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

[Slimy0233]

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?

Also, if an electron is moving with a velocity v, would it's total energy be
A fellow student said it's 1. $$E_{total} = mc^2 + \frac{1}{2} \frac{m_o v^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

Now, I think it's either
2. $E_{total} = m_o c^2 + \frac{1}{2} m_o v^2$ or 3. $E_{total} = mc^2$

Who is right?

 

[Ibix]

Who is right?

Once I've sorted through your notation, 3 is correct. But never use relativistic mass for anything - no modern source will use it because it just leads to confusion between different meanings of the word mass. Just use "mass" and mean the rest mass.

Using your $m_0$ for mass (by which I mean rest mass), total energy is $\gamma m_0c^2$, of which the rest energy is $m_0c^2$, leaving kinetic energy $(\gamma-1)m_0c^2$.

If you Taylor expand the $\gamma$ in the last expression you will find that the expression approximates $\frac 12mv^2$ when $v\ll c$.

 

[vanhees71]

There is no relativistic mass. It was a misconception in the very early writings on SRT due to a confusion about the correct description of point-particle mechanics, which was resolved by Planck already in 1906. Why still people insist in using this flawed concept, I don't know.

The correct definitions of energy and momentum of a massive particle with invariant mass $m$ are [EDIT: typo corrected in view of #17]
$$p^{\mu} = m \frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu}=m c u^{\mu}$$
with the propertime $\mathrm{d} \tau =\sqrt{1-\vec{\beta}^2} \mathrm{d} t$, leading to $u_{\mu} u^{\mu}=1$. Here $\vec{\beta}=(1/c) \mathrm{d} \vec{x}/\mathrm{d} t$. Thus you have the energy-momentum relation in covariant form
$$p_{\mu} p^{\mu}=m^2 c^2.$$
With $p^0=E/c$
$$E=c \sqrt{m^2 c^2 + \vec{p}^2}=\frac{m c^2}{\sqrt{1-\beta^2}}$$
and
$$\vec{p}=\frac{m}{\sqrt{1-\vec{\beta}^2}} \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
The advantage using this definition is that energy and momentum form the components of a four-vector. The energy includes the "rest energy", $$E_0=m c^2$$. The kinetic energy thus is
$$E_{\text{kin}}=E-E_0=m c^2 \left [\sqrt{1+\frac{\vec{p}^2}{(m^2 c^2)}}-1 \right].$$
The Newtonian limit is given for $|\vec{\beta}| = |c \vec{p}|/E<|\vec{p}|/(mc) \ll 1$. Then you can expand the square root in the previous formula
$$E_{\text{kin}}=mc^2 \left [\frac{\vec{p}^2}{2 m^2 c^2} + \mathcal{O}(\vec{p^2}^2/(m c)^4 \right ] \simeq \frac{1}{2m} \vec{p}^2.$$
At the same order you get
$$\vec{p} \simeq m \vec{v}=m c \vec{\beta}.$$

 

[Slimy0233]

Once I've sorted through your notation, 3 is correct. But never use relativistic mass for anything - no modern source will use it because it just leads to confusion between different meanings of the word mass. Just use "mass" and mean the rest mass.

Using your $m_0$ for mass (by which I mean rest mass), total energy is $\gamma m_0c^2$, of which the rest energy is $m_0c^2$, leaving kinetic energy $(\gamma-1)m_0c^2$.

If you Taylor expand the $\gamma$ in the last expression you will find that the expression approximates $\frac 12mv^2$ when $v\ll c$.

thank you very much! But, don't you think 2 would be right too? Where the rest energy is $m_0c^2$ and the added Kinetic energy would be $\frac{1}{2}mv^2$? I am sorry, I didn't understand why that was wrong.

But never use relativistic mass for anything - no modern source will use it because it just leads to confusion between different meanings of the word mass. Just use "mass" and mean the rest mass.

Please explain this in greater detail, I don't know what you mean here.

 

[Slimy0233]

@vanhees71 I must thank you greatly for the answer you have provided, but unfortunately, my math knowledge is not sufficient to understand it. I am bookmarking it and I will be able to read it within an year!

 

[PeroK]

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?

Also, if an electron is moving with a velocity v, would it's total energy be
A fellow student said it's 1. $$E_{total} = mc^2 + \frac{1}{2} \frac{m_o v^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

Now, I think it's either
2. $E_{total} = m_o c^2 + \frac{1}{2} m_o v^2$ or 3. $E_{total} = mc^2$

Who is right?

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

 

[vanhees71]

@vanhees71 I must thank you greatly for the answer you have provided, but unfortunately, my math knowledge is not sufficient to understand it. I am bookmarking it and I will be able to read it within an year!

What didn't you understand? It's all just very simple calculus (derivatives) and algebra.

 

[Slimy0233]

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

I am somewhat confused where to use relativistic mass and where not to.
1. Is mass in Kinetic energy equation, rest mass or relativistic mass?
2. I know the mass in $E = mc^2$ is relativistic mass

But yes, the rest were just to distinguish between rest mass and relativistic mass (or to better understand them).

 

[Slimy0233]

What didn't you understand? It's all just very simple calculus (derivatives) and algebra.

**cries in poor education**

Unfortunately sir, these equations look unfamiliar and when I was asked to derive $E = mc^2$ equation, I remember doing it in a very simple way (it still took two pages), but yes, not many of those equations are similar. You have done enough for me, It won't be wise or fair to ask you to explain those as the chances of me understanding them remains rather low the time being.

 

[Ibix]

Where the rest energy is $m_0c^2$ and the added Kinetic energy would be $\frac{1}{2}mv^2$?

That's the Newtonian formula for kinetic energy. It's only approximately correct, and the approximation gets worse as $v$ approaches $c$. The Taylor expansion I mentioned in my last post will show you that $\gamma mc^2=mc^2+\frac 12 mv^2+\ldots$, where the terms I haven't written are all small when $v\ll c$ but become large as you get to large fractions of $c$.

 

[Slimy0233]

That's the Newtonian formula for kinetic energy. It's only approximately correct, and the approximation gets worse as $v$ approaches $c$. The Taylor expansion I mentioned in my last post will show you that $\gamma mc^2=mc^2+\frac 12 mv^2+\ldots$, where the terms I haven't written are all small when $v\ll c$ but become large as you get to large fractions of $c$.

thank you very much!

 

[Ibix]

I am somewhat confused where to use relativistic mass and where not to.

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.

 

[Slimy0233]

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.

I am really sorry, but how can I not? I mean, when I say E = mc^2, I need relativistic mass, m here is relativistic mass which is different than rest mass, how can I not use relativistic mass?

[copied from a previous comment]
I am somewhat confused where to use relativistic mass and where not to.
1. Is mass in Kinetic energy equation, rest mass or relativistic mass?
2. I know the mass in $E = mc^2$ is relativistic mass

But yes, the rest were just to distinguish between rest mass and relativistic mass (or to better understand them).

I will check @levokun s profile, but meanwhile, who is he? also, You have been very helpful today, thank you! again!

 

[PeroK]

I am somewhat confused where to use relativistic mass and where not to.
1. Is mass in Kinetic energy equation, rest mass or relativistic mass?
2. I know the mass in $E = mc^2$ is relativistic mass

But yes, the rest were just to distinguish between rest mass and relativistic mass (or to better understand them).

Best to forget all about relativistic mass.

 

[Ibix]

I am really sorry, but how can I not? I mean, when I say E = mc^2, I need relativistic mass, m here is relativistic mass which is different than rest mass, how can I not use relativistic mass?

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.

I will check @levokun s profile, but meanwhile, who is he?

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

 

[vanhees71]

**cries in poor education**

Unfortunately sir, these equations look unfamiliar and when I was asked to derive $E = mc^2$ equation, I remember doing it in a very simple way (it still took two pages), but yes, not many of those equations are similar. You have done enough for me, It won't be wise or fair to ask you to explain those as the chances of me understanding them remains rather low the time being.

Well, I guess, you can understand it pretty easily, when I explain what you didn't understand. For that I'd have to know, what this is. Maybe you can point to some book/paper, where the approach is given, you used to derive $E=m c^2$.

The problem leading to many misunderstandings is the introduction of the "relativistic mass". There is no such thing in the modern understanding of relativity. Einstein clearly adviced not to use "relativistic mass" anymore but only "invariant mass" or "rest mass" and call that and only that quantity mass.

 

[Ibix]

Well, I guess, you can understand it pretty easily, when I explain what you didn't understand. For that I'd have to know, what this is. Maybe you can point to some book/paper, where the approach is given, you used to derive $E=m c^2$.

You have a typo in the first equation, ($dd\tau$ instead of $\frac{d}{d\tau}$) which may be confusing the OP.

 

[Nugatory]

I am somewhat confused where to use relativistic mass and where not to.

That's easy - never use relativistic mass

 

[PeroK]

That's easy - never use relativistic mass

It seems like we have a consensus on this!

 

[hutchphd]

Let me plug my stupid iidea: I think we should similarly define the "length" of an object as strictly rest length. Much confusion is engendered by allowing the length to be loosely defined when simultaneity is not well defined.

 

[PeroK]

Let me plug my stupid iidea: I think we should similarly define the "length" of an object as strictly rest length. Much confusion is engendered by allowing the length to be loosely defined when simultaneity is not well defined.

That's not a bad idea, but you would have to be rigorous in terms of solving specific problems. For example, in the rest frame of a muon travelling towards Earth, you could say nothing about the distance through the Earth's atmosphere. Having an undefined depth of the atmosphere is not necessarily helpful.

 

[vanhees71]

You have a typo in the first equation, ($dd\tau$ instead of $\frac{d}{d\tau}$) which may be confusing the OP.

Thanks for pointing this out. I've corrected it.

Maybe it helps to also define the space-time four vector
$$(x^{\mu})=\begin{pmatrix} c t \\ \vec{x} \end{pmatrix},$$
and it's derivative with respect to proper time,
$$\frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu}= \frac{\mathrm{d} t}{\mathrm{d} \tau} \frac{\mathrm{d}}{\mathrm{d} t} x^{\mu} = \left (\frac{\mathrm{d} \tau}{\mathrm{d} t} \right)^{-1} \frac{\mathrm{d}}{\mathrm{d} t} x^{\mu}.$$
From the definition of proper time, $\tau$ you get
$$\frac{\mathrm{d} \tau}{\mathrm{d} t} = \sqrt{1-\beta^2}, \quad \beta=\frac{1}{c} |\vec{v}|, \quad \vec{v}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}.$$
Then you get
$$\frac{\mathrm{d}}{ \mathrm{d} \tau} x^{\mu} = \frac{1}{\sqrt{1-\beta^2}} \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.$$
Now one can show that $\tau$ is the time measured by a clock carried on by the particle, i.e., $\mathrm{d} \tau$ is the time increment measured in the momentaneous rest frames of the particle. As such it is independent of the choice of the inertial reference frame, since
$$\mathrm{d} \tau=\frac{1}{c} \sqrt{c^2 \mathrm{d} t^2-\mathrm{d} \vec{x}^2},$$
is invariant under Lorentz transformations.

That's why it's plausible to define the momentum as
$$\vec{p}=m \frac{\mathrm{d}}{\mathrm{d} \tau} \vec{x}=\frac{m}{\sqrt{1-\beta^2}} \vec{v}.$$
Here, obviously $m$ is just the same quantity, called mass, as in Newtonian physics, and it doesn't depend on the motion of the particle, and it is invariant under Lorentz transformations. That's why it's called "invariant mass" and that's why one exclusively works ith this notion of mass when dealing with relativistic particles. Unfortunately there are many highschool textbooks, which still introduce the "relativistic mass", which is very unfortunate, because it introduces long outdated confusing concepts, which should be avoided, because unfortunately we remember best the things about a a subject, we learn first about it.

Thus you get a four-vector for momentum by just defining
$$p^{\mu} = m \frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu} = \frac{m}{\sqrt{1-\beta^2}} \begin{pmatrix} c \\ \vec{v} \end{pmatrix}.$$
To interpret the meaning of the time component, you can just calculate the non-relativistic limit for $|\beta|\ll 1$:
$$p^0=m c (1-\beta^2)^{-1/2}=mc \left (1+\frac{\beta^2}{2} + \mathcal{O}(\beta^4) \right) \simeq m c + \frac{m \vec{v}^2}{2 c},$$
i.e.,
$$E=c p^0 \simeq m c^2 + \frac{m}{2} \vec{v}^2.$$
This shows that it makes sense to interpret $p^0=E/c$, where $E$ is the energy, including the "rest energy" $E_0=m c^2$, i.e., you have
$$E_{\text{kin}}=E-E_0 = m c^2 \left (\frac{1}{\sqrt{1-\beta^2}}-1 \right),$$
for the exact relativistic kinetic energy of a particle, and you get also the usual non-relativistic limit
$$E_{\text{kin}} \simeq \frac{m}{2} \vec{v}^2,$$
but it's indeed only an approximation for particles moving with a speed much smaller than the speed of light (in vacuum).

 

[Dale]

I am somewhat confused where to use relativistic mass

Nowhere. There is nowhere that you should use relativistic mass.

where not to.

Everywhere. You should avoid using relativistic mass everywhere.

I mean, when I say E = mc^2, I need relativistic mass, m here is relativistic mass which is different than rest mass, how can I not use relativistic mass?

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.

 

[Vanadium 50]

I am somewhat confused where to use relativistic mass

That's easy. Never.

 

[Slimy0233]

@Dale Thank you very much for this!

 

[Herman Trivilino]

I think it's K.E.=12mov2 and my friend thinks it's K.E.=12mov21−v2c2

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 $(\gamma-1)mc^2$.

 

[PeterDonis]

I am somewhat confused where to use relativistic mass and where not to.

The best option is to never use it.

 

[PeterDonis]

2. I know the mass in $E = mc^2$ is relativistic mass

Not in modern notation. In modern notation, the full energy-momentum relation is $E^2 = p^2 c^2 + m^2 c^4$, where $m$ is the rest mass. $E = mc^2$ is only true for the special case $p = 0$, i.e., when the object is at rest. This is much better than trying to use relativistic mass.

 

[Herman Trivilino]

I know the mass in E=mc2 is relativistic mass

The correct equation is $E_o=mc^2$ where $E_o$ is the so-called rest energy.

 

[jtbell]

In modern notation, the full energy-momentum relation is $E^2 = p^2 c^2 + m^2 c^4$, where $m$ is the rest mass.

As a minor point, I prefer to write $E^2 = (pc)^2 + (mc^2)^2$, to reinforce that $E$, $pc$ and $mc^2$ all have the same units, namely units of energy.

When I was a graduate student in experimental elementary particle physics, long ago c. 1980, it was the universal practice among us to refer to mass using energy units (meaning $mc^2$) and likewise momentum (meaning $pc$), usually in MeV or GeV. (Our experiments didn't get to the TeV level, because they came before Fermilab upgraded to the Tevatron.)

In fact, it was and still is customary in HEP circles to use units with $c = 1$ so that numerically $E^2 = p^2 + m^2$.

 

[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.