Mechanics and Special Relativity Diagnostic

General relativity assumes that several earlier ideas are already usable together: stationary action and conservation in mechanics, invariant intervals and four-vectors in special relativity, and observer-based measurement. This diagnostic separates those capabilities so that one weak area does not force a review of all three. It supplies a concise review and then returns you to the GR route that sent you here.

You should leave able to formulate a mechanical problem variationally, reason with Minkowski causal geometry and energy–momentum, and distinguish coordinate components from quantities measured by a specified observer.

Mechanics and special-relativity diagnostic

1. A particle moves in a central potential $V(r)$ with $$L=\frac m2(\dot r^2+r^2\dot\phi^2)-V(r).$$ Identify the conserved angular momentum and derive the radial equation in terms of it.
2. In units $c=1$, events differ by $\Delta x^\mu=(5,3,0,0)$. Classify the separation, find the proper time between the events, and give the speed of the inertial frame in which they occur at the same place.
3. A particle has four-momentum $p^\mu=(5,3,0,0)$ and an observer has four-velocity $u^\mu=\gamma(1,v,0,0)$ with $v=1/2$. Compute the observer’s measured energy $E_{(u)}=-p\cdot u$ and check that it is positive.
4. The worldline $$t(\tau)=a^{-1}\sinh(a\tau), \qquad x(\tau)=a^{-1}\cosh(a\tau)$$ describes hyperbolic motion. Verify that $\tau$ is proper time and that the four-acceleration has constant magnitude $a$.

Hints and solutions

Hints for 1

$\phi$ is cyclic. Use its conjugate momentum to eliminate $\dot\phi$ from the Euler–Lagrange equation for $r$.

Solution to 1

Since $L$ has no explicit $\phi$ dependence,

$$\ell=\frac{\partial L}{\partial\dot\phi}=mr^2\dot\phi$$

is constant. The radial Euler–Lagrange equation is

$$m\ddot r=mr\dot\phi^2-V'(r) =\frac{\ell^2}{mr^3}-V'(r).$$

Equivalently $m\ddot r=-V_{\rm eff}'(r)$ for $V_{\rm eff}=V+\ell^2/(2mr^2)$. The $\ell\to0$ limit removes the centrifugal term, and every term has force dimension.

Hints for 2

With the site signature, compute $s^2=-(\Delta t)^2+(\Delta x)^2$. A timelike displacement has a rest frame with boost speed $v=\Delta x/\Delta t$.

Solution to 2

The interval is

$$s^2=-25+9=-16,$$

so the separation is timelike and the elapsed proper time is $\Delta\tau=\sqrt{-s^2}=4$. The boost $v=3/5$ gives $\Delta x'=\gamma(\Delta x-v\Delta t)=0$. Since $|v|\lt1$, this is a valid inertial frame.

Hints for 3

First find $\gamma=1/\sqrt{1-v^2}$. Remember the negative time component in the scalar product.

Solution to 3

For $v=1/2$, $\gamma=2/\sqrt3$. Therefore

$$E_{(u)}=-p_\mu u^\mu =\gamma(5-3v) =\frac{7}{\sqrt3}\gt0.$$

Both $p^\mu$ and $u^\mu$ are future-directed timelike vectors, so positivity was expected. The number is a scalar for the specified observer, although another observer measures another scalar.

Hints for 4

Differentiate once to obtain $u^\mu$ and twice to obtain $A^\mu$. Check $u^2=-1$, $u\cdot A=0$, and $A^2=a^2$.

Solution to 4

Differentiation gives

$$u^\mu=(\cosh a\tau,\sinh a\tau), \qquad A^\mu=a(\sinh a\tau,\cosh a\tau).$$

Thus

$$u^2=-\cosh^2(a\tau)+\sinh^2(a\tau)=-1,$$

so $\tau$ is proper time. Also $u\cdot A=0$ and

$$A^2=a^2\bigl[-\sinh^2(a\tau)+\cosh^2(a\tau)\bigr]=a^2.$$

The proper-acceleration magnitude is constant even though the coordinate acceleration is not.

Interpret the result

Use each prompt to check a different skill.

• Secure: the corresponding solution is set up invariantly and includes its limit or norm check.
• Uncertain: the right object is chosen but a sign, projection, or conservation argument needs a hint. Review only that capability and retry with changed data.
• Not yet: a cyclic coordinate is not connected to conservation, timelike and spacelike separations are confused, or measured energy is read directly from a coordinate component without naming an observer.

Question 1 samples mechanics; question 2 samples special-relativistic causal geometry; questions 3–4 sample observer measurement and four-vector kinematics. A weakness in one category does not imply a weakness in the others.

Focused review

Mechanics and variational reasoning

An action $S=\int L(q,\dot q,t)dt$ with fixed endpoints yields the Euler–Lagrange equations. A continuous symmetry gives a conserved quantity; a cyclic coordinate is the simplest case. Review the derivation and its boundary assumptions at Variational Mechanics and Symmetries. Effective potentials reorganize radial motion but do not add a new force.

Special-relativistic geometry

The invariant interval, not coordinate time alone, determines causal class. With $(-,+,+,+)$, timelike vectors have negative squared norm and future-directed four-velocities satisfy $u^2=-1$. Review Events, Intervals, and Minkowski Geometry and verify every boost by preserving the interval.

Observers and measured components

At an event, an observer is a future-directed unit timelike vector $u^\mu$. The measured energy of momentum $p^\mu$ is $-p\cdot u$, while the spatial momentum is its projection orthogonal to $u$. Coordinate components become readings only after a frame is specified. Observers, Rest Spaces, and Measured Components develops this construction.

Transfer task: the free relativistic particle

For one-dimensional motion in an inertial frame, take

$$L(v)=-m\sqrt{1-v^2}.$$

Derive the canonical momentum $p$, the Hamiltonian $H=pv-L$, and the identity relating $H$, $p$, and $m$. Then recover the Newtonian kinetic term from the small-$v$ expansion.

Hint 1

Differentiate $L$ with respect to $v$. Use $\gamma=(1-v^2)^{-1/2}$.

Hint 2

After finding $H$, eliminate $v$ using $p=m\gamma v$ and $\gamma^2(1-v^2)=1$.

Solution

The momentum is

$$p=\frac{\partial L}{\partial v} =\frac{mv}{\sqrt{1-v^2}}=m\gamma v.$$

The Legendre transform gives

$$H=pv-L=m\gamma v^2+\frac{m}{\gamma}=m\gamma.$$

Consequently

$$H^2-p^2=m^2\gamma^2(1-v^2)=m^2.$$

This is the mass-shell relation for energy $H$ and spatial momentum $p$. For $v\ll1$,

$$H=m\left(1+\frac12v^2+O(v^4)\right) =m+\frac12mv^2+O(v^4).$$

The rest energy is the constant $m$; the first velocity-dependent term is the Newtonian kinetic energy. The Lagrangian itself expands as $L=-m+mv^2/2+O(v^4)$, and the additive constant does not affect the equations of motion.

Common errors

• Calling stationary action “minimum action” without checking the second variation.
• Mixing the site’s $(-,+,+,+)$ norm with formulas copied from the opposite signature.
• Treating coordinate velocity, four-velocity, and proper acceleration as the same derivative.
• Saying energy is “not a scalar”: $-p\cdot u$ is a scalar once the observer is specified.

Return to your route

After reviewing only the topics you need and completing the transfer task, return to the Graduate General Relativity Self-Study Path or the First Graduate Course in General Relativity.