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Year 1 Sem 1/LC Classical Mechanics and Relativity 1/Lecture 08

From Ash's Lecture Notes

Lecture 8: Special Relativity IV and Mechanics V(a): Intro to Dynamics
Thu 23 Oct 2025 15:00

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The Relativistic Doppler Effect

For an emitted frequency $f_0$, emitted by an object moving with velocity (in 1D) $u$ relative to an observer, the received frequency $f$ is given by:

$$f = \sqrt{\frac{1 + u/c}{1 - u/c}} f_0$$

For a relativistic speed $u$. Note that we cannot use the standard Doppler formula for relativistic speeds. Also note the lack of $\pm$, as we encode this into $u$. If the object moves towards the observer, $u$ is positive, and if the object is moving away $u$ is negative.

If $u$ is non-relativistic, we assume that $u/c \ll 1$:

$$f = \left(1 + \frac{u}{c}\right)^{1/2} \left(1 - \frac{u}{c}\right)^{-1/2} f_0$$

And Taylor Series expanding:

$$= \left(1 + \frac{1}{2} \frac{u}{c} + \cdots\right)\left(1 - \frac{-1}{2} \frac{u}{c} + \cdots\right) f_0$$

Since $u/c$ is small, we ignore any quadratic, cubic etc terms of $u/c$, as these are very small.

$$= \left(1 + \frac{1}{2} \frac{u}{c} + \frac{1}{2} \frac{u}{c} \right) f_0$$

Hence:

$$f \approx \left(1 + \frac{u}{c}\right) f_0$$

Since we assume $u/c \ll 1$:

$$f = f_0 + \frac{u}{c} f_0$$
$$f - f_0 = \Delta f = \frac{u}{c} f_0$$
$$\frac{\Delta f}{f_0} = \frac{u}{c} \qquad \text{plus higher order terms we ignore}$$

This is the classical result that we’re familiar with, for non-relativistic speeds.

Lorentz Transformation

Say we have a reference frame $s^\prime$ which is moving along the $x$-direction relative to a static reference frame $s$.

An event in the $s$ frame has coordinates $(x, y, z, t)$ and the same event in the $s^\prime$ frame has coordinates $(x^\prime, y^\prime, z^\prime, t^\prime)$. We have the following transformations:

$$t^\prime = \gamma\left(t - \frac{u}{c^2}x\right)$$
$$x^\prime = \gamma \left(x - ut\right)$$

Noting that $y = y^\prime$, and $z = z^\prime$ as these are orthogonal to the direction of motion. We also have:

$$u^\prime_x = \frac{u_x - u}{1 - \frac{u}{c^2} u_x}$$

Please note that CMR1 does not include derivations of these equations (which collectively form the Lorentz Transformations), however for understanding’s sake I’ll include them here regardless.

Derivations

We want a transformation in time and space between a stationary frame $S$ and the moving frame $S^\prime$. We have three postulates to do this:

  • Linearity: The transformation must be linear, i.e. a straight line in $S$ must map to a straight line in $S^\prime$.
  • Standard rule for $c$: $c$ is invariant and has the same velocity in all frames, regardless of motion.
  • Inverse symmetry: The inverse transformation ($S^\prime \to S$) is the same, but with $u \to -u$, as in $S^\prime$’s reference frame, it is static with $S$ moving with speed $-u$.
Deriving $x^\prime$:

We derive the transformation in position using length contraction:

Imagine a ruler at rest in the moving frame $S^\prime$. It has one end on the origin $O^\prime$ and the right end at some coordinate $x^\prime$. The ruler therefore has proper length $L_0 = x'$.

From the stationary frame $S$:

  • The origin $O^\prime$ has moved distance $ut$ after some time $t$.
  • The ruler is moving, so has been length contracted to the observer in $S$. The ruler now appears to have length $x^\prime / \gamma$.

The total coordinate as seen by $S$ is therefore:

$$x = ut + \frac{x^\prime}{\gamma}$$

And rearranging gives:

$$x^\prime = \gamma(x - ut)$$
Deriving $t^\prime$:

We derive the transformation for time using the third postulate above:

If the transformation from $S \to S^\prime$ is:

$$x^\prime = \gamma(x - ut)$$

Then the transformation from $S^\prime \to S$ is:

$$x = \gamma(x^\prime + ut^\prime)$$

As the scenario is the same: $S'$ is moving with speed $u$ relative to $S$, if we instead consider $S^\prime$’s reference frame then it is static, and $S$ is moving in the opposite direction with the same magnitude of velocity, so the transformation must be the same with $u \to -u$.

Substituting the position transformation for $x^\prime$ into this:

$$x = \gamma[\gamma(x - ut) + ut^\prime]$$
$$\frac{x}{\gamma} = \gamma x - \gamma ut + u t^\prime$$

Rearranging for $t^\prime$:

$$t^\prime = \gamma t - \frac{x}{u}\left(\gamma - \frac{1}{\gamma}\right)$$

Using the identity $\gamma - 1/\gamma = \beta^2\gamma = \frac{u^2}{c^2}\gamma$:

$$t^\prime = \gamma t - \frac{x}{u}\left(\gamma \frac{u^2}{c^2}\right)$$
$$t^\prime = \gamma \left(t - \frac{u}{c^2}x\right)$$

Which gives us the time transformation.

Deiving $u_x^\prime$:

We find velocity in the moving frame as $dx' / dt'$.

The velocity in the moving frame $S^\prime$ is defined as $u^\prime_x = \frac{dx^\prime}{dt^\prime}$. Taking these derivatives

$$dx^\prime = \gamma(dx - u dt) \quad \text{and} \quad dt^\prime = \gamma\left(dt - \frac{u}{c^2}dx\right)$$

Substituting these into the definition of velocity:

$$u^\prime_x = \frac{\gamma(dx - u dt)}{\gamma\left(dt - \frac{u}{c^2}dx\right)} = \frac{dx - u dt}{dt - \frac{u}{c^2}dx}$$

Dividing through by $dt$:

$$u^\prime_x = \frac{\frac{dx}{dt} - u}{1 - \frac{u}{c^2}\frac{dx}{dt}}$$

And using $u_x = \frac{dx}{dt}$:

$$u^\prime_x = \frac{u_x - u}{1 - \frac{u u_x}{c^2}}$$

Dynamics

Kinematics is effectively looking at objects in motion. Dynamics is effectively “why” they move (Newton’s laws, static friction etc).

Newton’s Laws

We have three:

  1. If there is no resultant force acting upon an object, there are two possibilities:
    • The object was initially at rest, and stays at rest.
    • The object moves at a constant speed with no acceleration.
  2. $\vec{F} = m \vec{a}$. Force and acceleration are proportional with a constant of proportionality $m$, the “inertial mass”. We call it inertial mass because this is theoretically distinct from gravitational mass, however all experiments give them as having them same value.
  3. The Reaction Principle. If a body $A$ is producing a force $\vec{F}$ on a body $B$, then $B$ acts back upon $A$ with a force of the same magnitude but the opposite direction “Every action has an equal and opposite reaction”.

Superposition Principle

If we have some body with $N$ forces acting upon it, the final resultant force that acts upon an object is a vector sum of these forces:

$$\vec{F} = \vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_n = \sum_{i=1}^{N} \vec{F}_i = m \vec{a}$$

Example

Consider an object of mass $m$ hanging from the ceiling with an “ideal string”. This means that string is inextensible and is massless. The body is initially at rest.

Two forces act upon this body:

  • The weight force due to local gravitational acceleration: $w = m g$.
  • The force produced by the string (tension).

Since the body is at rest, the resultant force must be zero and:

$$T - mg = 0$$
$$T = mg$$

Example II

Consider a body on a horizontal surface. We laterally pull the object with force $F$. We have these forces:

  • Again a weight force: $w = mg$.
  • The normal force produced by the table acting back upon the body iaw Newton’s Third Law.
  • A frictional force acting in opposition to the direction of motion, $F_\text{fric}$

The frictional force is proportional to the Normal force with a coefficient depending on the materials used:

$$F_\text{fric} = \mu N$$

When moving an object there are two stages:

  • Attempting to take an object from stationary to actually moving.
  • Continuing the motion of the object once it’s moving (this is easier).

We therefore have multiple coefficients of friction. Here we consider the coefficient of static friction $\mu_s$ and the coefficient of kinetic friction $\mu_k$. There is also the coefficient of rolling friction $\mu_r$ seen in labs. Generally, $\mu_s > \mu_k$.


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