Richard Feynman: Deriving Lorentz Mass Transformation

[43arcsec]

TL;DR Summary

Cool derivation of Lorentz transformation for mass by Feynman seems to start with assuming mass is constant, thereby invalidating the derivation.

In section 3.8, Feynman does a derivation of the Lorentz transformation for mass starting from

$$\frac{d}{dt}E=F \cdot v \hspace{1cm}(1) $$

But is this a valid starting point if you are going to show mass changes with velocity?
He says (1) comes from chapter 13 of his Lectures which he derives by differentiating the Newtonian formula for kinetic energy

$$ \frac{d}{dt} \frac{1}{2} mv^2 = mav = Fv \hspace{1cm} (2)$$

But this assumes mass is constant. Feynman never misses, so I know I am missing some type of explanation as to why it's valid to start with (1) to prove

$$ m= \frac{m_0}{ \sqrt{1-\frac{v^2}{c^2} } } \hspace{1cm}(3)$$

 

[PeroK]

I think relativistic mass was something of a blind spot for the great man! It's not a useful concept - especially when you venture beyond elemntary SR:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[physicsworks]

In the words of Lev Okun

... even great scientists such as Landau and Feynman, when addressing nonscientists, have sometimes—though not always—used the equation $E = mc^2$. (Compare, for instance, The Feynman Lectures on Physics and Feynman's last published lecture "Thre Reason for Antiparticles".)

See his pedagogical articles in Physics Today, The Concept of Mass, and on the arxiv The Concept of Mass in the Einstein Year.

 

[Sagittarius A-Star]

I check the boards, and I just can't see what is wrong with this. The first instance of $$ latex-stuff $$ renders, then the 2nd and 3rd instances of $$ latex-stuff $$ doesn't. Very embarrassed, sorry.[/FONT]

The 2nd instance contains a wrong type of closing bracket:

\frac{d}{dt)

The 3rd instance misses a backslash before the sqrt and a closing bracket after the sqrt-function.

sqrt{1- \frac{v^2}{c^2} }

See also the chapter "Delimiting your LaTeX code" in the LaTeX guide linked on the left botton side below the input field.

 

[43arcsec]

The 2nd instance contains a wrong type of closing bracket:The 3rd instance misses a backslash before the sqrt and a closing bracket after the sqrt-function.See also the chapter "Delimiting your LaTeX code" in the LaTeX guide linked on the left botton side below the input field.

Thank you!

 

[Dale]

the Lorentz transformation for mass

Currently mass is considered to be an invariant scalar. So it doesn’t transform.

 

[Sagittarius A-Star]

In section 3.8, Feynman does a derivation of the Lorentz transformation for mass starting from
$$\frac{d}{dt}E=F \cdot v \hspace{1cm}(1) $$
But is this a valid starting point if you are going to show mass changes with velocity?
He says (1) comes from chapter 13 of his Lectures which he derives by differentiating the Newtonian formula for kinetic energy

This formula can be derived also from chapter 4 "Conservation of Energy" of his Lectures, that means without the Newtonian formula for kinetic energy:

The general principle is that the change in the energy is the force times the distance that the force is pushed, and that this is a change in energy in general:
$$ \begin{pmatrix}
\text {change in} \\
\text {energy}
\end{pmatrix} = \begin{pmatrix}
\text {force}
\end{pmatrix} \times \begin{pmatrix}
\text {distance force}\\
\text {acts through}
\end{pmatrix} \hspace{2 cm}(4.4)$$

Source:
https://www.feynmanlectures.caltech.edu/I_04.html

This can be written as:
$$ dE = \vec{F} \cdot d\vec{s} $$
$\Rightarrow$
$$ \frac{dE}{dt} = \vec{F} \cdot \frac{d\vec{s}}{dt} = \vec{F} \cdot \vec{v} $$

 

[Sagittarius A-Star]

... prove
$$ m= \frac{m_0}{ \sqrt{1-\frac{v^2}{c^2} } } \hspace{1cm}(3)$$

As others have mentioned, the lectures of Feynman are somewhat old and most of today's physicists don't use "relativistic mass". Actual standard is to use $m$ for the invariant mass, which is for massive particles equal to what Feynman called $m_0$. Instead of Feynman's relativistic mass, which he called $m$, today $E/c^2$ is used. That means, that he derived with a clever mathematical trick after his equation (15.14), in today's notation:
$$ E= \frac{E_0}{ \sqrt{1-\frac{v^2}{c^2} } } = \frac{mc^2}{ \sqrt{1-\frac{v^2}{c^2} } }$$
His starting point is $dE = \vec{F} \cdot d\vec{s}$ and $d\vec{p} = \vec{F} dt$ and then assuming, in today's notation: $\vec{p} = \frac{E}{c^2}\vec{v}$.

In SR, momentum and energy are combined to the four-momentum
$\mathbf P = \frac{E}{c^2} \frac {d}{dt} (ct, x, y, z) = (E/c, p_x, p_y, p_z)$

In the four-velocity, the $\gamma$-factor is already included:
https://en.wikipedia.org/wiki/Four-vector#Four-momentum

 

[43arcsec]

As others have mentioned, the lectures of Feynman are somewhat old and most of today's physicists don't use "relativistic mass". Actual standard is to use $m$ for the invariant mass, which is for massive particles equal to what Feynman called $m_0$. Instead of Feynman's relativistic mass, which he called $m$, today $E/c^2$ is used. That means, that he derived with a clever mathematical trick after his equation (15.14), in today's notation:
$$ E= \frac{E_0}{ \sqrt{1-\frac{v^2}{c^2} } } = \frac{mc^2}{ \sqrt{1-\frac{v^2}{c^2} } }$$
His starting point is $dE = \vec{F} \cdot d\vec{s}$ and $d\vec{p} = \vec{F} dt$ and then assuming, in today's notation: $\vec{p} = \frac{E}{c^2}\vec{v}$.

In SR, momentum and energy are combined to the four-momentum
$\mathbf P = \frac{E}{c^2} \frac {d}{dt} (ct, x, y, z) = (E/c, p_x, p_y, p_z)$

In the four-velocity, the $\gamma$-factor is already included:
https://en.wikipedia.org/wiki/Four-vector#Four-momentum

Very nice, thanks.

 

[vanhees71]

As others have mentioned, the lectures of Feynman are somewhat old and most of today's physicists don't use "relativistic mass". Actual standard is to use $m$ for the invariant mass, which is for massive particles equal to what Feynman called $m_0$. Instead of Feynman's relativistic mass, which he called $m$, today $E/c^2$ is used. That means, that he derived with a clever mathematical trick after his equation (15.14), in today's notation:
$$ E= \frac{E_0}{ \sqrt{1-\frac{v^2}{c^2} } } = \frac{mc^2}{ \sqrt{1-\frac{v^2}{c^2} } }$$
His starting point is $dE = \vec{F} \cdot d\vec{s}$ and $d\vec{p} = \vec{F} dt$ and then assuming, in today's notation: $\vec{p} = \frac{E}{c^2}\vec{v}$.

In SR, momentum and energy are combined to the four-momentum
$\mathbf P = \frac{E}{c^2} \frac {d}{dt} (ct, x, y, z) = (E/c, p_x, p_y, p_z)$

In the four-velocity, the $\gamma$-factor is already included:
https://en.wikipedia.org/wiki/Four-vector#Four-momentum

I'm also very puzzled by the fact that Feynman is so old-fashioned concerning the avoidance of the covariant formalism. For me it's also not to explain with any didactic argument, because this not manifestly covariant mish-mash between relativity and some kind of Newtonian ideas makes relativistic physics more complicated than the manifestly covariant formalism.

In fact relativistic point-particle mechanics (which can be formulated easily only for the motion of a particle in some external field; so let's concentrate on this case) can be heuristically understood by thinking about the non-relativistic laws as being valid in the momentaneous rest frame of the particle, but it's of course too cumbersome to use a non-relativistic equation of motion and always transform to the momentaneous restframe when integrating them, and thus it's more convenient to simply derive manifestly covariant equations of motion.

First of all this heuristics makes it clear that a "natural choice" for the parameter describing the particle's trajectory is not the coordinate time of the chosen inertial frame of reference but the proper time, because in the momentaneous (inertial!) rest frame of the particle by definition the increment of the coordinate time $\mathrm{d} t^*$ can be expressed in the fixed inertial reference frame ("lab frame") by $\mathrm{d} \tau =\mathrm{d} t \sqrt{\eta_{\mu \nu} \dot{x}^{\mu}} \dot{x}^{\nu}$ (where the dot here means the derivative wrt. the lab frame, $t$).

In Newtonian mechanics you have some equation of motion of the type
$$m \ddot{x} = \vec{F},$$
where $\vec{F}$ is more or less arbitrary, i.e., it has to be determined empirically (that's of course also the case in relativity, but as we'll see in a moment the possible forms of the force are somewhat more restricted).

So this equation should hold in the momentaneous rest frame, but it's very hard to guess from it the form in the lab frame, but no it's clear that the derivative wrt. the coordinate time $t^*$ in the of the momentaneous rest frame we can right away use the invariant proper time, and then it makes sense not only to look at the spatial components $\vec{x}^*$ but on all four components $x^{* \mu}=(t^*,\vec{x}^*)$ (using natural units with $c=1$), and we need time derivatives, i.e., we have for the components in the momentaneous rest frame
$$u^{* \mu} = \mathrm{d}_{t^*} x^{* \mu}=\mathrm{d}_{\tau} x^{* \mu} = (1,0,0,0).$$
Since $\tau$ is an invariant, these are components of a Minkowski four-vector, and in the lab frame its components simply are
$$u^{\mu} = \mathrm{d}_{\tau} x^{\mu},$$
and it fulfills the constraint
$$u_{\mu} u^{\mu}=\eta_{\mu \nu} u^{\mu} u^{\nu} =1. \qquad (*)$$
Now we look at the 2nd derivative. In the momentaneous rest frame again $t^*=\tau$, i.e., we can write
$$\mathrm{d_{t^*}}^2 x^{* mu} = \mathrm{d}_{\tau}^2 x^{* \mu} = \mathrm{d}_{\tau} u^{* \mu}.$$
Now it's plausible that $m$ as a parameter that characterizes the point particle is just the same parameter in relativistic physics as it is in Newtonian physics, and thus it should be an invariant quantity. Thus, it's again plausible to make the ansatz
$$m \mathrm{d}_{\tau} u^{\mu} = F^{\mu}$$
in the lab frame for the equation of motion since for the three spatial components of this equation in the momentaneous rest frame you get the Newtonian equation of motion.

Now because of (*) the time component is also fixed, because you must have
$$m u_{\mu} \mathrm{d}_{\tau} u^{\mu} =0 = F^{\mu} u_{\mu}=0. \qquad (**)$$
So in the momentaneous rest frame the EoM reads
$$m \mathrm{d}_{\tau} u^{* \mu} = \begin{pmatrix}0 \\ \vec{F}^* \end{pmatrix},$$
and from this you get the equation of motion in the general frame by a Lorentz boost from this momentaneous rest frame, but it's much easier to simply guess some forces which fulfill the constraint and see which equations of motion come out.

The most simple example is to choose
$$F^{\mu} = q F^{\mu \nu} u_{\nu}$$
with an antisymmetric tensor $F^{\mu \nu}$, because then (**) is trivially fulfilled. If you then split this in time and space components you see that what you get is the equation of motion for a charged particle (charge $q$) in an electromagnetic field (provided you assume that $F^{\mu \nu}$ is a function of the space-time variables $x^{\mu}$ only).

It also becomes clear that one has only three independent equations of motion as in Newtonian physics because of the constraint (**) which must be fulfilled for a force to make these equations consistent, i.e., if you have a solution for the three spatial components the temporal component of the covariant equation is automatically fulfilled. It takes the form of the work-energy theorem.