janm

[jtbell]

Yes, the Hyperphysics page has an error in the equation for {\vec P}_b. The immediately-following result for {\vec P}_a \cdot {\vec P}_b is nevertheless correct, so it's surely merely a transcription error.

I've seen worse in many textbooks!

[DaleSpam]

Thanks for letting me know. I have contacted the author and alerted them to the error.

[mathfeel]

Is there any reason why time-like interval are positive and not the other way around?

[Rasalhague]

Is there any reason why time-like interval are positive and not the other way around?

I think it's just a matter of convention (http://en.wikipedia.org/wiki/Sign_convention#Relativity), or rather competing conventions... I started a thread here (http://www.physicsforums.com/showthread.php?t=346920) listing some of the different approaches I've found and asking for general advice. Some people, e.g. most clearly Callahan, define the interval in such a way that it's always positive and real, unless it's zero. For others, it seems, a timelike interval is real and a spacelike interval imaginary (its square being negative). And for others, a timelike interval is imaginary (its square being negative) and a spacelike interval real.

[lightarrow]

PLanck: a general feature of physics is that as you learn more and more, different insights begin to emerge...for example Dalespam said above :his means that energy and momentum have the same relationship to each other as time and space do
That's enough to rock me back on my heels still and I've know about it for a while ...and while not smart enough to extrapolate that relationship further, I know enough to realize it's hinting at other things we likely don't understand fully yet. Just like E = Mc2...two pieces of a puzzle with many other pieces still missing....

I can't help but wonder if time and mass and energy and space all popped out, apparently togther in some sort of a bang, that we'll ultimately understand how they are all related at a fundamental level....maybe a fundamental constitutent from which these emerge, perhaps like quantum vacuum fluctuations...Ok, since you seem to like thinking about these things, now I give you something else on which think about, that it's not clear to me:
the 4-vectors (ct,x,y,z)=(ct,\mathbf{r}) and (E/c,\mathbf{p}) must be related in some way in QM, since the product of the time components' indeterminations of the two 4-vectors is the same as the product of the space components' indeterminations (every single component), by HUP. Furthermore, the scalar product of the two is \mathbf{p}\cdot \mathbf{r} -Et\ =\ \hbar\phi (phi is the phase) =\ S where S is the action, in the classical limit.
So, in some way, the two 4-vectors must be a sort of...the same one, but seen in different ways (of course I'm speculating here). But then, the interval (the first 4-vector's modulus) is related in some way to mass (the second 4-vector's modulus)? Their product is related to the action? And to the phase?:confused:

[Naty1]

I was reading Richard Feynman's SIX NOT SO EASY PIECES, excertps from pgs 88 and 89 and came across the following which blows the idea that relativistic mass is in some way inappropriate:

Astonishing as it may seem, in order for the conservation of momentum to work when two objects come together, the mass they form must be greater than the rest masses of the objects, even though the objects are at rest after the collision!....Of course we know from the conservation of energy that there is more kinetic energy (in moving objects)...but that does not affect the mass according to Newton's laws....We see that is impossible (in relativity) because the kinetic energy involved in a collision of two objects, the resulting object will be heavier, therefore, it will be a different object.

[DaleSpam]

I understand the quote, but not your use of it. The quote is about the invariant rest mass, not the relativistic mass. It does not support the use of relativistic mass at all.

Mathematically this is analogous to a statement that the norm of a sum of 2 vectors is never larger than the sum of the norms of the 2 vectors.

[lightarrow]

...or it is equivalent to say that any form of energy, when confined in a stationary (= not moving) region of space, has mass (invariant, of course).

[JANm]

I was reading Richard Feynman's SIX NOT SO EASY PIECES, excertps from pgs 88 and 89 and came across the following which blows the idea that relativistic mass is in some way inappropriate:
Hello Naty1
In every collision even a fully elastic one velocities change; so relativistic mass does. The heaviest one changes a little and the lightest one changes much. A very important thing is to know relativistic addition of velocities, because if that formula is known one can calculate center of gravity and subtract the velocity of the centre. Calculate the collision and a add the velocity of the centre again. Since at nonfully elastic collision heat is produced the relativistic mass diminishes if and when this heat is radiated; before this there has to be a formula predicting how much each of the colliding particles warm up.
Conclusion: the relativistic adding of velocities formula must keep the relativistic mass invariant.
Implication: even cooling bodies loose relativistic mass.
I didn't say it was going to be easy but it must remain exact.
examples with just: limits of velocitie to zero gives restmass is not sufficient alone!
greetings Janm

[DaleSpam]

...or it is equivalent to say that any form of energy, when confined in a stationary (= not moving) region of space, has mass (invariant, of course).Yes. The norm of the four momentum is:
m_0 ^2 c^2 = E^2/c^2 - \mathbf{p}^2
Which reduces to exactly what you said for \mathbf{p}=0.

[Rasalhague]

Mathematically this is analogous to a statement that the norm of a sum of 2 vectors is never larger than the sum of the norms of the 2 vectors.

Except that here the triangle inequality is reversed, isn't it? The mass of the sum of two energy-momentum vectors is never less than the sum of the masses of the same vectors.

And, as in Feynman's example, it may be greater: "when two objects come together, the mass (i.e. rest mass) they form must be greater than the rest masses of the objects".

In Spacetime Physics, Taylor and Wheeler give another example in their Figure 97:

"The sum of the rest masses of the fission fragments of plutonium is less than the rest mass of the original plutonium nucleus."

"The vector sum of two timelike 4-vectors has magnitude M (rest mass of Pu239 before fission), which is greater than the sum of magnitudes, m1 and m2, of the two individual 4-vectors (rest masses of fission products). Contrasts to Euclidean geometry in which the third side of a triangle always has a length that is less in value than sum of length of other two sides."

[DaleSpam]

Except that here the triangle inequality is reversed, isn't it?Precisely (for timelike four-vectors). That is due to the different signature for the Minkowski norm as opposed to the Euclidean one.

[Naty1]

Dalespam posted...

It does not support the use of relativistic mass at all.


so Feynman is commenting on the amazing situation that not only does relativistic mass change but so also does REST MASS....that seems almost beyond belief to me....rather incredible...or as one silly politician has said "requiring the willing suspension of disbelief.."...its exactly the kind of different perspective I like to keep in mind...

I inferred from Feynman's comments and surrounding discussion in his book just the kind of things JAN posted above in #30... like In every collision even a fully elastic one velocities change; so relativistic mass does.


Anyway, for those who would like a somewhat different perspective yet easily worded view of some fundamental physics, SIX NOT SO EASY PIECES is a FUN read....Time for me to skim it again...each time I do I see something else I like....

[Rasalhague]

so Feynman is commenting on the amazing situation that not only does relativistic mass change but so also does REST MASS....that seems almost beyond belief to me....rather incredible...or as one silly politician has said "requiring the willing suspension of disbelief.."...its exactly the kind of different perspective I like to keep in mind...

I'm fairly new to all this, and there's lots that still confuses me, but I think it's a case of strange but true... I think the idea is that mass (rest mass, the magnitude of the energy-momentum vector) is invariant under the Lorentz transformation (it doesn't depend on velocity; it's the same in all frames of reference), although it can be changed by actual physical interactions such as the collision Feynman refers to in that quote. This contrasts with energy (relativistic mass, the time component of the energy-momentum vector) whose value is changed by the Lorentz transformation and does depend on velocity.

[DrGreg]

It is perhaps worth mentioning two concepts, "conservation" and "invariance", which can sometimes be confused with each other.

A conserved quantity is something measured by a single observer that doesn't change over time; for example it has the same value before and after a collision, and typically it is the sum of several measurements, e.g. of multiple particles. Examples are energy (a 1D number), momentum (a 3D vector), four-momentum (a 4D vector), all when there are no external forces, of course. In Newtonian physics, mass is also conserved. In relativity, relativistic mass may be conserved (in the absence of any other form of energy) but rest mass isn't.

An invariant quantity is a single measurement whose value all observers agree upon, i.e. a frame-independent value. Examples are proper time, (scalar) proper acceleration, and rest mass. Or anything that can be expressed in the form g_{ab}U^aV^b (where U and V are genuine 4-vectors).

So, energy and momentum are both conserved but neither is invariant. In relativity, rest mass is invariant but not necessarily conserved across multi-particle interactions. (In Newtonian physics, mass is both conserved and invariant.)

[DaleSpam]

so Feynman is commenting on the amazing situation that not only does relativistic mass change but so also does REST MASS....that seems almost beyond belief to me....rather incredible...or as one silly politician has said "requiring the willing suspension of disbelief.."...its exactly the kind of different perspective I like to keep in mind.This kind of comment is unscientific. All you have to do is look at the evidence and see if it agrees with the theory, which it does. You don't need a "willing suspension of disbelief". This is not some summer action film, and although you find the conclusion "amazing" the universe did not give us any editorial control so we can't change it.

[JANm]

This means that a system's energy (timelike component of four-momentum), momentum (spacelike component of four-momentum), and mass ("length" of four-momentum) are also conserved and you get one conservation law which unifies three separate conservation laws from classical mechanics. To me it is one of the most elegant and compelling facets of relativity.
Hello Dalespam
You take the momentum four vector as (E/c, Vec(p)). How can you say that energy is the timelike part? I can see that momentum is the 3D vector part. But isn't it energy divided by c as timelike part. In the site and Wikipedia is spoken about (E, c*vec(p). Why is this difference in definitions?
greetings Janm

[DaleSpam]

Don't worry much about factors of c. Usually we work in units where c=1 so factors of c are not important, they are just there to make the units work out right and are frequently dropped entirely. In fact, there are even some equivalent conventions where the four-vector is (t,x,y,z) and the units are taken care of in the metric.

[Rasalhague]

When the equations are presented without c, as DaleSpam says, this just means that c has been taken to be 1, and other speeds expressed as fractions of c. In these "natural units" time and space are treated as having the same dimension, e.g. seconds of time and (light) seconds of space, or (light) metres of time and metres of space (the latter convention is followed by Taylor & Wheeler in Spacetime Physics), so that speed is dimensionless. In natural units, mass and energy also have the same dimension and so can both be measured in kilograms. You can convert between natural units and SI units by dimensional analysis, e.g.

c t_{seconds} = L \cdot T^{-1} \cdot T = L = t_{(light)metres}

\frac{E_{joules}}{c^{2}} = M \cdot L^{2} \cdot T^{-2} \cdot T^{2} \cdot L^{-2} = M = E_{kg}

\frac{p}{c} = M \cdot L \cdot T^{-1} \cdot L^{-1} \cdot T = M

where L is length, T is time, M is mass, and p is the magnitude of 3-momentum. Hence in SI units:

m^{2} = \frac{E^{2}}{c^{4}} - \frac{p^{2}}{c^{c}}

which can be rearranged to taste (and the right side multiplied by -1 if the opposite sign convention (http://en.wikipedia.org/wiki/Sign_convention#Relativity) is prefered), and which simplifies in natural units to

m^{2} = E^{2} - p^{2}.

Each of the components of the energy-momentum vector (also called momentum 4-vector or 4-momentum vector) can, for an object with mass, be defined as mass times the derivative of the coordinate with respect to proper time, e.g. mass times the derivative of position in the x direction, that's to say, mass times speed in the x direction:

m \frac{dx}{d \tau} = m v_{x}

Energy is the derivative of coordinate time with respect to proper time:

E_{kg} = m \frac{dt}{d \tau} = m \gamma

In the object's rest frame, the derivative of t with respect to tau = 1, since coordinate time equals proper time in that frame, thus rest energy--the energy of an object at rest--is equal to its (rest) mass.

[JANm]

Yes. The norm of the four momentum is:
m_0 ^2 c^2 = E^2/c^2 - \mathbf{p}^2
Which reduces to exactly what you said for \mathbf{p}=0.
Hello DaleSpam
Please do not forget that that E is total energy.
For kinetic energy there is: T=m*c^2-m_0*c^2
It is nice to have a thread here discussing the matter seriously.
Greatings Janm

[JANm]

m^{2} = \frac{E^{2}}{c^{4}} - \frac{p^{2}}{c^{c}}

which can be rearranged to taste (and the right side multiplied by -1 if the opposite sign convention (http://en.wikipedia.org/wiki/Sign_convention#Relativity) is prefered), and which simplifies in natural units to

m^{2} = E^{2} - p^{2}.


Hello Rasalhague
Thank you for explaining the c=1 matter. The m_0*c^2 is a huge energy of potential charakter.
I like to work with the kinetic energy T=E-m_0*c^2, or in the other language:T=E-m_0
with E=Sqrt(m^2+p^2)
in this new language T=srt(m^2+p^2)-m_0.
The thing I find strange is: if you put m=m_0/beta(v), which is the topic of this thread!
In relativity is stated that p=m_0*v/beta(v), why then not T=m_0*v^2/(2*beta(v))?
That question keeps me already occupied for years.
greetings Janm

[DaleSpam]

Please do not forget that that E is total energy.I didn't forget, did I say something to confuse you about that point?

[pervect]

A Lagrangian or a Hamiltonian approach will give you the correct expressions for p and E. There's more detail in textbooks like Goldstein's "Classical mechanics". If you're going to spend years on the problem, you have time to read up on the Hamiltonian and Lagrangian approaches are usually graduate level - I would assume you're probably not familiar with them?

wiki has an article at http://en.wikipedia.org/wiki/Hamiltonian_mechanics which may or may not be helpful


suppose we write

H(p,x) = sqrt(p^2+m^2) + V(x)

this is the hamiltonian, H, which is equivalent to the energy, as function of the momentum p, and some potential function V(x) which depends only on position.

H is always written as a function of momentum, and position. The position coordinates are "generalized coordinates", the only thing that's important is that giving the position coordinates gives the state of the system.

V(x) isn't really important to the problem, it drops out in the next step

then Hamilton's equations say that the velocity v is given by:
v = \partial H / \partial p
which gives

v = p/sqrt(p^2+m^2)

This can be solved to get p as a function of v

p = m*v/sqrt(1-v^2)

*add*

Let me motivate this from a simpler conservation of energy standpoint

The force on an object, is just dp/dt, the rate of change of its momentum

The work done on an object is force * distance, so force * velocity gives the rate at which work is done on an object, or the power.

rate at which work is done = dE/dt = Force * velocity = (dp/dt) * v


since dE/dt = v * (dp/dt), we must have

dE = v * dp

or v = dE/dp

so if we write the energy E as a function of momentum, p, we expect that
v = dE/dp

*end addition*

If we repeat this for clasical mechanics we have

H = p^2 / 2m. Note this is the first order approximation to H = sqrt(m^2+p^2) assuming p << m except for the constant factor of m. You can add any constant to the energy, or Hamiltonian, of a system, without changing the dynamics.

then
v = \partial H / \partial p

gives v = p/m, or p=mv

[Rasalhague]

Correction:

m^{2} = \frac{E^{2}}{c^{4}} - \frac{p^{2}}{c^{2}}

I accidentally typed cc instead of c2 in that last denominator on the right.

[Rasalhague]

Hello Rasalhague
Thank you for explaining the c=1 matter. The m_0*c^2 is a huge energy of potential charakter.
I like to work with the kinetic energy T=E-m_0*c^2, or in the other language:T=E-m_0
with E=Sqrt(m^2+p^2)

Sorry, I should have made it clearer that in all of the equations I posted, m stands for the magnitude of the energy-momentum vector (mass, i.e. "rest mass"), and E stands for its time component (energy, which some people have called "relativistic mass"). I followed the terminology and symbols used by Taylor & Wheeler in Spacetime Physics. They prefer to use the terms mass and energy rather than rest mass and relativistic mass, which they see as misleading. If you use the latter terms, representing them with the symbols m_0 (corresponeding to T&W's m) and m (corresponding to T&W's E), then the above equation would become, in natural units where c = 1,

E = \sqrt[]{m_{0}^{2} + p^{2}} = m

in this new language T=srt(m^2+p^2)-m_0.

If you want to use the letter m ("relativistic mass") for what Taylor & Wheeler call energy, E, in contrast to "rest mass" m_0, then this equation would have to be written

T = \sqrt[]{m_{0}^{2} + p^{2}} - m_{0} = m - m_{0}

In Taylor & Wheeler's terms:

T = \sqrt[]{m^{2} + p^{2}} - m

= E - m = m \frac{dt}{d \tau} - m = \frac{m}{\sqrt[]{1 - v^{2}}} - m

[JANm]

Hello Rasalhague
It is a pity that I cannot read your so nicely edited formulas. The background is black and the letters are miniaturally and there is not enough contrast...
greetings Janm

[DrGreg]

The background is black and the letters are miniaturally and there is not enough contrast...If you are able to do so, upgrade your browser to a more recent version, e.g. the latest version of Internet Explorer, or Firefox, or Safari.

[JANm]

If you want to use the letter m ("relativistic mass") for what Taylor & Wheeler call energy, E, in contrast to "rest mass" m_0, then this equation would have to be written

T = \sqrt[]{m_{0}^{2} + p^{2}} - m_{0} = m - m_{0}

In Taylor & Wheeler's terms:

T = \sqrt[]{m^{2} + p^{2}} - m

= E - m = m \frac{dt}{d \tau} - m = \frac{m}{\sqrt[]{1 - v^{2}}} - m
Hello Rasalhague
The dt/dtau I find very interesting. Do Taylor & Wheeler put the time dilation factor in the formulae at this moment? ..,or is it another t and tau I should have been aware of?
I understand that the term restmass is difficult to explain in relativistics, because standing still is not defined there. For that I made the sentence:

Just if you are standing still you can consider how much you move...

Greetings Janm

[DaleSpam]

I understand that the term restmass is difficult to explain in relativistics, because standing still is not defined there.Huh? Rest mass is easy to explain. It is simply the norm of the four-momentum. All reference frames agree on its value.

[JANm]

Huh? Rest mass is easy to explain. It is simply the norm of the four-momentum. All reference frames agree on its value.
Hello DaleSpam
Thanks for this answer. Have to look into it. Do you also have an answer to the dt/dtau?
greetings Janm

[DaleSpam]

I don't know what Taylor and Wheeler do with dt/dtau. Was that the question?

[JANm]

Rest mass is easy to explain. It is simply the norm of the four-momentum.
Hello DaleSpam
Momentum has to do with mass and velocity, while I expext that restmass has to do with mass and no velocity at all. So simple it is maybe for you; I have thought about what you wrote but I don't understand it at all...
greetings Janm

[DaleSpam]

Sorry about the confusion, that is my fault. I was being sloppy with my factors of c (conceptually working in units where c=1). The four-momentum is:
\left(\frac{E}{c}, \, p_x, \, p_y, \, p_z\right)
So it has units of momentum, but the timelike component is refered to as the total energy (even though it is actually total energy divided by c). The norm of this four-vector is:
m_0^2 c^2 = \frac{E^2}{c^2} - p_x^2 - p_y^2 - p_z^2
So it also has units of momentum, but is often refered to as the rest mass or invariant mass (even though it is actually invariant mass times c).

[Rasalhague]

Hello Rasalhague
The dt/dtau I find very interesting. Do Taylor & Wheeler put the time dilation factor in the formulae at this moment?

By \frac{\mathrm{d} t}{\mathrm{d} \tau}, Taylor & Wheeler denote the derivative of coordinate time with respect to proper time. This is the time dilation factor!

m_0c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau} = \frac{m_0c^2}{1 - \left ( \frac{v}{c} \right )^2} = m_0c^2 \gamma = m_0c^2 \cosh \left ( \phi \right ) = E = T + m_0c^2,

where E is the relativistic total energy (which some people in the past have called "mass"), the time component of the energy-momentum 4-vector (sometimes called the momentum 4-vector), and m_0 is rest mass (which I think is usually considered nowadays more deserving of the name "mass" and often written simply m).

Compare this with the space components of the energy-momentum 4-vector (the momentum part of it), which take the form:

m_0c \frac{\mathrm{d} x}{\mathrm{d} \tau} = m_0c \frac{\mathrm{d} x}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m_0cv \gamma = m_0c \ \sinh \left ( \phi \right ),

where x stands for a representative direction in space.

[Gregg]

Hello planck42
So you are easily satisfied. I think that to say that mass-velocity relation is outdated is not a scientific statement. Science should be timeless. Theories are prooved true or false.

Theories are proven to be false only.

[JANm]

The norm of this four-vector is:
m_0^2 c^2 = \frac{E^2}{c^2} - p_x^2 - p_y^2 - p_z^2
So it also has units of momentum, but is often refered to as the rest mass or invariant mass (even though it is actually invariant mass times c).
Hello DaleSpam
Where do these three minus signs come from? Shouldn't the norm be the thing with the plus signs? ||a+bi||=a^2+b^2 isn't it?
I found a formula for dynamic mass m=Sqrt(m_0+p^2/c^2)... I think this is the right relativistic formula. So the invariant mass you speak of is the mass with velocity and momentum; if the velocity is zero then p=0 and so dynamic mass m=m_0.
My proposal is use the inertial system orientated to the background, so the earth has velocity 370 km/sec. Things standing still in comparison to the background have velocity zero and dt/dtau is one. It is already known for years that dt/dtau in most common astronomical cases does not differ much from one.
Instead of using beta=sqrt(1-v^2/c^2) I am very fond of using the first order approximation: beta=1-v^2/(2*c^2).

So my mass velocity relation is m=m_0/(1-v^2/(2*c^2) and this mass velocity relation is realy very plausible...

greetings Janm

[edpell]

JANm how about we use \gamma = \frac{1}{\sqrt[]{1-\frac{v^2}{c^2}}}}
and say m = \gamma m_0 = m_0 + (\gamma -1)m_0 some people like to break it into these two parts the rest mass and the kinetic energy I think it is simpler to leave it as \gamma m_0 .

[edpell]

Theories are proven to be false only.

I think there are three separate things experimental physics, theoretical physics and mathematical physics. I would agree that with the first two a theory can only be "not false" or "false" given our current available data and calculation techniques.

But I sure get the impress that mathematical physics would like to say they are correct by construction in the same way a math proof is correct by construction. Not sure if they are physics or math or some new in between ground.

[DaleSpam]

Where do these three minus signs come from? Shouldn't the norm be the thing with the plus signs? ||a+bi||=a^2+b^2 isn't it?The minus signs are from the usual Minkowski norm in relativity.

I found a formula for dynamic mass m=Sqrt(m_0+p^2/c^2)... I think this is the right relativistic formula. So the invariant mass you speak of is the mass with velocity and momentum; if the velocity is zero then p=0 and so dynamic mass m=m_0.
Your "dynamic mass" is more commonly known as "relativistic mass". It is equivalent to the total energy E (of course, divided by c^2 in order to make the units correct). There are many discussions here contrasting "relativistic mass" and "invariant mass".

My proposal is use the inertial system orientated to the background, so the earth has velocity 370 km/sec. Things standing still in comparison to the background have velocity zero and dt/dtau is one.Sure, you could do all of that, but why bother? All refeence frames are equivalent so you can use any that is convenient.

[JANm]

Sure, you could do all of that, but why bother? All refeence frames are equivalent so you can use any that is convenient.
Hello DaleSpam
Here you relativate (=psychological indifference to stimuli smaller then threshold) the very fundament of relativity-theory.
In between Michelson with his experiment and Penzias and Wilson with their backgroundradiation: the relativity theory can survive with: benifit of the doubt. In common terms one does not know the absolute velocity of the earth through space.
Since Penzias and Wilson we and for that matter all creatures in all of space can find an unique inertial system. For everybody the same, everywhere the same, radiation of 3 Kelvin has to be isotropic in all space. Astronomers have found that our particular velocity is 370 km/sec to the sign Leo. That is now a physical fact.
Why bother, while we have a theory which states it doesn't matter?
greetings Janm

[DaleSpam]

In common terms one does not know the absolute velocity of the earth through space.In more precise terms there is no such thing as "absolute velocity" in any experimental sense.

Astronomers have found that our particular velocity is 370 km/sec to the sign Leo. That is now a physical fact.Certainly it is a fact. Geographers have similarly found that the velocity of the Nile is 4 knots to the north. That is also a physical fact, with equal physical significance. If you wish you may also construct a reference frame where the Nile is at rest and compute all of your physics (including relativistic mass) wrt that reference frame.

Again, why bother? What prediction do we get correct by doing it your way that we get wrong by doing it the easy way?

[vitruvianman]

I've read contrasting sources concerning the concept of an object increasing in mass at relativistic velocities. Some of my older calculus texts mention this as being accepted by physicists, while a newer(by comparison) textbook called Principles of Physics: A Calculus-Based Text by Serway and Jewett claims that relativistic mass is outdated. Normally, the newer book would be correct, but I don't think there is any other reasonable explanation of the momentum of photons. Can the good people at PF please shed some light on this matter?

I think it's never been outdated and will never be. Relativistic mass is calculated so:

for example, kinetic enerji can be calculated with KE=mc^2-m_0c^2
it's also KE=mc^2-\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2=mc^2-\frac{m_0c}{\sqrt{c^2-v^2}} and if we add momentum to this equ., it would be E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} where momentum isn't relativistic. Remeber that relativistic p is p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}

after we make momentum relativistic in KE equ., we'll find out the relativistic value of energy, which includes p. It's so: E=\frac{m_0^2c^3}{\sqrt{c^2v^2}}+\frac{p^2cm_0}{\s qrt{c^2-v^2}}=\frac{m_0c(m_0c^2+p^2}{\sqrt{c^2-v^2}}

------------

we can also say for pc when E=\sqrt{m_0^2c^4+p^2c^2} pc is pc={\sqrt{p_0^2-m_0^2c^4}} and so, we can say p^2c^2=\frac{m_0^2v^2c^2}{1-\frac{v^2}{c^2}}=\frac{m_0^2\frac{v^2}{c^2}c^4}{1-\frac{v^2}{c^2}}

after a quick calculation you'll find p^2c^2=\frac{m_0^2c^4[\frac{v^2}{c^2}-1]}{1-\frac{v^2}{c^2}}+\frac{m_0^2c^4}{1-\frac{v^2}{c^2}}=-m_0^2c^4+m^2c^4=(mc^2)^2\Rightarrow E=pc

You can find the basic ideas of this calculations (made by me :) ) at the notes part of the book "Relativity: The Special and General Theory" by Albert Einstein.

If you see the equations as text, you can see them as pic.s by copy-pasting them onto http://www.sitmo.com/latex/

[DaleSpam]

if we add momentum to this equ., it would be E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} where momentum isn't relativistic. Remeber that relativistic p is p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} Careful here. Your p is relativistic here. Remember m_0 and c are constant and E is unbounded so p must also be unbounded.

[vitruvianman]

Careful here. Your p is relativistic here. Remember m_0 and c are constant and E is unbounded so p must also be unbounded.

Well you're right I think I've gotten confused here. I made everything relativistic even it's already relativistic :)

But the relativistic value of energy calculated in my first part of calculations (before ---------), I mean E=\sqrt{m_0^2c^4+p^2c^2} is right I think. (am I right?)

After all, my primary aim was to show that E=pc is also right as the relativistic energy of photon.

And as you say p is unbounded because of the same feature of E.

[DaleSpam]

Yes, that is correct, and it is equal to the expression I wrote above for the rest mass as the norm of the four-momentum.

[JANm]

Yes, that is correct, and it is equal to the expression I wrote above for the rest mass as the norm of the four-momentum.
Hello DaleSpam
The term norm is defined in linear analysis. So we have to call that mathematical norm.
The norm you mention is realy different, so let us call that physical norm and that means the diagonal trace of a Minkowski matrix. As I pointed out a mathematical norm cannot have these minus signs because the thing could become zero in a case of a nonzero situation.
Norm is a partial ordening, so things can have the same norm while being different, but there is only one thing with norm zero and that is the zero element.

So one uses if you have || a-b||= 0 one may conclude a=b. With your physical norm this essential norm thesis is not true.
greetings Janm

[DaleSpam]

What you say is all correct, the Minkowski norm is not really a norm at all. It is technically "a nondegenerate, symmetric, bilinear form with the signature (+,-,-,-) defined on a four-dimensional real space". However, that is rather cumbersome to write so "Minkowski norm" is the conventional shorthand.

In any case, whether you call it "Minkowski norm" or "nondegenerate, symmetric, bilinear form with signature (+,-,-,-) defined on a four-dimensional real space" does not substantially alter anything I wrote above.

[JANm]

Hello DaleSpam
The things I wrote will not be altered either. I found an answer To shakespeare and Einstein at the same time:

Of the many things relative: they are; and all the others they aren't.

As I explained within relativity theory restmass cannot be defined. It involves mass and velocity and especially the special case that velocity is zero. A mass of an individual object needs velocity to calculate m = gamma * m_0.

The orientation to the CMB would bring joy to Michelson and actual re honouring to Galileo Galilei. I might even say that if Lorentz and Einstein knew of the possibility to orientate to cosmic background radiation and know the velocity of the earth through space, they would not have launched this cumbersome theory in the first place!

So hurray for Michelson and Hurray for Galileo Galilei and a little minus point for Lorentz and a little minus point for Einstein; let us say they did not know.

If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial system and by using the term of A.A. Robb; what can perfect translation of everything add to reality. Suppose you have everything in an encyclopedia in Greek language and you translate that perfectly to Latin, then the things sayd are exactly the same; so what use is there in translating and what possible effect can this translating have on the described physics in the firstplace: none.

Greetings Janm

[edpell]

... within relativity theory restmass cannot be defined.
... If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial system ...

Jan I do not understand either of these statements.

1) rest mass is the mass measured by one moving with the object.

2) why?

[DaleSpam]

As I explained within relativity theory restmass cannot be defined.Since I defined it just a couple of posts ago this comment seems disingenuous at best.

If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial systemThat doesn't make any sense at all. That is like saying that Mt. Everest should have the same velocity in all reference frames. The CMBR is not a law of nature.

[vitruvianman]

The CMBR is not a law of nature.

Right!

[vitruvianman]

A mass of an individual object needs velocity to calculate m = gamma * m_0.

But not everytime. Let v be 0, then m=m_0 so m\geq m_0

[vitruvianman]

JANm how about we use \gamma = \frac{1}{\sqrt[]{1-\frac{v^2}{c^2}}}}
and say m = \gamma m_0 = m_0 + (\gamma -1)m_0 some people like to break it into these two parts the rest mass and the kinetic energy I think it is simpler to leave it as \gamma m_0 .

Yes it's much simpler but not that effective. I mean you can't calculate what you want without using the original forms of equations.
For example the E^2=m_0^2c^4+p^2c^2 not only the famous E=mc^2 . Most of the time we "gotta" calculate momentum too... Simplifying doesn't work all the time. A quote from Einstein:

"Keep it simple, stupid — but never oversimplify."