Daily Archives: Tue, 2 June 2015

Acceleration under a constant force

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.15.

Under a constant force {F}, an object undergoes hyperbolic motion. In one dimension for a constant force we have

\displaystyle   \frac{dp}{dt} \displaystyle  = \displaystyle  F\ \ \ \ \ (1)
\displaystyle  p \displaystyle  = \displaystyle  Ft+C \ \ \ \ \ (2)

where {C} is a constant of integration. If the object starts at {t=0} at rest (in the lab frame), then {C=0}, and

\displaystyle  p=\frac{mu}{\sqrt{1-u^{2}/c^{2}}}=Ft \ \ \ \ \ (3)


which can be solved for the velocity {u} to give

\displaystyle  u=\frac{F}{m}\frac{t}{\sqrt{1+\left(Ft/mc\right)^{2}}} \ \ \ \ \ (4)

We can also get this formula by integrating the expression for the acceleration

\displaystyle  \mathbf{a}=\frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{\gamma mc^{2}}\left(\mathbf{u}\cdot\mathbf{F}\right) \ \ \ \ \ (5)

If {\mathbf{F}} is constant and acts on an object initially at rest, then {\mathbf{F}\parallel\mathbf{u}} and

\displaystyle   a \displaystyle  = \displaystyle  \frac{du}{dt}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{F}{\gamma m}\left(1-\frac{u^{2}}{c^{2}}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{F}{m}\left(1-\frac{u^{2}}{c^{2}}\right)^{3/2} \ \ \ \ \ (8)

To find {u\left(t\right)} we integrate:

\displaystyle   \int\left(1-\frac{u^{2}}{c^{2}}\right)^{-3/2}du \displaystyle  = \displaystyle  \frac{F}{m}\int dt\ \ \ \ \ (9)
\displaystyle  \frac{u}{\sqrt{1-u^{2}/c^{2}}} \displaystyle  = \displaystyle  \frac{F}{m}\left(t+t_{0}\right) \ \ \ \ \ (10)

If {u=0} at {t=0}, the constant of integration is {t_{0}=0} and we get 3 again.

From 3 we can get the inverse function

\displaystyle  t=\frac{mu}{F\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (11)

Thus as {u\rightarrow c}, {t\rightarrow\infty} so the object never quite reaches the speed of light.

Because of the velocity-dependent term in 5, the acceleration due to a constant force is not constant, but rather decreases as {u} increases. If we start with {a_{0}=F/m=g=9.8\mbox{ m s}^{-2}}, then the times required to reach various speeds are found from

\displaystyle  t=\frac{u}{9.8\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (12)

{\frac{u}{c}} {t}
0.9 {6.3\times10^{7}\mbox{ s}=2\mbox{ years}}
0.99 {2.14\times10^{8}\mbox{ s}=6.78\mbox{ years}}
0.999 {6.82\times10^{8}\mbox{ s}=21.6\mbox{ years}}
0.9999 {2.16\times10^{9}\mbox{ s}=68.4\mbox{ years}}
1.0 {\infty}

Relativistic acceleration in terms of force

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 4, Problem 4.14.

Ordinary force in relativity is given by

\displaystyle   \mathbf{F} \displaystyle  = \displaystyle  \frac{m}{\sqrt{1-u^{2}/c^{2}}}\left[\mathbf{a}+\frac{\left(\mathbf{u}\cdot\mathbf{a}\right)\mathbf{u}}{c^{2}-u^{2}}\right] \ \ \ \ \ (1)

To get a general expression for the acceleration {\mathbf{a}} in terms of the force, we take the dot product of both sides with the velocity {\mathbf{u}}:

\displaystyle   \mathbf{u}\cdot\mathbf{F} \displaystyle  = \displaystyle  \gamma m\left(\mathbf{u}\cdot\mathbf{a}\right)\left(1+\frac{u^{2}}{c^{2}-u^{2}}\right)\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{\gamma mc^{2}\left(\mathbf{u}\cdot\mathbf{a}\right)}{c^{2}-u^{2}}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \gamma^{3}m\left(\mathbf{u}\cdot\mathbf{a}\right)\ \ \ \ \ (4)
\displaystyle  \mathbf{u}\cdot\mathbf{a} \displaystyle  = \displaystyle  \frac{\mathbf{u}\cdot\mathbf{F}}{\gamma^{3}m} \ \ \ \ \ (5)

Substituting back into 1, we get

\displaystyle   \mathbf{a} \displaystyle  = \displaystyle  \frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{c^{2}-u^{2}}\frac{\mathbf{u}\cdot\mathbf{F}}{\gamma^{3}m}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mathbf{F}}{\gamma m}-\frac{\mathbf{u}}{\gamma mc^{2}}\left(\mathbf{u}\cdot\mathbf{F}\right) \ \ \ \ \ (7)

In the limit of small {\mathbf{u}}, this reduces to the familiar Newton’s law {\mathbf{F}=m\mathbf{a}}, but in the relativistic region, the acceleration depends on the object’s velocity. As a result, the acceleration isn’t parallel to the force unless {\mathbf{F}} is either parallel to {\mathbf{u}} or {\mathbf{F}\perp\mathbf{u}}; in the latter case {\mathbf{u}\cdot\mathbf{F}=0} and the second term is zero.