Why do we only detect Minkowski particles?
Answers
Answered by
0
I'm having trouble building intuition of the Unruh effect, altough I can follow the proof using Bogoliubov transformations.
When studying a massless scalar field in 1+1D QFT, we start from the classical Klein-Gordon equation. We decompose solutions as left and right moving ones (something like f(x+ct)f(x+ct) and g(x−ct)g(x−ct)). We then do a Fourier expansion of both solutions, so they are a sum of Minkowski normal modes: exp(−iω(x±vt))exp(−iω(x±vt)). To quantize the theory, we simply promote the Fourier coefficients to creation/annihilation operators, which create and destroy particles.
We could choose another expansion in terms of other modes, exp(−iωh(x±vt))exp(−iωh(x±vt)) where h is a new function, so a vacuum associated to these modes would contain particles according to Minkowski normal modes.
I don't understand then, what expansion should we choose. I've read that we should choose an expansion in terms of a null coordinate that describes the observer, so positive and negative frequencies don't mix. Is this choice unique?
To give an example, if we accelerate, we would see photons because the Rindler vacuum is different from the Minkowski vacuum. Does this mean that we could make photomultipliers that don't detect photons when accelerated, but will start clicking at rest?
It seems that a justification that we detect Minkowski particles is that all inertial observers agree on them, but I don't see how this is relevant
When studying a massless scalar field in 1+1D QFT, we start from the classical Klein-Gordon equation. We decompose solutions as left and right moving ones (something like f(x+ct)f(x+ct) and g(x−ct)g(x−ct)). We then do a Fourier expansion of both solutions, so they are a sum of Minkowski normal modes: exp(−iω(x±vt))exp(−iω(x±vt)). To quantize the theory, we simply promote the Fourier coefficients to creation/annihilation operators, which create and destroy particles.
We could choose another expansion in terms of other modes, exp(−iωh(x±vt))exp(−iωh(x±vt)) where h is a new function, so a vacuum associated to these modes would contain particles according to Minkowski normal modes.
I don't understand then, what expansion should we choose. I've read that we should choose an expansion in terms of a null coordinate that describes the observer, so positive and negative frequencies don't mix. Is this choice unique?
To give an example, if we accelerate, we would see photons because the Rindler vacuum is different from the Minkowski vacuum. Does this mean that we could make photomultipliers that don't detect photons when accelerated, but will start clicking at rest?
It seems that a justification that we detect Minkowski particles is that all inertial observers agree on them, but I don't see how this is relevant
Answered by
0
Hello Friend..
The answer of u r question is..✌️✌️
WHY DO WE ONLY DETECT MINKOWSKI PARTICLES????
Ans:✍️✍️✍️✍️✍️✍️✍️✍️✍️
We detect only MINKOWSKI PARTICLES ..
Reason..
If we accelerate we would see photos beacuse the rindler vacuum is different from the minkowski vacuum does this mean that we could make photomultiplier that don't detect water when acceleration but will thot clicking at rest it seems that a justification that we detect minkowski particles is that all inertial observers agree on them..
Thank you..⭐️⭐️⭐️
The answer of u r question is..✌️✌️
WHY DO WE ONLY DETECT MINKOWSKI PARTICLES????
Ans:✍️✍️✍️✍️✍️✍️✍️✍️✍️
We detect only MINKOWSKI PARTICLES ..
Reason..
If we accelerate we would see photos beacuse the rindler vacuum is different from the minkowski vacuum does this mean that we could make photomultiplier that don't detect water when acceleration but will thot clicking at rest it seems that a justification that we detect minkowski particles is that all inertial observers agree on them..
Thank you..⭐️⭐️⭐️
Similar questions