Does the Speed of Light Remain Constant in All Scenarios?

etcota
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Starting at the beginning and have been having some difficulty explaining the implications of the constancy of the speed of light. Seeking confirmation that I at least have it right. For all scenarios below the observer is Bob standing in the middle of a train car moving at speed v. Bob has a pull cord connected mechanically to each end of the train car such that when he pulls the cord the time to trigger an event at each end of the train car is identical

- Scenario 1) Bob’s cord is attached to guns pointed at his head. He pulls the chords and after identical response times the guns fire . Speed of bullet from the back end of the car is v(b) + v, speed of bullet from the front end of the car is v(b) – v. Bullets hit Bob, who is moving at speed v with the train car, at the same time.
- Scenario 2) Bob’s cord is attached to a mallet that hits a gong when pulled. He pulls the cord and after identical response times the gongs sound. Speed of sound from back of train is v(s) +v, speed of sound from front of train is v(s) – v. Sound reaches Bob, who is moving at speed v with the train, at the same time.
- Scenario 3) Bob’s cord is attached to flashlights pointed at him. He pulls the cord and after identical response times the flashlights turn on. Speed of light from the back of train is c and speed of light from front of train is c. Bob, moving at speed v with the train car sees the light from the front of the car first.

Is this the gist of it?
 
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etcota said:
Starting at the beginning and have been having some difficulty explaining the implications of the constancy of the speed of light. Seeking confirmation that I at least have it right. For all scenarios below the observer is Bob standing in the middle of a train car moving at speed v. Bob has a pull cord connected mechanically to each end of the train car such that when he pulls the cord the time to trigger an event at each end of the train car is identical

- Scenario 1) Bob’s cord is attached to guns pointed at his head. He pulls the chords and after identical response times the guns fire . Speed of bullet from the back end of the car is v(b) + v, speed of bullet from the front end of the car is v(b) – v. Bullets hit Bob, who is moving at speed v with the train car, at the same time.
- Scenario 2) Bob’s cord is attached to a mallet that hits a gong when pulled. He pulls the cord and after identical response times the gongs sound. Speed of sound from back of train is v(s) +v, speed of sound from front of train is v(s) – v. Sound reaches Bob, who is moving at speed v with the train, at the same time.
- Scenario 3) Bob’s cord is attached to flashlights pointed at him. He pulls the cord and after identical response times the flashlights turn on. Speed of light from the back of train is c and speed of light from front of train is c. Bob, moving at speed v with the train car sees the light from the front of the car first.

Is this the gist of it?
No, Bob will see the light from both ends of the train at the same time.

As far as the train and the air in the train and all the items in the train and Bob, traveling at v along the tracks is identical to being stationary on the tracks.

If you set up your scenarios in the Inertial Reference Frame (IRF) of the train and then used the Lorentz Transformation to transform them to the IRF of the tracks, you would find that the Coordinate Times and Distances would all be different, but it wouldn't change what Bob experiences.
 
etcota said:
He pulls the chords and after identical response times the guns fire .

...

He pulls the cord and after identical response times the gongs sound.

...

He pulls the cord and after identical response times the flashlights turn on.

"Identical response times" with respect to which frame? Bob's frame (the train frame) or the stationary frame? This is the key thing that makes the difference.

If, as I suspect, you mean "identical response times with respect to Bob's frame", then, as ghwellsjr pointed out, all three pairs of things arrive at Bob at the same time (where here, "time" is unambiguous because both items in the pair are at the same point in space as well): the pair of bullets, the pair of sounds, and the pair of light rays.

In fact, all of your reasoning involving the speed v of the train is irrelevant in this case, because, as ghwellsjr pointed out, as far as Bob and the bullets and sounds and light rays are concerned, Bob's frame is the "stationary" frame, and the fact that the train happens to be moving relative to the ground doesn't come into it.
 
Thank you both for your responses. Got it now.
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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