Is Relativistic Mass Plausible?

[Dale]

...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."

 

[Dale]

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.)

 

[Dale]

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

 

[Dale]

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 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 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

 

[Dale]

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 helpfulsuppose 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) * vsince 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 c^c 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

 

[Dale]

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

 

[Dale]

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

 

[Dale]

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 referred 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 referred 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 referred 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 really 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.

 

[Dale]

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.