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Consider the following toy problem:

A ship starts with velocity $v=0.8\text{c}$ from Earth along the $x$ axis. After it has reached some distance $d$, a light signal is sent after the ship with $c$. We equip Earth with a resting frame $\Sigma$ and the ship with a moving frame $\Sigma'$.

Now we want to know when the signal reaches the ship and we obviously will get different answers, depending on the coordinate system.

In the frame $\Sigma$ this is easy and doesn't actually need SR or Lorentz transformations: A signal with velocity $c$ following a receiver with velocity $v$ is equivalent to a stationary receiver and a signal velocity of $c-v$.

However, in the moving frame $\Sigma'$ we have to carefully calculate the space-time coordinates of the events $\texttt{SEND}$ and $\texttt{RECEIVE}$ via the Lorentz transform and then take $t_1'-t_0'$ to know when the ship receives the light signal.

One way of doing that is to say that at the moment of the broadcast, the two origins $O$ and $O'$ coincide, so in $\Sigma$ we have for $\texttt{SEND}$ $(t_0,x_0)=(0,-d)$, and in $\Sigma'$ we get $(\gamma \frac{v}{c^2}d,-\gamma d)$ from $t'=\gamma\left(t-\frac{v}{c^2}x\right)$ and $x'=\gamma\left(x-vt\right)$.

What if we were to disregard the notion of coinciding origins for the moment and just say that in $\Sigma$, we tag the event $\texttt{SEND}$ as $(0,0)$? From this, the Lorentz transformation gives us $(0,0)$ in $\Sigma'$, too, but that doesn't make sense: I expect the spatial coordinate $x'$ to be something like a length contracted $-d$, so $x'=-\gamma d$, since that is the distance the moving ship sees Earth at.

Comparing this with the general form of $x'$ from the transformation I see that I should have put $x=-d$ to get this result, which in turn means that my $t'$ should be $t'=\gamma \frac{v}{c^2} d,$ of course coinciding with the $\Sigma'$ time coordinate from the earlier case.

Now this inconsistency (the transformed coordinates of $\texttt{SEND}$ not taking the spatial seperation into account) suggests that it is important when (and probably where) the origins of the two frames coincide, since choosing $(0,0)$ as the $\Sigma$-coordinates of $\texttt{SEND}$ means that the origins coincided at the start of the ship.

Is this really the case? How does the time when the origins of a moving and resting coordinate system coincide enter the Lorentz transformation? Is it maybe something else entirely, like an implicit simultaneity assumption across different frames I'm overlooking, which usually leads to weird results in SR?

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You are looking for the difference between a Poincaré transformation, also known as an inhomogeneous Lorentz transformation and a (homogeneous) Lorentz transformation. The adjective is often omitted in the latter case.

In special relativity, dealing with inertial frames, we want to preserve the spacetime interval $$ds^2 = dt^2 - dx^2_1 - dx_2^2 - dx_3^2.$$ The homogeneous Lorentz transformations are the linear transformations that are realized by matrix multiplication: $$x' = Lx \tag 1$$ for some $4\times 4$-matrix. You have seen and used one of these transformations, your boost.

However, as you have realized, this does not give us all possible intertial frames, since the origins of the $\Sigma'$ and $\Sigma$ frames will always agree according to (1). The inhomogeneous Lorentz transformations, also called Poincaré transformations are all transformations of the type $$x' = Lx + a \tag 2$$ where $a$ is a constant 4-vector. We get a homogeneous Lorentz transformation by taking $a = 0$.

Because $a$ is a constant, the time between two events in the primed system is $$(x'-y')_0 = (Lx + a)_0 - (Ly + a)_0 = (L(x-y))_0 = \sum_{\mu=0}^3 L_{0\mu} (x_\mu - y_\mu). \tag 3$$ Clearly, it does not depend on $a$, so it does not matter if the origins of the two frames are aligned or not.

Concretely in your case (3) means that we shift $\text{SEND}$ to $(0, 0)$, but then to be consistent, we have to shift $\text{RECEIVE}$ to $(t, d)$.

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    $\begingroup$ Oh right, so either I stack the two frames on top of each other at $t_0$ and so have $a=0$ or at $t_0$ they are some distance apart and I have to encode that distance when changing coordinate systems by taking a nonzero $a$, in this case equal to $d$. Sounds reasonable. $\endgroup$ – Wojciech Morawiec Mar 18 '17 at 14:08

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