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Let's consider two cars moving with velocity $v$ and from the ground frame the separation between them to be equal to $l$.

In the reference frame fixed to the cars simultaneously 2 light bulbs are switched on, now I want to find the time interval between the switching on of the two bulbs for which I used two methods .

Lets say the events happen with coordinates $(0,0)$ and $(0,l')$ in the frame of the leftmost cart. Now using Lorentz transformation corresponding coordinates in ground frame will be $(0,0)$ and $(\frac{vl\gamma^2}{c^2},l)$.

When I work the other way out say from ground frame events happen at $(0,0)$, $(t,l)$. Now the same event happens in moving frame at $(0,0)$ , $(\gamma(t-\frac{vl}{c^2}),\gamma(l-vt)$. Now since the two event happen simultaneously this implies $t=\frac{vl}{c^2}$.

Why do the two methods give different results for time interval between flashing of two light bulbs in the ground frame?

($\gamma$ is the lorentz transformation factor $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$)

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    $\begingroup$ You must clarify the configuration of the systems : Are there two systems $\,\rm S\,$ the ground one and $\,\rm S'\,$ the moving one ??? if so, how the latter is moving with respect to the former ??? (velocity vector ???) What do you mean by ".... with a separation between them $\,=\ell\,$..." ???? What is $\,Y\,$??? $\endgroup$
    – Frobenius
    Oct 13, 2022 at 13:10
  • $\begingroup$ The two moving frames are at seperation l and Y is the lorentz factor gamma @Frobenius $\endgroup$ Oct 13, 2022 at 14:11
  • $\begingroup$ gamma $\gamma$ is \gamma between dollar-sign characters $\endgroup$
    – robphy
    Oct 13, 2022 at 14:24

1 Answer 1

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Let's consider two cars moving with velocity $v$ and from the ground frame the separation between them to be equal to $l$.

In the reference frame fixed to the cars simultaneously 2 light bulbs are switched on, now I want to find the time interval between the switching on of the two bulbs

In general, this type of problem requires substantial care. You have specified the separation in the ground frame but the simultaneity in the car frame. Luckily, it is easy to specify the separation in the car frame using the length contraction formula: $l'=l \gamma$.

Once you have everything specified in the car frame then it is far easier to start with the car frame and transform to the ground frame. However, even that must be done with care.

Lets say the events happen with coordinates $(0,0)$ and $(0,l')$ in the frame of the leftmost cart. Now using Lorentz transformation corresponding coordinates in ground frame will be $(0,0)$ and $(\frac{vl\gamma^2}{c^2},l)$.

Here is your first mistake. Let's call the event of the flash on the left car $a$ and the event of the flash on the right car $b$. So $a=a'=(0,0)$ and $b'=(0,l')$, but it is not correct that $b=(\frac{vl\gamma^2}{c^2},l)$. Even though the length is given as $l$ in the ground frame, you are forgetting that the car is moving in the ground frame, so the right car will not usually be at $l$.

The correct way is to actually fully transform into the ground frame to obtain $b=(\frac{l' v \gamma}{c^2},l' \gamma)$. Once you have that then you can substitute $l'=l\gamma$ to get $b=(\frac{l v \gamma^2}{c^2},l \gamma^2)$.

This is the correct answer. It really does not make sense to work it the other way since we have all of the relevant quantities in the car frame, but it can be done.

When I work the other way out say from ground frame events happen at $(0,0)$, $(t,l)$.

Here is your second error. In this approach $b=(t,l+vt)$. Again, you are making the same mistake, forgetting that the right car is moving in the ground frame so the position is not generally $l$. If you transform the corrected expression you get $b'=(t \gamma - \frac{v(l+vt)\gamma}{c^2},l \gamma)$. We can then set the first term to 0 and solve for $t$ to get $t=\frac{lv\gamma^2}{c^2}$ which matches the correct expression found above.

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  • $\begingroup$ why we have to take b position in ground frame to be $=l+vt$ we are just transforming the coordinates of an event in different frame $\endgroup$ Oct 15, 2022 at 11:53
  • $\begingroup$ @PrateekMourya the event we are transforming is an event on the right car. At time $t$ the right car is located at $l+vt$, not $l$. $\endgroup$
    – Dale
    Oct 15, 2022 at 12:02

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