What is m in Kinetic Energy? Relativistic mass or Rest mass?

[vanhees71]

Neither. And this is a perfect example of one of the reasons relativistic mass got removed from introductory college-level physics textbooks during the 1990's. It leads many people to think that relativistic mass is a genuine relativistic generalization of newtonian mass. So, for example, we take the newtonian expression for kinetic energy, $\frac{1}{2}mv^2$, replace $m$ with $\gamma m$ (where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}})$, the so-called relativistic mass, and the result is a relativistically valid expression.

It's not!

The correct expression is $(\gamma-1)mc^2$.

In addition it hinders to understand that the measure of inertia in relativistic physics is not mass but all kinds of energy. This also holds in general relativity, where the source of the gravitational field is the energy-momentum-stress tensor of "matter and radiation", and GR predicts (by assuming the validity of the Einsteinian "equivalence principle") that the coupling constant is universal (modulo some factors given by Newton's gravitational constant). Mass is only one contribution to this energy-momentum tensor. Famously this also predicts that light, which is described by the massless electromagnetic field, takes part in the gravitational interaction, i.e., it gets affected by gravitational fields, which was confirmed in 1919 by the British solar-eclipse excursion and made Einstein the first shooting star of science/physics. Also the em. field provides a source of the gravitational field although it has 0 mass.

 

[vanhees71]

BTW: also Einstein pretty early abandoned the idea of "relativistic mass" for the very reasons stated over and over in this thread:

https://faculty.washington.edu/seattle/physics544/Eismc2/1989 Okun Physics Today.pdf

 

[pervect]

TL;DR Summary: Need to know the kinetic energy of an electron moving with velocity v to find the total energy of an electron moving with velocity v

note:
m = relativistic mass
$m_o$ = rest mass
v = velocity of the objectQuestion 1: If a particle is moving at relativistic speeds what would it's kinetic energy be?
I think it's $K.E. = \frac{1}{2} m_o v^2$ and my friend thinks it's $K.E. = \frac{1}{2} \frac{m_o v^2}{\sqrt{1-\frac{v^2}{c^2}}}$
Who is right? Is it relativistic mass or rest mass?

I just wanted to reinforce Mr T"s observation that both answers were incorrect, and to repeat the correct answer using your preferred notation, even though there is only a very very slight notational difference.

For the situation you describe, the relativistic kinetic energy of the particle is $\left( \gamma-1 \right) m_0 c^2$ and the total energy is $\gamma m_0 c^2$, where $\gamma = \frac{1}{\sqrt{1 - {\left( v/c \right) }^2}}$. To show that in the low limit that the kinetic energy is given by $\frac{1}{2} m_0 v^2$ in the non-relativistic limit, you need to do something equivalent to taking the Taylor series expansion of $\gamma m_0 c^2$. You should get the following expresion for total energy E.

$$E = \gamma m_0 c^2 \approx m_0 c^2 + \frac{1}{2} m_0 v^2 + \frac{3}{8} m_0 v^4/c^2$$

plus terms of order greater than 6 in velocity. From this Taylor series expansion, you can see that when (v/c) << 1, the total energy approaches the rest energy $m_0 c^2$ plus the Newtonian kinetic energy $\frac{1}{2} m_0 v^2$.

The rationale for the Taylor series expansion is that (v/c) is a small number much less than 1. It might be clearer and easier to let $\beta = v/c$. So let's do it that way as an afterthought.

What we are saying is that the Taylor series expansion of $1/\sqrt{1-\beta^2}$ is given by
$$1 + \frac{1}{2} \beta^2 + \frac{3}{8} \beta^4$$

plus terms of order greater than 6 in beta. Using the fact that $\beta = v/c$, you should get the answer I gave earier.

If you take the advice of some of the other posters about not using relativistic mass, you may understand why they tend to automatically use different notation.

 

[jtbell]

For the situation you describe, the relativistic kinetic energy of the particle is $\left( \gamma-1 \right) m_0 c^2$ and the total energy is $\gamma m_0 c^2$, where $\gamma = \frac{1}{\sqrt{1 - {\left( v/c \right) }^2}}$.

I find it easiest to remember the relativistic kinetic energy by starting from the total energy (which is what unqualified "energy" usually means for a relativistic particle).

Total energy = rest energy + kinetic energy

Kinetic energy = total energy - rest energy

Kinetic energy = $\gamma m_0 c^2 - m_0 c^2$

Kinetic energy = $(\gamma - 1) m_0 c^2$

 

[Sagittarius A-Star]

Non-relativistic kinetic energy=
$\int \vec F \cdot d\vec s = \int (m\vec a) \, d\vec s = \int (m\frac{d\vec v}{dt}) \, d\vec s = m\int_0^V \vec v \cdot d\vec v = \frac{1}{2}mV^2$

With proper acceleration $\alpha=\gamma^3 a$ (follows from relativistic velocity addition), $m=$ invariant mass, and with force in direction of movement, the relativistic kinetic energy=
$\int F \cdot ds = \int m\alpha \cdot ds = m \int \gamma^3 a \cdot ds = m \int \gamma^3 \frac{dv}{dt} \cdot ds = m\int_0^V \gamma^3 v \cdot d v = mc^2(\gamma -1) = E-E_0$

Calculation

 

[Slimy0233]

Nowhere. There is nowhere that you should use relativistic mass.Everywhere. You should avoid using relativistic mass everywhere.No. The $m$ in $E=mc^2$ is the invariant mass since the formula only applies to systems at rest anyway. Although it is super-famous, $E=m c^2$ is not a very general expression. The general expression is $$ m^2 c^2 = E^2/c^2 - p^2$$ The famous equation is the special case of the general equation for $p=0$.

So, the famous equation is only true when the momentum of the system is 0. Thus the $m$ in the famous equation is the invariant or "rest" mass. This $m$ is also the $m$ that you find in tables listing the properties of particles, the $m$ that you use in the relativistic version of Newton's 2nd law, and the $m$ that you measure with a balance scale.

hey, thank god you had commented this, btw, my professor used $E = mc^2$ to calculate the velocity of a moving electron instead of the general expression $m^2 c^2 = E^2/c^2 - p^2$, so is he wrong here? Third line (I mean, I think he is wrong, but he is my tutor, so, I sought your help

edit: The kinetic energy of the electron is 0.34 MeV and thus the total energy will be 0.51MeV+0.34MeV = 0.85 MeVSo, I would say $({(0.85Mev)}^2 = (mc^2)^2 + (pc)^2$ and not the special case

 

[Ibix]

Third line (I mean, I think he is wrong, but he is my tutor, so, I sought your help

Yes and no. If he means relativistic mass by $m$ (which he apparently does) then his expression is correct - it's just that any use of relativistic mass is horribly open to misinterpretation and confusion. It's correct here because (for a particle of non-zero mass)$$\begin{eqnarray*}
E^2&=&m_0^2c^4+p^2c^2\\
&=&m_0^2c^4+\gamma^2m_0^2v^2c^2\\
&=&m_0^2c^2\left(c^2+\gamma^2v^2\right)\\
&=&m_0^2c^2\left(c^2+\frac{v^2}{1-v^2/c^2}\right)\\
&=&m_0^2c^2\left(\frac{c^2}{1-v^2/c^2}\right)\\
&=&\gamma^2m_0^2c^4
\end{eqnarray*}$$So for things that have non-zero mass (i.e. anything not travelling at the speed of light) there's nothing mathematically wrong with using $E=\gamma m_0c^2$. It's just a notational mess because you have to keep track of what $m$ means (and that only gets worse when you go beyond one dimensional motion because there are longitudinal and transverse relativistic masses that you need to use in different contexts). And it doesn't work at all for things like photons that have zero rest mass because they don't have a rest mass to start with so you have to have a special case explanation for what relativistic mass means in this case.

Much better not to use it.

If your course does use it (in any way other than mentioning that it exists and you have to keep in mind that some people mean $\gamma m$ when they write $m$) then I would consider picking a different course. If that is not an option, any time the prof writes $m$, you write $\gamma m_0$. And as soon as you're out of the course, join the 21st century and drop the subscript zero.

Incidentally, this kind of collision problem is much easier to solve (in both Newtonian and relativistic mechanics) if you work in the zero momentum frame.

 

[Dale]

hey, thank god you had commented this, btw, my professor used $E = mc^2$ to calculate the velocity of a moving electron instead of the general expression $m^2 c^2 = E^2/c^2 - p^2$, so is he wrong here? Third line (I mean, I think he is wrong, but he is my tutor, so, I sought your help

edit: The kinetic energy of the electron is 0.34 MeV and thus the total energy will be 0.51MeV+0.34MeV = 0.85 MeVSo, I would say $({(0.85Mev)}^2 = (mc^2)^2 + (pc)^2$ and not the special case
View attachment 329678

Ugh, that is a bad sign. It means the professor is decades out of date with their source material. I would solve this using the four-momentum as follows

The four momentum is $P=(E/c,p_x,p_y,p_z)$, for convenience I will use units where $c=1$ and $m$, $E$, and $p$ are all measured in $\mathrm{MeV}$. We can ignore $y$ and $z$ here, so I will just drop those. The four-momentum is conserved, so setting the four-momentum before the scattering equal to the four-momentum after the scattering we get $$P_{photon,before} + P_{electron,before}=P_{photon,after}+P_{electron,after}$$$$(0.51,0.51) + (0.51,0.00) = (E_{photon,after},p_{photon,after}) + (E_{electron,after},p_{electron,after})$$ This gives us two equations in four unknowns, so we use the general equation $m^2=E^2-p^2$ to get two more equations $$0=E^2_{photon,after} - p^2_{photon,after}$$$$0.51^2=E^2_{electron,after}-p^2_{electron,after}$$
Solve to get $$E_{photon,after}=0.17$$$$p_{photon,after}=-0.17$$$$E_{electron,after}=0.85$$$$p_{electron,after}=0.68$$ and we can easily get the velocity of the electron with $$v_{electron,after}=\frac{p_{electron,after}}{E_{electron,after}}=0.8$$ and the kinetic energy of the electron with $$KE_{electron,after}=E_{electron,after}-m_{electron} = 0.34$$

 

[Vanadium 50]

the professor

Is there one? I thought he was studying on his own, although there has been some mention of a tutor.

 

[Nugatory]

hey, thank god you had commented this, btw, my professor used $E=mc^2$ to calculate the velocity of a moving electron instead of the general expression $m^2c^2 =E^2/c^2-p^2$ so is he wrong here? Third line (I mean, I think he is wrong, but he is my tutor, so, I sought your help

Not exactly wrong, because with a bit of algebra and an understanding of the four-momentum you can get from that more general frame-independent relationship to the frame-dependent $E=\gamma mc^2$ - and that is indeed the easiest way of calculating the total energy of a particle with known velocity. It's bad pedagogy though, because it tempts people to think of $\gamma m$ as a "mass" which it really isn't, and because in most realistic particle physics problems we know the momentum more accurately than the velocity.

 

[romsofia]

Since your course uses relativistic mass, there is some luck for you. Rindler (a great relativist) has some notes in which he uses that notation, which can be found here: http://www.scholarpedia.org/article/Special_relativity:_mechanics

 

[robphy]

Here's an energy-momentum diagram (energy upwards) that visualizes @Dale's calculation above

• The numbers are nice enough to be drawn on "rotated graph paper"
  by taking 1 mass-diamond unit to be 0.17MeV
  so that the electron-mass is approximately 3 mass-diamonds (0.51 MeV).
• The electron has initial 4-momentum OM.
• The given incident photon has 4-momentum MZ.
• The conservation of 4-momentum is OM+MZ = \stackrel{?}{ON}+\stackrel{?}{NZ}, where the state N is unknown.
• As @Dale says, we need more relations to locate N.
  • ON is a timelike 4-momentum for the electron...
    so N lies on the "electron mass-shell", the hyperbola of radius 3 mass-diamonds centered at O.
    (fancy: The causal diamonds OM and ON have area 3^2=9.)
  • NZ is a lightlike 4-momentum for the scattered photon.
    [so N lies on the hyperbola of radius 0 mass-diamonds centered at O].
    (fancy: The causal diamonds MZ and NZ have area 0.)
• So, trace a (past-directed) lightlike ray from state Z until one intersects the mass-shell at state N.
  (M and N are the intersections of the two hyperbolas from O and Z, of radii 3 and 0, respectively.
  MN is the 4-impulse (the increment in 4-momentum) gained by the electron.)
• Once you have located N, you can determine the 4-momenta of the final products.
  • ON has energy-component 5 mass-diamonds =(5*0.17MeV) and momentum-component 4 mass-diamonds =(4*0.17 MeV).
  • NZ has energy-component 1 mass-diamond =(0.17 MeV) and momentum-component -1 mass-diamond = -(0.17 MeV).
  • The velocities are gotten from the slopes of the 4-momenta.
  • The electron's final kinetic energy is the final energy-component minus the mass-energy = ((5-3)*0.17 MeV) [drawn in green].

 

[Sagittarius A-Star]

Question 1: If a particle is moving at relativistic speeds what would it's kinetic energy be?
I think it's $K.E. = \frac{1}{2} m_o v^2$ and my friend thinks it's $K.E. = \frac{1}{2} \frac{m_o v^2}{\sqrt{1-\frac{v^2}{c^2}}}$
Who is right? Is it relativistic mass or rest mass?

None. My impression is, that your friend wrongly thinks, switching to relativistic mass would make any formula from classical, non-relativistic mechanics valid in SR. That is a good example for the didactic disadvantage of the concept of relativistic mass.

 

[PeroK]

Why are you using relativistic mass? Is it in your course notes or textbook?

@Slimy0233 a straight answer to this question would be helpful.

 

[Slimy0233]

Ugh, that is a bad sign. It means the professor is decades out of date with their source material. I would solve this using the four-momentum as follows

The four momentum is $P=(E/c,p_x,p_y,p_z)$, for convenience I will use units where $c=1$ and $m$, $E$, and $p$ are all measured in $\mathrm{MeV}$. We can ignore $y$ and $z$ here, so I will just drop those. The four-momentum is conserved, so setting the four-momentum before the scattering equal to the four-momentum after the scattering we get $$P_{photon,before} + P_{electron,before}=P_{photon,after}+P_{electron,after}$$$$(0.51,0.51) + (0.51,0.00) = (E_{photon,after},p_{photon,after}) + (E_{electron,after},p_{electron,after})$$ This gives us two equations in four unknowns, so we use the general equation $m^2=E^2-p^2$ to get two more equations $$0=E^2_{photon,after} - p^2_{photon,after}$$$$0.51^2=E^2_{electron,after}-p^2_{electron,after}$$
Solve to get $$E_{photon,after}=0.17$$$$p_{photon,after}=-0.17$$$$E_{electron,after}=0.85$$$$p_{electron,after}=0.68$$ and we can easily get the velocity of the electron with $$v_{electron,after}=\frac{p_{electron,after}}{E_{electron,after}}=0.8$$ and the kinetic energy of the electron with $$KE_{electron,after}=E_{electron,after}-m_{electron} = 0.34$$

I really appreciate your help! I will try and use this way instead. I don't want to kick the can down the road just because my professor uses it. Also, this answer of yours helped a lot, thank you!

 

[Slimy0233]

@Slimy0233 a straight answer to this question would be helpful.

I am sorry for not having answered it, we don't really have a textbook per se. Professor probably doesn't use it when he is doing actual problem, but he does use it when he is teaching us, he probably thinks this is easier to grasp than the alternative.He did suggest a number of books as course material and he strictly mentioned to not waste your time completely reading them as we have a lot to study in a very short time. Instead he asked us to use them for reference. i.e., if we don't understand something we can refer that book for greater detail and depth.

I am pretty sure all the books he has mentioned don't use relativistic mass. But he does use it for the reasons mentioned before.

PS: I use Quantum Mechanics by Griffiths (hardly ever, but for reference) and Modern Physics by Krane if I need something cleared up.

 

[Slimy0233]

And as soon as you're out of the course, join the 21st century and drop the subscript zero.

Thank you!! I shall do that.

 

[Slimy0233]

Is there one? I thought he was studying on his own, although there has been some mention of a tutor.

Hello again sir, I am studying alone somewhat, I do have a tutor, but I take online classes and I don't consider him to be a reliable professor (he's good, but I can't run upto him and ask questions) he is not available most of the times.

 

[Sagittarius A-Star]

Just don't use it. It was a mis-step that has bred decades of confusion like you are experiencing. @levokun made a strong push in the 1990s to stamp it out, but it clings on in some quarters.

The following collection of letters contains a discussion of Lev Okun's article THE CONCEPT OF MASS with comments by W. Rindler and others, and replies and remarks to his own article by L. Okun:
https://faculty.washington.edu/seattle/physics544/Eismc2/1990 Rindler Physics Today.pdf

 

[vanhees71]

Well, although I believe that for sure Rindler understands the concepts well, he's wrong in claiming that "energy has mass-like properties". It's rather the opposite "mass has energy-like properties", because the mass term is one contribution to energy in special relativity not vice versa. It's in fact total energy that's a measure of inertia in relativity and not only mass. This even to a certain extent works as a heuristics in GR, where it is not mass that is the source of gravity but energy (or more precisely energy, momentum, and stress of matter and radiation).

I couldn't agree more to Vandyck and Morugesan.

 

[Vanadium 50]

The problem here is that the OP isn't trying to learn physics, but to score well enough in an exam so he can start graduate school and THEN learn some physics. Hopefully learning physics and doing well on the exam are correlated, but probably not 100%. In particular, if relativistic mass is on the test, he better learn it.

Of course this cuts both ways - if it's not on the test, he's not just wasting his time, he's actually making negative progress. While he can not afford.

 

[vanhees71]

I know, today it's more important to earn some "credit points" and have other idiosyncratic burocracy fulfilled than learning the actual physics. In Europe it's the "Bologna process", turning our universities to schools...

 

[Slimy0233]

The problem here is that the OP isn't trying to learn physics, but to score well enough in an exam so he can start graduate school and THEN learn some physics. Hopefully learning physics and doing well on the exam are correlated, but probably not 100%. In particular, if relativistic mass is on the test, he better learn it.

Of course this cuts both ways - if it's not on the test, he's not just wasting his time, he's actually making negative progress. While he can not afford.

You have very accurately described my situation.

 

[Dale]

if relativistic mass is on the test, he better learn it

Yes, unfortunately that is true.

 

[Herman Trivilino]

Professor probably doesn't use it when he is doing actual problem, but he does use it when he is teaching us, he probably thinks this is easier to grasp than the alternative.

He is probably using an older book that is no longer in print. See if you can find out what it is and buy a used one somewhere. This happened to me once and about half way through the semester we found an old copy of the textbook. It was like having a light come on, illuminating the course material!

 

[lightarrow]

Don't say that, then. Use $E=mc^2$ to refer to the rest energy of something and $E=\gamma mc^2$ for the total energy (which reduces to $mc^2$ when $v=0$ and hence $\gamma =1$). There is nothing expressed in terms of the relativistic mass that cannot be expressed in terms of the rest mass times the Lorentz gamma factor. And then it is always crystal clear what $m$ means - the rest mass.

The rest mass is also an invariant quantity, which makes it much easier to deal with when you use more sophisticated tools.

Lev Okun (https://en.wikipedia.org/wiki/Lev_Okun). Sadly he passed away a few years ago, but he posted here a few times.

In the spirit of Okun's good article, better to write
$E_0=mc^2$
to avoid confusion.

--
lightarrow