- #1

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- 915

## Main Question or Discussion Point

I've been wondering if there is an expression for the magnitude of the proper acceleration of an observer in terms of three-velocities and three-accelerations.

I didn't find any, so I took a stab at calculating it. However, it was not at all convenient to do the calculations without 4-vectors, so I used them in my caculations. Even so perhaps the results could be mildly useful to someone who asks a question about acceleration in special realtivity (of which we get large numbers ) without a background in 4-vectors (of which there are also large numbers. The point is that one can see that given the 3-acceleration and 3-velocity of an observer, one can compute the invariant magnitude of the 4-acceleration which is just the magnitude of the acceleration that is experienced by an instantaneously co-moving observer.

Onto the calculations.

We start with defining the components of the 3-velocity as vx, vy, vz, and also the 3-acceleration ax=dx/dt, ay=dy/dt, az=dz/dt. I used geometric units, with c=1, though I'll attempt to put the necessary factors of c back in in the final result.

We can write the components of the 4-velocity as:

$$u^t = \frac{1}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^x=\frac{vx}{\sqrt{1-vx^2-vy^2-vz^2}}$$$$ u^y=\frac{vy}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^z=\frac{vz}{\sqrt{1-vx^2-vy^2-vz^2}} $$

Then we can compute the 4-accerlation ##\alpha## as:

$$\alpha^t =\frac{d}{d\tau} u^t = \frac{du^t}{dt}\frac{dt}{d\tau} = u^t \,at \quad \alpha^x = \frac{d}{d\tau} u^x = \frac{du^x}{dt} \frac{dt}{d\tau} = u^t\,ax$$

$$\alpha^y = u^t \, ay \quad \alpha^z = u^t \, az$$

Then the square magnitude of the proper acceleration, ##A^2## is giving by

$$A^2 = (\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2$$

Much computer algebra later we get the following expression for ##A^2##:

$$\frac{(1-\beta_y^2-\beta_z^2)\,ax^2 + (1-\beta_x^2-\beta_z^2)\,ay^2+(1-\beta_x^2-\beta_y^2)\,az^2 + 2\beta_x\beta_y\,ax\,ay +2\beta_x\beta_z\,ax\,az+2\beta_y\beta_z\,ay\,az}{ (1-\beta_x^2-\beta_y^2-\beta_z^2)^3}$$

where we have introduced ##\beta_x =\frac{vx}{c} \quad \beta_y = \frac{vy}{c} \quad \beta_z = \frac{vz}{c}##.

Hopefully this will serve as a motivation for why it's much easier to use 4-vectors to calculate proper accelerations, as the 4-vector expession ##(\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2## is much simpler than the 3-vector expression.

As far as whether or not I made any errors - it's hard to say, but it seems to pass a few quick tests, namely the 1d case where only ##\beta_x## is nonzero, and the 2d case of the sliding block.

I didn't find any, so I took a stab at calculating it. However, it was not at all convenient to do the calculations without 4-vectors, so I used them in my caculations. Even so perhaps the results could be mildly useful to someone who asks a question about acceleration in special realtivity (of which we get large numbers ) without a background in 4-vectors (of which there are also large numbers. The point is that one can see that given the 3-acceleration and 3-velocity of an observer, one can compute the invariant magnitude of the 4-acceleration which is just the magnitude of the acceleration that is experienced by an instantaneously co-moving observer.

Onto the calculations.

We start with defining the components of the 3-velocity as vx, vy, vz, and also the 3-acceleration ax=dx/dt, ay=dy/dt, az=dz/dt. I used geometric units, with c=1, though I'll attempt to put the necessary factors of c back in in the final result.

We can write the components of the 4-velocity as:

$$u^t = \frac{1}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^x=\frac{vx}{\sqrt{1-vx^2-vy^2-vz^2}}$$$$ u^y=\frac{vy}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^z=\frac{vz}{\sqrt{1-vx^2-vy^2-vz^2}} $$

Then we can compute the 4-accerlation ##\alpha## as:

$$\alpha^t =\frac{d}{d\tau} u^t = \frac{du^t}{dt}\frac{dt}{d\tau} = u^t \,at \quad \alpha^x = \frac{d}{d\tau} u^x = \frac{du^x}{dt} \frac{dt}{d\tau} = u^t\,ax$$

$$\alpha^y = u^t \, ay \quad \alpha^z = u^t \, az$$

Then the square magnitude of the proper acceleration, ##A^2## is giving by

$$A^2 = (\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2$$

Much computer algebra later we get the following expression for ##A^2##:

$$\frac{(1-\beta_y^2-\beta_z^2)\,ax^2 + (1-\beta_x^2-\beta_z^2)\,ay^2+(1-\beta_x^2-\beta_y^2)\,az^2 + 2\beta_x\beta_y\,ax\,ay +2\beta_x\beta_z\,ax\,az+2\beta_y\beta_z\,ay\,az}{ (1-\beta_x^2-\beta_y^2-\beta_z^2)^3}$$

where we have introduced ##\beta_x =\frac{vx}{c} \quad \beta_y = \frac{vy}{c} \quad \beta_z = \frac{vz}{c}##.

Hopefully this will serve as a motivation for why it's much easier to use 4-vectors to calculate proper accelerations, as the 4-vector expession ##(\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2## is much simpler than the 3-vector expression.

As far as whether or not I made any errors - it's hard to say, but it seems to pass a few quick tests, namely the 1d case where only ##\beta_x## is nonzero, and the 2d case of the sliding block.