# I Position as a function of time in Special Relativity

Recently I have come into Special Relativity and specifically Lorentz transformation. Lets assume two frames A and B moving relative with speed $v$. The position of a particle moving with respect to B is given by $x′=f(t′)=3t′$. What is the function of position $x=f(t)$ of the particle for the observer A? Thanks in advance. As I said before I am new to Special Relativity.

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#### PeterDonis

Mentor
The position of a particle moving with respect to B is given by $x′=f(t′)=3t′$.
What units are you using? In the natural units of relativity, where $c = 1$, the worldline $x' = 3 t'$ is impossible; it would mean that the particle would be moving at 3 times the speed of light.

What units are you using? In the natural units of relativity, where $c = 1$, the worldline $x' = 3 t'$ is impossible; it would mean that the particle would be moving at 3 times the speed of light.
Thanks for the reply. I am using the "classical" units where c=3*10^8 m/s. I was just trying to figure out if given a position function for one (A) reference frame we can deduce another position function for another reference frame (B) moving relative with A.

#### PeterDonis

Mentor
I am using the "classical" units where c=3*10^8 m/s.
So that means $x' = 3 t'$ implies a speed in the primed frame of 3 m/s? Ok, good.

was just trying to figure out if given a position function for one (A) reference frame we can deduce another position function for another reference frame (B) moving relative with A.
Yes, we can, but for the simple function you give, you don't need the full Lorentz transformation to figure it out. You just need the velocity addition law: in the unprimed frame, the function will be $x = w t$ where

$$w = \frac{v + 3}{1 + \frac{3 v}{c^2}}$$

#### Dale

Mentor
Recently I have come into Special Relativity and specifically Lorentz transformation. Lets assume two frames A and B moving relative with speed $v$. The position of a particle moving with respect to B is given by $x′=f(t′)=3t′$. What is the function of position $x=f(t)$ of the particle for the observer A? Thanks in advance. As I said before I am new to Special Relativity.
This is just straight algebra. You have $x’=f(t’)$ so just substitute $x’=\gamma\ (x-vt)$ and $t’=\gamma\ (t-vx/c^2)$. Then solve for $x=g(t)$

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So that means $x' = 3 t'$ implies a speed in the primed frame of 3 m/s? Ok, good.

Yes, we can, but for the simple function you give, you don't need the full Lorentz transformation to figure it out. You just need the velocity addition law: in the unprimed frame, the function will be $x = w t$ where

$$w = \frac{v + 3}{1 + \frac{3 v}{c^2}}$$
Thanks so much. I was trying to figure it out through position and time transformations.

#### Ibix

Thanks so much. I was trying to figure it out through position and time transformations.
That's how you derive the velocity transformation, so it ought to have worked.

#### pervect

Staff Emeritus
Just a comment. The transformation of x' = f(t') into x = f(t) winds up to be rather ugly. t and t' are different coordinates, due to the relativity. Thus, as other posters have mentioned, one needs to transform both x' and t'. The transformation x' = f($\tau$) into x = f($\tau$) is much simpler. Here $\tau$ denotes proper time. Because $\tau$ is a world scalar, it doesn't need to be transformed, thus one only needs to use the Lorentz transform to transform x' to x, making it a significantly easier calculation.

Sometimes you do want to express things in terms of coordinate time t rather than proper time $\tau$, but it's often worth looking at the alternative.

"Position as a function of time in Special Relativity"

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