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.
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2What kind of reasoning is this? What, on your understanding, does the symbol $\Delta t$ represent in the energy-time uncertainty relation? http://physics.stackexchange.com/q/53802/ – Alfred Centauri Nov 04 '14 at 21:48
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A change in time. – Goodies Nov 04 '14 at 22:10
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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.
Prof. Legolasov
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The question is correct in that the relativistic (or lorentz-invariant) time-energy uncertainty relation is a bit different (but still there is)
For example here is a pre-print Lorentz-Invariant Time-Energy Uncertainty Relation for Relativistic Photon, arxiv
Abstract:
The time-energy uncertainty relation is discussed for a relativistic massless particle. The Lorentz-invariant uncertainty relation is obtained between the root-mean-square energy deviation and the scatter of registration time. The interconnection between this uncertainty relation and its classical analogue is established.
