M∝TxCyhz
M1L0T0=(T1)x(L1T−1)y(M1L2T−1)z
[M1L0T0]=[MzLy+2zTx−y−z]]
⟹z=1
y+2z=0,x−y−z=0
solving
y=−2
and x=−1
M⇒[T−1C−2h1]
Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:
Held on 8 Apr 2017 · Verified 6 Jul 2026.
[M]=[T−1C−2h]
[M]=[T−1C2h]
[M]=[T−1C−2h−1]
[M]=[TC−2h]
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