Does anybody know what the formula is for the graph you see in text books, of speed against time for a constant force applied to an object. The one where speed increases with time, and then levels of as it approaches C.

I know how to figure it out:

 

1) Find out what the momentum after time t is, when a constant force is applied

2) Find out how the velocity of an object depends on its momentum (and its mass/energy).

3) Plug it together.

For non reletavistic speeds I think the formula is f=ma, which when re-arranged would give t=mv/f. and for a force of 1N applied to a mass of 1kg t=v. So after 10 seconds the mass would be doing 10m/s,after 20s 20m/s.

 

But at speeds close to see we know that this is not the case, so does anybody know the formula that matches the graph I mentioned?

sorry, I meant to post this in 'Relativity'. Can anyone move it please?

You should go via momentum. Sadly, no one else bothered replying here and I will probably not have time to guide you to the equation within the next weeks. So I will just give you the formula I suppose you are looking for:

 

[math] F = \frac{dp}{dt} \Rightarrow p = Ft, v = \frac{pc^2}{E} = \frac{pc^2}{\sqrt{m^2c^4 + p^2c^2}} \Rightarrow v = \frac{Ftc^2}{\sqrt{m^2c^4 + F^2 t^2 c^2}} [/math]

 

The difference is in part 2) of the points I previously mentioned. In classical mechanics, the relation of velocity and momentum is p = mv. In relativistic mechanics, the relation is different (the one I used above).

You should go via momentum. Sadly, no one else bothered replying here and I will probably not have time to guide you to the equation within the next weeks. So I will just give you the formula I suppose you are looking for:

 

[math] F = \frac{dp}{dt} \Rightarrow p = Ft, v = \frac{pc^2}{E} = \frac{pc^2}{\sqrt{m^2c^4 + p^2c^2}} \Rightarrow v = \frac{Ftc^2}{\sqrt{m^2c^4 + F^2 t^2 c^2}} [/math]

 

The difference is in part 2) of the points I previously mentioned. In classical mechanics, the relation of velocity and momentum is p = mv. In relativistic mechanics, the relation is different (the one I used above).

 

Thanks a lot, that works perfect.