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The formula for length contraction is:

$$ l' = \frac{l}{ \gamma} $$

So if the distance between earth and an asteroid is 2 lightyears. I travel with speed 0.5c, then my lorentz factor is 1.155. So the distance from my (l') perspective is 2/1.55 = 1.73c. So it will take 1.73 / 0.5 = 3.46 years.

So let us calculate that with the time dilation equation:

$$ t' = \gamma t $$

So logically, l' would be the same frame as t'. t = 2 lightyears / 0.5 = 4 years, and the lortentz factor is 1.155.

So from my (t') perspective, the time to travel that distance would be 1.155 * 4 = 4.62 years. Wait what. That doesn't match up.

Why isn't t' defined as:

t' = t / lorentz, just like the length contraction formula.

I'm guess I went wrong somewhere... but I'd like to know where. Am I wrong to assume that t' and l' are from the same perspective?

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Am I wrong to assume that t' and l' are from the same perspective?

In both equations given, the unprimed quantities are proper quantities, i.e., $l$ is the proper length and $t$ is the proper time.

See that, understood this way, the proper length (rest length) is the greatest length observed from any inertial reference frame (length contraction) while the proper time ('wristwatch' time) is the smallest time observed (time dilation).

What is the proper length in your scenario? It is the length of a 'ruler' that extends between the Earth and destination. Clearly, this ruler is at rest with respect to the Earth so $l$ in your length contraction formula must be the length in the Earth's frame.

What is the proper time in you scenario? It is elapsed time according to a clock that moves from the Earth to the destination and so $t$ in the time dilation formula must be the time according to the clock on the spaceship.

It would probably better to clearly denote proper quantities like so:

$$l_\gamma = \frac{\Lambda}{\gamma},\qquad t_\gamma = \gamma \tau$$

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Normally the mistakes is this kind of problems are that it's not easy to identify the reference frames. Let's consider $(x',t')$ our astronaut, who's moving at $v=0.5c$, and $(x,t)$ the person who stays at Earth. The distance between the Earth and the asteroid is $x=2$ ligthyears, so the timetravel is $t=\frac{x}{v}=4$ years. We have now the parameters of the person who is in the Earth, and now we can use the boost to calculate the time that will spend our astronaut doing his travel. $t'=\gamma(t-v·x)=\gamma(t-v·v·t)=\gamma·t(1-v^2)=\frac{t}{\gamma}$, comparing with the equation that you're showing your $t'$ refers to the Earth's frame and $t$ is for the astronaut. It's better to use the boost for not doing this mistakes and you can be sure when you solve this problems always thinking that the person who is moving will spend less time and will move less distance than the person whos stays at the Earth.

So the results are:

For the Earth $(x,t)=(2$ lightyears$,4$ years$)$

For the astronaut $(x',t')=(v·t',t')=(1.73$ lightyears$, 3.46$ years$)$

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As others have mentioned, a common problem is understanding the meaning of the variables in the formula. [Which is primed? and unprimed? Maybe this notation is ambiguous.]

In my opinion, a spacetime diagram is good way to clarify the meaning of the different variables because one has something more tangible than "carefully-chosen but maybe not-well-understood strings of words or symbols". ("A spacetime diagram is worth a thousand words.")

The answer I offer here elaborates on the earlier answers by using the spacetime diagram, together with rapidities (Minkowski-angles), which suggests helpful analogies with ordinary Euclidean trigonometry. [In addition, given the Euclidean geometry version of this problem, would one use rotation matrices?]

spacetime diagram with rapidities

With time running upwards, we have drawn the worldlines of the earth, the asteroid [at rest with respect to the earth], and the ship that leaves the earth at event O and arrives at the asteroid at event Q.

With the ship velocity $v=(1/2)c$,
the gamma-factor is $\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{2}{\sqrt{3}}=1.1547...$ [in units where $c=1$].
Using rapidities, where $v=\tanh\theta$, we have $\gamma=\cosh\theta$. So, $\gamma=\cosh(\rm arctanh(v))=\cosh(\rm arctanh(0.5))=1.1547...$ (with help from WolframAlpha). The Rapidity-angle is angle between timelike worldlines of the earth and ship, and between their spacelike-directions in this plane.

  • Consider Minkowski-right-triangle OPQ (with hypotenuse $OQ$ and legs $PQ \perp OP$, along the earth's space and time axes). The velocity is $v=\frac{PQ}{OP}$.
    Time-Dilation compares the measured-durations between events O and Q by comparing the (apparent-duration) timelike-leg OP to the (proper-time) timelike-hypotenuse OQ].
    The time-dilation factor arises from $\gamma=\cosh\theta=\frac{ADJ}{HYP}=\frac{OP}{OQ}$. So, $$(\mbox{apparent-duration } OP)=\gamma(\mbox{proper-time }OQ).$$

  • Consider Minkowski-right-triangle QPX (with hypotenuse $QX$ and legs $PX \perp QP$, along the earth's time and space axes).
    Length-contraction compares the measured-distances between two parallel timelike-worldlines by comparing the (proper-length) spacelike-leg PQ to the (apparent-length) spacelike-hypotenuse XQ.
    The length-contraction factor arises from $\gamma=\cosh\theta=\frac{ADJ}{HYP}=\frac{QP}{QX}$. So, $$(\mbox{proper-length }QP)=\gamma(\mbox{apparent-length }QX),$$ or, more traditionally, $$(\mbox{apparent-length }QX)=\frac{(\mbox{proper-length }QP)}{\gamma}.$$

  • With $v=(1/2)$ and $(PQ)=2$,
    $$(QX)=\frac{(PQ)}{\gamma}=\frac{2}{\cosh({\rm{arctanh\ }0.5})}=\frac{2}{2/\sqrt{3}}=\sqrt{3}$$ $$(OP)=\frac{(PQ)}{v}=\frac{2}{1/2}=4$$ $$(OQ)=\frac{(OP)}{\gamma}=\frac{4}{2/\sqrt{3}}=2\sqrt{3},$$ in accord with the answers given earlier.
    Refer to the diagram for a physical interpretation.

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