Probably Fairly Simple Special Relativity Calculation

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SUMMARY

The discussion focuses on deriving the velocity approximation for an ultra-relativistic particle, specifically the expressions v ≈ c[1 - (1/2)(mc²/E)²] and 1/v ≈ (1/c)[1 + (1/2)(mc²/E)²]. The user struggled with missing the factor of 1/2 and not applying the ultra-relativistic condition correctly. A key insight was provided by a participant suggesting the use of a Taylor series expansion to approximate the square root of 1 - x, which helped the user resolve their confusion and complete the derivation successfully.

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Astrofiend
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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] <br /> v \approx c \left[1-\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]<br /> [/tex]

and[tex] <br /> \frac{1}{v} \approx \frac{1}{c} \left[1+\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]<br /> [/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!
 
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I didn't look over the rest of your derivation, but this step:
Astrofiend said:
[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]
...doesn't follow. (You forgot to take the square root of the RHS!)

Hint: Use a Taylor series expansion to approximate the square root of 1 - x, where x << 1.
 
"(You forgot to take the square root of the RHS!)"

Der! Must be tired - or just plain dumb...

Thanks a lot Doc Al - I got it out easily after that little hint. I appreciate it a lot!
 

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