- #1
vinven7
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HI : Consider a cylinder of length L and volume V that contains one mole of an ideal gas. The familiar ideal gas law states that:
PV = RT
Now, if the cylinder were to move with velocity v parallel to the length direction, special relativity requires the length to contract given by:
L' = L √(1-(v^2/c^2))
As the length contracts, the volume must also go down by the same relation:
V' = V√(1-(v^2/c^2))
However, the ideal gas law must continue to hold ( I presume) - which can only be true if the pressure P increases by the same factor. Thus, simply because of it's motion, the pressure experienced the cylinder seems to have gone up by
P' = P /√(1-(v^2/c^2))
if this argument is correct, does this then mean that pressure is not a relativistic invariant but must depend on the frame of reference? A vessel that does not experience any pressure in one frame of reference might be under pressure in another one?
Could anyone confirm or refute this argument? Thanks!
PV = RT
Now, if the cylinder were to move with velocity v parallel to the length direction, special relativity requires the length to contract given by:
L' = L √(1-(v^2/c^2))
As the length contracts, the volume must also go down by the same relation:
V' = V√(1-(v^2/c^2))
However, the ideal gas law must continue to hold ( I presume) - which can only be true if the pressure P increases by the same factor. Thus, simply because of it's motion, the pressure experienced the cylinder seems to have gone up by
P' = P /√(1-(v^2/c^2))
if this argument is correct, does this then mean that pressure is not a relativistic invariant but must depend on the frame of reference? A vessel that does not experience any pressure in one frame of reference might be under pressure in another one?
Could anyone confirm or refute this argument? Thanks!