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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.

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  • $\begingroup$ This question is very similar to this one: Time Dilation + Length Contraction Scenario, though it's worded a little differently. Check out my answer there. $\endgroup$
    – Philip
    Aug 22 '20 at 6:17
  • $\begingroup$ What you write is not a Lorentz transformation. $\endgroup$
    – my2cts
    Aug 22 '20 at 7:25
  • $\begingroup$ Thank you so much Philip for typesetting and upgrading the language of question. Surely I’ll learn and use MathJax in future. $\endgroup$
    – rim
    Aug 22 '20 at 8:31
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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.

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  • $\begingroup$ 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/𝛾 & second’=secondX 𝛾? $\endgroup$
    – rim
    Aug 22 '20 at 7:56
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    $\begingroup$ @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? $\endgroup$
    – Philip
    Aug 22 '20 at 7:59
  • $\begingroup$ 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. $\endgroup$
    – rim
    Aug 22 '20 at 8:27
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    $\begingroup$ 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. $\endgroup$
    – Philip
    Aug 22 '20 at 8:30
  • $\begingroup$ 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/𝛾^2)?! I may be measuring speed of light as c m/s but for you I must be measuring speed of light as c/𝛾^2 m/s, don’t you think? Please read “measure of unit” as scale in my previous comment. $\endgroup$
    – rim
    Aug 22 '20 at 8:55
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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.

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