Physics and Astronomy, University of Kent
In this and the following lectures we will discuss further implications of special relativity on momentum and energy. Since both momentum and kinetic energy depend on velocity, it’s perhaps not surprising that our definitions for these will need to be modified. However, first we must study the relativistic equivalent of the velocity transformations which we looked at in Lecture 1.
Just as we replaced the Galilean position transformations with the Lorentz transformations, we must now replace the velocity transformations with their relativistic equivalents. A little care is needed here because there is one critical difference between position and velocity. For the position transformations, co-ordinates in which the reference frames aren’t moving (\(y\) and \(z\)) experience no relativistic correction (i.e. \(y=y'\) and \(z=z'\)). This isn’t the case for velocity - the y and z velocities, \(v_y\) and \(v_z\) do experience a correction. This is because velocity is a function of time, and since the time co-ordinate does needs a relativistic correction, it follows that all velocities will also experience a correction, regardless of the direction.
The relativistic velocity transformations between frames \(S\) and \(S'\), where \(S\) is moving with velocity \(u\) along the x-axis with respect to \(S'\) are: \[v_x = \frac{v_x' + u}{1 + uv_x'/c^2}, \qquad v_y = \frac{v_y'}{\gamma(1 + uv_x'/c^2)}, \qquad v_z = \frac{v_z'}{\gamma(1 + uv_x'/c^2)}\] Similarly, we have the inverse transforms \[v_x' = \frac{v_x - u}{1 - uv_x/c^2}, \qquad v_y' = \frac{v_y}{\gamma(1 - uv_x/c^2)}, \qquad v_z' = \frac{v_z}{\gamma(1 - uv_x/c^2)}\]
As for the Lorentz transformation, it should be the case that these reduce to our familiar Galilean velocity transformation for speeds far below the speed of light (\(u << c\)). This is indeed true, because \(uv_x'/c^2\) is then approximately equal to 0 and \(\gamma \approx 1\) and so \(v_x \approx v_x' + u\), \(v_y \approx v_y'\) and \(v_z \approx v_z'\).
It must also be the case that a velocity of \(c\) is one reference frame transforms into a velocity of \(c\) into another, since this is one of the postulates on which the Lorentz transformations are based. We can confirm this for light moving along the x-axis by substituting in \(v_x'=c\), so that:
\[\begin{aligned} v_x &= \frac{c+ u}{1 + uc/c^2} \\ &= \frac{c+ u}{1 + u/c} \\ &= \frac{c(c+ u)}{c + u} \\ & = c \end{aligned}\]
For light moving along the y-axis, we can show the same thing. If we simply substitute \(v_y = c\) into the equation for \(v_y'\) we get \(v_y' = c/\gamma\). However, we have to remember that there will be both an \(x\) and \(y\) component to the transformed velocity (you can verify this by substituting \(v_x = 0\) into the equation for \(v_x'\), it can be seen that \(v_x = u\)).
The total velocity is then \(\sqrt{v_x^2 +
v_y^2}\), which can be shown to be \(c\). This is an exercise on the problem
sheet.
Also note the following:
The equations for \(v_y\) and \(v_z\) contain \(v_x'\) in the denominator. That is, the transformation of the \(y\) and \(z\) velocities depends on the \(x\) velocity.
Don’t confuse the \(v\)s with \(u\). \(u\) is the velocity of the frame of reference \(S'\) relative to \(S\) (along the \(x\) direction), while the \(v\)s and \(v'\)s are the velocity of some object as measured in frames of reference \(S\) and \(S'\), respectively.
The relativistic velocity transformations can be derived from the Lorentz transformations. To make this simple, we first need to recognise that the Lorentz transformations are equally valid if we replace the \(x\), \(y\), \(z\) and \(t\) co-ordinates and their primed counterparts with intervals (or differences), \(\Delta x\), \(\Delta y\), \(\Delta z\) and \(\Delta t\). We can make these intervals infinitesimally small, calling them, \(dx\), \(dy\), \(dz\), and \(dt\). Then we also note that the x, y and z components of velocity can be written \(v_x = dx/dt\), and so on. So, starting with the velocity in the x-direction, we have the following to work with: \[dx = \gamma(dx' + udt'), \qquad dt = \gamma\bigg(dt' + u\frac{dx'}{c^2}\bigg),\] and \[v_x = \frac{dx}{dt}, \qquad v_x' = \frac{dx'}{dt'}.\] Now, plug in the expressions for \(dx\) and \(dt\) into the equation for \(v_x\), giving us \[\begin{aligned} v_x &= \frac{dx}{dt} \\ &= \frac{\gamma(dx' + udt')}{\gamma(dt' + udx'/c^2)} \end{aligned}\] The \(\gamma\) factors cancel, \[v_x = \frac{dx' + udt'}{dt' + udx'/c^2}.\] We want to end up with an equation containing velocities, \(v_x'\), so let’s divide through every term by \(dt'\) \[\begin{aligned} v_x &= \frac{dx'/dt' + u}{1 + udx'/c^2dt'} \end{aligned}\] Now, we can simply substitute in \(v_x' = dx'/dt'\) to get the equation for \(v_x\) in terms of \(v_x'\). \[\boxed{v_x = \frac{v_x' + u}{1 + uv_x'/c^2}.}\] To derive the velocity in the \(y\) direction, \(v_y\), we similarly start with \[dy = dy', \qquad dt = \gamma(dt' + u\frac{dx'}{c^2}),\] and \[v_y = \frac{dy}{dt}, \qquad v_y' = \frac{dy'}{dt'}.\] We perform a similar substitution to get \[\begin{aligned} v_y = \frac{dy}{dt} = \frac{dy'}{\gamma(dt' + udx'/c^2)}. \end{aligned}\] Notice it looks a bit different to the \(v_x\) case, we no longer have a \(\gamma\) on the top, because \(dy = dy'\). Again, we divide through by \(dt'\) to get velocities: \[v_y = \frac{dy'/dt'}{\gamma(1 + udx'/c^2dt)},\] and so \[\boxed{v_y = \frac{v_y'}{\gamma(1 + uv_x'/c^2)}.}\] The derivation for \(v_z\) is the same, replacing \(z\) every time you see a \(y\).
Relative Velocity of Two Rockets: Two rockets are heading straight towards each other, both moving at \(c/2\) relative to the Earth. What does one rocket measure for the velocity of the other?
Solution: Common sense (and the Galilean velocity transformations) would tell us \(c\). But we know that’s not right. Instead, let’s use the velocity transformation: \[\begin{aligned} v_x' = \frac{v_x - u}{1 - uv_x/c^2} \end{aligned}\]
We need to decide what \(v_x\) and \(u\) are. The velocities of the rockets are measured to be \(c/2\) in the Earth’s reference frame (\(S\)). So, for example, one of the rockets (call it rocket A) has \(v_x = -c/2\) in the Earth’s frame. Now we want to transform this velocity to the frame of rocket B (\(S'\)), so that we know what velocity rocket B measures rocket A to have. The reference frame of Rocket B is moving relative to Earth at \(u=c/2\), and so: \[\begin{aligned} v_x' &= \frac{v_x - u}{1 - uv_x/c^2} \\ v_x &= \frac{-c/2 - c/2}{1 - (-c/2)(c/2)/c^2} \\ v_x &= \frac{-c}{1 + 1/4} \\ v_x &= -0.8c \end{aligned}\] Which, as we would expect, is less than \(c\).