Is Relativistic Mass Plausible?

[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.

 

[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

 

[Dale]

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: $$E=\frac{m_0^2c^3}{\sqrt{c^2v^2}}+\frac{p^2cm_0}{\sqrt{c^2-v^2}}=\frac{m_0c(m_0c^2+p^2}{\sqrt{c^2-v^2}}$$

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

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/

 

[Dale]

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.

 

[Dale]

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 really 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

 

[Dale]

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?