change in relativistic momentum

[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?

[DW]

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?
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.

[Nenad]

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}

[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.

[pmb_phy]

Is it alright to say that force = rate of change of relativistic momentum..Yes. Force is defined as

\bold F = \frac{d\bold p}{dt}


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?That is the average force. The instantaneous force is F = dp/dt.

Pete

[DW]

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
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.