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Suppose two objects collide and combine into a single object, will the total relativistic momentum and relativistic mass stay the same? I read in my grade 12 physics text book that relativistic momentum can be defined as $|\text{relativistic momentum}| = \frac{\text{rest mass} \,\times\, v}{\sqrt{1 - \frac{v^2}{c^2}}}$ and relativistic mass can be defined as $\text{relativistic mass} = \frac{\text{rest mass}}{\sqrt{1 - \frac{v^2}{c^2}}}$. The total rest mass on the other hand doesn't necessarily have to stay the same. For example, the collision could heat up the combined system and the faster vibrating atoms could have stronger relativistic effects giving the combined system more inertia at the macroscopic level so we define it to have a higher rest mass.

Let's assume that any any frame of reference, two non-spinning object of any positive rest mass any subluminal velocity if they combine at such an angle that the combined system doesn't spin either, will always combine into an object with the same rest mass and velocity. Furthermore, let's assume that relativistic momentum and relativistic mass are a function of rest mass and subluminal velocity such that:

  • For any subluminal velocity, relativistic momentum and relativistic mass vary linearly with rest mass.
  • Relativistic momentum is in the direction of the velocity for nonzero subluminal velocity.
  • Relativistic mass and magnitude of relativistic momentum don't vary between any two velocities with the same magnitude.
  • The relativistic mass of an object at zero velocity is its rest mass and for any rest mass, as its velocity approaches zero, its relativistic momentum divided by its velocity approaches its rest mass.
  • When ever two non-rotating objects combine into a single non-rotating object, the total relativistic mass and relativistic momentum are conserved.

After I did a lot of work, I figured out a mathematical proof in my head that in more than one dimension, the only solution to these criteria is:

  • $\text{relativistic mass} = \frac{\text{rest mass}}{\sqrt{1 - \frac{v^2}{c^2}}}$ where $v$ is the speed of the object, not the velocity and $c$ is the speed of light.
  • the relativistic momentum at zero velocity is zero and for nonzero subluminal velocity, the relativistic momentum is in the direction of the velocity and $|\text{relativistic momentum}| = \frac{\text{rest mass} \,\times\, v}{\sqrt{1 - \frac{v^2}{c^2}}}$
  • For 2 nonrotating objects of any positive rest mass and subluminal velocity that combine into a single nonratating object, the rest mass and velocity of the system can be determined as follows: compute the relativistic mass and relativistic momentum of each object then add them to get the relativistic mass and relativistic momentum of the combined system then compute the rest mass and velocity of the combined system from its relativistic mass and relativistic momentum using the formulae for relativistic mass and relativistic momentum. Solving these equations, we get $\text{rest mass} = \sqrt{\text{relativistic mass}^2 - \frac{|\text{relativistic momentum}|}{c}^2}$ and $\text{velocity} = \frac{\text{relativistic momentum}}{\text{relativistic mass}}$ for nonzero relativistic momentum and $\text{rest mass} = \text{relativistic mass}$ for zero relativistic momentum.

How to we know the result I just proved as actually correct? I didn't prove that the original criteria I deduced this result from are correct. That problem can be solved by deciding that by definition, special relativity defines relativistic mass and relativistic momentum to follow the function they were shown to follow in the third last and second last bullet point and predicts that when ever two nonrotating objects combine into a single nonrotating object, the rest mass and velocity of the combined system actually is determined the way it is according to the last bullet point. Also using math, we can show that this prediction according to the theory is in fact a solution to the original criteria and it's actually easier to show that it is a solution than that only it can be a solution. I might not show a complete mathematical proof that it is a solution because it might be too confusing but I actually did figure out a mathemetical proof myself in my head that it is a solution also.

It's easy to show that in special relativity, subluminal velocities can be represented by points on a hyperbolic plane. Now if you have a Minkowski space and for an object of a given rest mass and any subluminal velocity, you plot a point in that Minkowski space where the time coordinate represents its relativistic mass and the spacial coordinates represent its relativistic momentum, you get a function from subluminal velocities to a plane in that Minkowski space that all have the same distance from the origin and therefore has hyperbolic geometry and each velocity exactly corresponds to the point in that hyperbolic plane that it represents as described earlier. From this, it's easy to show that in any frame of reference, two nonrotating objects of a given rest mass and velocity that combine into a nonrotating object will combine into an object of the same rest mass and velocity as they would in a stationary frame of reference.

Now in the real world, is it the case for any one specific material, the hotter it is, the more rest mass it has per atom and for a nonvibrating spinning object, its rest mass is defined as the triple integral of the relativistic density of each part in the frame of reference where the object has no overall movement and the relativistic density of each part is defined as $\text{relativistic density} = \frac{\text{rest density}}{\sqrt{1 - \frac{v^2}{c^2}}}$ and its relativistic mass in any frame of reference is defined the same way and it also turns out that even for a spinning object, $\text{relativistic mass} = \frac{\text{rest mass}}{\sqrt{1 - \frac{v^2}{c^2}}}$? Is it also the case that when ever two objects of a given rest mass and velocity if they combine whether or not they're spinning before or after the collision will always combine into an object of the rest mass and velocity predicted earlier regardless of the source of how much of its rest mass comes from thermal energy and how much of it comes from spinning kinetic energy? Maybe a quantum theory that doesn't include gravity can predict whether or not that's the case.

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    $\begingroup$ Markup wise, you should use \text for typesetting plain text statements in mathematical expressions (but you'll notice that it is not visually distinguished from non-math text and so doesn't work well in-line), and note that the \sqrt takes an argument surrounded by curly braces {} to tell the program how big/long to make the expression. $\endgroup$ – dmckee Jun 12 '18 at 19:28
  • $\begingroup$ @dmckee I didn't know how to make the square root sign longer so I made it so that brackets would appear visually in the question to indicate which part of the expression the square root operation was intended to operate on. Now I know how to make the square root sign itself go over the part of the expression I want it to go over. $\endgroup$ – Timothy Jun 12 '18 at 19:34
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    $\begingroup$ Now, physics-wise I'll note that these kinds of questions are a lot easier in the modern nomenclature where there is only one mass and we don't give $\gamma m$ a name. See physics.stackexchange.com/q/133376 (and many other questions on the site) for some discussion on the matter $\endgroup$ – dmckee Jun 12 '18 at 19:36
  • $\begingroup$ @dmckee Don't we need to use the concept of rest mass to prove conservation of energy. This question proves conservation of rest mass and if we insist that extra relativistic mass and Newtonian kinetic energy are the same thing, we can derive $E = mc^2$. It turns out that an electron and positron also annihilate into photons of energy $mc^2$ even though that derivation doesn't prove they do. I'm guessing that as a result of all that discovery, we now define relativistic energy as $\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$ $\endgroup$ – Timothy Jun 12 '18 at 19:54
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    $\begingroup$ "Don't we need to use the concept of rest mass to prove conservation of energy." Absolutely not. Despite it's prominence in pop-sci treatments relativistic mass is a both unnecessary and an invitation to erroneous thinking. Understand that there are always several ways to talk about physical theories that have the same underlying math, and they can all be internally consistent, but you can get into trouble when mixing the nomenclatures unless you are very careful. In modern nomenclature $E = \sqrt{(mc^2)^2 + (\vec{p}c)^2} = \gamma m c^2$ (here $m$ is the invariant mass AKA the rest mass). $\endgroup$ – dmckee Jun 12 '18 at 20:07
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"Suppose two objects collide and combine into a single object, will the total relativistic momentum and relativistic mass stay the same?"

The answer is "yes", or rather their sums over the system of bodies will stay the same, but I would counsel you to stop using the term relativistic mass. It's going out of use for a number of good reasons that I won't bore you with now.

$m \gamma c^2$ in which m is the body's mass (formerly called 'rest mass') represents the sum of the body's internal energy, $mc^2$ and its kinetic energy ($\gamma m - m)c^2$. So for a closed system its sum over the bodies of the system is conserved in collisions, elastic or inelastic. Think of $\gamma m$ as the body's total energy, expressed in mass units.

The beauty of it is that a body's total energy (divided by the mere constant, c) and the three components of its momentum $(m \gamma u_x, m \gamma u_y, m \gamma u_z)$ make up a 4-component vector (or 4-vector): $(m \gamma c, m \gamma u_x, m \gamma u_y, m \gamma u_z)$. So for a closed system, despite collisions elastic or inelastic, the vector sum of these vectors is conserved, that is the sums of each component separately. One conserved 4-vector deals with conservation of energy and conservation of momentum.

Note also that the modulus of the 4-vector, defined as $\sqrt{(m \gamma c)^2 - (m \gamma u_x)^2 - (m \gamma u_y)^2 - (m \gamma u_z)^2}$, is simply $mc$, the body's mass multiplied by the mere constant, c.

$m$ is a constant for the body (provided we don't tamper with the body, e.g. by changing its internal energy!) and doesn't vary from frame to frame. It is a Lorentz invariant. [Beware: the sum of the masses (rest masses) of bodies in a system has no obvious significance; it is certainly not the mass (rest mass) of the system!]

I've gone on longer than I should have done. It's all so wonderful. A classic introduction to Special Relativity, first rate on concepts, is Spacetime Physics by Taylor and Wheeler.

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  • $\begingroup$ If I drop the use of the phrase "relativistic mass" and replace the equations describing it with the following equations describing relativistic energy: relativistic energy is a function of rest mass and velocity and is conserved when two objects collide and combine into one; for any given rest mass, relativistic energy doesn't vary between two velocities with the same magnitude; for any velocity, relativistic energy varies linearly with rest mass; and change in relativistic energy approximates to change in $\frac{1}{2}mv^2$ for low speeds; then we can derive $E = mc^2$ for an object at rest. $\endgroup$ – Timothy Jun 14 '18 at 2:42
  • $\begingroup$ Yes, I think what you've said is right up to "change in relativistic energy approximates to change in $\frac{1}{2}mv^2$ for low speeds". Energy could change because of change in rest mass. What you $can$ say is that relativistic KE , ($(\gamma -1)mc^2$), approximates to $\frac{1}{2}mv^2$ at low speeds. As for deriving $E=mc^2$ for an object at rest, there are several ways to come at this. Above all I would strongly urge you to read a book like Taylor/Wheeler. The four dimensional space-time approach makes it all so much easier and more natural. $\endgroup$ – Philip Wood Jun 14 '18 at 7:31
  • $\begingroup$ I was thinking a change in relativistic energy approximates to a change in $\frac{1}{2}mv^2$ for low speeds if the rest mass doesn't change but then forgot to write that. I think I ended up not having enough characters to write that but that's not why I didn't write it. I know how to prove that there's exactly one solution to how momentum and energy vary with rest mass and velocity and what rest mass and subluminal velocity 2 objects of a given rest mass and subluminal velocity will combine into. $\endgroup$ – Timothy Jun 14 '18 at 15:47
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Yes, conservation laws are what you define energy (relativistic mass) and momentum based on -- in relativity as well as in non-relativistic mechanics. Energy is defined to as a conserved scalar quantity of motion, and momentum as a conserved vector quantity of motion.

Rest mass is not conserved, because the norm of a sum is not the same as the sum of norms. $\sqrt{(v^\mu+w^\mu)(v_\mu+w_\mu)}\neq\sqrt{v^\mu v_\mu}+\sqrt{w^\nu w_\nu}$.

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