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This question is about special relativity.

Assumptions

1. Speed of light is constant to all observers.

Context:

Let us imagine a train traveling in deep space. Now, let us suppose an observer 'A' inside the train and an observer 'B' outside the train. The train is traveling at 50% speed of light relative to observer 'B'. Now let us provide observer 'A' with a torch light. If 'A' opens the torch at time t=0 for B and we take a duration t=1sec outside the train.

Relative to B,

Distance covered by train = 0.5c as (d=vt),
Distance covered by light beam = c,
Total distance covered by light = c+0.5c,
According to this,
Relative to B light travels at 1.5c per second,

But, this breaks 2 rules,

  1. Universal Speed Limit
  2. Our 1st assumption

Now, let us assume time is not absolute but relative. In that case to satisfy our 1st assumption,
If, train travels 0.5c distance than light must travel 0.5c distance as well for total distance covered to be c in 1 sec. But, light can't slow down again from our 1st assumption. So, let us slow down time. To satisfy out assumption, how much must the time passed inside the train be? Well, let's calculate

Distance covered by train = 0.5c,
Distance to be covered by light = 0.5c,
Since,
time = ${\frac d v}$
t = ${\frac {0.5c} {c}}$
t = 0.5s,

However, from special relativity formulae,
${\triangle}$t' = ${\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}}$
According to this formula, for v = 0.5c & t=1
${\triangle}$t'=1.15,

I don't think my results defy Einstein. Could someone help me find what the mistake I committed here is. According to me, time inside the train should be twice as slow. AND, I have one other question, if ${\triangle}$t'>1 for t=1, how is time slowed inside the train. Does this mean time inside the train has greater duration? That is, 1.15s inside the train is equivalent to 1s outside it. This means 1s inside the train has a time duration less than duration of 1 sec outside the train. Thank you for any help.

Note I am not even an undergraduate in physics, so please don't get angry if I have committed grave errors in my question. I beleive my question is clear. Thank you for any help

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  • $\begingroup$ Take into account length contraction. $\endgroup$
    – Bill Watts
    May 4, 2021 at 3:56
  • $\begingroup$ @BillWatts Could you explain the answer? I am not an expert in this field. I know only that length contracts in the direction of motion but how does that conspire to give the same result? My understanding is as follows: As length contracts, it would seem as if light covered a lesser amount of distance thus the slowed down time must speed up. Thank you for the help $\endgroup$ May 4, 2021 at 4:06
  • $\begingroup$ Light is traveling a shorter distance to the outside observer than you are assuming. $\endgroup$
    – Bill Watts
    May 4, 2021 at 4:19
  • $\begingroup$ Oh right, thanks a lot. $\endgroup$ May 4, 2021 at 4:21
  • 2
    $\begingroup$ Any effort to understand this stuff in Special Relativity without the Lorentz transformation will go in vain. $\endgroup$
    – Frobenius
    May 4, 2021 at 4:33

1 Answer 1

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Relative to B,
Distance covered by train = 0.5c as (d=vt)
Distance covered by light beam = c
Total distance covered by light = c+0.5c,

The last line is incorrect. The distance covered by light is c.
It wont cover additional distance because of motion of train.
Once, light leaves the torch, the motion of train does not "add" to the motion of the light.
Hence, in 1 sec, the light will cover distance c.

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  • $\begingroup$ I have developed this intuitive idea because of some analogies in special relativity. One of them is the idea of the photon-clock. If the clock moves forward, why does the photon within it travel forward as well? I do not even know if my question makes sense so sorry for any errors as I said I am just curious, I haven't gotten a formal degree. $\endgroup$ May 4, 2021 at 12:47
  • $\begingroup$ I guess that ought to do it. My main assumption was wrong. But could you please explain the solution to the problem in my previous comment @silverrahul $\endgroup$ May 4, 2021 at 13:01
  • $\begingroup$ @SamipGyawali The clock does not start to move forward and the photon does not start to travel forward, . In the frame in which , the clock is moving, the clock is always moving and hence the photon always has forward velocity $\endgroup$ May 4, 2021 at 13:59

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