GRE Physics Subject Test common mistakes
GRE Physics Subject Test · Understand the traps and how to avoid them.
Computing the speed of a rolling or sliding object by setting the entire gravitational potential energy equal to translational kinetic energy, ignoring the rotational term (1/2)Iω², so a solid sphere rolling down an incline is reported as moving faster than it actually does.
The translational energy equation is memorized first and applied by reflex; the moment of inertia is only invoked when the word 'rotation' appears explicitly. Rolling without slipping is treated as a kinematics detail rather than an energy constraint, and the factor of 2/5 or 1/2 is easily forgotten under time pressure.
What to do instead
Before writing any energy equation, list every moving mass element and ask whether it both translates and rotates. For a rigid body rolling without slipping, write E = 1/2 mv² + 1/2 Iω² and substitute ω = v/R immediately so the v² cancels. Drill the standard moments of inertia (solid sphere 2/5 mR², cylinder 1/2 mR², hoop mR²) and check the limiting case: the faster the body, the smaller its effective inertia coefficient.
Treating the electric potential at a point as a scalar sum of the potentials from each charge but then also adding the field vectors as if they contributed to potential, or conversely computing potential by vector addition and getting zero where the field is not zero (or vice versa).
Field and potential are both produced by the same charge distribution and are introduced back to back, so the computational habits blur. Candidates remember that potential is a scalar and field a vector but do not reliably carry that distinction into sign conventions, especially for dipoles where the potential on the perpendicular bisector is zero while the field is not.
What to do instead
State the object's type before computing: potential is a scalar with sign from the charge sign, field is a vector with direction from the charge sign. For a dipole, verify the perpendicular-bisector case explicitly: V = 0 there but E points opposite to the dipole moment. Use E = −∇V as a consistency check by computing the gradient of your potential and comparing direction and magnitude with the direct field sum.
Applying the thin-lens or mirror equation with a sign convention chosen inconsistently between object, image, and focal length, then reporting a virtual image as real (or a magnification with the wrong sign) when the setup involves a diverging element.
Multiple sign conventions circulate (real-is-positive versus Cartesian), and the exam rarely restates one. Under timing pressure candidates plug numbers into |1/f| = 1/do + 1/di without deciding in advance what a negative image distance means for the particular element, so the arithmetic is right but the physical interpretation is inverted.
What to do instead
Commit to one convention and write it at the top of your scratch work: for lenses, real object distance positive, real image distance positive on the far side, converging f positive, diverging f negative; magnification m = −di/do. Sanity-check every result with a ray sketch: a diverging lens always gives a virtual, upright, diminished image of a real object. If the computed image is on the same side as the incoming light for a lens, it is virtual.
Using the ideal-gas relation PV = nRT with the wrong R (8.314 J/mol·K versus 0.0821 L·atm/mol·K) or plugging Celsius temperatures directly into a ratio, producing answers off by orders of magnitude or by the 273 offset.
Both gas constants are legitimate and the choice depends on the pressure and volume units in the problem, which are often given in atm and liters for chemistry-flavored items but pascals and cubic meters for physics-flavored ones. Candidates also shortcut temperature ratios in adiabatic or Carnot expressions using Celsius values because the numbers are smaller and 'look right'.
What to do instead
Convert everything to SI (Pa, m³, K, mol) before substituting, then R = 8.314 J/mol·K is the only constant needed. For ratios such as T₂/T₁ or the Carnot efficiency, convert to kelvin first; a Celsius ratio is never valid. As a check, one mole of ideal gas at STP occupies about 22.4 L, so any computed volume far from that scale signals a unit error.
Solving the infinite square well or harmonic oscillator with the wrong quantum number, for example using n = 0 as the ground state of the particle in a box, or computing the energy spacing of the oscillator as if it were proportional to n rather than uniform.
The n = 1 ground state of the box conflicts with the harmonic oscillator's n = 0 ground state, and both formulas are memorized as bare expressions (n²h²/8mL² and (n + 1/2)ħω) without the boundary conditions that generate them. Candidates also assume all bound-state spectra have growing gaps.
What to do instead
Re-derive the box quantization from the boundary condition sin(kL) = 0, which gives kL = nπ with n = 1, 2, 3, …, so n = 0 is excluded because it gives a zero wavefunction. For the oscillator, note the spectrum is evenly spaced by ħω with ground-state energy ħω/2. When a problem asks for a transition energy, subtract the two level energies rather than quoting a single level.
Predicting the fine-structure or Zeeman splitting pattern from the wrong selection rules or from ignoring the spin-orbit coupling sign, so the number of allowed spectral lines or their ordering is wrong.
Selection rules (Δl = ±1, Δm = 0, ±1, Δs = 0) are listed as facts but not connected to the angular momentum algebra, and the anomalous Zeeman effect requires the Landé g-factor, which candidates often omit and treat as g = 1 for all states. The sign of the spin-orbit term also depends on whether the shell is more or less than half full.
What to do instead
For any transition, write the initial and final term symbols and check Δl and Δm explicitly before counting lines. For the anomalous Zeeman effect, compute the Landé g-factor g = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)] and use ΔE = g μ_B B Δm. Remember the normal Zeeman triplet only appears for singlet states (S = 0).
Adding velocities with the Galilean rule u = u′ + v for speeds close to c, or applying the time-dilation factor γ in the wrong frame, so a relativistic momentum or lifetime problem comes out larger than physically allowed.
Everyday experience and Newtonian mechanics train the additive rule, and the Lorentz transformation is recalled as a formula but not as a change of which frame measures proper time and proper length. Candidates also forget that γ depends on the relative speed between the two frames, not on the speed of the object in the lab.
What to do instead
Use the velocity-addition formula u = (u′ + v)/(1 + u′v/c²) whenever any speed exceeds about 0.1c, and check that the result never exceeds c. Identify proper time as the interval measured by a clock at rest with respect to the events (the shortest one) and proper length as the rest-frame length (the longest one); then apply γ as a stretch to time and a contraction to length. For momentum, use p = γmv, not mv.
Reading an uncertainty or error bar as a fixed correction and reporting a measured value without combining statistical and systematic uncertainties in quadrature, or quoting a result with more significant figures than the data support.
Introductory labs emphasize a single 'percent error' against a textbook value, so candidates treat uncertainty as a grade rather than a property of the measurement. The quadrature rule for independent uncertainties is learned for one case and then applied incorrectly to correlated or purely systematic errors, and calculator output is copied verbatim.
What to do instead
Separate statistical from systematic contributions, combine independent ones in quadrature (σ_total = √(σ_stat² + σ_syst²)), and treat a systematic shift as a bias, not a random scatter. Report the result as value ± uncertainty with the uncertainty to one or two significant figures and the value rounded to the same decimal place. Check that the quoted uncertainty is consistent with the spread of repeated trials.
Assuming a small-angle pendulum or a mass-spring system is exactly simple harmonic and using T = 2π√(L/g) or T = 2π√(m/k) outside the regime where the approximation holds, or confusing the angular frequency ω with the frequency f.
The SHM formulas are the first oscillation results learned and are applied to any periodic motion, including large-amplitude pendulums, physical pendulums with extended mass distributions, and damped systems. The factor of 2π between ω and f is a frequent slip because both are called 'frequency' in casual speech.
What to do instead
State the approximation before using it: the pendulum formula requires small amplitude (roughly under 10°) and a point mass on a massless string; for an extended body use T = 2π√(I/(mgd)) with I about the pivot. Convert deliberately between ω = 2πf and f = 1/T, and check units: ω has units of rad/s, f of Hz. For damping, confirm whether the problem asks for the damped frequency or the natural frequency.
Confusing the thermodynamic potentials and applying the wrong one to a process: using ΔU = Q − W with the wrong sign for work, or treating enthalpy, Helmholtz, and Gibbs free energies as interchangeable when deciding spontaneity or available work.
The sign convention for work (work done by versus on the system) differs between physics and chemistry texts, and the potentials are defined by Legendre transforms that are memorized as symbols rather than as statements about which variables are held constant. Candidates also assume ΔG < 0 applies to any process rather than to constant temperature and pressure.
What to do instead
Write the first law in one convention and keep it: ΔU = Q − W with W the work done by the system. Then attach each potential to its natural variables: U(S, V), H(S, P), F(T, V), G(T, P). Use ΔG < 0 for spontaneity at constant T and P, ΔF < 0 at constant T and V, and ΔS_universe > 0 as the general criterion. For an ideal gas isothermal process, ΔU = 0 and Q = W, which is a quick consistency check.