Relativistic Momentum Help With Equation Reduction

In summary, the conversation discusses a problem with simplifying an equation from a wiki article about special relativity and momentum. Despite attempts at different approaches, the person is unable to reach the desired equation. After receiving help, they realize their mistake and are able to solve the problem. The final question is about choosing the appropriate solution for the equation.
  • #1
jimbobian
52
0
Hi, so basically have been looking at http://en.wikibooks.org/wiki/Special_Relativity:_Dynamics#Momentum" and working my way through the maths for myself. However I have hit this point and can't get past it:

\begin{align}
u = \frac{v - u}{1-\frac{uv}{c^2}}
\end{align}
Which should be able to become:
\begin{align}
u = \frac{c^2}{v(1-\sqrt{1-\frac{v^2}{c^2}})}
\end{align}
I have tried and tried but can't seem to get it to work. Can anyone help me out on this one please?!

Thanks
James
 
Last edited by a moderator:
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  • #2
jimbobian said:
Hi, so basically have been looking at http://en.wikibooks.org/wiki/Special_Relativity:_Dynamics#Momentum" and working my way through the maths for myself. However I have hit this point and can't get past it:

\begin{align}
u = \frac{v - u}{1-\frac{uv}{c^2}}
\end{align}
Which should be able to become:
\begin{align}
u = \frac{c^2}{v(1-\sqrt{1-\frac{v^2}{c^2}})}
\end{align}
I have tried and tried but can't seem to get it to work. Can anyone help me out on this one please?!

Thanks
James

That wiki entry is not being clear with it's order of operations!

What you should be getting is:

[tex] u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})[/tex]
 
Last edited by a moderator:
  • #3
G01 said:
That wiki entry is not being clear with it's order of operations!

What you should be getting is:

[tex] u = \frac{c^2}{v}(1-\sqrt{1-\frac{v^2}{c^2}})[/tex]

Hi G01, thanks for your reply. Have tried aiming for that equation instead and still can't get there. I have tried all sorts of different approaches, such as getting it in the form of a quadratic or dividing the top and bottom of the original fraction by c^2, but I just end up nowhere. Could you perhaps give me a hand in the right direction, maybe the first step or two - but don't make it too easy for me!
 
  • #4
Show us exactly what you've tried, and where you get stuck, and tell us why you're stuck.
 
  • #5
Ok, have shut my computer down now and am replying on my phone. I will post my attempts tomorrow.
Thanks,
James
 
  • #6
Well, brute force use of the quadratic formula leads directly to the desired answer.
 
  • #7
Haha, went back over my general quadratic attempt after PAllen's suggestion and realized that I had made a rather fundamental cross multiplication error! Fixed that and got to the desired equation. Lovely!

Last question, the equation that I get is:
\begin{align}
u = \frac{c^2}{v}(1\pm\sqrt{1-\frac{v^2}{c^2}})
\end{align}
I assume that we choose the negative version of the equation, because the positive version would yield a value for u which is greater than c?

Cheers
 

FAQ: Relativistic Momentum Help With Equation Reduction

What is relativistic momentum and how is it different from classical momentum?

Relativistic momentum is a concept in physics that describes the motion of an object that is moving at speeds close to the speed of light. It is different from classical momentum in that it takes into account the effects of special relativity, such as time dilation and length contraction, which are not accounted for in classical mechanics.

How is the equation for relativistic momentum reduced?

The equation for relativistic momentum, p = mv/√(1-v^2/c^2), can be reduced using the Lorentz factor, γ = 1/√(1-v^2/c^2), to give the simplified equation p = γmv. This equation takes into account the effects of special relativity on an object's mass, velocity, and momentum.

What is the significance of the speed of light, c, in the equation for relativistic momentum?

The speed of light, c, is a fundamental constant in the equation for relativistic momentum. It represents the maximum speed at which any object can travel in the universe. As an object approaches this speed, its momentum increases exponentially, making it impossible to accelerate to the speed of light.

How does the concept of relativistic momentum apply to particle accelerators?

In particle accelerators, particles are accelerated to speeds close to the speed of light, making the effects of special relativity significant. The equation for relativistic momentum is used to calculate the momentum of these particles and predict their behavior in the accelerator.

Can relativistic momentum be used for objects with massless particles, such as photons?

Yes, the equation for relativistic momentum can be applied to massless particles, such as photons. In this case, the mass, m, is equal to 0 and the equation reduces to p = E/c, where E is the energy of the photon. This equation is used to describe the momentum of photons in electromagnetic radiation and other phenomena.

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