Energy and Speed

  • Thread starter DR13
  • Start date
  • #1
150
0

Main Question or Discussion Point

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 dont really have a clue.

Thanks for the help,
DR13
 

Answers and Replies

  • #2
28,903
5,165
E=mc^2 is the energy of a mass at rest, so v=0 in your first equation.
 
  • #3
K^2
Science Advisor
2,469
28
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.

[tex]KE = \left(\sqrt{\frac{c^2}{c^2-v^2}} - 1\right)mc^2[/tex]

For small values of v, the following approximation works.

[tex]\sqrt{\frac{c^2}{c^2-v^2}} \approx 1 + \frac{1}{2}\frac{v^2}{c^2}[/tex]

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.

[tex]E = \sqrt{\frac{c^2}{c^2-v^2}}mc^2[/tex]

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.

[tex]E^2 = p^2c^2 + m^2c^4[/tex]

[tex]KE = E - mc^2[/tex]

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

[tex]p=\sqrt{\frac{c^2}{c^2-v^2}mv[/tex]

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.

[tex]E = pc[/tex]

[tex]KE = pc - mc^2 = pc[/tex]

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.
 
  • #4
150
0
Oh ok. Now I get it. Thanks guys!
 

Related Threads for: Energy and Speed

Replies
1
Views
563
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
21
Views
5K
Replies
4
Views
6K
  • Last Post
Replies
2
Views
468
Replies
2
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
6
Views
833
Top