# Why geometrically does the orientation of a scalar particle plane wave relate to its velocity

+ 0 like - 0 dislike
43 views

In this question I am trying to think about quantum field theory geometrically. I abbreviate "in the zero speed direction" to "vertically" and "in the equal-time plane" to "horizontally".

I base my question of the following understandings. A correction of any of these would (helpfully) pre-empt my question:

• A quantum field of a specific particle variety in a single particle state may be in a plane wave state for that particle (idealised situation where the whole field contains one particle and it is in an exactly known abstract (reference frame independent) momentum state).
• The field mode of such particle may be geometrically transformed via Lorentz boost, yielding another plane wave, to get its appearance in another inertial frame.
• The rest mass/energy corresponds to the frequency of the field in the frame in which the frequency is least. Equivalently the vertical spacing (period) is most. In this frame the wavefronts are oriented horizontally. The higher the frequency / smaller the vertical spacing, the greater the energy. If the particle is massless there is no such frame.
• The total energy in a frame corresponds to the frequency / closeness of vertical spacing in that frame.
• The momentum in a frame corresponds to the closeness of horizontal spacing in that frame ('horizontal version of the energy').

That in itself seems fine to me. However I can't reconcile these definitions of energy and momentum with velocity. In a simple sense the velocity of something is how much spatial translation per unit time translation is required to keep it invariant. So I would expect a vertically (in a spacetime sense) aligned object to be stationary. A sloped configuration to have a faster speed the more the slopes deviate from vertical. But this is the opposite situation from the one implied above. Instead of vertical features a zero momentum has horizontal features (wave fronts). I understand the relativistic energy-velocity-momentum relationship is v proportional to p / E. Or equivalently v proportional to period / wavelength. Again this is the opposite of what I would expect, which would be for increasing wavelength that stretches the phenomena horizontally to increase speed and the opposite for increasing period.

asked May 21, 2018

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.