# Cutoff Point for Relativistic Effects

1. Dec 26, 2009

### pzona

I'm looking over some common physics equations as a review for the course I'll be taking next semester, and for each equation, I'm coming across two equations for each, a relativistic and a non relativistic. The class I'll be taking will be on mechanics, inertia, etc., so I figure I'll only need to know the nonrelativistic equations. I'm just curious though, at what velocity do you start taking relativistic effects into account? Is there an accepted cutoff point, or do you just judge it based on the problem? Thanks in advance for any help with this

2. Dec 26, 2009

### Nabeshin

I generally use .1c as the cutoff point for relativistic corrections. At this velocity, the difference between the result a relativistic and non-relativistic formula will give starts to be within the number of significant digits of the problem.

3. Dec 26, 2009

### tiny-tim

Yes, there's no way you'll need relativistic equations for the mechanics of bodies.

You'ld only need it for particles (eg a photon colliding with an electron).

4. Dec 26, 2009

### espen180

You can use the Lorenz factor $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ to check how prominent relativistic effects are. For velocities way below c, it is very close to 1, and it will say around 1 for velocities up to maybe $$3\cdot10^7\frac{m}{s}$$, like Nabeshin suggested. For example, at v=0.1c, the Lorentz factor is about 1.005.

5. Dec 26, 2009

### arunma

I think most people usually refer to the "extreme relativistic limit" as the point where a particle's momentum is much greater than its mass, and the "non-relativistic limit" as when the momentum is much smaller than the mass. This comes from the relativistic energy formula:

$$E = \sqrt{(pc)^2+(mc^2)^2}$$

We can see that when $pc>>mc^2$, it just becomes $E = pc$. It's often convenient to work in this regime when doing quantum or stat mech calculations, because you've got a Hamiltonian that goes linearly with momentum. When $pc<<mc^2$, we can do a Taylor expansion on p,

$$E \approx mc^2 + \dfrac{p^2}{2m}$$

You can see that the only two terms left are the rest energy term and the Newtonian kinetic energy, so now you can just use non-relativistic mechanics. This, by the way, is one of the ways to get the famous formula $E = mc^2$. When you use relativity to get a non-relativistic limit, you end up with something left over besides the kinetic energy.

When the momentum and mass are approximately the same, you can't really use any approximations, and this is when all the usual relativistic formulas apply.

6. Dec 26, 2009

### KrisOhn

I have a question: What does the Lorentz Factor mean? I know that it comes out to be a dimensionless constant, but that role does that constant play? What does it mean?

7. Dec 26, 2009

### arunma

It's the factor that you put in front of many mechanical quantities to adjust for relativistic effects. For example, the length contraction formula is $L=\gamma L'$, the time dilation formula is $dt = \gamma dt'$, the relativistic momentum is $p = \gamma mv$, etc. It also appears in the Lorentz transformations. For even large speeds, the gamma factor is extremely close to 1. Only at speeds very close to the speed of light does it become important, and at speeds above .99c, even small increases in velocity correspond to very large changes in the gamma factor. Basically it's a measure of how important relativity is.

8. Dec 26, 2009

### KrisOhn

Thank you, that makes sense.

9. Dec 27, 2009

### pzona

Now that you mention the Lorenz Factor, it was showing up in a few different places (relativistic momentum, etc.) while I was trying to review equations, and I didn't know what it was. That makes sense that it can be considered a factor itself though. Thanks for the help everyone

10. Dec 28, 2009