Mass in Relativity: Explaining How Velocity Does Not Increase Mass

In summary: Yes, the definition of mass is the same in all of these fields--mass is just a scalar quantity. The principle of equivalence states that the laws of physics should be the same for all observers moving at the same speed, and this is why invariant mass is the norm in SR and QM. In contrast, the principle of relativistic equivalence states that the laws of physics should be the same for all observers moving at the same speed with respect to each other, and this is why relativistic mass is needed in GR.
  • #1
AlbertEinstein
113
1
[Mentor's note: this is a continuation of a thread in the Quantum Physics forum:

https://www.physicsforums.com/showthread.php?t=144296

which turned into a discussion of the various kinds of mass in relativity.]Originally Posted by Los Bobos
"It is actually more consistent to say, that from SR we know that mass does not increase with velocity. Here you would have to think what is momentum in SR. And of course the biggest problem of elementary QM, the difference in the handling of spatial and temporal coordinates."

Well, no. I haven't studied either of QM or SR in real depth.I'm still in high school.:( How mass does not incrase with velocity? I will be very thankful if you clearly explain it in bit detail.

Thanks
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
The increase of mass with velocity is just an evil pedagogical plan and a historical misunderstanding (this was included in the Lorentz-Fitzgerald theory (if I remember correclty, transverse masses and so on)).

This was a good text about the subject
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html.

But nevertheless we have only one mass which is invariant and anything else makes very little physical sence.
 
  • #3
AlbertEinstein said:
How mass does not incrase with velocity?

When people talk about "mass" in SR, they are usually referring to one of two different kinds of "mass": invariant mass (also often called rest mass) which does not change with velocity, and relativistic mass which does change with velocity. Relativistic equations involving mass can be written using either kind of mass. Of course the equations look different depending on which kind of mass is being used, but if you use them properly they predict the same results for physical experiments.

It's ultimately a matter of convention or preference which kind of mass people prefer to talk about, and there have been repeated arguments about this here. Most popular-level books and articles about relativity talk about relativistic mass. In serious introductory textbooks (college/university level) there has been a trend in the last couple of decades away from relativistic mass and towards invariant mass, which some people disagee with. In elementary particle physics, people generally use invariant mass. (I worked in that field as a grad student, and I personally do not remember anyone ever using relativistic mass, among the people that I worked with or among the papers that I read.)

Further discussion of this issue really belongs in the relativity forum.
 
Last edited:
  • #4
That's fine. Thanks Los Bobos and jtbell.This means that invariant mass and relativistic mass are the two forms of one single entity. What is that "entity"? hope my question is not silly:) But one question popping out. What about the gravitational mass? How is it related to any of these two ?
 
  • #5
AlbertEinstein said:
That's fine. Thanks Los Bobos and jtbell.This means that invariant mass and relativistic mass are the two forms of one single entity. What is that "entity"? hope my question is not silly:) But one question popping out. What about the gravitational mass? How is it related to any of these two ?

The only physical mass is the invariant mass, the other one is just a teaching tool.
 
  • #6
The term "physical mass" is rather vague. I usually stress the fact that independence of the frame of reference is a highly desirable quality, and it is this independence of frame that invariant mass has (at least for isolated systems), and that relativistic mass lacks.
 
  • #7
AlbertEinstein said:
That's fine. Thanks Los Bobos and jtbell.This means that invariant mass and relativistic mass are the two forms of one single entity. What is that "entity"? hope my question is not silly:)

That "entity" is that particle's momentum-4vector [tex]p^a[/tex].
Up to conventional signs and factors of c,
Invariant mass is the norm of that 4vector: [tex]\sqrt{p_ap^a}[/tex].
Relativistic mass is the temporal-component of that 4vector for some observer with unit 4velocity [tex]u^a[/tex]: [tex]p_au^a[/tex].

Note that both are scalars, which are invariant...the same for all observers [with unit 4velocity [tex]v^a[/tex]]. The relativistic mass always requires specification of an associated observer with unit 4velocity [tex]u^a[/tex].
 
  • #8
Originally Posted by"Robphy"

"That "entity" is that particle's momentum-4vector .
Up to conventional signs and factors of c,
Invariant mass is the norm of that 4vector: .
Relativistic mass is the temporal-component of that 4vector for some observer with unit 4velocity : .

Note that both are scalars, which are invariant...the same for all observers [with unit 4velocity ]. The relativistic mass always requires specification of an associated observer with unit 4velocity ."

By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?

Thanks.
 
  • #9
mass?

jtbell said:
When people talk about "mass" in SR, they are usually referring to one of two different kinds of "mass": invariant mass (also often called rest mass) which does not change with velocity, and relativistic mass which does change with velocity. Relativistic equations involving mass can be written using either kind of mass. Of course the equations look different depending on which kind of mass is being used, but if you use them properly they predict the same results for physical experiments.

It's ultimately a matter of convention or preference which kind of mass people prefer to talk about, and there have been repeated arguments about this here. Most popular-level books and articles about relativity talk about relativistic mass. In serious introductory textbooks (college/university level) there has been a trend in the last couple of decades away from relativistic mass and towards invariant mass, which some people disagee with. In elementary particle physics, people generally use invariant mass. (I worked in that field as a grad student, and I personally do not remember anyone ever using relativistic mass, among the people that I worked with or among the papers that I read.)

Further discussion of this issue really belongs in the relativity forum.
That is a very lucide explanation. When we start deriving the bazic equations of relativistic dynamics we start with the invariant m0. Volens-nolens we arive at a given point of the derivation to gammam0, Physicists, as well trained name givers call it relativistic mass. Others multiply it with c^2 and call it energy. Is there more to say for a beginner?
Regards
 
  • #10
Los Bobos said:
The increase of mass with velocity is just an evil pedagogical plan and a historical misunderstanding (this was included in the Lorentz-Fitzgerald theory (if I remember correclty, transverse masses and so on)).


.
Please have a critical look at

Physics, abstract
physics/0605203
Relativistic dynamics without conservation laws
We show that relativistic dynamics can be approached without using conservation laws (conservation of momentum, of energy and of the centre of mass). Our approach avoids collisions that are not easy to teach without mnemonic aids. The derivations are based on the principle of relativity and on its direct consequence, the addition law of relativistic velocities.

Regards
 
  • #11
But my question reamains unanswered! As in post #8
"By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?"

At least please answer my questions.
 
  • #12
AlbertEinstein said:
But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?"

At least please answer my questions.

Mass is the norm of the momentum 4-vector. It is the same and the one and the only everywhere.

Your second question I don't understand.
 
  • #13
Please read posting #3 again.

There is no single quantity that everyone agrees on calling simply "mass" in relativity. There is no single quantity in relativity that has all the properties of classical mass and serves all the purposes of classical mass.
 
  • #14
jtbell, you are completely right in the statistical sense :smile: , but for me there is only one mass.
Relativistic mass causes some logical inconsistensies rising from incorrect use of simultaneity (at least it seems so to me). Relativity is not classical and therefore it makes no sense to me in explaining it in classical terms.

Is "in explaining it in" english? :).
 
  • #15
Los Bobos said:
Relativity is not classical and therefore it makes no sense to me in explaining it in classical terms.

That's why a mathematical formulation (as I gave #7) helps clarify what is going on. Of course, one may still have to interpret what is going on... but at least we have a mathematical structure and rules of mathematics to guide us. Certainly, if the mathematics leads us somewhere that is inconsistent with the physics, we need a better mathematical model.
 
  • #16
Yes, but for me there is no logic in calling the temporal part of the 4-momentum "mass" as it is not mass which is the norm of this 4-vector :).
I would like to see for example the lagrangian of QED written with this weird mass.

"The road to emitter theory is paved with relativistic masses" ;).
 
Last edited:
  • #17
Los Bobos said:
Yes, but for me there is no logic in calling the temporal part of the 4-momentum "mass" as it is not mass which is the norm of this 4-vector :).
I would like to see for example the lagrangian of QED written with this weird mass.

"The road to emitter theory is paved with relativistic masses" ;).

I'm not necessarily advocating its use.
I'm advocating a less-ambiguous mathematical language of what has already been defined by others... so as to avoid "loose talk" using overused (overloaded in computer-speak) terms inspired by classical thinking.

Once given a mathematical formulation, its physical meaning [especially what it depends on, etc] is clarified. Then one can judge how useful or not it is in whatever context you want to apply it in.
 
  • #18
Ok then. what should be the correct answer to my question posted -
"Can we mix SR and uncertainty principle? for ex. from uncertainty principle we have [tex]\Delta x \Delta p \geq h/4\pi [/tex]Books then write [tex]m \Delta x \Delta v \geq h/4\pi [/tex]
with an assumption that m can be measured accurately.
However form SR we know that mass depends on velocity; with mass increasing with velocity. Now if we can't measure v exactly how can we measure m exactly?"
I am bit confused with all these discussions. Please help me out.
 
  • #19
Maybe it will help in this case to think of the momentum in terms of the invariant (rest) mass instead of the relativistic mass:

[tex]p = \frac{m_0 v} {\sqrt{1 - v^2 / c^2}}[/tex]

Then we have

[tex]\Delta x \Delta p = m_0 \Delta x \Delta \left( \frac{v} {\sqrt{1 - v^2 / c^2}} \right)[/tex]

and we can express that complicated [itex]\Delta[/itex] on the right in terms of [itex]\Delta v[/itex]. I'm in a bit of a hurry so I won't do that last step right now.
 
  • #20
AlbertEinstein said:
But my question reamains unanswered! As in post #8
"By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?"

At least please answer my questions.
After reading your posts, I resent your choice of name.
I bet you aren't really who you say you are.
 
  • #21
jtbell said:
Maybe it will help in this case to think of the momentum in terms of the invariant (rest) mass instead of the relativistic mass:

[tex]p = \frac{m_0 v} {\sqrt{1 - v^2 / c^2}}[/tex]

Then we have

[tex]\Delta x \Delta p = m_0 \Delta x \Delta \left( \frac{v} {\sqrt{1 - v^2 / c^2}} \right)[/tex]

and we can express that complicated [itex]\Delta[/itex] on the right in terms of [itex]\Delta v[/itex]. I'm in a bit of a hurry so I won't do that last step right now.

i'll do it.

[tex] \frac{d}{dv} \left( \frac{v} {\sqrt{1 - \frac{v^2}{c^2}}} \right) = \frac{d}{dv} \left( v \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \right)[/tex]
[tex] = -\frac{1}{2} v \left(1 - \frac{v^2}{c^2} \right)^{-3/2} \frac{-2 v}{c^2} + \left(1 - \frac{v^2}{c^2} \right)^{-1/2}[/tex]
[tex] = \frac{v^2}{c^2} \left(1 - \frac{v^2}{c^2} \right)^{-3/2} + \left(1 - \frac{v^2}{c^2} \right)^{-1/2}[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \frac{v^2}{c^2}\left(1 - \frac{v^2}{c^2} \right)^{-1} + 1 \right)[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \left(\frac{c^2}{v^2} - 1 \right)^{-1} + 1 \right)[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \frac{1+\left(\frac{c^2}{v^2} - 1 \right)}{\frac{c^2}{v^2} - 1} [/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \frac{1 }{1 - \frac{v^2}{c^2} } \right)[/tex]
[tex] =\left(1 - \frac{v^2}{c^2} \right)^{-3/2}[/tex]so

[tex] \Delta x \Delta p = \left(1 - \frac{v^2}{c^2} \right)^{-3/2} \Delta x \Delta (m_0 v) [/tex]

looks sort of ugly. don't know if it's right.
 
Last edited:
  • #22
AlbertEinstein said:
Ok then. what should be the correct answer to my question posted -
"Can we mix SR and uncertainty principle? for ex. from uncertainty principle we have [tex]\Delta x \Delta p \geq h/4\pi [/tex]Books then write [tex]m \Delta x \Delta v \geq h/4\pi [/tex]
with an assumption that m can be measured accurately.
However form SR we know that mass depends on velocity; with mass increasing with velocity. Now if we can't measure v exactly how can we measure m exactly?"
I am bit confused with all these discussions. Please help me out.

I'm not quite sure what your question is. Standard undergraduate quantum mechanics (the Schrodinger wave equation, for instance) is not compatible with special relativity, i.e. its not Lorentz invariant.

To get compatibility with SR, you either need to do quantum field theory rather than quantum mechanics (the most useful and most accurate choice but also the most demanding), or you need to do things like replace the Schrodinger equation with the Klein-Gordon equation (and if you have particles with spin, you need to do even more than that, the Klein-Gordon equation is the QM formulation of a spin 0 particle.)

Is your problem purely quantum mechanical, or are you also having problems with the classical aspects of mass? If so, you might want to stick to the classical issues and forget about QM for the time being.

It seems to me from reading your posts that you are "stuck" on the Newtonian idea that momentum = mass * velocity. All you really need to do is to unlearn this, then learn how SR defines momentum (mass * velocity * gamma, where mass is invariant mass), and how quantum mechanics defines momentum - as an operator - (which is a much trickier topic) - you might want to start with

http://en.wikipedia.org/wiki/Momentum#Momentum_in_quantum_mechanics

And if you want to tackle QFT, it's got an even different notion of momentum. (Actually, I suppose it might be better to say it's got a different notion of particles. The wikipedia article on QFT explains it reasonably well.)

You also might find it helpful to take a side-trip down classical mechanics into the Lagrangian formulation, where you'll learn that momentum is the partial derivative of the Lagrangian with respect to the velocity. This leads to the Hamiltonian formulation of classical mechanics, which is where more serious treatments of quantum mechanics begin.
 
Last edited:
  • #23
Originally Posted by Pervect
Is your problem purely quantum mechanical, or are you also having problems with the classical aspects of mass? If so, you might want to stick to the classical issues and forget about QM for the time being.

It seems to me from reading your posts that you are "stuck" on the Newtonian idea that momentum = mass * velocity. All you really need to do is to unlearn this, then learn how SR defines momentum (mass * velocity * gamma, where mass is invariant mass), and how quantum mechanics defines momentum - as an operator - (which is a much trickier topic) - you might want to start with

This is exactly what I wanted to know that how momentum is exactly defined in QM.Thanks guys I suppose I have got an answer to the question.
 
  • #24
Quantum mechanics is weird (IMO) - a lot weirder than relativity.

But it turns out that if you've got a wavefunction [itex]\Psi(x,t)[/itex], the momentum can be represented by a vector

[tex]
[px,py,pz] = -i \hbar \left[\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right] \Psi
[/tex]

which is the momentum operator that I was talking about. See for instance

http://scienceworld.wolfram.com/physics/MomentumOperator.html

This defintion of momentum doesn't involve velocity at all (which as you point out isn't really defined in quantum mechanics)- just partial derivatives of the wave function. There might be someone in the QM forums who can explain in detail why this works, but it's not me. I just use the results when I need to. I'll add that

http://www.missioncollege.org/depts/physics/P4poe/P4D/Schrodinger.htm

does a somewhat reasonable job of motivating this result (they don't talk about momentum operators specifically but are mainly interested in the Schrodinger equation) starting from the de-Broglie relationship.
 
Last edited by a moderator:

Related to Mass in Relativity: Explaining How Velocity Does Not Increase Mass

1. What is the concept of mass in relativity?

The concept of mass in relativity is based on Albert Einstein's theory of special relativity, which states that the mass of an object increases as its velocity approaches the speed of light. This means that the mass of an object is not constant and can change depending on its velocity.

2. How does velocity affect an object's mass?

As an object's velocity increases, its mass also increases. This is due to the fact that energy and mass are equivalent in relativity, and as an object approaches the speed of light, its kinetic energy increases, which in turn increases its mass.

3. Does an object's mass actually increase as it approaches the speed of light?

No, an object's mass does not actually increase as it approaches the speed of light. Instead, its relativistic mass increases, which is a measure of the object's energy content. The object's rest mass, which is its mass at rest, remains constant.

4. How does this concept impact our understanding of mass?

The concept of mass in relativity challenges the traditional understanding of mass as a constant and unchanging property of an object. It shows that mass is not an absolute quantity, but rather depends on the object's velocity.

5. What are some real-life examples of mass changing with velocity?

One example of mass changing with velocity is the behavior of particles in accelerators, such as the Large Hadron Collider. As particles are accelerated to near the speed of light, their mass increases, allowing for the study of high-energy collisions. Another example is the increase in mass of an object as it approaches the speed of light in space travel, which can impact fuel requirements and overall mission planning.

Similar threads

  • Special and General Relativity
Replies
11
Views
2K
  • Special and General Relativity
Replies
15
Views
2K
  • Special and General Relativity
Replies
5
Views
2K
  • Special and General Relativity
Replies
4
Views
853
  • Special and General Relativity
2
Replies
38
Views
3K
  • Special and General Relativity
Replies
4
Views
1K
  • Special and General Relativity
Replies
14
Views
4K
  • Special and General Relativity
Replies
14
Views
2K
  • Special and General Relativity
Replies
10
Views
2K
Replies
2
Views
4K
Back
Top