Why is a relativistic calculation needed unless p c is much smaller than the rest energy of a particle?After introducing the de Broglie wavelength equation, my textbook gives a rather simple example where it asks to find the kinetic energy of a proton whose de Broglie wavelength is 1 fm. In the solution to this problem, it states that "A relativistic calculation is needed unless p c for the proton is much smaller than the proton rest energy."Could someone please explain why this is the necessary condition? I'm not sure what the quantity ' p c' represents or means here. I know that for massless particles like photons, the total energy E is equal to p c. I'm not sure what it means for particles having rest mass like protons.

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Why is a relativistic calculation needed unless   p  c is much smaller than the rest energy of a particle?After introducing the de Broglie wavelength equation, my textbook gives a rather simple example where it asks to find the kinetic energy of a proton whose de Broglie wavelength is 1 fm. In the solution to this problem, it states that &quot;A relativistic calculation is needed unless   p  c for the proton is much smaller than the proton rest energy.&quot;Could someone please explain why this is the necessary condition? I&#039;m not sure what the quantity &#039;  p  c&#039; represents or means here. I know that for massless particles like photons, the total energy   E is equal to   p  c. I&#039;m not sure what it means for particles having rest mass like protons.

reflam2kfnr · Accepted Answer

In relativity, the energy   E of a particle is related to its mass   m and momentum   p by  E  =            m      2              c      4        +          p      2              c      2      Now let&#039;s think about the non-relativistic limit,   c  →  ∞. To do this, we will expand the square root  E  =  m      c    2        1    +                  p        2                              m          2                          c          2                      =  m      c    2    +            p      2              2      m        +  ⋯where the   ⋯ refer to terms that vanish in the limit   c  →  ∞The first term is the famous equation   E  =  m      c    2  . In non-relativistic physics, the mass is a constant, so this is just a constant term in the energy we can ignore.The second term   E  =            p      2              2      m       is the non-relativistic expression for kinetic energy.You can see that non-relativistic physics is a good approximation to relativistic physics when we can ignore the higher order terms. Thinking back to how we expanded the square root, this amounts to the condition   p  c  ≪  m      c    2

lurtzslikgtgjd · Accepted Answer

The momentum of a particle of mass   m moving at   v is   γ  m  v where   γ is given by the following.  γ  =      1          1      −                        v          2                          c          2                    Let&#039;s always take   v to be non-negative. If you need a direction then put it in later. Since   v  &amp;lt;=  c,   γ is a real number greater than or equal to 1. For   v very small compared to   c it is only slightly larger than 1.The rest energy of a proton is just   m      c    2  . So the comparison they are making is the following.  γ  m  v  c  &amp;lt;&amp;lt;  m      c    2  So dividing out the common factors you get the following.  γ  v  &amp;lt;&amp;lt;  cAnd for   v small compared to   c, this is just   v  &amp;lt;&amp;lt;  c

Why is a relativistic calculation needed unless p c is much smaller than the rest energy of

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