# How to calculate the minimum kinetic energy (special relativity)?

• Efeguleroglu
In summary, the conservation of energy states that the total energy of the system remains the same after the collision. The conservation of momentum states that the momentum of the particles after the collision is the same.
Efeguleroglu
Homework Statement
A moving electron collides to an stable electron and in conclusion additionaly 1 electron and 1 positron occur. At the end if all the 4 particles have the same velocity, we can say the kinetic energy required for this process is minimum. Show that KE(min)=6mc^2 (m is the rest mass of an electron)
Relevant Equations
KE=γmc^2-mc^2
Sadly, that's what all I could do.

Last edited:
I find it's better to work with energy and momentum, rather than writing things in terms of velocities.

You've used conservation of energy. What relationship do you get if you use conservation of momentum?

I still can't get the answer. I thought I didn't understand the question and calculated for both inital kinetic energy and final kinetic energy ( I know the change in kinetic energy is -mc^2 from the conservation of energy) but couldn't get the answer. There shouldn't be any calculation mistakes because I used a calculator.

That's a lot of work!

1) What can you say about the total momentum after the collision?

2) What can you say about the momentum of each particle after the collision?

The momentum of the particles are equal. And that's the momentum which must be conserved.

Efeguleroglu said:
The momentum of the particles are equal. And that's the momentum which must be conserved.
View attachment 245271

You've gone wrong, I'm sorry to say. Best to start again. Here are some general hints:

a) Find the total energy of the moving electron. You can easily get the KE from this. Splitting the energy into ##T + mc^2## too soon just complicates the equations.

b) You have 3 key equations:

Conservation of total energy
Conservation of total momentum
The energy-momentum-mass equation for each particle (##E^2 = p^2c^2 + m^2c^4##).

Concentrate on those.

There must be a problem in this question. It's ridiculous but V(i) is less than it should be thus I found V(f) is imaginary. I didn't spot any mistake. Sorry for bothering you with this :(

Again, it's difficult to see where you are going wrong. This should only be 3-4 lines of algebra using the energy and momentum equations. It's always messy when you try to solve for ##v##.

I would start as follows:

Let ##E_0, p_0## be the energy and momentum of the moving electron; and, ##E_1, p_1## be the energy and momentum of each of the particles after the collision.

Conservation of energy gives:

##E_0 + mc^2 = 4E_1 \ \ \ (1)##

Conservation of momentum gives:

##p_0 = 4p_1 \ \ \ (2)##

And, we have:

##E_0^2 = p_0^2c^2 + m^2c^4 \ \ \ (3a)##
##E_1^2 = p_1^2c^2 + m^2c^4 \ \ \ (3b)##

The next obvious step is to square equation (1), and then do some substitutions. It should come out in a couple of steps.

You made an error pretty much from the start. The fourth line is wrong because you did something akin to ##(a+b)^2 = a^2 + b^2##.

As I suggested yesterday and as @PeroK has again suggested, work in terms of energy and momentum.

PeroK
vela said:
You made an error pretty much from the start. The fourth line is wrong because you did something akin to ##(a+b)^2 = a^2 + b^2##.

As I suggested yesterday and as @PeroK has again suggested, work in terms of energy and momentum.

Ahh yes, I saw it. I applied the momentum-energy equation from particles to the set of particles wrongly. Thanks, you helped me to spot that.

PeroK said:
Again, it's difficult to see where you are going wrong. This should only be 3-4 lines of algebra using the energy and momentum equations. It's always messy when you try to solve for ##v##.

I would start as follows:

Let ##E_0, p_0## be the energy and momentum of the moving electron; and, ##E_1, p_1## be the energy and momentum of each of the particles after the collision.

Conservation of energy gives:

##E_0 + mc^2 = 4E_1 \ \ \ (1)##

Conservation of momentum gives:

##p_0 = 4p_1 \ \ \ (2)##

And, we have:

##E_0^2 = p_0^2c^2 + m^2c^4 \ \ \ (3a)##
##E_1^2 = p_1^2c^2 + m^2c^4 \ \ \ (3b)##

The next obvious step is to square equation (1), and then do some substitutions. It should come out in a couple of steps.

Thank you! It appeared quickly.

BvU and PeroK

## 1. How do I calculate the minimum kinetic energy in special relativity?

The minimum kinetic energy in special relativity can be calculated using the formula E = (γ - 1)mc2, where γ is the Lorentz factor, m is the rest mass of the object, and c is the speed of light.

## 2. What is the significance of calculating the minimum kinetic energy in special relativity?

The minimum kinetic energy in special relativity is important because it helps us understand the behavior of objects moving at high speeds. It is also a fundamental concept in the theory of relativity and has implications in various fields such as particle physics and astrophysics.

## 3. How does special relativity affect the calculation of minimum kinetic energy?

In special relativity, the concept of mass is relative and depends on the observer's frame of reference. This means that the minimum kinetic energy of an object will also vary depending on the observer's perspective. Therefore, the formula for calculating minimum kinetic energy takes into account the object's rest mass and its velocity relative to the observer.

## 4. Can the minimum kinetic energy ever be zero in special relativity?

According to the formula E = (γ - 1)mc2, the minimum kinetic energy can only be zero if the velocity of the object is zero, i.e. the object is at rest. However, in special relativity, the concept of rest mass is also relative, so it is possible for an object to have zero minimum kinetic energy in one frame of reference but not in another.

## 5. How is the minimum kinetic energy related to the speed of light in special relativity?

In special relativity, the speed of light is considered to be the maximum speed that any object can attain. This means that as an object approaches the speed of light, its minimum kinetic energy will approach infinity. This is because the Lorentz factor, which is used to calculate minimum kinetic energy, increases exponentially as the velocity of the object approaches the speed of light.

Replies
8
Views
1K
Replies
2
Views
327
Replies
33
Views
1K
Replies
21
Views
2K
Replies
3
Views
1K
Replies
1
Views
903
Replies
9
Views
1K
Replies
1
Views
945
Replies
14
Views
1K
Replies
4
Views
875