Comparing relativistic momentum to classical

[PsychonautQQ]

EDIT: Okay I don't expect an answer for this because of my crappy attempt at LaTex, i'll work on making it look prettier sorry

Homework Statement

If the kinetic energy of a particle is equal to twice its rest energy, what percentage error is made by using p = mu for the magnitude of its momentum?

Homework Equations

$$\[E_i=mc^2\]$$
$$\[p = \frac{mu}{(1-\frac{u^2}{c^2})^\frac{1}{2}}\]$$
$$\[p = mu\]$$
$$\[K=\frac{p^2}{2m}\]$$

The Attempt at a Solution

I set K = 2mc^2 and then solved for the relative velocity u and ended up with u=(2/sqrt(5))*c

I then set K = 2mc^2 again but this time the momentum term was non relativistic, and solving for u I got u = 2c

now I'm lost

 

[Chestermiller]

Try using your two equations for momentum to get the ratio of the relativistic momentum to the pre-relativistic momentum. If you did the algebra right so far, you should get the right answer.

 

[vanhees71]

Just perform a Taylor expansion in the quantity $p/(mc)$, which is small for non-relativistic motion. Here $m$ is the invariant (rest) mass of the particle. One should not use any other masses in relativistic physics anymore. That's outdated since 1908 when Minkowski figured out the mathematical structure of special-relativistic space-time!

 

[Chestermiller]

Another equation you might consider using is K=(γ-1)mc². This would greatly simplify the analysis.

Chet

 

[Nickie]

I would like to point out that the concept of relativistic momentum is necessary to accurately describe the behavior of particles at high speeds. In classical physics, momentum is defined as the product of mass and velocity, but in the realm of relativistic physics, this definition is not sufficient. The correct equation for momentum in relativistic physics is given by:

\[p = \frac{mu}{(1-\frac{u^2}{c^2})^\frac{1}{2}}\]

where m is the mass of the particle, u is its velocity, and c is the speed of light. This equation takes into account the effects of time dilation and length contraction at high speeds.

In this scenario, we are given that the kinetic energy of the particle is equal to twice its rest energy, which can be expressed as:

\[K = 2mc^2\]

Using the above equation for relativistic momentum, we can solve for the velocity of the particle:

\[u = c\sqrt{1-\frac{1}{2}} = c\sqrt{\frac{1}{2}} = \frac{c}{\sqrt{2}}\]

Now, let's compare this to the non-relativistic equation for momentum:

\[p = mu\]

If we substitute the value of u we just found into this equation, we get:

\[p = m(\frac{c}{\sqrt{2}}) = \frac{mc}{\sqrt{2}}\]

As you can see, there is a clear difference between the relativistic and non-relativistic equations for momentum. Using the non-relativistic equation would result in a significant error when describing the momentum of a particle at such high speeds.

To calculate the percentage error, we can use the following formula:

\[Error = \frac{|p_{relativistic}-p_{non-relativistic}|}{p_{relativistic}} * 100\]

Substituting the values we calculated earlier, we get:

\[Error = \frac{|\frac{mc}{\sqrt{2}} - \frac{mc}{\sqrt{5}}|}{\frac{mc}{\sqrt{2}}} * 100 = \frac{\frac{mc}{\sqrt{10}}}{\frac{mc}{\sqrt{2}}} * 100 = \frac{\sqrt{2}}{\sqrt{10}} * 100 \approx