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I'm just starting to learn special relativity, and I'm having trouble with the following concept:

In relativity, units of length and time of moving frame are related to that of stationary one through $$x’=\frac{x}{\gamma}\quad \quad \text{ and }\quad \quad t’=t\times \gamma$$ respectively, where $\gamma$ is Lorentz Factor.

Does this also mean that units of velocity or speed, i.e. length/time are related as $$v’=\frac{x’}{t’}=\frac{x}{t}\times\frac{1}{\gamma^2}=\frac{v}{\gamma^2}?$$

Note: By unit, I mean scale of axes in a respective coordinate system and I am not asking about addition or subtraction of velocities, I am enquiring about mutual “scale” difference between the quantity called velocity as measured in two different frames in uniform relative motion to each other.

Why scale of length contracts and not expands while that of time dilates, i.e. expands when the two are symmetrical for Lorentz transformations! Only if length expands with dilation of time can the “scale” of velocity or speed in general and speed of light in particular can remain truly invariant, I guess.

rim
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2 Answers2

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No. Velocities do transform in a non-intuitive way in special relativity, but not in the way you're describing. This is sadly a very common misunderstanding due to the fact that Relativity is usually first introduced using time-dilation and length contraction, without actually explaining under which conditions they are applicable. The best way to begin understanding the subject (and to also avoid all these "paradoxes") is to work with the Lorentz Transformations. In one dimension, if the frame $S'$ is moving with respect to $S$ with a velocity $v$, then

\begin{aligned} x' &= \gamma\left( x - vt \right)\\ t' &= \gamma \left( t - \frac{v}{c^2}x\right) \end{aligned}

provided that $x=x'=0$ when $t=t'=0$. Remember, though, that these are coordinates, not intervals. To find how intervals of length and intervals of time are related, we need to take differences of the coordinates, and since $v$ and $\gamma$ are constants, it's easy to show that the intervals satisfy similar equations:

\begin{aligned} \Delta x' &= \gamma\left( \Delta x - v \Delta t \right)\\ \Delta t' &= \gamma\left( \Delta t - \frac{v}{c^2}\Delta x\right) \end{aligned}

We can use the above equations to easily calculate how the velocities transform between the frames $S$ and $S'$. Remember that an observer in $S$ will calculate the velocity of an object to be $$u = \frac{\Delta x}{\Delta t},$$ and one in $S'$ will calculate it to be $$u' = \frac{\Delta x'}{\Delta t'}.$$

We can now divide $\Delta x'$ by $\Delta t'$ to show that:

$$u' = \frac{\Delta x'}{\Delta t'} = \frac{u - v}{1 - \frac{uv}{c^2}}.$$

Why doesn't your argument work?

Length contraction and time dilation are special cases of the general formulae that I have given above. They hold when certain conditions are satisfied, and these conditions are more certainly not satisfied simultaneously. Which means dividing the equations is not going to give you anything sensible. In special relativity, it is best to think in terms of "events" which occur at spacetime points $(t, x)$ to avoid such false "paradoxes". My answer here, and the links at the end, should explain it in more detail.

Philip
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  • First of all, thanks Philip for the reply, but I am afraid I am still unable to grasp the concept. The point of confusion is not the numerical aspect but the unit aspect. Correct me if I am wrong, any dimensional value has two things, a number and a unit, right! Now in one coordinate system velocity may be v meters/second and in other it may be v’ meter’/second’ with v & v’ related as per relation you have given. My confusion is how meter’/second’ be related to meter/second. Is it as per the relation in my question since meter’=meter/ & second’=secondX ? – rim Aug 22 '20 at 07:56
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    @rim I see, I didn't understand that from the question. But that's really not a problem at all! $\gamma$ has no units (it's dimensionless, just a pure number). So $x$ and $x'$ have the same units (just as $t$ and $t'$ have the same units). All physical quantities (including velocities) have the same units in both frames. Does that help? – Philip Aug 22 '20 at 07:59
  • Again thanks for reply Philip but sorry for my inability to understand, If units remain same between different coordinate systems then why do we say that length has contracted or time has dilated in the first place? i.e. why one meter measured by an observer is different from one meter measured by another observer in relative motion with former! During transformation of coordinates three things can happen, either number of unit can change or measure of unit itself can change or both. I am unable to grasp second & third aspect of Lorentz transformations, I guess. – rim Aug 22 '20 at 08:27
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    A metre continues to be a metre, since it's just a measure of length. Let's take an example: you are in $S'$ moving with respect to me in $S$ with a speed $v$. Now, we both have metre scales in our hands. I measure my scale to be 1 metre in my frame, but I will measure your scale to be contracted to $1/\gamma$ metres. Similarly, you measure your own scale to be 1 metre, but you will measure mine to be $1/\gamma$ metres. Does this help? The unit itself doesn't change, it is the numerical value that does. – Philip Aug 22 '20 at 08:30
  • It seems I am equally bad at conveying as at understanding! Doesn’t that mean “scale” of velocity or speed also changes, i.e. forget the numerical part, since my scales of meter and second seems different to you, won’t my scale for a “unit of” velocity (meter/second) will seem different to you (and that too by 1/^2)?! I may be measuring speed of light as c m/s but for you I must be measuring speed of light as c/^2 m/s, don’t you think? Please read “measure of unit” as scale in my previous comment. – rim Aug 22 '20 at 08:55
  • I'm not sure I understand what you mean by "scales". If you want to ignore the numerical part, from your own comment, both of us will in fact be measuring speed as being m/s. (You do understand that $\gamma$ has no units, yes?) If you don't want to ignore the numerical part, then you should use the formula that I have provided in my answer to find the numerical part, and you will see that both of us measure the an object moving at the speed $c$ to be $c$. It is true that for other velocities, we would not agree, but that's not strange since the same thing happens in "everyday" physics. – Philip Aug 22 '20 at 09:05
  • Actually my query is rooted in a far more basic issue, i.e. why in actual measurements length contract and time dilates when the Lorentz transformations seems symmetrical. Even in minkowski diagrams, scales of both axes in a respective coordinate system are drawn equal then why while transforming we differentiate between time & space? Is this difference apparent, i.e. time scale appears to expand & length scale appears to contract because of apparent difference between time (in continuous flowing state) & space (in rest state) themselves!? – rim Aug 22 '20 at 09:11
  • Let me make another attempt at clearing my query, why both space and time don’t contracts or expands scale wise while transforming from one coordinate system to another when Lorentz transformations are symmetrical, x4(time) can be exchanged with x1(space) without loosing form of Lorentz transformations! Because only then effect of scale change will cancel out and we will get same speed of light in all frames of reference, numerically & scale wise! Sorry for being such a dumb but I am really curious to understand this fact! – rim Aug 22 '20 at 09:21
  • I understand is dimensionless but suppose I measure 1 of your meter as 0.1 of my meters and 1 of your second as 10 of my seconds then won’t I be measuring 1 of your meter per second as 0.01 of my meter per second? Phew, This is my query. – rim Aug 22 '20 at 09:27
  • Ok, there are many different questions here, and I think this comment thread is getting a little too long. I can't see the option to move this to chat automatically, so I've created a chat room here, you may join if you wish to continue the conversation there. – Philip Aug 22 '20 at 09:29
  • Sorry Philip I don’t have enough “reputation“ to join you on chat, if you are willing then just please try answer my last query/comment on the subject. I am really glad for your patience with a dumb head like me! Thank you so much! – rim Aug 22 '20 at 09:37
  • Ok. Relativity is not an obvious concept, so don't be too worried if you can't understand it completely. I have actually dealt with this in my answer. If you notice, I have actually calculated the velocity that different observers see using the fact that $u = \frac{\Delta x}{\Delta t}$ and $u' = \frac{\Delta x'}{\Delta t'}$. However, you will notice that I did not use $\Delta x' = \gamma \Delta x$ and $\Delta t' = (1/\gamma) \Delta t$. The reason for this is because these two formulae are not applicable here. Their ratio does not describe the velocity of any specific object. – Philip Aug 22 '20 at 09:46
  • The point is, even if I measure 1 of your metres to be 0.1 m, and one of your seconds to be 10 s, I will not measure your speed (or the speed of light) to be their ratio. Speeds are not just ratios of these quantities. To calculate the speed correctly, you need to use the Lorentz Transformations I've described above. I can't seem to find a succinct way to explain it here, so perhaps you can modify the question (or ask another one) with this idea. Sorry :/ – Philip Aug 22 '20 at 09:52
  • Now getting the grasp of it a bit, thank you so much Philip. Just one last query, let’s decide to measure space time interval in meters. Now space time interval is invariant, right. Let suppose for some events it is say 10 meters. That means for me it will be 10 meters and for you too it will be 10 meters. Right? Now considering you to be in relative motion w.r.t. me let my meter scale be 10 times of your meter scale, so will that mean I must deduce space time interval length to be of 1 meter from your point of view?! I guess not, but why? – rim Aug 22 '20 at 10:21
  • The spacetime interval is an interval between two events, say $(t_1,x_1)$ and $(t_2,x_2)$. If you use the Lorentz Transformations that I have given you in my answer, you will find that while $(x_2 - x_1)$ will not be the same as $(x'_2 - x'_1)$, and $(t_2 - t_1)$ will not be the same as $(t'_2-t'_1)$, they change in exactly the right way so that the interval is constant. We will still both measure it to be 10m. Additionally, I think this question is pretty much the same thing you're asking in your original question. – Philip Aug 22 '20 at 11:02
  • Agreed but my meter scale measures 10 times value of your meter scale, so when you say that you have deduced space time interval of 10 by your meter scale, then wouldn’t that mean for me you are suggesting value of 10/10=1 meter by my meter scale! I mean numbers in various equations of SR are alright but what about unit scales that are used in them, e.g. which of the following represents transformations x’=f(x) correctly, (a) x’ unit’ = f(x unit) with unit’=g(unit) or (b) x’ unit (or unit’)= f(x unit (or unit’)) with unit’=unit but then how my meter stick will read different value from yours?! – rim Aug 22 '20 at 14:44
  • Thank you so much Philip, with your insights and the explanations provided by MinutePhysics YouTube Channel, I now, am finally able to comprehend what’s going on in Relativity. And yes, Black Holes don’t exist! ;P – rim Aug 23 '20 at 11:06
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Along with Philip’s answer, this video ( https://youtu.be/-NN_m2yKAAk ) brought clarity to the problem.

Basically what I understood from the two is that time dilation and length contraction are not similar or symmetrical things (as I was assuming in original question) but by nature, we are able to measure time only using one and length using another method respectively.

In fact, there are concepts of length dilation and duration contraction that are not generally discussed but should be.

rim
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