# Do time dilation and length contraction cancel out?

I've recently been getting into special relativity by myself. I noted something interesting: We know that the factor by which time dilates and lengths contract is the Lorentz factor

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

where $$c$$ is the speed of light and $$v$$ is the relative speed of some inertial frame of reference. We can then calculate the "slowing" of time:

$$t'=\frac{t}{\gamma}=t\sqrt{1-\frac{v^2}{c^2}}$$

and length contraction:

$$L'=\frac{L}{\gamma}=L\sqrt{1-\frac{v^2}{c^2}}$$

where $$t'$$ and $$L'$$ are the dilated time and contracted length

Since the velocity of an object is defined as $$v=\frac{s}{t}$$, we can say that for any velocity in the moving frame of reference

$$v'=\frac{s'}{t'}=\frac{s\sqrt{1-\frac{v^2}{c^2}}}{t\sqrt{1-\frac{v^2}{c^2}}}=\frac{s}{t}$$

which would imply that we would observe the same velocity as in our frame of reference. Is this true? I had always thought that only the speed of light is the same in all inertial frames of reference, but not other speeds. I am also not sure about whether one can use the formulas for time dilation and length contraction to do such calculations or whether one has to use the Lorentz transformations which I haven't learned yet.

So: If an object has some velocity in a frame of reference that is moving relative to our frame of reference, would we observe it to have the same velocity as if the object were in our frame of reference?

• Comments are not for extended discussion; this conversation has been moved to chat. – rob Dec 14 '20 at 5:49

There is a lot of confusion here. Also in Special Relativity is fundamental to be clear about what we mean when we write variables, especially when dealing with multiple frames of reference.

Let's treat things properly: given two frames of reference (so two observers if you like this name better) $$O$$ and $$O'$$, in relative motion with respect to each other with constant velocity $$v$$, from the point of view of $$O$$ the equation regarding time dilation and length contraction are: $$t'=\frac{t}{\gamma}$$ $$l'=l\gamma$$ but these equations by themselves do not mean anything! We have to really understand what $$l,t,l',t'$$ are to make use of these equations.
Well: $$t,l$$ are measurements of time and length respectively, made by $$O$$ on an object that is still in the frame of reference of $$O'$$. On the other hand $$t',l'$$ are measurements made by $$O'$$ on the same object. So, to get the picture in your head: $$O$$ sees the object being measured as moving, while $$O'$$ doesn't.

You can easily remember these equations by the names of the phenomena that they describe: time dilation roughly means that the time elapsed from the point of view of $$O$$ ($$t$$) is greater compared to the time elapsed from the point of view of $$O'$$ ($$t'$$), and we know that the proportionality constant should be $$\gamma$$, but since $$\gamma$$ is always greater than one we immediately see that: $$t>t' \ \Rightarrow \ t'=\frac{t}{\gamma}$$ but be careful: this is not at all a proof! It is simply an useful mnemonic technique to remember the formula.
You can of course do the same with length contraction, but this time $$l'>l$$.

Keep in mind also another thing: online or on books you will definitely find sources reporting seemingly the exact opposite equations that I reported, but just remember to pay attention to the meaning behind the letters and all the incongruencies should vanish.

Also remember that this phenomena work in both ways, $$O$$ sees $$O'$$ moving, but from the point of view of $$O'$$ is $$O$$ that moves, so the phenomena are completely symmetrical! This usually seems really strange at first, but with time you will get the hang of it. Hope to have spared you some confusion.

Now you should be able to see why your reasoning is incorrect.

• Thank you answering! It seems in fact like my confusion arised from different uses of $'$ (for example, the reference I used used $t'$ for the time measured on, lets say a spaceship). However, shouldn't $l'<l$ since we observe lengths to contract, i.e. being smaller? – Jonas Dec 13 '20 at 14:44
• As stated before $l$ is the measurement made by $O$ on a body that is moving for him, with the same velocity as $O'$, lenght contraction then implies that $O$ will see the body contract, so shorter in the direction of the motion. Meanwhile $O'$ sees the body at rest in his frame, so $O'$ sees the body without contraction. So it must be that $l'>l$. – Noumeno Dec 13 '20 at 14:48
• Even with the correct time dilation and length contraction equations, OP's argument will still be incorrect. – TaeNyFan Dec 14 '20 at 3:00
• @TaeNyFan Yes, I am aware of this. For this reason I stated that there was a lot of confusion. But anyway for one reason or for the other OP's argument is surely incorrect; let's not rub it in too much :) – Noumeno Dec 14 '20 at 8:51

Let's consider the usual setup with frame O' moving to the right with speed v with respect to frame O. Let's assume the two frames coincide at t=t'=0 and x=x'=0. So suppose we have an object traveling in O between x=0 and x=L in time T. Thus, (x$$_1$$, t$$_1$$)=(0,0) and (x$$_2$$,t$$_2$$)=(L,T). The speed of the object in O is therefore u=L/T.

To find the corresponding (x$$_1'$$, t$$_1'$$) and (x$$_2'$$, t$$_2'$$) we use the Lorentz transformation and we find $$(x_1', t_1')=(0,0),(x_2', t_2')=(\gamma(L-vT),\gamma(T-vL/c^2)$$ Thus, the observer in O' would measure for the speed of the object u'=x$$_2'$$/t$$_2'$$ = (L-vT)/(T-vL/c$$^2$$).

Note that in the limit v/c -> 0 we get u' = (L/T) -v as expected (the relative velocity).

I believe what you are calculating is the relative velocity of the two frames, which is the same to both observers. Check out my answer here: Measuring relative speeds in SR

I will present the formulas, diagrams, and interpretation of time-dilation and length-contraction in a less ambiguous notation. (Prime and unprimed notations often lead to confusions.)

At the end,
I will attempt to repeat your calculation that cancels the $$\gamma$$-factor, and demonstrate an error in your physical interpretation.

I use a spacetime diagram on rotated graph paper so that we can immediately read off the ticks along segments. (The "clock diamonds" are generated by the spacetime paths of light-signals in a light-clock. By Lorentz invariance, the area of all clock diamonds are equal.)

We draw $$\color{red}{\mbox{Alice's (RED)}}$$ spacetime diagram , where $$\color{blue}{\mbox{Bob (BLUE)}}$$ moves inertially with $$(v/c)=6/10$$.

Time dilation involves Minkowski-right triangle OTQ,
measuring the adjacent-component OT of the timelike-displacement hypotenuse-OQ.

Length contraction involves Minkowski-right triangle OLD,
measuring the spacelike hypotenuse OD (the apparent length of Bob's ladder), where OL is the "distance between parallel lines" (the proper length of the Bob's ladder), which is the adjacent side of triangle OLD, which is Minkowski-perpendicular to DL along the worldline of the front of Bob's ladder.

(These triangles are numerically similar. One involves the rapidity (the Minkowski-angle between two timelike lines) and the other involves the Minkowski-angle between two spacelike-lines, which are Minkowski-orthogonal to the timelike lines. All of these vectors are coplanar.) (For the relative velocity $$(v/c)=\displaystyle\frac{(6)}{(10)}$$, we have $$\gamma=\displaystyle\frac{1}{\sqrt{1-(v/c)^2}}=\frac{(5)}{(4)}$$.)

Here is Time Dilation:
$$\gamma=\frac{(adj)}{(hyp)}=\frac{OT}{OQ}$$ so \begin{align} (OT) &=\gamma(OQ)\\ \large\Delta t^A_{OT} &\large =\gamma\ \Delta t^B_{OQ}\\ \left( \begin{array}{c} \mbox{duration of OQ}\\ \mbox{according to Alice} \end{array} \right) &= \gamma \left( \begin{array}{c} \mbox{duration of OQ}\\ \mbox{according to Bob} \end{array} \right)\\ \Large \Delta t^A_{OQ} &\Large =\ \gamma\ \Delta t^B_{OQ}\\ (10) & \stackrel{\checkmark}{=} \left(\frac{5}{4}\right) (8) \end{align}

Here is Length Contraction:
$$\gamma=\frac{(adj)}{(hyp)}=\frac{OL}{OD}$$ so \begin{align} (OD) &=\frac{(OL)}{\gamma}\\ \large\Delta x^A_{OD} &\large =\ \frac{\Delta x^B_{OL} }{\gamma}\\ \left( \begin{array}{c} \mbox{distance between Bob's lines}\\ \mbox{according to Alice} \end{array} \right) &= \frac{ \left( \begin{array}{c} \mbox{distance between Bob's lines}\\ \mbox{according to Bob} \end{array} \right) } { \gamma } \\ \Large L^A_{\scriptsize\mbox{ B's ladder}} &\Large = \frac{ L^B_{\scriptsize\mbox{ B's ladder}} }{\gamma}\\ \Delta x^A_{OD} & = \frac{\Delta x^B_{OD}}{\gamma}\\ (4) & \stackrel{\checkmark}{=} \frac{ (5) }{ \left(\frac{5}{4}\right) } \end{align}

You may want to check with your formulas with primes and "unprimes".

Now let's attempt your calculation to cancel the $$\gamma$$-factors.

\large \begin{align} \frac{ \Delta x^A_{OD} }{ \Delta t^B_{OQ} } &= \frac{\displaystyle \frac{ \Delta x^B_{OD} }{\gamma} } { \frac{ \displaystyle \Delta t^A_{OQ} }{\gamma} } = \frac{\Delta x^B_{OD} }{\Delta t^A_{OQ}} \end{align}

The important feature to note is that neither side is a velocity
since neither side is a slope measured by Alice or Bob
since neither side has the form $$\left( \begin{array}{c} \mbox{velocity of something}\\ \mbox{according to Alice} \end{array} \right)= \frac{ \left( \begin{array}{c} \mbox{spatial component of a hypotenuse}\\ \mbox{according to Alice} \end{array} \right) } { \left( \begin{array}{c} \mbox{temporal component of a hypotenuse}\\ \mbox{according to Alice} \end{array} \right) }$$ By counting diamonds, we have $$\frac{(4)}{(8)} = \frac{ (5)}{(10)}$$ which suggests no immediate physical interpretation of velocity.
By choosing a longer elapsed time along OQ or a longer ladder along OL, the ratios will change by the same factor. In any case, there is still no immediate physical interpretation.

They are just two things wrong with your argument:

Firstly, the claim that $$s=\sqrt{1-\frac{v^2}{c^2}} s'$$ where $$s$$ is the distance traveled in frame $$S$$ and $$s'$$ is the distance traveled in frame $$S'$$ is incorrect.

We have to take into account that the frame $$S'$$ is moving with speed $$v$$ relative to frame $$S$$, so the correct equation should be

$$s=\sqrt{1-\frac{v^2}{c^2}} s'+vt,$$ where $$t$$ is the elapsed time as measured in S. (Note that this is similar to $$x=x'+vt$$ in Galilean transformations, but with length $$x'$$ contracted.)

Secondly, the claim that $$t=\sqrt{1-\frac{v^2}{c^2}} t'$$ where the $$t$$ is the time taken to travel distance $$s$$ in $$S$$ and $$t'$$ is the time taken to travel distance $$s'$$ in $$S'$$ is incorrect.

The derivation of the time dilation formula is such that it is only valid for a clock that is fixed at a particular point in frame $$S'$$. To measure the time taken to travel a distance in $$S'$$, we have to use two clocks in $$S'$$ that are situated at two different points in $$S'$$. Thus we cannot use the time dilation formula.

To relate $$t$$ and $$t'$$, we take the perspective of frame $$S'$$. According to frame $$S'$$, frame $$S$$ is moving away with a speed of $$-v$$. For a distance $$s$$ moved in frame $$S$$, the distance $$s'$$ moved in frame $$S'$$ will be $$s'=\sqrt{1-\frac{v^2}{c^2}} s-vt'.$$

Relating this to the equation $$s=\sqrt{1-\frac{v^2}{c^2}} s'+vt,$$ we can solve that $$t'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(t-\frac{v}{c^2}s)=\gamma{(t-\frac{v}{c^2}s)}.$$

Now that we have related $$s$$, $$s'$$, $$t$$ and $$t'$$ correctly to each other, you may proceed to show that in general $${s \over t}\neq{s'\over t'}$$.

Note: The relationship between $$s\over t$$ and $$s' \over t'$$ will be Einstein's velocity addition rule.

References:

1. Griffiths, Introduction to Electrodynamics, Chapter 12