Why is this incorrect (special relativity time dilation)

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Homework Statement



Anna is on a railroad flatcar moving at 0.6c relative to Bob. (Their clocks read zero as Anna's center of mass passes Bob's) Anna's arm is outstretched in the direction the flatcar moves, and in her hand a flashbulb. When the flashbulb goes off, the reading on Anna's clock is 100ns. The reading on Bob's differs by 27 ns. How long is Anna's arm (i.e., from her hand to her center of mass)?

Homework Equations



d=vt

The Attempt at a Solution



Okay I know the answer is d=0.6c*100E-9 but I don't understand the reasoning behind that. My train of thought on this problem (...no pun intended) was as follows: Anna sees the light go off at 100ns, which means the light took that long to get to her so that she could see it, and she doesn't see herself as moving so the distance the light traveled to her eyes was d=ct=c*100E-9---> why is this incorrect?
 

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  • #2
Ibix
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Usually, problems of this type assume that observers correct for the lightspeed delay. That means that the flashbulb goes off at 100ns and Anna actually sees the flash a bit later. Anna was kind enough to subtract off the "bit later", so you know when the flash went off, not when Anna saw it.

Your answer is wrong because you are assuming that the flash went off at t=0 and Anna saw it at t=100ns. That's not what the question says.

Bob thinks the flashbulb went off at a different time because he is in a different reference frame. What equations link positions and times in different frames?
 
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Usually, problems of this type assume that observers correct for the lightspeed delay. That means that the flashbulb goes off at 100ns and Anna actually sees the flash a bit later. Anna was kind enough to subtract off the "bit later", so you know when the flash went off, not when Anna saw it.

Your answer is wrong because you are assuming that the flash went off at t=0 and Anna saw it at t=100ns. That's not what the question says.

Bob thinks the flashbulb went off at a different time because he is in a different reference frame. What equations link positions and times in different frames?
So for in the future, how can I tell that they mean an adjusted time as opposed to the time observed? I understand what you're saying but it still messes with my head a bit...I don't see how we're supposed to know to assume that Anna can know the light has gone off without her actually being able to see the light.
 
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Ibix
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It's not so confusing. Think about it like this: you see a lightning flash and hear the thunder five seconds later. When would you say the sound was emitted?

You need to read the question carefully. It does say that the clock reads 100ns when the pulse was emitted. It also doesn't say where Anna's eyes are. Is she the same height as Bob? The question would need to specify all that and more if it were talking about the times Anna and Bob saw the flash.

In other words, this question only talks about the pulse being emitted, not being received.

Does that help?
 
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vela
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Homework Statement



Anna is on a railroad flatcar moving at 0.6c relative to Bob. (Their clocks read zero as Anna's center of mass passes Bob's) Anna's arm is outstretched in the direction the flatcar moves, and in her hand a flashbulb. When the flashbulb goes off, the reading on Anna's clock is 100ns. The reading on Bob's differs by 27 ns. How long is Anna's arm (i.e., from her hand to her center of mass)?

Homework Equations



d=vt

The Attempt at a Solution



Okay I know the answer is d=0.6c*100E-9 but I don't understand the reasoning behind that. My train of thought on this problem (...no pun intended) was as follows: Anna sees the light go off at 100ns, which means the light took that long to get to her so that she could see it, and she doesn't see herself as moving so the distance the light traveled to her eyes was d=ct=c*100E-9---> why is this incorrect?
Are you sure that's the right answer? That would make her arm 18 meters long.
 

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