# Energy-Time Uncertainty Principle and Photons

Heisenberg's uncertainty principle states that: $$\Delta E \cdot \Delta t \ge \frac{\hbar}{2}$$ It is clear that this has nothing to do with the accuracy of our measurements, but rather is a fundamental 'law' in the quantum world. Now, we also know that photons do not experience time because of the Lorentz transformation time dilation: $$\gamma t = \frac{t}{1-\frac{v^2}{c^2}} = \frac{t}{1-\frac{c^2}{c^2}} = \frac{t}{1-1} = \frac{t}{0} = \text{undefined}$$ Is the uncertainty principle not relativistic? Thanks.

• What kind of reasoning is this? What, on your understanding, does the symbol $\Delta t$ represent in the energy-time uncertainty relation? physics.stackexchange.com/q/53802 – Alfred Centauri Nov 4 '14 at 21:48
• A change in time. – Goodies Nov 4 '14 at 22:10

The relation says that it would take about $\Delta t$ time to measure the energy with an error of order $\Delta E$. $\Delta t$ is not the photon time, it is the time in the observer's (laboratory) frame of reference.