relativistic scratch

[CarwilB]

To account for relativistic effects, we use the relativistic formula for velocity under constant force. The relativistic version of Newton's second law is:

[ F = \frac{dp}{dt} ]

where ( p ) is the relativistic momentum given by:

[ p = \gamma m v ]

and

[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} ]

Given:

• ( F = 100 ) N,
• ( m = 1 ) kg,
• ( c = 3.0 \times 10^8 ) m/s,
• ( v = 0.9c = 2.7 \times 10^8 ) m/s.

We need to find the time ( t ) it takes to reach ( v = 0.9c ).

First, we express the force in terms of the rate of change of velocity:

[ F = \frac{d}{dt} (\gamma m v) ]

Since ( m ) is constant:

[ F = m \frac{d}{dt} (\gamma v) ]

We need to differentiate ( \gamma v ) with respect to ( t ):

[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} ]

Let ( \beta = \frac{v}{c} ), then:

[ \gamma = \frac{1}{\sqrt{1 - \beta^2}} ]

[ \frac{d(\gamma v)}{dt} = \frac{d}{dt} \left( \frac{v}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \right) ]

Using the chain rule, we get:

[ \frac{d(\gamma v)}{dt} = \frac{d}{dv} \left( \frac{v}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \right) \cdot \frac{dv}{dt} ]

[ \frac{d(\gamma v)}{dv} = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} + v \cdot \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \right) ]

[ \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \right) = \frac{v/c^2}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} ]

Thus:

[ \frac{d(\gamma v)}{dv} = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} + \frac{v^2/c^2}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} ]

[ \frac{d(\gamma v)}{dv} = \frac{1}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} ]

So:

[ F = m \frac{d(\gamma v)}{dt} = m \frac{1}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} \cdot \frac{dv}{dt} ]

[ 100 = 1 \cdot \frac{1}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} \cdot \frac{dv}{dt} ]

[ \frac{dv}{dt} = 100 \left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2} ]

Separate the variables and integrate:

[ \int_0^{0.9c} \frac{dv}{\left(1 - \left(\frac{v}{c}\right)^2\right)^{3/2}} = 100 \int_0^t dt ]

Let ( v = \beta c ), then ( dv = c d\beta ):

[ \int_0^{0.9} \frac{c d\beta}{\left(1 - \beta^2\right)^{3/2}} = 100 t ]

[ c \int_0^{0.9} \