Explaining the Speed Limit: Energy & Speed

[DR13]

Please tell me why I am wrong here. I know I am, I just want to know why.

K=.5mv^2
E=mc^2
So if all of the energy is kinetic then
mc^2=.5mv^2
Solving for v
v=sqrt(2)*c

However, the "speed limit" is c. How would this be explained? My guess is that the equation for K does not apply here for some reason but I don't really have a clue.

Thanks for the help,
DR13

 

[Dale]

E=mc^2 is the energy of a mass at rest, so v=0 in your first equation.

 

[K^2]

Your equation for kinetic energy is not quite correct. It only works for values of v that are very small compared to c. The correct equation is quite a bit more complicated.

$$KE = \left(\sqrt{\frac{c^2}{c^2-v^2}} - 1\right)mc^2$$

For small values of v, the following approximation works.

$$\sqrt{\frac{c^2}{c^2-v^2}} \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$$

If you substitute this into the above, you get kinetic energy equation that you are familiar with.

The total energy also needs to be corrected. As Dale points out, your equation only works for v=0. In general, total energy is given by this formula.

$$E = \sqrt{\frac{c^2}{c^2-v^2}}mc^2$$

Now, if you set this energy to be equal to KE, you'll find that the only solution is v infinitely close to c. (v=c causes you to divide by zero, but you can get arbitrarily close.) However, this also makes E and KE go to infinity.

The way this works out for real particles is that mass is zero. That, unfortunately, makes above equations undefined. So there is yet another set of equations you can use.

$$E^2 = p^2c^2 + m^2c^4$$

$$KE = E - mc^2$$

Here, p is momentum. For a particle with m>0, momentum is given by the following.

$$p=\sqrt{\frac{c^2}{c^2-v^2}mv$$

If you substitute this in, you'll get exactly the same equations for kinetic and total energy as before. The added benefit is that you can use the equations with momentum even when m=0. Then.

$$E = pc$$

$$KE = pc - mc^2 = pc$$

This is exactly what you have been looking for.

Yes, that's a lot of equations, but you really only need to know the one giving E in terms of p. All of the other equations can be derived from it.

 

[DR13]

Oh ok. Now I get it. Thanks guys!

 

[PJC01]

First of all, it is important to note that the equation K=.5mv^2 applies to objects in classical mechanics, while E=mc^2 applies to objects in relativistic mechanics. These two equations cannot be directly compared or used interchangeably.

The speed limit, or the maximum speed that an object can reach, is determined by the theory of special relativity. According to this theory, the speed of light (c) is the maximum speed that any object in the universe can attain. This means that no object, regardless of its mass or energy, can ever travel at a speed greater than the speed of light.

In the equation E=mc^2, c represents the speed of light, which is a constant in the universe. This means that the energy (E) of an object is directly proportional to its mass (m). This equation shows that as the mass of an object increases, its energy also increases, but it can never reach or exceed the speed of light.

On the other hand, in the equation K=.5mv^2, the variable v represents the velocity or speed of an object, which can theoretically reach any value. However, as an object approaches the speed of light, its mass increases and its energy becomes infinite, making it impossible to reach the speed of light.

In summary, the speed limit is determined by the theory of special relativity and the constant speed of light. The equations for kinetic energy and energy do not directly apply to this concept, as they are based on different principles and theories.