# Why don't these conversions of time between reference frames seem consistent? [closed]

I am a student and due to school closures I am reading ahead in physics. I have been learning about relativity and I made up a problem for myself. You can see a diagram that I drew below: The diagram shows car X travelling at $$0.7c$$ with respect to car B and $$0.5c$$ with respect to car A. Car A is travelling at $$0.2c$$ with respect to car B, where $$c$$ is the speed of light.

This means, using Lorentz transform, 1 year on car X is equivalent to:

$$\frac{1}{\sqrt{1-0.5^2}}=1.15$$

1.15 years on car A.

And 1 year on car X is equivalent to:

$$\frac{1}{\sqrt{1-0.7^2}}=1.40$$

1.40 years on car B.

This means that 1.15 years on car A should be the same as 1.40 years on car B.

So if we were to use the Lorentz transform we should get the same answer but instead we get:

$$\frac{1.15}{\sqrt{1-0.2^2}}=1.17$$

We get 1.17 years on car B is the same as 1.15 years on car A. I don't understand what I have done wrong. Could you help me why this problem arises?

It is also worth noting that if I was to use a different method it would work. If I was to convert the time for car X into the time for car A then convert the time for car A into the time for car B, it would work. But I don't understand why this method works. In fact, I believe that method shouldn't work. This is because car X will be moving at 0.7c with respects to B still so 1 years on car X should be the same as 1.40 years on car B.

Could you help me with this?

The contradiction is in the setup. It is impossible that those three relative velocities can simultaneously be true.

You can't just add relative velocities together directly in special relativity. If an observer moving at velocity $$v$$ measures, in their own frame, a velocity $$v'$$ of some passing object, then a stationary observer would measure the velocity

$$u=\frac{v+v'}{1+\frac{vv'}{c^2}}$$

If car A is moving at $$0.2c$$ with respect to car B, and car X is moving at $$0.5c$$ in the same direction with respect to car A, then car X must be moving at roughly $$0.64c$$ with respect to car B, not $$0.7c$$.

• Using velocity-composition alone does not resolve the key issue, which is that time-dilation factors aren’t multiplicative. May 19, 2020 at 19:29
• @robphy The fact that time-dilation factors aren't multiplicative can be directly derived from the velocity-composition law. And that doesn't address what's actually wrong with the question's reasoning, which is that it has assumed, at the start, conditions which are impossible. May 19, 2020 at 20:46
• Couldn't those relative speeds be possible if they aren't all going the same direction? That is, if A is moving relative to X in a manner such that a component of its velocity is in the opposite direction to the decomposition of B's velocity with respect to X along the same axes then the speeds given become possible. In fact, there's probably exactly two possible velocity configurations in 2 dimensions that work. May 20, 2020 at 15:31
• @Pleasestopbeingevil This is why I specifically said "relative velocities", not speeds. The direction does indeed matter. May 20, 2020 at 15:58
• @probably_someone For sure, but that's a pretty sneaky addition to the premises, since the question itself seems to treat them only as speeds, even if the diagram does admittedly make it look like they are all parallel. I just feel like you might want to call out that assumption of parallelism somewhere more clearly. May 20, 2020 at 16:15

The diagram shows car X travelling at 0.7c with respects to car B and 0.5c with respects to car A. Car A is travelling at 0.2c with respects to car B. Where c is the speed of light.

This is impossible. If car X is travelling at a velocity $$0.7c$$ w.r.t. car B and at a velocity $$0.5c$$ w.r.t. car A then according to the relativistic addition of velocities, car A would be travelling at a velocity $$> 0.2c$$.

If the velocities of two objects are $$u$$ and $$v$$ w.r.t. an observer then the relative velocity of the first w.r.t. the second is given by

$$\frac{u-v}{1-\frac{uv}{c^2}}$$

In your example, car A and car B are moving with velocities $$-0.5c$$ and $$-0.7c$$ respectively in the frame of car X. Thus, the velocity of A w.r.t. B will be given by

$$\frac{-0.5c+0.7c}{1-{(0.5)(0.7)}}$$

which would be obviously not $$0.2c$$ like you assume.

The morale of the story is that the Newtonian expectations of how velocities get combined don't remain valid in relativity and you need to be careful as to whether your assumptions are internally inconsistent in the purview of relativity.

• As the answer by @probably_someone shows, you can also take the relation between the velocities of (X and A) and (A and B) to be true and show that your assumption about the velocity between (X and B) won't hold. The point is that the three assumptions are inconsistent as given, you'd have to modify at least one of them to be consistent with relativity.
– user87745
May 18, 2020 at 23:07
• Using velocity-composition alone does not resolve the key issue, which is that time-dilation factors aren’t multiplicative. May 19, 2020 at 19:30
• @robphy Good point, I didn't really look closely what OP was doing afterwards. Thanks for pointing it out.
– user87745
May 19, 2020 at 19:40

Although the OP mistakenly used galilean velocity-composition ($$v_{AC}=v_{AB}+v_{BC}$$)
instead of special-relativisitic velocity composition ($$v_{AC}=\displaystyle\frac{v_{AB}+v_{BC}}{1+v_{AB}v_{AC}}$$),
using special-relativisitic velocity-composition alone
does not resolve the key issue in the OP.

The key issue is

## time-dilation factors are not multiplicative

Although $$\displaystyle\frac{0.5+0.2}{1+(0.5)(0.2)}=\frac{7}{11}\approx0.6363...$$, $$\def\GAM#1{\displaystyle\frac{1}{\sqrt{1-{#1}^2}}} \def\BGAM#1{\displaystyle\frac{#1}{\sqrt{1-{#1}^2}}} \GAM{0.5}\GAM{0.2}\neq \GAM{\left(\frac{7}{11}\right)}$$

Instead, $$\GAM{0.5}\GAM{0.2} + \BGAM{0.5}\BGAM{0.2} = \GAM{\left(\frac{7}{11}\right)}$$
Here it is in a form that you can copy-paste into wolframalpha [using implied multiplication]:
(1/sqrt(1-0.5^2))(1/sqrt(1-0.2^2))+(0.5/sqrt(1-0.5^2))(0.2/sqrt(1-0.2^2))-(1/sqrt(1-(7/11)^2))

Or, with an alternate set of velocities...

Although $$\displaystyle\frac{0.7-0.5}{1-(0.7)(0.5)}=\frac{4}{13}\approx0.307692...$$, $$\def\GAM#1{\displaystyle\frac{1}{\sqrt{1-{#1}^2}}} \def\BGAM#1{\displaystyle\frac{#1}{\sqrt{1-{#1}^2}}} \GAM{0.5}\GAM{(\frac{4}{13})}\neq \GAM{ 0.7 }$$

Instead, $$\GAM{0.5}\GAM{(\frac{4}{13})} + \BGAM{0.5}\BGAM{(\frac{4}{13})} = \GAM{0.7}$$
Here it is in a form that you can copy-paste into wolframalpha [using implied multiplication]:

(1/sqrt(1-0.5^2))(1/sqrt(1-(4/13)^2))+(0.5/sqrt(1-0.5^2))((4/13)/sqrt(1-(4/13)^2))-(1/sqrt(1-(0.7)^2))

More generally,
time-dilation factors are not multiplicative--there is an extra factor $$\gamma_{AC}=\gamma_{AB}\gamma_{BC}(1+v_{AB}v_{BC}).$$

These are easier to recognize if you work with rapidities (Minkowski angles),
where $$v_{AC}=\tanh\theta_{AC}$$, $$\gamma_{AC}=\cosh\theta_{AC}$$, $$v_{AC}\gamma_{AC}=\sinh\theta_{AC}$$, etc...

Since rapidities are additive $$\theta_{AC}=\theta_{AB}+\theta_{BC},$$ we have \begin{align*} v_{AC}=\tanh(\theta_{AC}) &=\tanh(\theta_{AB}+\theta_{BC})\\ &=\frac{\tanh\theta_{AB}+\tanh\theta_{BC}}{1+\tanh\theta_{AB}\tanh\theta_{BC}}\\ &=\frac{v_{AB}+v_{BC}}{1+v_{AB}v_{BC}}, \end{align*} which is the velocity-composition formula.

So, the time-dilation-factor is \begin{align}\gamma_{AC}=\cosh(\theta_{AC}) &=\cosh(\theta_{AB}+\theta_{BC})\\ &=\cosh\theta_{AB}\cosh\theta_{BC}+\sinh\theta_{AB}\sinh\theta_{BC}\\ &=\cosh\theta_{AB}\cosh\theta_{BC}(1+\tanh\theta_{AB}\tanh\theta_{BC})\\ &=\gamma_{AB}\gamma_{BC}(1+v_{AB}v_{BC}), \end{align} which is the formula used in the two numerical examples above.
Again, the key issue is that time-dilation factors are not multiplicative.

A: the Doppler factors: $$k_{AB}=\displaystyle\sqrt{\frac{1+v_{AB}}{1-v_{AB}}}=e^{\theta_{AB}}$$.)