How to prove the relativistic momentum? [duplicate]

Question

As far as I know, the relativistic momentum of a particle is given by the equation: $$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $m_0$ is the mass of the particle and $v$ is the velocity of the particle.

But why does the number $\sqrt{1-\frac{v^2}{c^2}}$ appear? How can I derive it from the special theory of relativity?

I have not enrolled in University yet, and I couldn't find the proof for the equation above anywhere online.

But if I found the proof for the equation of time dilation $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$ (which I did), can I prove the relativistic momentum equation above?

(Sorry for my English. If the question has been asked anywhere else, or if this question is inappropriate, please let me know. Any answers or comments will be appreciated)

Answer 1

I feel like you are saying that you have enough information to derive the Lorentz transform and you wish to see how we can use the reference-frame independence of conservation of momentum to derive this form $\gamma_\mathbf v m \mathbf v$, which is a very nice elementary question that deserves an answer.

A basic collision

Suppose you see two masses $m$ come in from opposite directions with velocities $\pm \mathbf v$ and speed $v = \|\mathbf v\|$, and they collide and stick together. Someone else sees this happen in a reference frame which regards you as traveling with velocity $\mathbf u$, and hence you see them traveling with velocity $-\mathbf u$.

Let’s firm up coordinates by saying that $\mathbf u = u \hat x$ so that this transformation of reference frames happens in the $x$-direction. Then let us say that $\mathbf v$, making a second line, also defines the $xy$-plane by lying within it: WLOG $$\mathbf v = \begin{bmatrix}v_x\\ v_y\end{bmatrix}.$$ Then due to the ways velocities add in relativity, the second frame must see velocity components for these two incoming particles’ vectors as $$\mathbf v_\pm' = \frac{1}{1 \pm u v_x/c^2} \begin{bmatrix}u \pm v_x\\ \pm v_y \sqrt{1 - u^2/c^2}\end{bmatrix}.$$

In fact these expressions are very easy to derive from the Lorentz transform if you take the following convention: make the event-of-collision your spacetime-origin point $w = ct=0, x=0, y=0$ for both frames. Then you can find $v_x$ by the simplified expression $c~x/w$ and $v_y$ by the simplified expression $c~y/w$. Well, after the Lorentz transform you have the same formulas, $v_x'=c~x'/w' = c(x - \beta w)/(w - \beta x)$ and $v_y' = c~y'/w' = c y/(\gamma w - \gamma \beta x),$ and in this case $\beta = -u/c.$ So that is why you get this $1/\gamma_u$ term on the $v_y$ component and these $1/(1 \pm u v_x/c^2)$ terms on both components otherwise.

Failure of diagonal momentum conservation

So our first problem is that we have these two $y$-components of velocity coming into our collision, and the outcome of the collision is something moving like $v_x' = u, v_y' = 0$, but the two $y$-components are not canceling out to zero due to this $\pm$ term in the denominator! Rather if I were to add them I would find$$ (\mathbf v_+')_y + (\mathbf v_-')_y = \sqrt{1 - \frac{u^2}{c^2}}\left(\frac{-2 u v_x v_y / c^2}{1 - u^2 v_x^2/c^4}\right).$$

As a consequence if we want momentum to be conserved in the primed frame in all of its various directions, we need to find something other than just $\mathbf p = m\mathbf v$ as this can work perhaps for perpendicular and parallel motion to the reference frame disagreement, but never for diagonal motion $v_x \ne 0, v_y\ne 0.$

And it is easy to see what I have to multiply by; somehow I have to multiply $v^\pm$ terms by something looking like $1 \pm u v_x/c^2$. So I need an expression for this, preferably one without explicit reference to $u$ or $v_x$ as they depend on how I choose my $x,y$-axes and I need to know what to do in general, not just in this particular scenario.

Fixing the problem

Seeing this term occur in both $(\mathbf v_\pm')_{x,y}$ components, and knowing that the length of my vector does not depend on my choice of $x,y$-axes, it makes sense to simply take this squared-length, $$\begin{align} \mathbf v_\pm'\cdot\mathbf v_\pm' &= \frac{ (u \pm v_x)^2 + v_y^2 (1 - u^2/c^2)}{(1 \pm u v_x/c^2)^2} \\ &= \frac{ u^2 \pm 2 u v_x + v_x^2 + v_y^2 - v_y^2 u^2/c^2}{1 \pm 2 u v_x/c^2 + u^2 v_x^2/c^4}. \end{align}$$ So we again have two $\pm$ terms, but one of them, in the numerator, is very simple to remove because the shape is exactly identical to the shape in the denominator. So we look at the combination, $$ \begin{align} \mathbf v_\pm'\cdot\mathbf v_\pm' - c^2 &= \frac{ u^2 \pm 2 u v_x + v_x^2 + v_y^2 - v_y^2 u^2/c^2}{1 \pm 2 u v_x/c^2 + u^2 v_x^2/c^4} - \frac{c^2 \pm 2 u v_x + u^2 v_x^2/c^2}{1 \pm 2 u v_x/c^2 + u^2 v_x^2/c^4}\\ &=\frac{ u^2 + v_x^2 + v_y^2 - v_y^2 u^2/c^2 -c^2 - u^2 v_x^2 / c^2}{1 \pm 2 u v_x/c^2 + u^2 v_x^2/c^4}\\ &=\frac{ v_x^2 + v_y^2 -c^2}{(1 \pm u v_x/c^2)^2}\left(1 - \frac{u^2}{c^2}\right) = - c^2 \frac{(1 - \|\mathbf v\|^2/c^2)(1 - u^2/c^2)}{(1 \pm u v_x/c^2)^2}. \end{align}$$ where in the last line we just happened to recognize that suddenly our numerator was divisible by $(1 - u^2/c^2)$ and factored that term out, and then saw the term $(1 - v^2/c^2)$ emerge as well. Regardless we now have removed our $\pm$ from the one side and discovered that actually $$\gamma_\pm = \frac{1}{\sqrt{1 - \|\mathbf v_\pm'\|^2/c^2}}= \gamma_u\gamma_v~(1 \pm u v_x/c^2),$$ allowing us to remove this $1/(1 \pm u v_x/c^2)$ in the denominator. So we can find that $$(\gamma_+ \mathbf v_+')_y + (\gamma_- \mathbf v_-')_y = 0$$ and in fact given the derivation above it is very likely the only coordinate-independent choice having this property, as there are no other convenient coordinate-indpendent scalars to be be formed here other than the length of $\mathbf v_\pm '.$ So the only choice which allows momentum to cancel in the $y$-direction in the diagonal case is $$\mathbf p = \frac{m \mathbf v}{\sqrt{1 -v^2/c^2}} = \gamma_v ~m \mathbf v.$$

Failure of mass conservation

As a result of the above derivation we will also find that $$ (\gamma_+~m~\mathbf v_+')_x + (\gamma_-~m~\mathbf v_-')_x = \frac{2 ~m~u}{\sqrt{(1 - u^2/c^2)(1 - \|\mathbf v\|^2/c^2)}},$$ rather than the semiclassical expression $2~m~u/\sqrt{1 - u^2/c^2}$ that we would have expected from our new definition of momentum. In other words, we might say, mass has somehow not been conserved; beforehand I had $2m$ mass and now I have $2m\gamma_v.$

However we can immediately also see that this expression has a very nice interpretation as $\gamma_+ + \gamma_- = 2 \gamma_u \gamma_v.$ This was the basic motivation behind early relativistic physicists regarding $\gamma m_0$ as the “relativistic mass;” you find that this mass is conserved when other mass is not.

Today with the language of 4-vectors we do not really regard this the same way; we would say that it is, instead, a relativistic energy (and we would say that it is a “rest mass” energy $m_0c^2$ , plus a “kinetic energy” $(\gamma - 1)m_0c^2$). So we would just say straight-up that mass is not conserved in relativity. And we would also say that the momentum 4-vector has a $w$-component (where $w=ct$ which is $E/c$ where $E$ is the energy, and is related to the 4-velocity by $p^\mu = m_0 v^\mu.$ So since the 4-velocity $v^\bullet = (\gamma~c, \gamma \mathbf{v})$ transforms correctly as a 4-vector under the Lorentz group, this 4-momentum does too as long as everyone agrees on the value of the mass $m_0$; as a consequence conservation of momentum naturally should be promoted to a conservation of 4-momentum including an expression for the conservation of this energy $E = \gamma m_0 c^2.$

Answer 2

If you are just curious as to how they came up to the value: $$\sqrt{1-\frac{v^{2}}{c^{2}}}$$ The basis of the calculation is on time dilation.

The time interval between ticks is the proper time $t_0$ and the time needed for the light pulse to travel between the mirrors at speed of light $c$ is $\frac{t_0}{2}$. Hence, $\frac{t_0}{2} = \frac{L_0}{c}$ and $$t_0 = \frac{2L_0}{c}$$

The time interval between ticks is $t$. Because the clock is moving, the light pulse, as seen from the ground follows a zigzag path. On its way from the lower mirror to the upper one in the time $\frac{t}{2}$, the pulse travels a horizontal distance of $v(\frac{t}{2})$ and a total distance of $c(\frac{t}{2})$. Since $L_0$ is the vertical distance between mirrors,

$$(\frac{ct}{2})^2 = L_0^2 + (\frac{vt}{2})^2$$ $$(\frac{t^2}{4})(c^2 - v^2) = L_0^2$$ $$t^2 = \frac{4L_0^2}{c^2 - v^2} = \frac{(2L_0^2)}{c^2(1 - \frac{v^2}{c^2})}$$ $$t = \frac{\frac{2L_0}{c}}{\sqrt{1-\frac{v^2}{c^2}}}$$

Time dilation now is

$$t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

Thus,

$$\sqrt{1-\frac{v^2}{c^2}} = \frac{t_0}{t}$$

Where:

$t_0$ = time interval on clock at rest relative to an observer = proper time

$t$ = time interval on clock in motion relative to an observer

$v$ = speed of relative motion

$c$ = speed of light

Answer 3

One cannot "prove" that momentum is equal to the quantity you quoted, just as nobody can really prove that classical momentum is given by $p = mu$. These formulas are mostly definitional and we always try to stick with useful definitions.

With that being said, the equation $$\mathbf{p} = m \mathbf{u}$$ is simply not useful in the context of special relativity. Why? The main reason is that this quantity could not be conserved in all reference frames. To see this explicitly, let's calculate the total such 'momentum' of objects $A, B, C$ and $D$ in a particular reference frame. Suppose that the momentum calculation turns out to be as follows: \begin{equation} m_{A} u_{A}+m_{B} u_{B}=m_{C} u_{C}+m_{D} u_{D}. \end{equation} Now, lets move to a reference frame that is moving uniformly with some velocity $v$. The transformation of each velocity is given by: \begin{equation} u_{i} \rightarrow \frac{\bar{u}_{i}+v}{1+\left(\bar{u}_{i} v / c^{2}\right)} \end{equation} and hence our momentum calculation yields: \begin{equation} m_{A} \frac{\bar{u}_{A}+v}{1+\left(\bar{u}_{A} v / c^{2}\right)}+m_{B} \frac{\bar{u}_{B}+v}{1+\left(\bar{u}_{B} v / c^{2}\right)}=m_{C} \frac{\bar{u}_{C}+v}{1+\left(\bar{u}_{C} v / c^{2}\right)}+m_{D} \frac{\bar{u}_{D}+v}{1+\left(\bar{u}_{D} v / c^{2}\right)}. \end{equation} This equation, in general, forces us to conclude that \begin{equation} m_{A} \bar{u}_{A}+m_{B} \bar{u}_{B} \neq m_{C} \bar{u}_{C}+m_{D} \bar{u}_{D} \end{equation} so momentum conservation is now dependent on the reference frame.

We could just throw our hands and forsake momentum conservation in all frames. It just so happens that clever physicists have shown that the definition \begin{equation} \mathbf{p}=\frac{m_{0} \mathbf{u}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{equation} yields a definition of momentum which is conserved in all reference frames. This is just a more useful definition of momentum, and so we stick with that.