Light Speed In Forward and Backward Directions

• B
• AlienGrey
In summary, the conversation discusses a thought experiment involving a moving frame and the discrepancy in the speed of light for forward and backward light beams as viewed from a stationary frame. The experiment shows that the time taken for the beams to reach the end points is different in the moving frame compared to the stationary frame. This is due to the relativity of simultaneity and can be explained using the Lorentz transforms. The speed of light is found to be consistent at 300,000 km/s in both frames. This experiment is a basic concept in special relativity and can be easily understood by competent physics students.
AlienGrey
TL;DR Summary
Thought experiment showing discrepancy in the speed of light for forward and backward light beams in a moving frame as viewed from a stationary frame
In the illustration below, a laser beam is emitted, and the time recorded, from the rear of a frame which is 150,000 km long and is moving at velocity 0.5 c to the right. When the beam reaches a half silvered mirror at the front of the frame, the frame being shown as a rectangle and using different colors for three different positions along its path of travel, half of the beam goes through the mirror to a sensor which records its time of arrival and the other half bounces from the mirror to the rear of the frame from whence it was emitted, striking a second sensor which also records the time. The frame was already up to speed when it emitted the beam at position one, shown in blue, so no acceleration effects are involved.

The time taken for the beam to reach the mirror, at position two shown in red, is 1 second in our stationary frame, having traveled 300,000 km, and the time taken for the reflected beam to reach the rear sensor is 0.33 second, the beam having traveled 100,000 km rearward (left) while the frame having moved ahead (right) 50,000 km by the time the beam hit the sensor, totaling the full length of the frame, 150,000 km, the frame's position shown in green. The one-way speed of light would be difficult to gauge in the real world, but this a thought experiment so it's a lot easier to do.

How do observers within the moving frame record the time taken for both the forward and rearward beams to be 0.5 second, as would be the case if the frame had been standing still? Or would they record the same times as we did in the stationary frame? For the time to be 0.5 second in both directions, their time would have to pass at 50% the rate of ours while the beam traveled forward, and it would have to pass at 150% the rate of ours on its rearward trip. How do you reach that outcome using a single time dilation and/or length contraction factor, considering how different they are? If you can explain how light speed can be recorded as 300,000 km/s in both directions in both frames, I think you deserve a Nobel Prize, because I suspect that Einstein wouldn't be able to do it using his equations.

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weirdoguy
AlienGrey said:
TL;DR Summary: Thought experiment showing discrepancy in the speed of light for forward and backward light beams in a moving frame as viewed from a stationary frame

AlienGrey said:
How do observers within the moving frame record the time taken for both the forward and rearward beams to be 0.5 second, as would be the case if the frame had been standing still?
They measure the fixed distance between the end points and the time between the emission and detection events. The result is predicted by SR to be ##c## (as the speed of light is independent of the motion of the source). That experiment, in some shape or form, is an important test of relativity.

topsquark
In your analysis you appear to have neglected the relativity of simultaneity (which is the error in about 99% of cases when people get weird answers from analysing simple experiments).

The general recipe is to write down the ##x,t## coordinates of the various emission and reflection events in one frame, and use the Lorentz transforms to get the ##x',t'## coordinates in the other frame (not just time dilation, which is a special case of them with certain assumptions that aren't applicable here, which is why your attempt goes wrong). You can then compare the transformed distances and times and you will find that the speed of light is ##c## in both frames.

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Dale, Nugatory and topsquark
AlienGrey said:
If you can explain how light speed can be recorded as 300,000 km/s in both directions in both frames, I think you deserve a Nobel Prize, because I suspect that Einstein wouldn't be able to do it using his equations.

Do you realise that this type of scenarios are analyzed even in high schools? It's that basic. I don't know what's the purpose of that comment.

russ_watters, malawi_glenn and PeroK
AlienGrey said:
If you can explain how light speed can be recorded as 300,000 km/s in both directions in both frames, I think you deserve a Nobel Prize.
These days, any competent physics student can learn SR in about 2-3 months. That's unlikely to lead to a Nobel Prize.

russ_watters and malawi_glenn
AlienGrey said:
In the illustration below, a laser beam is emitted, and the time recorded, from the rear of a frame which is 150,000 km long and is moving at velocity 0.5 c to the right.
Your picture depicts a 150000 km object in the 'stationary' frame, which means its proper length would be 173200 km and any clock on the object at the emission point would record about 1.15 seconds for the round trip.
If the proper length of the object is really 150000 km, then you need to redraw the diagram showing its contracted length to be about 1300000 km which affects all the times in your description.

If there is a clock at the right side of this object in sync with the left clock in the object's frame, it will record the reflection time as half the round trip, or 0.575 seconds.

topsquark
How would the length of the object change the the one-way speed of light differently depending on direction, forward or rearward? To correct both of them, the length would need to change so it's contracted when the beam travels in one direction and lengthened when it travels in the other. Since that would be implausible, the discrepancy would still exist, I would just have to do a bunch of extra calculations to change the values to conform to your speculation about objects contracting.

I illustrated the actual length of the object and the distance it traveled. You can use your theory that things contract arbitrarily if you want. You stated that the occupants would record the reflected beam's travel time as being half of the round trip time, but you could obtain the same figure for half of the round trip time using numerous combinations of different one-way trip times, that's the point of me stating that there were two sensors and each recorded the one-way trip time in the two opposite directions. So what would each one-way trip time be, using whatever object length you want to choose arbitrarily?

AlienGrey said:
How would the length of the object change the the one-way speed of light differently depending on direction, forward or rearward?
It wouldn't. It's the relativity of simultaneity that does that, although the length contraction is a necessary part of the calculation too.

Instead of ranting about "how can this beeee!?!?", why not look up the Lorentz transforms (Wikipedia has them correctly) and do the calculation correctly? Nothing more complex than a square root needed.

Dale and berkeman
AlienGrey said:
How would the length of the object change the the one-way speed of light differently depending on direction, forward or rearward?
Length contraction doesn't really address the main problem with your OP. Yes, there is length contraction in the picture, but your assumptions about the relationship between the two frames are more fundamentally flawed.
AlienGrey said:
To correct both of them, the length would need to change so it's contracted when the beam travels in one direction and lengthened when it travels in the other. Since that would be implausible, the discrepancy would still exist, I would just have to do a bunch of extra calculations to change the values to conform to your speculation about objects contracting.
This is not what's being proposed.
AlienGrey said:
I illustrated the actual length of the object and the distance it traveled.
What you showed was that SR is incompatible with Newtonian ideas of space and time. That's true. SR is fundamentally a theory of spacetime. You have to use the SR model of spacetime in order to say anything about it.

Dale
Weirdoguy wrote:

"Do you realise that this type of scenarios are analyzed even in high schools? It's that basic. I don't know what's the purpose of that comment."

I find that unlikely. I think it more likely that they used a setup involving the two-way trip time of a mirror reflected beam and divided it in two. I could get the same two-way trip time of 4 seconds with the combinations of one-way trip times 2 and 2, 1 and 3, and 0 and 4. So does anybody have figures for the trip times in each direction, rather than a round trip time divided by 2? Haven't seen that yet.

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weirdoguy
AlienGrey said:
Haven't seen that yet.
I've told you twice how to calculate it. This is textbook stuff.

Dale
AlienGrey said:
I find that unlikely.

I'm a high school teacher and I do teach this stuff.

Dale, russ_watters and PeroK
weirdoguy said:
I'm a high school teacher and I do teach this stuff.
Okay, what are the times for each of the trips of the beams, forward and rearward? Should be easy for you, right? Go.

weirdoguy
weirdoguy said:
I'm a high school teacher and I do teach this stuff.
Same here.
AlienGrey said:
Okay, what are the times for each of the trips of the beams, forward and rearward? Should be easy for you, right? Go.
Just use the Lorentz transformation, or draw a Minkowski diagram.

russ_watters
Ibix said:
I've told you twice how to calculate it. This is textbook stuff.
Yeah but I want YOU to do it and explain the procedure, that's why I'm here. I already did calculate it, you dispute my answer so the onus is on you to show how I'm wrong.

malawi_glenn said:
Same here.

Just use the Lorentz transformation, or draw a Minkowski diagram.
Please stop telling me to do stuff, it's getting tedious. Now please show me the trip times that YOU get using whatever procedure you want, show your work.

AlienGrey said:
that's why I'm here
It can also be that you have this as a homework assignement and are looking for someone to write a solution.
AlienGrey said:
I already did calculate it, you dispute my answer so the onus s on you to show how I'm wrong.
That is not how it works "prove that it's wrong"... the one who is making the claim that something is correct should be the one who is presenting the proof.
AlienGrey said:
Without calculations, we can not show you where you are wrong. Use the Lorentz transformation.

Dale
AlienGrey said:
Yeah but I want YOU to do it and explain the procedure, that's why I'm here. I already did calculate it, you dispute my answer so the onus is on you to show how I'm wrong.
Not really how it works here - the idea is that you learn how to do it, not that I do something I have known how to do for nearly three decades. I already told you what you did wrong and how to do it right. If Google or the necessary algebra is beyond you then say so and we can help. If you just don't want to solve your own problems then you're probably out of luck here.

Dale, russ_watters and malawi_glenn
Ibix said:
Not really how it works here - the idea is that you learn how to do it, not that I do something I have known how to do for nearly three decades. I already told you what you did wrong and how to do it right. If Google or the necessary algebra is beyond you then say so and we can help. If you just don't want to solve your own problems then you're probably out of luck here.
I could go to Wikipedia for that. The whole point is that I don't believe that SR can actually correct the trip times in two directions in the same moving frame. This is not the homework section, I'm not asking for help, I'm asking if anyone here can show how my calculation is erroneous by using SR equations and getting equal times for both one-way trips of the beam. Should be simple enough, right? Please proceed.

Dale
AlienGrey said:
Yeah but I want YOU to do it and explain the procedure, that's why I'm here. I already did calculate it, you dispute my answer so the onus is on you to show how I'm wrong.
I doubt this is homework.

In the moving frame, the path is ##0.5## light seconds long. The beam takes ##0.5## seconds to reach the far end and ##0.5## seconds to return. That's just the distance divided by the speed of light.

In the ground frame, the path is only ##d = \frac{0.5}{\gamma}## light seconds long. The time to reach the far end is $$t_1 = \frac d {c - 0.5c} = \frac{2d}{c} = \frac {1}{\gamma} = \frac{\sqrt 3}{2} \approx 0.87 \ \text{seconds}$$ (as the far end is moving away from the light beam in this frame).

For the return journey the beam takes $$t_2 = \frac d {c + 0.5c} = \frac{2d}{3c} = \frac {1}{3\gamma} \frac{\sqrt 3}{6} \approx 0.29 \ \text{seconds}$$ (as the near end is moving towards the near end in this case.)

Here the gamma factor is ##\gamma = \frac 1 {\sqrt {1 - v^2/c^2}} = \frac{\sqrt 3}{2}## and gives the length contraction as mentioned above. Note that I assumed that the apparatus was ha;lf a light second long in its rest frame.

Dale
PS the total round trip time in the moving frame is ##1## second. The total time in the ground frame is ##1.15## seconds. This can be seen as an example of the time dilation between frames as
$$t_{ground} = \gamma t_{moving} = \frac{\sqrt 3}{2} \approx 1.14 \ \text{seconds}$$

PeroK said:
I doubt this is homework.

In the moving frame, the path is ##0.5## light seconds long. The beam takes ##0.5## seconds to reach the far end and ##0.5## seconds to return. That's just the distance divided by the speed of light.

In the ground frame, the path is only ##d = \frac{0.5}{\gamma}## light seconds long. The time to reach the far end is $$t_1 = \frac d {c - 0.5c} = \frac{2d}{c} = \frac {1}{\gamma} = \frac{\sqrt 3}{2} \approx 0.87 \ \text{seconds}$$ (as the far end is moving away from the light beam in this frame).

For the return journey the beam takes $$t_2 = \frac d {c + 0.5c} = \frac{2d}{3c} = \frac {1}{3\gamma} \frac{\sqrt 3}{6} \approx 0.29 \ \text{seconds}$$ (as the near end is moving towards the near end in this case.)

Here the gamma factor is ##\gamma = \frac 1 {\sqrt {1 - v^2/c^2}} = \frac{\sqrt 3}{2}## and gives the length contraction as mentioned above. Note that I assumed that the apparatus was ha;lf a light second long in its rest frame.
Is this the homework section? No, so you doubting it being homework seems irrelevant. You didn't explain how the trip times were both 0.5 seconds in the moving frame, you simply stated that they were, based on nothing. You say in our frame it takes 0.87 seconds going to the right and 0.29 seconds going left. How did you get 0.5 seconds both ways in the moving frame starting from that?

AlienGrey said:
Is this the homework section? No, so you doubting it being homework seems irrelevant. You didn't explain how the trip times were both 0.5 seconds in the moving frame, you simply stated that they were, based on nothing.
The length of the apparatus id ##0.5## light seconds, so it takes light ##0.5## seconds to move the length of the apparatus. That's just ##t = \frac d c##.

Dale and russ_watters
PS the speed of light is ##c## (invariant) in all inertial reference frames. That's a postulate of SR.

russ_watters
PeroK said:
The length of the apparatus id ##0.5## light seconds, so it takes light ##0.5## seconds to move the length of the apparatus. That's just ##t = \frac d c##.
How do we transform 0.87 seconds and 0.29 seconds to 0.5 and 0.5? We're not starting from the assumption that the trip takes the same time in both directions in the moving frame. We're starting from the assumption that it doesn't. If you dispute that assumption then you would have to explain why, which you have yet to do.

PeroK said:
That's a postulate of SR.

Backed up by zilions of experiments, in case you want to dismiss it.

russ_watters
weirdoguy said:
Backed up by zilions of experiments, in case you want to dismiss it.
Just focus and try to stay on topic. How do you get 0.5 and 0.5 starting from 0.87 and 0.29?

Dale
AlienGrey said:
Just focus and try to stay on topic. How do you get 0.5 and 0.5 starting from 0.87 and 0.29?
Through the Lorentz transforms that you can't be bothered to look up.

Dale, russ_watters and malawi_glenn
AlienGrey said:
How do we transform 0.87 seconds and 0.29 seconds to 0.5 and 0.5?
You could get that using the Lorentz transformation. Or, the way I did, by simple kinematics in each frame.
AlienGrey said:
We're not starting from the assumption that the trip takes the same time in both directions in the moving frame. We're starting from the assumption that it doesn't. If you dispute that assumption then you would have to explain why, which you have yet to do.
Yes, that assumption is dead wrong.

Dale, russ_watters and malawi_glenn
PeroK said:
You could get that using the Lorentz transformation. Or, the way I did, by simple kinematics in each frame.

Yes, that assumption is dead wrong.
It seems like a very logical assumption to me. You're the one who seems to think it's not, so show how you get 0.5 and 0.5 from 0.87 and 0.29 using whatever transformation you want. It is rather obvious that the same factor can't correct for both, which is why you won't demonstrate it doing that, isn't it?

@AlienGrey For someone who only has posts in this thread, and joining today, you have a "special" attitude
AlienGrey said:
It seems like a very logical assumption to me.
That is why you are wrong, and so many others, when it comes to relativity. An assumption is not logical. An assumption is an assumption and from that you draw conclusions and statements using logic. SR starts with the assumption that the speed of light is same in all inertial frames, from which you predict things which can be observed and measured.
Perhaps you should take some time and actually learn SR and then re-visit this setup of yours?

dextercioby, Dale and russ_watters
malawi_glenn said:
@AlienGrey For someone who only has posts in this thread, and joining today, you have a "special" attitude

That is why you are wrong, and so many others, when it comes to relativity. SR starts with the assumption that the speed of light is same in all inertial frames, from which you predict things which can be observed and measured.
Perhaps you should take some time and actually learn SR and then re-visit this setup of yours?
So you can't show how to get 0.5 and 0.5 from 0.87 and 0.29 using one time dilation/length contraction factor? Can't say I'm surprised, since it's obviously impossible.

weirdoguy
AlienGrey said:
It seems like a very logical assumption to me.
It is a rationale assumption. It's part of Newtonian mechanics. But, it's not a valid assumption in the relativistic universe we have. The assumption ultimately fails experimentally. That's why in high-energy particle physics we use SR and not Newtonian physics.

All you've proved is that SR is not Newtonian.

russ_watters
PeroK said:
It is a rationale assumption. It's part of Newtonian mechanics. But, it's not a valid assumption in the relativistic universe we have. The assumption ultimately fails experimentally. That's why in high-energy particle physics we use SR and not Newtonian physics.

All you've proved is that SR is not Newtonian.
Show how to get 0.5 and 0.5 from 0.87 and 0.29 using the same SR factor. We both know you can't, isn't that right?

weirdoguy
AlienGrey said:
So you can't show how to get 0.5 and 0.5 from 0.87 and 0.29 using one time dilation/length contraction factor? Can't say I'm surprised, since it's obviously impossible.
It is impossible to teach anyone anything they do not want to learn. If you are not interested in learning SR, then there is no point to the discussion.

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