- #1

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assuming of course that all events lie along the x axis which as also the direction of motion.

I am not asking how to calculate the interval. I know that. I'm asking how to calculate how a different observer will see it.

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- Thread starter granpa
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- #1

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assuming of course that all events lie along the x axis which as also the direction of motion.

I am not asking how to calculate the interval. I know that. I'm asking how to calculate how a different observer will see it.

- #2

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assuming of course that all events lie along the x axis which as also the direction of motion.

I am not asking how to calculate the interval. I know that. I'm asking how to calculate how a different observer will see it.

If according to observer A the difference in between the two events is [itex]( \Delta t, \Delta d) [/itex] then the the difference between the two events according to observer B is [itex](\Delta t*g , \Delta d/g) [/itex] ..Simple as that ;)

Please don't get confused by the Lorentz transformation equations

where setting x to zero in [tex]t^{\prime} = g \left( t - \frac{vx}{c^2} \right)[/tex] appears to give [tex]t^{\prime} = g* t [/tex]

and setting t to zero in [tex]x^{\prime} = g (x - vt)[/tex] appears to give [tex]x^{\prime} = g*x[/tex]

..where [tex]g = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

That is not the correct way to analyse the situation.

The correct way can be found here for time :http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c2 and here for distance http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c1

There is a small error in the hyperphysics calculation for time transformation but they get the right final result in the end due to a correcting error. See if you can spot it ;)

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- #3

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that cant be. in the frame of a stationary observer the distance between 2 events might be zero and the time nonzero. to a moving observer the distance between those same 2 events would not be zero.

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- #4

jtbell

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[tex]x_1^{\prime} = \gamma (x_1 - v t_1)[/tex]

[tex]t_1^{\prime} = \gamma (t_1 - v x_1 / c^2)[/tex]

[tex]x_2^{\prime} = \gamma (x_2 - v t_2)[/tex]

[tex]t_2^{\prime} = \gamma (t_2 - v x_2 / c^2)[/tex]

Subtracting pairs of equations gives

[tex]\Delta x^{\prime} = x_2^{\prime} - x_1^{\prime} = \gamma ((x_2 - x_1) - v (t_2 - t_1)) = \gamma (\Delta x - v \Delta t)[/tex]

[tex]\Delta t^{\prime} = t_2^{\prime} - t_1^{\prime} = \gamma ((t_2 - t_1) - v (x_2 - x_1) / c^2) = \gamma (\Delta t - v \Delta x / c^2)[/tex]

That is, the Lorentz transformation applies to [itex]\Delta x[/itex] and [itex]\Delta t[/itex] just as it does to x and t, because the transformation is linear.

- #5

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that cant be. in the frame of a stationary observer the distance between 2 events might be zero and the time nonzero. to a moving observer the distance between those same 2 events would not be zero.

Exactly. See my reply to your question about proper velocity.

Imagine you are on a train and you have a flashlight. You flash it once and few seconds later you flash it again. If you consider yourself (and the train) to be stationary, then the two flashes occured in the same place according to you. Another observer on the side of the track says the two flashes did not happen in the same place relative to his reference frame (The railway track).

In this case, what you call the "moving observer" is the observer on the trackside. Is he really moving? He is stationary as far as he is concerned, because he not moving relative to the railway embankment. Is the observer on the train really stationary or moving? Stationary and moving are relative concepts and that is sort of the whole point of relativity. All you can say is that object A is moving relative to object B or observer A is moving relative to observer B. To say observer A is moving or stationary without reference to something else is meaningless.

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- #6

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[tex]x_1^{\prime} = \gamma (x_1 - v t_1)[/tex]

[tex]t_1^{\prime} = \gamma (t_1 - v x_1 / c^2)[/tex]

[tex]x_2^{\prime} = \gamma (x_2 - v t_2)[/tex]

[tex]t_2^{\prime} = \gamma (t_2 - v x_2 / c^2)[/tex]

Subtracting pairs of equations gives

[tex]\Delta x^{\prime} = x_2^{\prime} - x_1^{\prime} = \gamma ((x_2 - x_1) - v (t_2 - t_1)) = \gamma (\Delta x - v \Delta t)[/tex]

[tex]\Delta t^{\prime} = t_2^{\prime} - t_1^{\prime} = \gamma ((t_2 - t_1) - v (x_2 - x_1) / c^2) = \gamma (\Delta t - v \Delta x / c^2)[/tex]

That is, the Lorentz transformation applies to [itex]\Delta x[/itex] and [itex]\Delta t[/itex] just as it does to x and t, because the transformation is linear.

there must be an error in the second equation. it doesnt work for the trivial case of a particles position relative to itself (ie 0,0). are you sure you dont divide by v? then it would work(but only if c=1)

at time=zero the particle is at (0,0). at time=t it is at (t,tv). changing to the frame of the particle itself it should always be at (0,0)

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- #7

atyy

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at time=zero the particle is at (0,0). at time=t it is at (t,tv). changing to the frame of the particle itself it should always be at (0,0)

the formula is probably right, it gives:

at t=0 the particle is at (t=0,x=0),(t'=0,x'=0).

at t=t it is at (t=t,x=vt),(t'=t(1-v

t' is not zero, since time will pass for in the particle's frame. it should be less than t, because of time dilation, but not so slow until time has stopped, unless it is a photon.

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- #8

jtbell

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at time=zero the particle is at (0,0). at time=t it is at (t,tv). changing to the frame of the particle itself it should always be at (0,0)

In the rest frame of the particle, its position is always zero, but time still elapses for it (although more slowly because of time dilation).

Letting [itex]x_2 = vt[/itex] and [itex]t_2 = t[/itex] we get

[tex]x_2^{\prime} = \gamma (x_2 - v_2 t) = \gamma (vt - vt) = 0[/tex]

[tex]t_2^{\prime} = \gamma (t_2 - v x_2 / c^2) = \gamma (t - v^2 t / c^2) = \gamma (1 - v^2 / c^2) t = \gamma (1 / \gamma^2) t = t / \gamma[/tex]

- #9

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seems obvious now

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