# Position of Particle in Special Relativity (A vs B)

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In summary: In this case, x' = f(t') does indeed give the desired velocity in terms of coordinate time.In summary, the position of a particle moving with respect to B is given by ##x′=f(t′)=3t′##. The function of position ##x=f(t)## is just 3 t' in the classical units.
Recently I have come into Special Relativity and specifically Lorentz transformation. Let's 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.

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.

PeterDonis said:
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.

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

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

Recently I have come into Special Relativity and specifically Lorentz transformation. Let's 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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PeterDonis said:
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.

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.

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.

Dale and Nugatory

## 1. What is the difference between a particle's position in special relativity compared to classical mechanics?

In classical mechanics, a particle's position is described by its coordinates in three-dimensional space. However, in special relativity, the position of a particle is described by its coordinates in four-dimensional spacetime, which includes time as the fourth dimension.

## 2. How does the position of a particle change in special relativity when it approaches the speed of light?

In special relativity, as a particle approaches the speed of light, its position in space appears to contract in the direction of motion. This phenomenon is known as length contraction and is a consequence of the time dilation effect.

## 3. Can a particle's position in special relativity be measured in the same way as in classical mechanics?

No, the measurement of a particle's position in special relativity requires the use of four-dimensional coordinates and the concept of spacetime. This is different from the traditional three-dimensional coordinates used in classical mechanics.

## 4. How does the position of a particle in special relativity affect its energy and momentum?

In special relativity, the energy and momentum of a particle are not independent quantities but are related to each other through the famous equation E=mc². This means that as a particle's position changes in spacetime, its energy and momentum will also change accordingly.

## 5. Is the concept of a fixed position in space and time still applicable in special relativity?

No, in special relativity, the concept of a fixed position in space and time no longer holds. This is because the position of a particle is relative to the observer's frame of reference and can change depending on the observer's relative motion. Therefore, there is no absolute position in spacetime in special relativity.

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