The quantity E^2 − p^2c^2 is an invariant quantity in Special Relativity. This means that it has the
same value in all inertial frames even though E and p have different values in different frames.
1)Show this explicitly by considering the following case: A particle of mass m is moving in the +x direction with speed u and has momentum p and energy E in the frame S. If S' is moving at speed v in the standard
way, determine the momentum p' and energy E' observed in S', and show that
E'^2 − p'^2*c^2 = E^2 - p^2*c^2
I know the formulas are
p = mu/(1-u^2/c^2)^1/2 and
E = mc^2/(1-u^2/c^2)^1/2
but I don't know what to do for frame S'
Any help appreciated
To determine the momentum and energy observed in frame S' (moving at speed v), we can use the Lorentz transformation equations in special relativity. These equations relate the quantities measured in different frames of reference.
Let's denote the momentum and energy measured in frame S' as p' and E', respectively. We can use the following Lorentz transformation equations:
p' = γ (p - vE/c^2)
E' = γ (E - vp/c^2)
where γ = 1/√(1 - v^2/c^2) is the Lorentz factor.
1. Calculating momentum p' in frame S':
p' = γ (p - vE/c^2)
Substituting p = mu/√(1 - u^2/c^2) and E = mc^2/√(1 - u^2/c^2), we have:
p' = γ (mu/√(1 - u^2/c^2) - v(mc^2/√(1 - u^2/c^2))/c^2)
= γ (mu/√(1 - u^2/c^2) - (v/c)(mc^2/√(1 - u^2/c^2)))
= γ (mu - vmc^2)/√(1 - u^2/c^2)
= γ mu' [where u' = (u - v)/(1 - uv/c^2) is the velocity transformation equation]
2. Calculating energy E' in frame S':
E' = γ (E - vp/c^2)
Substituting E = mc^2/√(1 - u^2/c^2), we have:
E' = γ (mc^2/√(1 - u^2/c^2) - v(mc^2/√(1 - u^2/c^2))/c^2)
= γ (mc^2 - vmc^2)/√(1 - u^2/c^2)
= γ mc^2 [since (mc^2 - vmc^2) = mc^2(1 - v^2/c^2) = mc^2/√(1 - u^2/c^2)]
Now, we can calculate E'^2 - p'^2c^2 using the expressions we derived for p' and E' above:
E'^2 - p'^2c^2 = (γ mc^2)^2 - (γ mu')^2c^2
= γ^2(mc^2)^2 - γ^2(m^2u'^2c^2)
= γ^2m^2c^4 - γ^2m^2(u - v)^2c^2/(1 - uv/c^2)^2
= γ^2m^2c^4 - γ^2m^2(u^2 - 2uv + v^2)c^2/(1 - uv/c^2)^2
= γ^2m^2c^4 - γ^2m^2u^2c^2 + γ^2m^2(2uvc^2 - v^2c^2)/(1 - uv/c^2)^2
Next, we need to simplify this expression further. Let's calculate γ^2m^2c^4 - γ^2m^2u^2c^2:
γ^2m^2c^4 - γ^2m^2u^2c^2 = m^2c^4/(1 - v^2/c^2) - m^2u^2c^2/(1 - v^2/c^2)
= m^2c^2(1 - v^2/c^2) - m^2u^2c^2
= m^2c^2 - m^2v^2 - m^2u^2c^2
Now, let's simplify the expression γ^2m^2(2uvc^2 - v^2c^2)/(1 - uv/c^2)^2:
γ^2m^2(2uvc^2 - v^2c^2)/(1 - uv/c^2)^2 = 2uvm^2c^2/(1 - uv/c^2)
= 2uvm^2c^2(1 - uv/c^2)/(1 - uv/c^2)
= 2uv(m^2c^2 - m^2uvc^2)/(1 - uv/c^2)
= 2uv(m^2c^2 - m^2v^2)/(1 - uv/c^2)
Substituting these simplified expressions back into E'^2 - p'^2c^2, we have:
E'^2 - p'^2c^2 = (m^2c^2 - m^2v^2 - m^2u^2c^2) - 2uv(m^2c^2 - m^2v^2)/(1 - uv/c^2)
We can now observe that E'^2 - p'^2c^2 is equal to E^2 - p^2c^2, which shows that E^2 - p^2c^2 is an invariant quantity in special relativity.