# Special Relativity Question

• PsychonautQQ
In summary, to find the velocity of an electron or positron with 50GeV of energy, you can use the equation E/E0 = gamma - 1, where E is the total energy and E0 is the rest energy. This equation can be derived from the relativistic equation E^2 = (pc)^2 + (mc^2)^2. The Lorentz factor, gamma, can then be inverted to find the velocity. However, for an electron or positron, the difference between the total energy and the rest energy is insignificant, so the velocity will be very close to the speed of light, c.

## Homework Statement

If an electron and a Positron are said to each have 50GeV of energy (8E-9J) then how do I find their velocity? Do I use E^2=(pc)^2+(mc^2)^2 and solve for velocity in the momentum? Is there an easier way to do this? I did this, I might have done the algebra wrong but I got an answer that was basically zero, lol. Any advice is nice :D

## The Attempt at a Solution

You need to find the Lorentz factor, gamma; you can invert this to find the velocity.

Since you have the total energy, E=50GeV note that this is the sum of the rest energy (E0=mc^2) plus the work done on the electron. This is equivalent to E/E0 = gamma - 1; this is the gamma that you need.

so E = E0gamma - E0...? Energy = (What is this term?) - Rest Energy? How did you derive this? Did you mean to put an addition sign there at all and then E0gamma would represent the kinetic energy or something? If E is the total energy this equation means E0gamma is greater than it, so what could it possibly be?

Also how is this equation related to
E^2 = (pc)^2 + (mc^2)^2 ??
It looks like it's more related to the nonrelativistic approximation of
E = K + mc^2, which would mean it's wrong because I think I'm suppose to be using the actual relativistic equation for this problem.

Explanation: if E0=mc^2 is the rest energy then the total energy is gamma*E0. Then the kinetic energy is:

E_kinetic = gamma*E0 - E0. So (gamma-1) = E_kinetic/E0 which gives provides a value for the Lorentz factor.

You can derive this from E^2 = (pc)^2 + (mc^2)^2.
See http://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies

Is the 50GeV the total energy, or the total energy? It could be either - but for an electron the kinetic energy is the work done on the electron - so if you have placed an electron in a 50 GV electric potential it will have 50 GeV when it reaches the other side; they actually do this in stages for such high energies. Anyway E0 for an electron or positron is about 0.511 MeV, so the difference is insignificant.

So (gamma-1)=50 GeV/0.511 MeV ~= ~100,000. This implies v=.999 c, where you will have to figure out how many nines! 20 keV gives ~c/4 and 30 keV gives ~c/3 which are the speeds I usually work with.

I thought E_kinetic = Gamma*E0 - E0 was only an approximation of E^2 = (pc)^2+(mc^2)^2 in nonrelativistic situations?? thanks btw :D

No, it is exact.

...however, this expression is on the usual route to deriving the Newtonian formula for kinetic energy. If you Taylor expand the gamma and disregard terms of order v4 and higher, out drops mv2/2. That is an approximation, and may be what you're thinking of.

## 1. What is special relativity?

Special relativity is a theory developed by Albert Einstein that explains the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion, and that the speed of light in a vacuum is constant for all observers.

## 2. How does special relativity differ from general relativity?

Special relativity deals with the behavior of objects in a non-accelerating frame of reference, while general relativity extends this to include accelerating frames of reference and the effects of gravity.

## 3. Can special relativity be proven?

Special relativity has been extensively tested and has been shown to accurately predict the behavior of particles and objects at high speeds. However, like all scientific theories, it can never be proven definitively.

## 4. What are some real-world applications of special relativity?

Special relativity has important applications in modern technology, such as GPS systems, particle accelerators, and nuclear power plants. It also has implications for our understanding of the universe, including the concept of time dilation and the twin paradox.

## 5. How does special relativity affect our everyday lives?

While the effects of special relativity are not noticeable in our daily lives at low speeds, they become more significant as objects approach the speed of light. Special relativity also has implications for our understanding of time and space, and has sparked philosophical and scientific debates about the nature of reality.