Derive u from v with Lorentz Transform

[mntb]

derivating a from v (lorentz transform)
u is the velocity in the +x direction
u=(u-v)/(1-vu/c^2)
du? how do you du the right side of the eq?

 

[selfAdjoint]

derivating a from v (lorentz transform)
u is the velocity in the +x direction
u=(u-v)/(1-vu/c^2)
du? how do you du the right side of the eq?

What are you taking the derivative with respect to? Note that the u-time in the observed frame is not the same as the v-time in the reference frame. If you want your derivative to be Lorentz invariant you have to differentiate with respect to proper time.

 

[quicknote]

To derive u from v using the Lorentz Transform, we can use the equation:

u = (u-v)/(1-vu/c^2)

where u is the velocity in the +x direction and v is the velocity in the +x direction in the original reference frame.

To differentiate the right side of the equation with respect to v, we can use the quotient rule:

du/dv = [(1-vu/c^2)(du/dv) - (du/dv)(u-v)]/(1-vu/c^2)^2

Since we are differentiating with respect to v, we can treat u as a constant. Thus, du/dv = 0.

Plugging this into the equation, we get:

du/dv = [(1-vu/c^2)(0) - (0)(u-v)]/(1-vu/c^2)^2

du/dv = 0

Therefore, the derivative of u with respect to v is 0. This means that u is independent of v and remains constant regardless of the velocity in the original reference frame.