psyche

Not sure if this is correct, but I thought this was quite good (done while procrastinating about doing Mr Miller’s Cosmology prep):

Consider an object whose initial speed, u or v₀, is zero. A constant force F then acts on it (and continues acting on it). I will (attempt) to find an expression for the speed v(t) of the object. I will use u’ to mean du/dt. Please ignore terrible interchanges between functions of stuff and such like - don’t really know what I’m doing.

Remember, v₀ is zero.

First, use the crunched version Newton’s second law: F = ma, so a = F/m, so v’ = F/m. Integrating both sides with respect to t gives v(t) = Ft/m + k. Since v(0) = 0, k = 0 so v = tv’ (cf. suvat stuff, v = u + at when u = 0… it’s all fine so far).

Now consider the relativistically tweaked version of Newton II: F = ma/sqrt(1-(v/c)^2). Using the same juggling as before, this gives:

v’ = Fsqrt(1-(v/c)^2)/m

Rearrange and use separation of variables/whatever to get:

int(1/sqrt(1-(v/c)^2))dv = int(F/m)dt

which gives (remembering the boundary condition v(0) = 0)

csin^-1(v/c) = Ft/m

sin^-1(v/c) = Ft/mc

v/c = sin(Ft/mc)

finally:

v(t) = csin(Ft/mc)

This looked promising (note the maximum value of v is c)…

To try to convince myself that this was correct, consider the sin small angle approximation: sin(x) is approx. equal to x for small values of x. Well, Ft/mc is certainly small for most F and t when you consider how large c is, so: for values of F that are small compared to c (and small t but not sure what that means physically), we get v(t) is approx. equal to c*Ft/mc = Ft/m (which = at for small v as this means we can ignore the Lorentz factor), which is the Newtonian one… which seems quite interesting. Tell me where I’ve messed up.

I almost did it with a non-zero initial speed but I saw a bunch of sqrt(1-(v₀/c)^2) blah (because of the trig addition formulae and Pythagorean identities) in there so I avoided that and then gave up entirely (good times).

Pax