- #1
Astrofiend
- 37
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Homework Statement
I am trying to show that the velocity of an ultra-relativistic particle can be approximated by the following expressions:
[tex]
v \approx c \left[1-\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]
[/tex]
and[tex]
\frac{1}{v} \approx \frac{1}{c} \left[1+\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]
[/tex]
...but I'm struggling. I know it can't be that hard, but I just can't quite get there. It seems I can get close, but I miss the factor of 1/2 out the front of the mass/energy term. I'm definitely doing something wrong, so any help would be much appreciated. Here's how I went about trying to obtain the first expression:
The Attempt at a Solution
So we have our usual relativistic expression:
[tex]
E^2 = p^2 c^2 + m^2 c^4
[/tex]
Then, I divided through by m^2 c^4 to get:
[tex]
\frac{E^2}{(mc^2)^2}-1 = \frac{p^2 c^2}{(mc^2)^2}
[/tex]
Now, using a relation that I pulled from somewhere - [tex] (pc)^2 = E^2 \frac{v^2}{c^2}[/tex]
I simply sub that in for (pc)^2, getting
[tex]
\frac{E^2}{(mc^2)^2}-1 = \frac{E^2}{(mc^2)^2}. \frac{v^2}{c^2}
[/tex]
and dividing through by [tex]\frac{E^2}{(mc^2)^2}[/tex], get
[tex]
\frac{v^2}{c^2} = 1 - \frac{(mc^2)^2}{E^2}
[/tex]
so
[tex]
\frac{v}{c} = 1 - \frac{(mc^2)^2}{E^2}
[/tex]
i.e
[tex]
v = c\left[1 - \frac{(mc^2)^2}{E^2}\right]
[/tex]
Hmmmm - the problem is that a) this is an exact equality not an approximation, b) I'm missing the damn factor of 1/2 out the front of the mass-energy fraction, and c) nowhere have I used the fact that the particle is highly relativistic. These facts are almost certainly all related, but I can't see how and I've been staring at it for some time now.
Any help or suggestions, or more ideas for more rigorous derivations would be exceedingly welcome!