Suppose we have a reference $$R$$ and a reference $$R'$$ with velocity $$v \ne 0$$ respect to $$R$$ then we know that:

$$\Delta t = \gamma \Delta t'$$

but respect to $$R'$$ the reference $$R$$ has also velocity $$v$$ (in opposite direction) then we have:

$$\Delta t' = \gamma \Delta t$$

so by substitution we have $$\Delta t = \gamma^2 \Delta t$$ which implies $$\gamma^2 = 1$$ and then $$v=0$$ which is a contradiction to the hypothesis. Where is the error in this argument?

• Time dilation is one of the things that helped to popularize SR, and as you can see, leads to confusion when it is oversimplified. As a good PSE citizen you may want to learn what proper time is and find out that each timelike space time interval has the same proper time in every reference frame. That's the essence of it that will make the twin paradox disappear also. Mar 13 at 18:06
• What does $\Delta t$ mean? Mar 13 at 18:59
• The two references need to compare their clocks locally in order to make this judgement. So one observer would have to turn around, and go back to the other. If you take this acceleration into account, all paradoxes disappear. Mar 14 at 2:59

The mistake is that you cannot substitute the deltas, as they refer to different quantities in each of the two equations. In the first equation, delta t refers to a time interval between two events that occur in different places in the unprimed frame, while in the second equation it refers to a time interval between two events that occur in the same place in the unprimed frame. Since they are referring to different quantities, you cannot substitute one for the other.

If you are confused by that, try the following analogy. Suppose you have two people, R and R', who are each of the same height, standing a long way apart, so that from the perspective of R, the other person R' seems to have shrunk by a half because they are so far away. Then if h represents the apparent height of R and h' represents the apparent height of R', you would have, from the perspective of R

h = 2h'

From the perspective of R' it is R who seems to have shrunk by a half, so you have

h' = 2h

Those two equations seem completely contradictory if you make the mistake of assuming that h in one of the equations means the same thing as h in the other. They don't. h in the first equation means the height of R as seen by himself, while h in the second equation means the height of R as seen by R' standing a long way away.

• Nice analogy. Reminds me of Father Ted: "Ok, one last time. These are small, but the ones out there are far away. Small... Far away". Mar 14 at 10:09
• For those who have not seen this, here is a 27 second clip: youtube.com/watch?v=vh5kZ4uIUC0 Mar 14 at 10:13
• @MarcoOcram. So, if I'm not wrong, the formula $\Delta t = \gamma \Delta t'$ is valid for an event in same place in R' and I can't use it in reverse because if event is in same place in R' is not in same place in R. So the second formula is not valid. Is this correct?
– asv
Mar 14 at 10:24

A Lorentz transformation (for simplicity: 1 spatial dimension) is generally given by $$\left(\begin{matrix}ct'\\x'\end{matrix}\right)=\left(\begin{matrix}\gamma&-\beta\gamma\\-\beta\gamma&\gamma\end{matrix}\right)\left(\begin{matrix}ct\\x\end{matrix}\right)\quad\text{with}\quad\beta=\frac{v}{c},\,\gamma=\frac{1}{\sqrt{1-\beta^2}}\,,$$ where in Minkowski space the coordinates themselves can be treated like vectors, as well as differences in coordinates. Setting $$x'=0$$, we see that in this definition the origin of the system $$\Sigma'$$ travels with velocity $$+v$$ in the system $$\Sigma$$. We can invert the relation to find $$\left(\begin{matrix}ct\\x\end{matrix}\right)=\left(\begin{matrix}\gamma&\beta\gamma\\\beta\gamma&\gamma\end{matrix}\right)\left(\begin{matrix}ct'\\x'\end{matrix}\right)\,.$$ See how composing the two matrix multiplications gives the identity, which resolves your initial question. Therefore you have two consider the complete Lorentz transformation at once.

Extending the question just a little, imagine two events $$A=(0,0)$$ and $$B=(ct_B,x_B)$$ in the $$\Sigma$$ inertial frame. Their temporal distance is obviously $$\Delta t=t_B$$. In the $$\Sigma'$$ inertial frame we find $$A'=(0,0)\quad\text{and}\quad B'=(\gamma(ct_B-\beta x_B),\gamma(x_B-\beta ct_B))\,.$$ This nicely shows the essence of special relativity: Whereas all events with arbitrary $$x$$ but same $$t=t_B$$ happen "at the same time" in $$\Sigma$$, they surely happen at different times $$t'(\equiv t'(t_B,x_B))$$ in $$\Sigma'$$ and simultaneity is not absolute anymore.

• So, if I'm not wrong, the formula Δt=γΔt′ is valid for an event in same place in R' and I can't use it in reverse because if event is in same place in R' is not in same place in R. So the second formula is not valid. Is this correct?
– asv
Mar 14 at 10:34
• @asv yes that is correct. Mar 14 at 11:19
• Thank you very much
– asv
Mar 14 at 14:41