Special relativity - photograph of a moving meter stick

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Homework Statement


A meter stick, pointing in the x-direction moves along the x-axis with speed v=0.8c, with its midpoint passing through the origin at time t=0. A stationary observer situated at the origin takes a photograph of the stick at time t=0. At what poisition does the photograph show the front and back end of the stick to be at t=0?


Homework Equations



x'=γ(x-ut)


The Attempt at a Solution


I'm confused here by the use of photograph. I've tried to use the Lorentz contraction but do I need to do it twice, once for the light going out and once for it coming back to the photographic plate? Very confused and all help/guidance gratefully accepted.
 
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Answers and Replies

  • #2
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I think the photograph is just a means of measuring points on both sides of the observer simultaneously.

Think of it as two points that are a meter apart (or a half meter in front of and behind the center of the meter stick) in the reference frame of the meter stick. At t=0, the observer will agree that the center of the stick is at 0 m. Where does that observer measure as the position of the two points?
 
  • #3
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Hi

Thanks for answering Fewmet.

The first part of the question was where in the observer's frame are the end points of the meter stick and I have successfully calculated that to be +/- 0.3m. Now I'm stuck on the next bit which I've posted. Just not sure how to tackle this next bit.
 
  • #4
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The first part of the question was where in the observer's frame are the end points of the meter stick and I have successfully calculated that to be +/- 0.3m. Now I'm stuck on the next bit which I've posted. Just not sure how to tackle this next bit.

OK: I am a little confused. You said you got an answer for the the observer taking the photograph. It seems to me that is what this
A meter stick, pointing in the x-direction moves along the x-axis with speed v=0.8c, with its midpoint passing through the origin at time t=0. A stationary observer situated at the origin takes a photograph of the stick at time t=0. At what poisition does the photograph show the front and back end of the stick to be at t=0?
asked for. What's the next bit?
 
  • #5
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No. My answer was for where in the observer's frame of reference is the stick. I calcualted +/-/0.3m. Now I have to think about the position on the photograph. I suspect that you have to take into account not only the contraction but also the finite speed of light?
 
  • #6
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No. My answer was for where in the observer's frame of reference is the stick. I calcualted +/-/0.3m. Now I have to think about the position on the photograph. I suspect that you have to take into account not only the contraction but also the finite speed of light?

The original problem refers to a meter stick moving at 0.8c and a "stationary observer" taking a photograph. I read that as "moving 0.8c with respect to x=0" and "stationary with respect to x=0".

Am I right that you are reading the question differently?

In my interpretation, the photographer sees the meter stick as shortened. An observer moving with the meter stick is in its reference frame and sees nothing unusual: the is 1 m long.

Are we in agreement on that?
 
  • #7
Doc Al
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No. My answer was for where in the observer's frame of reference is the stick. I calcualted +/-/0.3m. Now I have to think about the position on the photograph. I suspect that you have to take into account not only the contraction but also the finite speed of light?
That's exactly right. The photograph doesn't record where the ends of the stick are at time t = 0, but where they were when the light from the ends of the stick was emitted.
 
  • #8
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Thanks for the answers.

I think what I need to do is work out what additional length will be added to the -x direction to allow for the finite speed of light and what length needs to be taken off the +x direction? Not sure how I should do that though?

Am I right in thinking that the length on the photo must still be .6m?
 
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  • #9
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Thanks for the answers.

I think what I need to do is work out what additional length will be added to the -x direction to allow for the finite speed of light and what length needs to be taken off the +x direction? Not sure how I should do that though?

Am I right in thinking that the length on the photo must still be .6m?

I am not sure about Doc Al's reply. Yes the finite speed of light is taken into account, but I think that is incorporated in the equations. You don't need to make any further calculations

If the meter stick moves with respect to the photographer, the photographer measures the stick as being 0.6 m long using the same light that the photograph records. The photo looks just like what he sees.

If I am wrong about this, I'd be delighted if someone corrected me.
 
  • #10
Doc Al
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I am not sure about Doc Al's reply. Yes the finite speed of light is taken into account, but I think that is incorporated in the equations. You don't need to make any further calculations

If the meter stick moves with respect to the photographer, the photographer measures the stick as being 0.6 m long using the same light that the photograph records. The photo looks just like what he sees.

If I am wrong about this, I'd be delighted if someone corrected me.
The photo records the light that hits it at t=0. That light takes time to reach it, so the photo records the position of the ends at some earlier time. It does not record their position at t = 0.

That's why there are two parts to the question: (1) Where are the ends at t = 0 (according the observer) and (2) What would a photograph show.
 
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  • #11
Doc Al
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I think what I need to do is work out what additional length will be added to the -x direction to allow for the finite speed of light and what length needs to be taken off the +x direction?
Exactly.
Not sure how I should do that though?
Use a little algebra. You know the speed of the stick and the speed of light.

Am I right in thinking that the length on the photo must still be .6m?
I would make no such assumption. :wink:
 
  • #12
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The photo records the light that hits it at t=0. That light takes time to reach it, so the photo records the position of the ends at some earlier time. It does not record their position at t = 0.

That's why there are two parts to the question: (1) Where are the ends at t = 0 (according the observer) and (2) What would a photograph show.

Thanks for addressing that, Doc Al. To clarify for myself, am I correct that you are drawing the distinction between what is subjectively measured at t=0 and the objective reality at that instant that can be calculated (assuming uniform velocity of the stick)?
 
  • #13
Doc Al
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Thanks for addressing that, Doc Al. To clarify for myself, am I correct that you are drawing the distinction between what is subjectively measured at t=0 and the objective reality at that instant that can be calculated (assuming uniform velocity of the stick)?
You could put it that way, yes. To measure the 'actual' length of the stick at time t=0 you would have to measure the location of each end at exactly t=0. A single photograph at x=0 will not do that.
 
  • #14
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You could put it that way, yes. To measure the 'actual' length of the stick at time t=0 you would have to measure the location of each end at exactly t=0. A single photograph at x=0 will not do that.

I don't like putting it that way because it feels like it's violating the first postulate, and think that is the source of all my qualms here.

Please check my reasoning from there: someone moving with the meter stick sees it being a meter long and having the 50 cm mark at x=0, the leading edged at xL=0.5 m and the trailing edge at xT=-0.5 m. I could calculate what the photographer sees (which is the same as what the photograph documents) as xL' and xT' by using the Lorentz coordinate transformations. But that, you are saying, is not where the ends "really" are...

Doesn't that lead us to the ends "really" being at -0.5 m and 0.5 m, and that the observer moving with the stick makes the "correct" measurements. Isn't that in violation of the first postulate?

I know that cannot be the case, but what am I missing here?
 
  • #15
Doc Al
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Please check my reasoning from there: someone moving with the meter stick sees it being a meter long and having the 50 cm mark at x=0, the leading edged at xL=0.5 m and the trailing edge at xT=-0.5 m.
I would say that observers moving along with the meter stick would 'see' the ends at ±0.83 m on the coordinate grid that the photographer is using. One must put 'see' in quotes because we really mean 'calculate and measure' not literally 'see'. (The meter stick observers 'see' the other frame as contracted.)
I could calculate what the photographer sees (which is the same as what the photograph documents) as xL' and xT' by using the Lorentz coordinate transformations.
The Lorentz transformations will tell you what observers in the 'stationary' frame would measure as the positions of the ends of the stick at t=0: ±0.3 m.
But that, you are saying, is not where the ends "really" are...
I'm saying just the opposite. At t=0, observers in the 'stationary' frame will measure the ends of the stick to 'really be' at ±0.3 m.

But what is captured on a photograph is not where the ends 'really are', but it is distorted by the travel time of the light reaching the film. The Lorentz transformations deal with observations after all such effects have been taken into account; they deal with 'reality'.

Doesn't that lead us to the ends "really" being at -0.5 m and 0.5 m, and that the observer moving with the stick makes the "correct" measurements. Isn't that in violation of the first postulate?
Each set of observers is entitled to their own measurements and there is no contradiction.

I know that cannot be the case, but what am I missing here?
The subtlety is the difference between 'reality' and what is recorded on the film.

What makes this example a bit silly is that the stick is only one meter long and moving fast. Lots of luck measuring that! A better example might be something like this. Imagine the stick is a train that is one light year long. At the second the middle of the train passes the origin, an observer there who is able to view the ends of the train through telescopes records where he sees them. How do his raw observations through the telescopes (analogous to what was recorded on the film in the original example) compare with what he would calculate as the positions of those ends?
 
  • #16
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I would say that observers moving along with the meter stick would 'see' the ends at ±0.83 m on the coordinate grid that the photographer is using. One must put 'see' in quotes because we really mean 'calculate and measure' not literally 'see'. (The meter stick observers 'see' the other frame as contracted.)

That's interesting: I've not seen the distinction between coordinate grids before. So the values of -0.5 m and 0.5 m are on the coordinate grid of the meter stick. The observer at rest with respect to the meter stick sees the photographer's coordinate grid as shortened, so the ends of the stick have larger coordinates. (That's just me thinking out loud...or perhaps thinking "in print".)

Thanks so much for your patience and persistence. These ideas are elusive and this is being fantastically helpful for me. I would like to go one more step to cement my understanding (if you are willing). To that end I have solved your new and improved wording of the problem from both reference frames.

Case 1: From the perspective of an observer in the train, the ends of the train are at xL=0.5 ly and xT=-0.5 ly, and the length of the train is 1 ly. That observer would use the Lorentz coordinate transformations and predict that the observer on the platform (formerly known as the photographer) would calculate based on his observations that xL’= 0.833 ly and xT’=-0.833 ly.

Case 2: From the perspective of the of the observer on the platform, xL’=0.5 ly and xT’=-0.5 ly and (from the Lorentz coordinate transformations) xL=0.3 ly and xT=-0.3 ly.

If those values look right, I'll take a stab at the interpretation each observer has of these calculated positions.
 
  • #17
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Hello and thanks for the very helpful discussion which has clarified the ideas in my mind.

I'm still having difficulty with getting the right answers out though. According to 'French' they are +0.27 and -0.34. The light left them before t=0 and took the time difference to arrive at the camera. We know that the final position is +0.3. So for the for front end, the time the light takes to travel x = the time the stick takes to travel d, where x +d =0.3. Which clearly doesn't result in the right answer.
 
  • #18
Doc Al
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Case 1: From the perspective of an observer in the train, the ends of the train are at xL=0.5 ly and xT=-0.5 ly, and the length of the train is 1 ly. That observer would use the Lorentz coordinate transformations and predict that the observer on the platform (formerly known as the photographer) would calculate based on his observations that xL’= 0.833 ly and xT’=-0.833 ly.
The first sentence is fine, but the second has me confused. The ± 0.833 ly are where on the platform the ends of the train would be at t = 0 according to the observers on the train. (The platform observers disagree, of course, as you state below.)

Case 2: From the perspective of the of the observer on the platform, xL’=0.5 ly and xT’=-0.5 ly and (from the Lorentz coordinate transformations) xL=0.3 ly and xT=-0.3 ly.
Yes, at t = 0 the ends of the train are at positions ± 0.3 ly according to the platform observers. Of course, the train observers disagree.

The trick to understanding how they can disagree on where the ends of the train are at t=0 (when the midpoint of the train passes the origin) is in the relativity of simultaneity. That, and other effects, are built into the Lorentz transformations.
 
  • #19
Doc Al
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Hello and thanks for the very helpful discussion which has clarified the ideas in my mind.

I'm still having difficulty with getting the right answers out though. According to 'French' they are +0.27 and -0.34. The light left them before t=0 and took the time difference to arrive at the camera. We know that the final position is +0.3. So for the for front end, the time the light takes to travel x = the time the stick takes to travel d, where x +d =0.3. Which clearly doesn't result in the right answer.
Those answers do not look right to me. What book is this problem from? (Maybe I have it in my library.) Hopefully I haven't misinterpreted the question.
 
  • #20
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The first sentence is fine, but the second has me confused. The ± 0.833 ly are where on the platform the ends of the train would be at t = 0 according to the observers on the train. (The platform observers disagree, of course, as you state below.)

Here's how I got ± 0.833 ly.

For the observer in the train,
xT= - 0.5 ly
xL= 0.5 ly
v = -0.8c

Using those values to calculate what the observer on the platform measures
xT'= [- 0.5 ly -(-.8c*0)]/0.6 = -0.833 ly
Similarly xL'= 0.833 ly

Could it be that my interpretation of xT' is wrong? Isn't it what an observer on the train calculates the observer on the platforms sees?

Is the problem my confusion over whose coordinate grid is being used?
 
  • #21
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Those answers do not look right to me. What book is this problem from? (Maybe I have it in my library.) Hopefully I haven't misinterpreted the question.

Hi

They are from French, Special Relativity (MIT). Thanks for your help. I've been struggling with this one and just want to make sure that I can use the right technique in my exam next week.
 
  • #22
Doc Al
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Here's how I got ± 0.833 ly.

For the observer in the train,
xT= - 0.5 ly
xL= 0.5 ly
v = -0.8c

Using those values to calculate what the observer on the platform measures
xT'= [- 0.5 ly -(-.8c*0)]/0.6 = -0.833 ly
Similarly xL'= 0.833 ly

Could it be that my interpretation of xT' is wrong? Isn't it what an observer on the train calculates the observer on the platforms sees?

Is the problem my confusion over whose coordinate grid is being used?
I see what you did and that's perfectly OK. (As long as you realize that the platform observers will disagree about the times at which the train ends were at ±0.833 ly.)

Also, to follow standard convention I would have called the platform frame the unprimed frame and the train frame the primed frame. In the unprimed frame, the primed frame moves in the direction of the +x axis at a speed of v.
 
  • #23
Doc Al
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They are from French, Special Relativity (MIT). Thanks for your help. I've been struggling with this one and just want to make sure that I can use the right technique in my exam next week.
I have that book. Are you talking about problem 4-3? Realize that that's not quite the problem you posed here. For one, the observer is not at the origin, but at x = 0 and y = 1 m. Also, it asks for the apparent position of the end points when the observer sees the midpoint pass the origin, not at t=0.
 
  • #24
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Yes, of course. That was very careless of me.

Can you let me know whether my previous analysis was right and what the correct answer would be so that I can try to work towards it?

Thanks again for your help.
 
  • #25
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Hi, sorry for the late bump, I'm also looking at this question (as in the question presented in the original post) and I'm still struggling. I understand that to calculate the length on the photograph, the time it takes for light to reach the camera has to be taken into account but I'm still not sure how this can be done? Any help or hints would be appreciated.
 

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