# Change in relativistic momentum

1. Aug 21, 2004

### kurious

Is it alright to say that force = rate of change of relativistic momentum

F = [ m0 v2 / (1 - v2^2/c^2)^1/2 - m0 v1/(1 - v1^2/c^2)^1/2 )] / (t2 - t1)

and can this relation be used to get sensible results for particles?

Last edited: Aug 21, 2004
2. Aug 21, 2004

### DW

Thats close to ordinary force f. To be precise ordinary force f involves the limit of that as t2-t1 becomes infinitesimal dt in a calculus limit. Also, you shouldn't subscript the mass with a zero as it is invariant.

3. Aug 21, 2004

DW is right, your notation is wrond. You must let t2-t1 approach 0, it must be alimit.
Here is the wau you want to write it:

$$F = \frac {d(\frac{mv}{\sqrt{1-v^2 / c^2}})}{dt}$$

4. Aug 21, 2004

### krab

Yes, but in general, you gain nothing in writing the differential equation in terms of velocity. Just leave it in momentum; equations are far simpler.

5. Aug 21, 2004

### pmb_phy

Yes. Force is defined as

$$\bold F = \frac{d\bold p}{dt}$$
That is the average force. The instantaneous force is F = dp/dt.

Pete

6. Aug 24, 2004

### DW

Just to be clear to you, in light of the notation having been used here for a while, it is an expression for ordinary force f, not the four-vector force F.