Explain Paradox of Light Moving at c in All Frames

[DaveC426913]

I wonder what distance the light would be in front of Bob after one of his seconds?

The light would be one light-second in front of him. It's traveling at c, right? In one second, light travels on light-second.

Is it about the same distance that light travels in one second in Alice's frame?

Light travels one light-second in one second in Alice's frame too.

What if Bob points a flashlight out the back window of his ship.

One of things you must come to terms with is the fact that Bob is effectively not moving at all. There is no experiment Bob can do (locally) to demonstrate that he is anything but stopped dead in space. Yes, he could look out the window and watch Earth fly past but, other than doing this, he is effectively not moving.

This means that when he shines his beam of light out the front, the back or the side of his spaceship it will behave in every way as if he is not moving at all.

This light would move at c for Alice I presume.

In one of Alice's seconds, she would see the light move one light-second to the "left" while Bob's ship moves .99ls to the "right" for a toal separation of 1.99ls.

What would it move at for Bob? At the end of one second on Bob's ship, the light would be almost 200 of Alice's light-seconds of distance from Bob, wouldn't it.

How does Bob know how many seconds have passed for Alice? This is key.

You cannot assume they both know what each other's clocks do.

If true, this is a much further distance behind Bob than the front light has traveled in front of Bob.

Alice would see this, yes.

Is this asymetric distance really happening?

This is no "really". Both are real. The answer you get depends on who you ask.

And the light speed for Bob would be close to 2c, which is not allowed.

Wrong calc.

It seems to me that the light should move from Bob the same distance in every direction in a given amount of time.

Correct.

Every frame should have the same behavior.

No. The timing of events is relative, depending on who you ask. That is the "relative" in relativity.

 

[Dick]

The strange transformations of space and time were concocted to explain the embarrassing experimentally observed fact that the speed of light is c in any inertial frame. Why don't we observe strange effects as a result of the Earth's motion around the sun? It's also kind of awkward that Maxwell's equations for electromagnetism only assume a simple form in one frame. So why can we use them anyway if we are not in the privileged frame? You aren't convinced, are you? I didn't think so. The Lorentz transformations are an internally consistent view of the world. If you choose not to believe in it in spite of experimental evidence, then that's really your choice, isn't it?

 

[stever]

Alice can "see-know-think" that the front and back lights have attained different distances from Bob. How is it that Bob thinks these lights have attained the same distance from him at each moment?

I admit this is a re-asking of the same old question, but for me it is the main problem I have with understanding the process.

 

[Dick]

Alice can "see-know-think" that the front and back lights have attained different distances from Bob. How is it that Bob thinks these lights have attained the same distance from him at each moment?

I admit this is a re-asking of the same old question, but for me it is the main problem I have with understanding the process.

I think the only way to really do this is to figure out how the Lorentz transformations work. Then get a couple of pieces of graph paper. Assign the x,t coordinates in Alice's frame. Then work out the x',t' coordinates in Bob's frame. You find that the fact they have different notions of when two things are 'simultaneous' explains it.

 

[DaveC426913]

That graph is already done, http://en.wikipedia.org/wiki/Relativity_of_simultaneity" .

The trick is interpreting it and extrapolating it to stever's scenario. It's not easy. I need to mull over it.

 

[vela]

Assume the standard setup: two frames S and S'; origins coincide at t=t'=0; and S at rest and S' moves with speed β in the +x direction. For concreteness, let β=0.6, so γ=5/4.

At time t=0, two light pulses leave the origin. One moves in the +x direction and the other in the opposite direction. At time t=1 s, the forward-moving pulse will be at x=1 ls, and the other one will be at x=-1 ls. This is what Alice observes.

Alice also observes Bob moves with speed β=0.6, so the distance between Bob and the forward-moving pulse at t=1s is 0.4 ls and the distance between Bob and the backward-moving pulse is 1.6 ls.

What does Bob see? The spacetime event at (t,x)=(1 s, 1 ls) corresponds to the point (t',x')=(0.5 s, 0.5 ls) in S' coordinates, and the event at (t,x)=(1 s, -1 ls) to the point (t',x')=(2 s, -2 ls). So Bob also sees that the backward moving pulse is farther away because it has propagated for 2 s whereas the forward-moving pulse only propagated for 0.5 s.

So for those two events, Alice ascribes the asymmetry to Bob's motion. Bob sees the same asymmetry but for him it's because of a difference in propagation times. The two events which were simultaneous for Alice aren't simultaneous for Bob.

 

[DaveC426913]

What does Bob see? The spacetime event at (t,x)=(1 s, 1 ls) corresponds to the point (t',x')=(0.5 s, 0.5 ls) in S' coordinates, and the event at (t,x)=(1 s, -1 ls) to the point (t',x')=(2 s, -2 ls). So Bob also sees that the backward moving pulse is farther away because it has propagated for 2 s whereas the forward-moving pulse only propagated for 0.5 s.

You lost me. Bob is shining a flashlight out his front and rear windows simultaneously (in his frame). Why would propagation time be different for each?

 

[vela]

You lost me. Bob is shining a flashlight out his front and rear windows simultaneously (in his frame). Why would propagation time be different for each?

In Alice's frame, the two events are at the same spatial distance from the origin, so the light pulses reach those points at the same time. The same two events in Bob's frame aren't at the same spatial distance from his origin. Using the Lorentz transformations, you find one is 0.5 ls away and the other is 2 ls away. So the light leaves Bob's ship simultaneously, but since the distances are different, the propagation times will be different.

You could look at where the light pulses are at, say, t'=1 s as seen by Bob. He'd say one is at x'=1 ls and the other at x'=-1 ls. From Alice's point of view, however, those two events occur at x=2 ls and x=-0.5 ls, so in this case, she'd see different propagation times.

 

[DaveC426913]

In Alice's frame, the two events are at the same spatial distance from the origin, so the light pulses reach those points at the same time. The same two events in Bob's frame aren't at the same spatial distance from his origin. Using the Lorentz transformations, you find one is 0.5 ls away and the other is 2 ls away. So the light leaves Bob's ship simultaneously, but since the distances are different, the propagation times will be different.

You could look at where the light pulses are at, say, t'=1 s as seen by Bob. He'd say one is at x'=1 ls and the other at x'=-1 ls. From Alice's point of view, however, those two events occur at x=2 ls and x=-0.5 ls, so in this case, she'd see different propagation times.

No, I've realized where the problem is. Bob cannot simply "know" how fast the light rays are traveling away from him. There must be some method for him to do so, such as the light bounces off an asteroid in front of him and an asteroid behind him. Bob must know how far away the two asteroids are in order to measure the speed of the beams of light.

And measruing how far way these two asteroids are from Bob is non-trivial. Spatial distances are distorted along Bob's line of travel. Are the two asteroids equidistant in Bob's frame? Or in Alice's frame?

 

[Gear.0]

If I may attempt to address the original question, I think I can highlight the problem assumptions the OP is making.

You say that in your (rest) frame of reference (relative to the light sources) that the two photons will hit your friend at the same time.
But what does that mean if your friend is moving? It means that the only way this is possible is if the emitter ahead of her be positioned farther from her and the emitter behind her must be positioned closer to her. If this were not so then it would not be possible for the photons to hit her at the same time, so it must be so.
Then let's say that the emitters fire at the same time and at the instant that she is passing right in front of you (so you are also standing offset from the center of the emitters).

Which should all make sense so far as it requires no relativity yet.

Ok, so now we jump into your friends frame of reference, from her frame we know that at the instant she is passing by you she must agree that she is not directly in the middle (she doesn't agree on the exact distances, but she does agree that she is closer to one detector than the other).
These are fairly obvious facts I don't think you should have trouble with...

Now the problem is that you are saying:
when she passes by you both emitters fire right? So then if she is not in the middle of the emitters there should be a contradiction because the light will reach her at different times...

However, the thing that reconciles this (and if you think about it, it's the only possible way it can be reconciled) is that in her reference frame the emitters do not go off at the same time. She would claim that although she was not in the middle, the detector that was farther away shot first, then (she sees the detectors moving) at a later time the detector that was closer fires. When this happens she does in fact agree that the two photons hit her at the same time.

 

[stever]

Gear.o, please refer to my original post, the first in this thread, to see the exact conditions of the puzzle, which are different from what you are using. I'm not going to try to deal with the changed conditions you made, although it might be interesting. There is also the Bob and Alice thought experiment under discussion. Two of these thinking exercises are enough to keep me overworked.

 

[Gear.0]

I strongly disagree stever. The conditions are exactly the same.

The only thing I can think of that you would say I "changed" is the fact that I placed your friend not in the center of the detectors... But, if you think that I changed a condition by doing that then you are wrong.. Because if she were in the center of the emitters when they fired, then it is not physically possible for YOU to observe the photons hitting her at the same time... nevermind what your friend sees.
So because you said that you observe the photons hitting her at the same time, then she MUST be closer to the detector that is behind her. This is your condition NOT mine, I didn't change it, I simply brought that fact to light and I think your overlooking of that hidden condition is causing your confusion.
I explained this in the post above, please read it more carefully.

Were there any other "changed conditions" that I didn't catch? Please let me know.

 

[weaselman]

You have to specify whose reference frame you are talking about when you define the moment at which the emitters fire before figuring out where she was at that moment. Because in different reference frame it would be a different moment(s), and therefore, since she is moving, she will be in different places.

For example, if the emitters fire simultaneously in your reference frame just when your friend passes you, standing in the middle between two emitters, then your friend will not only not be in the middle when they fire, but (more importantly), the emitter behind her will fire earlier, than the one in front in her reference frame (so, obviously, because she is moving, she won't be in the middle when at least one of them fires). However, in this case the signals from emitters will not reach your friend at the same time. They will reach you at the same time, but your friend will not be there anymore.

If however, the light emitters are moving with your friend, and she is sitting in the middle between them, then stever is right - she is in the middle when they fire, by definition. (They don't really have to be moving, what's important is that in here reference frame they fire simultaneously, and she is in the middle between them at that moment).
If emitters fire simultaneously in her reference frame, she will receive their signals at the same time. But you will have to conclude that in your reference frame the emitter behind her fired earlier, than the one in front, because the light had to cover greater distance to catch up with her.

I think, the question of whether or not she is in the middle when the emitters fire, is really misleading. First, because, as I said above, "the time when emitters fire" is actually not one moment of time, in all reference frames, except at most one. But also, because the answer to this question depends on the unspecified condition of whether or not the emitters are moving.
For example, if the emitters are attached to the front and the back of the car your friend is traveling in, and she is sitting in the middle between them, then, of course, she will be in the middle between them when they fire, as well as at any other moment.
If the emitters are stationary, but fire simultaneously in her reference frame, and she receives the signals at the same time, then she is in the middle the moment they fire too (but not at any other point in time).

Basically, since it's given that she receives the signals at the same time, we can say, that if emitters fire simultaneously in her reference frame, then she must be in the middle at that moment. If they don't fire simultaneously, then there is more than one moment to consider, but, if she receives the signals at the same time, she must be either moving with respect to the emitters, or at different distance to them, or both (if she is moving, the distance would have to be different too, because they fire at different time, and her position changes with time).

So, I recommend, that instead of asking where she was when the light was emitted, focus on the question of simultaneity - if the signals are emitted simultaneously in her frame, and she receives them at the same time (it automatically means, that she was at the middle when they were emitted, but it does not matter), then in your reference frame, the emitter at the back fired earlier, than the one in front by exactly the right amount of time to allow the signals going with constant speed "c" reach the moving detector at the same time.
This is the answer to your original question, stever. This is how you reconcile the discrepancy you pointed out.

 

[weaselman]

Also, in regards to "Alice and Bob" problem. Alice sees the light travel at "c" with respect to her, not to Bob. In one second (by her clock) the light will be one light-second away from the point where Bob was when the light was emitted, not from where Bob is now. That distance will be respectively larger/smaller than the current distance of Bob's ship depending on whether he flashes his flashlight forward or backwards.

 

[stever]

"So the light leaves Bob's ship simultaneously, but since the distances are different, the propagation times will be different."

Vela, it is very interesting to conceive that Bob sees the distance difference between the two opposite-going pulses as a time difference. By definition perhaps, the problem of speed differences suddenly seems to disappear. And you indicate it works out mathematically. What seems strange is that Bob would not notice what Alice notices: that light has two speeds, in relation to Bob, from Alice's perspective. Alice might see, if she could, a photon go very quickly from the front to the back of Bob's ship while a photon going from the back to the front would take much longer. Bob just can't experience that, or can he? Is there no evidence of Alice's way of seeing it?

What if Bob went to the back wall of his ship and observed the front wall of his ship, and then compared this to standing at the front wall and observing the back wall? Would he not see a difference in the length of his ship from the two viewpoints? Of course he doesn't, does he?

 

[DaveC426913]

What seems strange is that Bob would not notice what Alice notices: that light has two speeds, in relation to Bob, from Alice's perspective. Alice might see, if she could, a photon go very quickly from the front to the back of Bob's ship while a photon going from the back to the front would take much longer.

This is not "light having two speeds".

Alice observes a light beam head West at c and a light beam East at c. She observes Bob's ship heading East at .99c, so it is closely following the Eastward beam of light.

i.e.:
Alice sees light moving at c. Full stop.
Alice sees a spaceship moving at .99c. Full stop.
Nothing out of the ordinary here.

Bob just can't experience that, or can he? Is there no evidence of Alice's way of seeing it?

Certainly he could. If he watched Alice shine a beam of light East and West from Earth, he'd see a beam of light moving East at c and a beam of light moving West at c. And he'd see the Earth flash by Westward at .99c. So Earth is very closely following the beam of light.

What if Bob went to the back wall of his ship and observed the front wall of his ship, and then compared this to standing at the front wall and observing the back wall? Would he not see a difference in the length of his ship from the two viewpoints? Of course he doesn't, does he?

Bob always observes light moving at c in his own frame of reference (which, as far as he's concerned, is at rest).

 

[stever]

"What seems strange is that Bob would not notice what Alice notices: that light has two speeds, in relation to Bob, from Alice's perspective. Alice might see, if she could, a photon go very quickly from the front to the back of Bob's ship while a photon going from the back to the front would take much longer."

Alice measures light at c in her frame, but when she looks at light going by a moving Bob, she could imagine that Bob would experience various speeds of light depending on his movement in relation to her. If she knows SR she might not bother to imagine such things, as Bob will measure light at c only, but if she doesn't know SR she might easily conclude that the speeds of light and material objects are additive/ subtractive. We don't really know if Alice knows SR.

 

[DaveC426913]

"What seems strange is that Bob would not notice what Alice notices: that light has two speeds, in relation to Bob, from Alice's perspective. Alice might see, if she could, a photon go very quickly from the front to the back of Bob's ship while a photon going from the back to the front would take much longer."

Please, use the quote feature.

Alice measures light at c in her frame,

Alice can only measure light moving in her frame.

but when she looks at light going by a moving Bob, she could imagine that Bob would experience various speeds of light depending on his movement in relation to her. If she knows SR she might not bother to imagine such things, as Bob will measure light at c only, but if she doesn't know SR she might easily conclude that the speeds of light and material objects are additive/ subtractive. We don't really know if Alice knows SR.

What difference does it make what she possibly-mistakenly thinks is going on with Bob? She can surmise all she wants about what Bob might or might not be seeing. She could be right, she could be wrong. Is that what this question is about?