Recent content by jollyredgiant

Ah, ok. I think its still basically the same problem. The stationary observer will claim the time delay between detections is different than what the observer on the ship claims. However, they will also claim the light traveled different lengths. They will both disagree on the time interval and...

I think you're forgetting that simultaneity is relative. According to the stationary observer, the light will hit the beam splitter, then hit the back of the ship, then the front. However, according to the ship's frame of reference, the laser will hit the beam splitter and then both ends at the...

A set of vectors is linearly independent if for any vector, v_i , in the set of vectors, \{ v_1, v_2, ..., v_n \} , it cannot be written as a linear combination of the other vectors. So what this means is that if v_i = c_1 v_1 + ... c_{i-1} v_{i-1} + c_{i+1} v_{i+1} + ... + c_n v_n (where...

Wait a minute... When the Hamiltonian is written as \hat{H}_D = -i\gamma^0 \gamma \cdot \nabla + \gamma^0 m then \gamma \cdot \nabla is a four vector product?? So that \gamma \cdot \nabla = -\vec{\gamma} \cdot \vec{\nabla} . I think that's it. Is that what you meant, Bill? So. Much...

I thought I did take that into account. Perhaps I should've include more steps between eq 1 and 2, so here they are: Start w/ dirac eq (i \gamma^{\mu} \partial_{\mu} - m) \psi = (i \gamma^0 \partial_t -i \vec{\gamma} \cdot \nabla - m) \psi = 0 Keep time derivative on the left, move...

Nah, its the standard +--- metric.

Alright. So the Dirac Eq is (i \gamma^{\mu} \partial_{\mu} - m) \psi = 0 or putting the time part on one side with everything else on the other and multiplying by \gamma^0 , i \partial_t \psi = (i \gamma^0 \vec{\gamma} \cdot \nabla + \gamma^0 m) \psi I would think that this is the...