Relativistic mass and energy

In summary: Relativistic mass is not outdated, but it is an unnecessary concept now that everything is based on invariant mass. The equation E = mc^2 still holds true, but it is now interpreted as the energy of an object at rest being equivalent to its invariant mass multiplied by the speed of light squared. The concept of relativistic mass complicates things and is no longer needed in modern physics.
  • #1
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well, I'm very new to this, but I'm interested in learning. recently i followed a link from these forums that lead me here http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html which states:

If you see "relativistic mass" in your first-year physics textbook, complain! There is no reason for books to teach obsolete terminology.
...
The "relativistic mass" of an object is really just the same as its energy, and there isn't any reason to have another word for energy: "energy" is a perfectly good word.

well, I'm a little confused now, because if energy and relativistic mass are the same thing then why do we have the equation E = mc^2? in the equation, E is the energy and m is the relativistic mass, right? that equation seems to say to me that there is a propper use for the term "relativistic mass" because while it may be related to energy by a constant, it's not the same thing. could anyone shed some light on this for me?

thanks
 
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  • #3
nwall said:
well, I'm a little confused now, because if energy and relativistic mass are the same thing then why do we have the equation E = mc^2? in the equation, E is the energy and m is the relativistic mass, right? that equation seems to say to me that there is a propper use for the term "relativistic mass" because while it may be related to energy by a constant, it's not the same thing. could anyone shed some light on this for me?
If m is the relativistic mass, then E = mc^2 will be valid for any object regardless of the velocity. But if m is just the rest mass, then E=mc^2 only works for an object at rest, and for an object with nonzero velocity you'd use this equation:

[tex]E^2 = m^2 c^4 + p^2 c^2[/tex]

Where p is the relativistic momentum [tex]p = m v / \sqrt{1 - v^2/c^2}[/tex]
 
  • #4
nwall said:
well, I'm a little confused now, because if energy and relativistic mass are the same thing then why do we have the equation E = mc^2? in the equation, E is the energy and m is the relativistic mass, right? that equation seems to say to me that there is a propper use for the term "relativistic mass" because while it may be related to energy by a constant, it's not the same thing. could anyone shed some light on this for me?
You are correct. It is quite wrong to say that they are the same thing. If one quantity is proportional to another quantity that doesn't mean that they're the same thing. E.g. E = hf where E is the energy of a photon, h is Planck's constant and f is the frequency of the photon. E is therefore proportional to f. However this obviously doesn't mean that E an f are the same thing. E and m have a different definition and in some, but not all, cases they are proportional.

For an example where E does not equal to mc^2 please see

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Pete
 
  • #5
An object at rest has energy given by E_{rest}=mc^2.
This relation is sometimes considered as an "equivalence" of mass and energy,
but it is not a true equivalence. The energy and mass of an object are closely related, but not equivalent. To within factors of c, the energy is the time-like component of the momentum four-vector, while the mass is the invariant length of the four-vector. The energy in the rest system equals mc^2, but while in motion, the energy varies with velocity.

In early expositions of special relativity, and still in some popularizations,
the mass is considered to vary as m=m_0/\sqrt{1-v^2/c^2} with m_0 called the "rest mass". Then E=mc^2 even in motion. However this interpretation goes against the principle that the mass, as an intrinsic property of an object, should be the same in all Lorentz frames. In the modern interpretation, m is the "invariant mass" of an object, and the equation E=mc^2 holds only in the rest system.
 
  • #6
Meir Achuz said:
An object at rest has energy given by E_{rest}=mc^2.
That only applies when the object has no forces acting on it.

Pete
 
  • #7
thanks pmb_phy and meir_achuz. that was a very helpful explanation. so i guess while the article was wrong in saying that relativistic mass and energy are the same, it was correct in saying that relativistic mass is an outdated concept because now everything is based on invariant mass? that makes much more sense. thanks again.
 
  • #8
part of the answer is that in standard units used by relativistic physicists c = 1
so E =m and m = m(rest mass)/sqrt(1-v^2/c^2).
 
  • #9
nwall said:
thanks pmb_phy and meir_achuz. that was a very helpful explanation. so i guess while the article was wrong in saying that relativistic mass and energy are the same, it was correct in saying that relativistic mass is an outdated concept because now everything is based on invariant mass? that makes much more sense. thanks again.
No. Not everything is based on invariant mass. And relativistic mass is not outdated. The author is speaing about point particles. Since these particles don't have any internal structure then the time component of the 4-momentum will be mc.

So why is it you like energy rather than mass?

Pete
 
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  • #10
pmb_phy said:
And relativistic mass is not outdated.

Pete

Monty Python and the Relaitivistic Mass Question:

It's not outdated MUCH, said the crowd. "Maybe a bit", said sir Bevedere.
"A bit! A bit! A bit!" chanted the crowd.

"This new learning amazes me, Sir Bevedere. Explain to me once again
how ram's bladders may be employed to prevent earthquakes", aksed King Arthur.
 
  • #11
pmb_phy said:
You are correct. It is quite wrong to say that they are the same thing. If one quantity is proportional to another quantity that doesn't mean that they're the same thing. E.g. E = hf where E is the energy of a photon, h is Planck's constant and f is the frequency of the photon. E is therefore proportional to f. However this obviously doesn't mean that E an f are the same thing. E and m have a different definition and in some, but not all, cases they are proportional.
In the case of E=hf, we have independent experimental techniques for measuring the frequency and the energy of a particle. But do we have independent experimental techniques for measuring the rest mass of an object and measuring its energy at rest? How do you measure rest energy?
 
  • #12
JesseM said:
In the case of E=hf, we have independent experimental techniques for measuring the frequency and the energy of a particle. But do we have independent experimental techniques for measuring the rest mass of an object and measuring its energy at rest? How do you measure rest energy?
That's not right. They're are different methods to measure E and m.

Pete
 
  • #13
pmb_phy said:
That's not right. They're are different methods to measure E and m.
I didn't say there weren't, I just asked what they were.
 
  • #14
Meir Achuz said:
In early expositions of special relativity, and still in some popularizations,
the mass is considered to vary as m=m_0/\sqrt{1-v^2/c^2} with m_0 called the "rest mass". Then E=mc^2 even in motion. However this interpretation goes against the principle that the mass, as an intrinsic property of an object, should be the same in all Lorentz frames. In the modern interpretation, m is the "invariant mass" of an object, and the equation E=mc^2 holds only in the rest system.
Can you give an example (in numbers) of the value of E changing as v approaches c? I just need some illustration to get my mind around it.
 
  • #15
LindaGarrette said:
Can you give an example (in numbers) of the value of E changing as v approaches c? I just need some illustration to get my mind around it.

A handy value to learn is that at 86.6% the speed of light*, the dilation factor

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

takes the value .5. And energy at a relative velocity of v is rest energy [tex]mc^2[/tex] over [tex] \gamma[/tex]. So at 86.6% of c, the energy is doubled over the rest energy.


*Actually it's [tex]\frac{v}{c} = \frac{\sqrt{3}}{2} \approx .866[/tex]. Work it out.
 
  • #16
JesseM said:
I didn't say there weren't, I just asked what they were.
Ah. Sorry my good Sir. I applogize. :redface:

Let me get back to you in the morning.

Pete
 
  • #17
JesseM said:
In the case of E=hf, we have independent experimental techniques for measuring the frequency and the energy of a particle. But do we have independent experimental techniques for measuring the rest mass of an object and measuring its energy at rest? How do you measure rest energy?
The concept of rest energy is pre-relativity. Relativity did not prove that a body at rest has energy. Its a postulate on which to derive the relation between E and m. You can actually chose what the value of the rest energy is since energy is defined up to an arbitrary constant. Choose the constant to make life easy. The details of the rest will depend on specifics.

Pete
 
  • #18
I suppose one method of measuring a particle's rest energy would be to collide it with the corresponding antiparticle and measure the energy released...
 
  • #19
JesseM said:
I suppose one method of measuring a particle's rest energy would be to collide it with the corresponding antiparticle and measure the energy released...

I've been thinking about Tachyons lately. The theory of tachyons does not assume [itex]E^2 - (pc)^2 = m_0^2 c^4[/itex] to be valid. So if you get a measured value for which thid relation does not hold then how were E and p measured and what does that say for the definitions of E and m0 ?? :confused:

Pete
 
  • #20
nwall said:
well, I'm very new to this, but I'm interested in learning. recently i followed a link from these forums that lead me here http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html which states:



well, I'm a little confused now, because if energy and relativistic mass are the same thing then why do we have the equation E = mc^2? in the equation, E is the energy and m is the relativistic mass, right? that equation seems to say to me that there is a propper use for the term "relativistic mass" because while it may be related to energy by a constant, it's not the same thing. could anyone shed some light on this for me?

thanks

I am almost positive E=mc^2 stated that relativistic mass DOES equal energy but rest mass doesnt. since the speed of light squared is constant, does that have to meen that energy does equal relativistic mass?
 
  • #21
It does,they mean and are the same thing (especially in Heaviside -Lorentz units used in QFT),just as long you don't consider macroscopical bodies,for which,that formula will not hold.

Daniel.

P.S.There are no such problems,if people work only with pointlike particles.QFT is a theory of pointlike particles.
 
  • #22
thank you! that's what i thought but everyone on the first page of this forum doesn't seem to think the same
 
  • #23
Incidentally,in (elementary) particle physics,one never uses this concept "relativistic mass",but only "energy".

Daniel.
 
  • #24
pmb_phy said:
I've been thinking about Tachyons lately. The theory of tachyons does not assume [itex]E^2 - (pc)^2 = m_0^2 c^4[/itex] to be valid.
I think this equation is still assumed to hold for tachyons, with the assumption that tachyons have imaginary rest mass--on this page from John Baez's site, that equation is used to get the conclusion that the energy of a tachyon is a negative real, as is E^2 - (pc)^2.
 
  • #25
dextercioby said:
Incidentally,in (elementary) particle physics,one never uses this concept "relativistic mass",but only "energy".

Daniel.
exactley! which is was nwall was asking!
 
  • #26
The 4D force-free relativistic Lagrangian *) for a particle in space-time is a constant of motion, which (at least that is my interpretation) means that from a 4D perspective the energy of the particle is no longer a function of the spatial velocity but a constant. It lacks therefor the odd infinities that arise in the approach according to 3D relativistic dynamics. This is a strong indication that these infinities are merely a result of our limited 3D observations that may lead to "projections" of the 4D values. Compare e.g. the shadow of a stick in the ground with sunset. Its length also tends to infinity.

*) see e.g. Herbert Goldstein, Classical Mechanics. He derives the equation [tex]\Lambda=\frac{1}{2}mu_{\nu}u_{\nu}[/tex], where [tex]u_{\nu}[/tex] is the Minkowski 4-velocity which is equal to [tex]c[/tex] and [tex]m[/tex] is the rest mass. I must add the remark that also Goldstein warns that active research on relativistic Lagrangians is still ongoing. Conclusions may be premature although I have seen several articles using it.
 
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  • #27
Mortimer said:
The 4D force-free relativistic Lagrangian *) for a particle in space-time is a constant of motion, which (at least that is my interpretation) means that from a 4D perspective the energy of the particle is no longer a function of the spatial velocity but a constant.
In which frame is this energy to be measured? The energy of a particle is frame dependent, and not, in general in GR, conserved. However the energy equivalent of its 'rest' mass is conserved as it is measured in the rest frame of the particle.

Garth
 
  • #28
Garth said:
In which frame is this energy to be measured?

Goldstein (and others) use a 4D reference frame where the progress of the particle along its worldline is measured using a 5th, independent parameter. I quote the text:
"We shall assume the choice of some Lorentz-invariant quantity [tex]\Theta[/tex] with no further specification than that it be a monotone function of the progress of the world point along the particle's world line"
In this frame the velocity of any particle is always [tex]c[/tex], in fact being equal to the conventional Minkowski 4-velocity. So essentially it assumes kind of an absolute 4D rest frame where all particles move at [tex]c[/tex].
Another article of Hans Montanus (http://www.kluweronline.com/article.asp?PIPS=346166&PDF=1) is based on Euclidean relativity but uses the same principles.
 
  • #29
It's much more clever to use the einbein approach for the free relativistic particle

[tex]S^{L}_{0}\left[x^{\mu},e\right]=-\frac{1}{2}\int_{\tau_{1}}^{\tau_{2}} d\tau \ \left(\frac{\dot{x}^{\mu}\dot{x}_{\mu}}{e}+m^{2}e\right) [/tex]

,where all variables involved are bosonic.Dot means derivative wrt [itex]\tau[/itex].

Daniel.
 
  • #30
This is because you're in the different view.relativistic mass is relativistic mass,and energy is energy.they are connectted by M-E eqoution.They are equal.
 
  • #31
Meir Achuz said:
In the modern interpretation, m is the "invariant mass" of an object, and the equation E=mc^2 holds only in the rest system.



is the rest system to viewer?
 
  • #32
dreamfly said:
is the rest system to viewer?

I'm pretty sure Meir meant "the rest system of the object in question."

In the usual modern terminology the energy of an object is

[tex]E = \frac {mc^2}{\sqrt {1 - u^2 / c^2}}[/tex]

where [itex]u[/itex] is the speed of the object in whatever reference frame you're working in, and [itex]m[/itex] is the invariant mass, which is often called the "rest mass". This reduces to [itex]E = mc^2[/itex] when [itex]u = 0[/itex], i.e. when the object is at rest.
 
  • #33
jtbell said:
I'm pretty sure Meir meant "the rest system of the object in question."

In the usual modern terminology the energy of an object is

[tex]E = \frac {mc^2}{\sqrt {1 - u^2 / c^2}}[/tex]

where [itex]u[/itex] is the speed of the object in whatever reference frame you're working in, and [itex]m[/itex] is the invariant mass, which is often called the "rest mass". This reduces to [itex]E = mc^2[/itex] when [itex]u = 0[/itex], i.e. when the object is at rest.



Thank you!
 

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