2
$\begingroup$

Let's say we have a pair of flywheels rotating in opposite directions and they're inside a black box so that maybe we don't know that's what's inside, and we accelerate the box in the direction of their axii. Let's say it's a special flywheel where the energy stored on it is a substantial portion of the energy in the black box, including E=m*c^2. To an external observer, their radii do not change because length contraction is along their axii, but their velocities get smaller due to time slowing up. The structure's non-rotating rest mass increases as expected, but the initial rotating energy on the flywheel does not increase by the same proportion because it's a function of velocity squared, which has reduced because time slowed up but meters in the direction of rotation did not get shorter. The initial rest mass of the black box included the fly wheel's energy, but that portion of the black box's mass-energy did not increase as much as we expect because the energy on the flywheel is a function of m*v^2. Is this reasoning correct? The mass will not increase via relativity as much as we expect, giving that it's a black box and didn't know a flywheel is inside?

To make the question clear: initial energy is (after the flywheels are given a velocity $v$ but before the entire box is given a velocity $w$ perpendicular to the two components of $v$):

$E = (m_o+m_1+m2)c^2 + \frac{1}{2}(m_1 + m_2)v^2$

$m_0 = $ mass of the box, not including the point masses $m_1$ and $m_2$ which are the flywheels, i.e. point masses on two strings going in opposite directions.

Kinetic energy comes from relativistic energy so this is the same as

$E = \left(m_o + \cfrac{m_1+m_2}{\sqrt{1-(v/c)^2}}\right)c^2 $

My question is asking why my reasoning applied to the first equation comes out different from the 2nd equation.

My reasoning applied to the first equation when adding a velocity $w$ perpendicular to $v$'s two vector components gives:

$E_w = \cfrac{(m_o+m_1+m2)}{\sqrt{1-(w/c)^2}}c^2 + \frac{1}{2}\cfrac{(m_1 + m_2)}{\sqrt{1-(w/c)^2}}\left(v*\sqrt{1-(w/c)^2}\right)^2$

But the correct answer should be

$E_w = \left(m_o + \cfrac{m_1+m_2}{\sqrt{1-(v/c)^2}}\right)c^2 * \cfrac{1}{\sqrt{1-(w/c)^2}} $

Doing some math, they would be the same only if the following were equal, which is not the case:

$\cfrac{1-(w/c)^2}{2}v^2 \ne \left(\cfrac{1}{\sqrt{1-(w/c)^2}} - 1\right)c^2$

The question could be simplified to a single point mass flywheel that has zero "excess" mass for its support structure. That's not realistic, but correct reasoning for the single mass can be extended to each point mass in a real system.


I believe my mistake is that I can't just jump from 0 to $w$ and expect the kinetic energy equation to incorporate relativity effects. The derivation that shows $\frac{1}{2}mv^2$ is from special relativity depends on an integral where the mass (more precisely momentum as Dale mentions) increases as velocity increases, in each instant. Integrating each point mass on the flywheel as the box velocity increases from 0 to $w$ in the x direction (parallel to the flywheel axis) is not merely a $w$ velocity, but also involves the flywheel's velocity component.

$\endgroup$
  • 1
    $\begingroup$ Relativistic energy is not a function of the velocity squared. $\endgroup$ – safesphere Sep 2 '18 at 15:35
  • 1
    $\begingroup$ You should not add your answer the the question. Add it as a separate answer and accept it. Additionally, do not make your post contain edit tables. See this meta post. Edit histories can be viewed easily by those interested, and are hence superfluous. $\endgroup$ – user191954 Sep 3 '18 at 13:08
  • $\begingroup$ As @safesphere says, KE is not equal to $1/2\;mv^2$. So your first equation is just wrong and will necessarily not produce the same answer as any correct method. In relativity the KE is $m/\sqrt{1-v^2}-m$ using units where c=1. Your second equation is correct, the first is not $\endgroup$ – Dale Sep 3 '18 at 19:44
2
$\begingroup$

Relativistic mass is a deprecated concept by professional physicists: http://www.stat.physik.uni-potsdam.de/~pikovsky/teaching/stud_seminar/einstein_okun.pdf However, your question can easily be asked and answered in terms of total energy instead.

So for simplicity and clarity, let’s use units where c=1, and consider the flywheel’s axis along x and examine a single point mass (rest mass m) on the flywheel with a velocity v in the y direction in the rest frame of the box. Now this mass has a total energy $E_0=m/\sqrt{1-v^2}$ in the box frame.

Now, let’s consider a frame where the box is moving with velocity u. In this frame the total energy of the mass is $E_u=m/(\sqrt{1-v^2}\sqrt{1-u^2})$. So it gains as much total energy as expected in relativity.

How is this possible given your reasoning above? Note that you suggested that the velocity would decrease due to time dilation. But don’t forget that velocity is a vector. If you compute the velocity in the moving frame you get $v_u=(u,v\sqrt{1-u^2},0)$. So the y component does indeed decrease as you mentioned, but the x component increases. The total energy thus increases by exactly the usual amount.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.