Speed of light constant

  • Thread starter Adrian07
  • Start date

ghwellsjr

Science Advisor
Gold Member
5,120
146
Basically what I'm getting from the first 2 sections of the paper is that In order to synchronize 2 clocks the time it takes light to travel from A to B must equal the time it takes for light to be reflected from B and arrive back at A. Also, that the distance from A to B and B to A (round trip distance) divided by the time it takes light to originate from A, travel to B, reflect off of B, and travel back to A equals the speed of light. This makes sense if and only if the the distance between points A and B constantly remains the same.
That is correct.
My question is that if these are true then, wouldnt that technically mean that no 2 clocks in motion relative to each other could ever be synchronized?
That is correct.
And if the speed of light is constant, since v = d/t wouldnt that inherently mean that time and distance would end up being dependent on each other instead of independent variables? when referring to points A and B?
Yes, that's the importance of Special Relativity. Time is relative, space is relative. We now talk about spacetime where space and time are not independent of each other.
 
12
0
That is correct.

That is correct.

Yes, that's the importance of Special Relativity. Time is relative, space is relative. We now talk about spacetime where space and time are not independent of each other.
So as you go faster lengths get smaller? Or does time flow faster? or both?
 

ghwellsjr

Science Advisor
Gold Member
5,120
146
So as you go faster lengths get smaller? Or does time flow faster? or both?
Once you use Einstein's definition of a Reference Frame, any rigid object that is traveling is length contracted along its direction of motion and it's clocks take longer to tick which means they run slower. But if you transform everything to a different Reference Frame moving at some speed with respect to the first one, the speeds of those objects can be different, meaning their length contractions and time dilations can be different and you can always transform to a frame in which any given object is at rest in which case there will be no length contraction or time dilation.

Also, be aware that any object or ruler that is traveling with an object will have the same length contraction and time dilation so any measurement that is performed on a "moving" object with identically moving rulers and clocks will come out the same no matter which Reference Frame is used.
 
12
0
Once you use Einstein's definition of a Reference Frame, any rigid object that is traveling is length contracted along its direction of motion and it's clocks take longer to tick which means they run slower. But if you transform everything to a different Reference Frame moving at some speed with respect to the first one, the speeds of those objects can be different, meaning their length contractions and time dilations can be different and you can always transform to a frame in which any given object is at rest in which case there will be no length contraction or time dilation.

Also, be aware that any object or ruler that is traveling with an object will have the same length contraction and time dilation so any measurement that is performed on a "moving" object with identically moving rulers and clocks will come out the same no matter which Reference Frame is used.
ok, so if i'm travelling at let's say minutely just under the speed of light. To me, from my perspective, nothing changes. Time and space go on as normal? Or is it actually possible to turn on a flashlight and watch the beam of light slowly extend outward in front of me? Wouldn't time technically stop if i were to reach the speed of light? From my perspective in a fast spaceship that reached the speed of light, would i just instantaneously skip over the speed of light "gap" or interval of time until my speed was reduced to that less than the speed of light?

man this is some freaky stuff
 

ghwellsjr

Science Advisor
Gold Member
5,120
146
ok, so if i'm travelling at let's say minutely just under the speed of light. To me, from my perspective, nothing changes. Time and space go on as normal?
What you're saying is that in some particular Reference Frame, you're traveling at just under the speed of light and yes, everything is normal for you. But, of course, whatever is at rest or traveling at slow speeds in that frame will be traveling at just under the speed of light relative to you so it's not like you can't tell that you are traveling at a high speed.
Or is it actually possible to turn on a flashlight and watch the beam of light slowly extend outward in front of me?
How do you watch a beam of light? It's not like watching some kind of projectile traveling away from you for which you shine light on it and the reflected light off the surface of the projectile is what you are actually seeing, correct? Since projectiles travel at a very small fraction of the speed of light, you simply ignore the additional time it takes for the reflected light to get back to you and you approximately "see" the projectile moving away from you.

But you can't do that with light. What you have to do instead is have some portion of the light beam itself reflect off of other things placed at increasing distances away from you but now you can't ignore the additional time it takes for the reflected light to get back to you. And what will it look like? Since the light is making a round trip, when you turn on the flashlight, it will look like the beam is traveling at one-half the speed of light, do you understand that?

Now because you are traveling in the same direction that the beam is traveling, and light is defined to travel at c in the selected Reference Frame, the beam will be traveling very slowly away from you and then after it hits one of the objects out in front of you, the reflected light will travel back to you almost instantly. Remember how you can't tell if the light takes the same amount of time to go away as it does to come back? This is an example of how you can't tell. As far as you are concerned, it won't look any different than when you are at rest in the Reference Frame and the light takes the same amount of time to go away as it does to come back (by definition, not by observation).
Wouldn't time technically stop if i were to reach the speed of light?
Didn't you start off your post by saying "To me, from my perspective, nothing changes. Time and space go on as normal?" To which I agreed. So even if you had accelerated from being at rest in the Reference Frame and you spent an enormous amount of energy getting to your high rate of speed, it would be just like you were at rest and you would have to start all over again to get back to your high rate of speed. You could repeat this as often as you wish and you'd be no closer to the speed of light than before you started.

But technically, you can't reach the speed of light, so there is no meaning to your question of what would happen if you were to reach it.
From my perspective in a fast spaceship that reached the speed of light, would i just instantaneously skip over the speed of light "gap" or interval of time until my speed was reduced to that less than the speed of light?
Another meaningless question for the same reason.
man this is some freaky stuff
Not to me and hopefully not to you some day.
 
12
0
What you're saying is that in some particular Reference Frame, you're traveling at just under the speed of light and yes, everything is normal for you. But, of course, whatever is at rest or traveling at slow speeds in that frame will be traveling at just under the speed of light relative to you so it's not like you can't tell that you are traveling at a high speed.

How do you watch a beam of light? It's not like watching some kind of projectile traveling away from you for which you shine light on it and the reflected light off the surface of the projectile is what you are actually seeing, correct? Since projectiles travel at a very small fraction of the speed of light, you simply ignore the additional time it takes for the reflected light to get back to you and you approximately "see" the projectile moving away from you.

But you can't do that with light. What you have to do instead is have some portion of the light beam itself reflect off of other things placed at increasing distances away from you but now you can't ignore the additional time it takes for the reflected light to get back to you. And what will it look like? Since the light is making a round trip, when you turn on the flashlight, it will look like the beam is traveling at one-half the speed of light, do you understand that?

Now because you are traveling in the same direction that the beam is traveling, and light is defined to travel at c in the selected Reference Frame, the beam will be traveling very slowly away from you and then after it hits one of the objects out in front of you, the reflected light will travel back to you almost instantly. Remember how you can't tell if the light takes the same amount of time to go away as it does to come back? This is an example of how you can't tell. As far as you are concerned, it won't look any different than when you are at rest in the Reference Frame and the light takes the same amount of time to go away as it does to come back (by definition, not by observation).

Didn't you start off your post by saying "To me, from my perspective, nothing changes. Time and space go on as normal?" To which I agreed. So even if you had accelerated from being at rest in the Reference Frame and you spent an enormous amount of energy getting to your high rate of speed, it would be just like you were at rest and you would have to start all over again to get back to your high rate of speed. You could repeat this as often as you wish and you'd be no closer to the speed of light than before you started.

But technically, you can't reach the speed of light, so there is no meaning to your question of what would happen if you were to reach it.

Another meaningless question for the same reason.

Not to me and hopefully not to you some day.
ok got it. So hypothetically, suppose we get the technology needed to travel to Proxima Centauri. It's about 4 light years away. We go there and travel from point A (being earth) to point B (being Proxima Centauri) at half the speed of light. Going off what Einstein said, me in the spaceship would perceive time to be travelling like normal, but it wouldnt take me 8 years to get there from MY time. Back on earth though it would be an 8 year wait (technically 12 years since the first transmission sent out once the ship reaches Proxima Centauri would take 4 years to get back to earth). This sorta doesnt make sense. The distance between here and Proxima Centauri (for the purposes of the example) doesnt change. The velocity also does not change (suppose we got a running head start and only wanted to do a flyby of point B). Yet on the ship, both the distance and the time have changed? Is this correct? Could we actually calculate how long the trip would "feel" like on the ship?
 

ghwellsjr

Science Advisor
Gold Member
5,120
146
ok got it. So hypothetically, suppose we get the technology needed to travel to Proxima Centauri. It's about 4 light years away. We go there and travel from point A (being earth) to point B (being Proxima Centauri) at half the speed of light. Going off what Einstein said, me in the spaceship would perceive time to be travelling like normal, but it wouldnt take me 8 years to get there from MY time.
That is correct. It would take you 6.9282 years to get there according to your clock.
Back on earth though it would be an 8 year wait (technically 12 years since the first transmission sent out once the ship reaches Proxima Centauri would take 4 years to get back to earth).
That is correct. In the common earth/Centauri rest frame, it takes 8 years to get there but it takes 12 years for the earthlings to see you get there.
This sorta doesnt make sense. The distance between here and Proxima Centauri (for the purposes of the example) doesnt change.
In your rest frame, the distance that Proxima Centauri is away from you is not 4 light years but rather 3.4641 light years.
The velocity also does not change (suppose we got a running head start and only wanted to do a flyby of point B). Yet on the ship, both the distance and the time have changed? Is this correct? Could we actually calculate how long the trip would "feel" like on the ship?
Yes, that is correct. Your trip will take 6.9282 years and it will feel like 6.9282 years.
 
441
8
harrylin #3
Now, if you do what you seem to suggest - take your physical system out of your lab and in your car, and measure the one-way speed of entering light rays while you are driving - then you may not find the same value. However, after you re-synchronize your on-board clocks at that velocity, then you'll have again a standard reference system if you keep approximately an inertial course. Subsequently you'll find again the standard value.
The 1-way, 2-way,...n-way speed of light is c.
The relative light speed v/c is the variable, but that's what can't be measured. The observer can't receive the same unidirectional signal he sends (unless he can move faster then light). If he uses a 2nd observer, a 2nd clock is needed and the synchronization problem appears. Despite being unable to measure his absolute speed (he has one, otherwise v/c would be variable and unreliable), he can still achieve a relative synchronization. This is the familiar parallelogram with the skewed axis, seen in Hermanns (aka Minkowski spacetime diagrams,...who needs verboseness). The synchronization does not alter the value of c, it makes the outbound and inbound paths equal, per Einsteins 'stipulation', i.e, it's not a rule of physics, as he clearly states, but a definition (physics by decree). Since the observer can only be coincident with the emission and detection of the reflected signal, only the round trip time is measurable. The time and location of the reflection event is speculation, thus the skewed spatial axis is bogus.
Do you think it's possible to alter distance by setting a clock?
 
3,871
88
harrylin #3
[..] The observer can't receive the same unidirectional signal he sends (unless he can move faster then light). If he uses a 2nd observer, a 2nd clock is needed and the synchronization problem appears. [..] Do you think it's possible to alter distance by setting a clock?
Sorry but I can't follow you. A person who has set up a physical reference system uses as many clocks and detectors as he wants (the system that I referred to uses two clocks, typically as detailed in post #7). That has nothing to do with "altering distances".
 
84
1
Have been looking at the special relativity link in post 12 particlarly the diagram at the end.
Quote The observer on the spaceship will measure the blip of light to be traveling at c relative to the spaceship, the observer on the ground will measure the same blip to be traveling at c relative to the ground. That is the unavoidable consequence of the Theory of Relativity.
While the speed of light remains the same the measured speed must be different as it would take longer to travel between the sensors.
If we add to the diagram the spaceship moving at c and a blip of light fired from the back of the ship towards the front as it passes the stationary light source, from outside both blips move at the same speed, inside the blip would seem as not moving, if it was seen as moving inside the ship it would be seen as moving at a different speed to the outside one from outside the ship, so how do we keep the measured speed of light as constant?
Please keep replies in laymans terms as simple as possible please.
 

Erland

Science Advisor
735
135
If we add to the diagram the spaceship moving at c.
A ship cannot move at c. It cannot, even in theory, be accelerated up to c. That is a consequence of SR. Therefore, it is not meaningful to speculate of how things will be perceived inside such a ship.
 

ghwellsjr

Science Advisor
Gold Member
5,120
146
Have been looking at the special relativity link in post 12 particlarly the diagram at the end.
Quote The observer on the spaceship will measure the blip of light to be traveling at c relative to the spaceship, the observer on the ground will measure the same blip to be traveling at c relative to the ground. That is the unavoidable consequence of the Theory of Relativity.
While the speed of light remains the same the measured speed must be different as it would take longer to travel between the sensors.
If we add to the diagram the spaceship moving at c and a blip of light fired from the back of the ship towards the front as it passes the stationary light source, from outside both blips move at the same speed, inside the blip would seem as not moving, if it was seen as moving inside the ship it would be seen as moving at a different speed to the outside one from outside the ship, so how do we keep the measured speed of light as constant?
Please keep replies in laymans terms as simple as possible please.
As Erland pointed out, the spaceship cannot travel at c so your following description is not meaningful but aside from that, you can still make your point, and it's a good one, with the spaceship traveling at one half c.

The point is that the diagram is illustrating that when you want to measure the speed of light, it is always a round-trip measurement, although the author seems blissfully unaware of that because he links it to Einstein's second postulate when it is actually a consequence of the first postulate.

Here's the author's explanation of how the measurements take place:
The setup is as follows: on a flat piece of ground, we have a flashlight which emits a blip of light, like a strobe. We have two photocells, devices which click and send a message down a wire when light falls on them. The photocells are placed 10 meters apart in the path of the blip of light, they are somehow wired into a clock so that the time taken by the blip of light to travel from the first photocell to the second, in other words, the time between clicks, can be measured. From this time and the known distance between them, we can easily find the speed of the blip of light.
He doesn't say where the clock is or how the wires go but let's assume that the clock is located next to the first photocell. We will assume that the messages from the photocells are sent down the wires to the clock at the speed of light.

As soon as the blip of light hits the first photocell, it instantly registers the time on the clock since the wire is very short. Now we wait for the blip to reach the second photocell and for the message to travel down the wire back to the clock. Both observers are making the same measurement separately with their own clock, photocells and wires, although they are measuring the same blip of light. Let's assume that light actually travels at c with respect to the light source and the ground observer.

Then for the ground observer, the time it takes the light to travel between the two photocells will be equal to the time it takes for the signal to travel from the second photocell back to the clock (colocated with the first photocell). So if the ground observer is blissfully unaware (like the author) of what's going on, he will say that his measurement of the speed of light is actually one half of c. If he realizes that the message takes additional time to travel down the wire, he might be inclined to double his calculation of the measured speed of light and then he'd get the right answer.

But would the same thing happen for the spaceship observer? Well it better or Einstein's first postulate would be violated. So how can they both get the same measurement when obviously it will take a lot longer for the blip to get from the first photocell to the second one which I think is your point? Well, what if the two photocells were closer together than the spaceship observer measured them to be? And what if his clock is running slower than he thinks it is? Then his measurement can come out the same as for the ground observer. We don't have to go into the details because you're a layman but I think you can see how this would allow the spaceship observer to get the same answer as the ground observer.

Unfortunately, the diagram doesn't show this and they have a note saying they ignored this salient fact but that is the explanation.

So in conclusion, if this measurement really was of the one-way speed of light like the author claims, then both observers would get a different answer because the amount of length contraction between the photocells and the amount of time dilation of the clock that explains the same answer for the round-trip measurement would not also work for the one-way measurement.

I hope I have hit the nail on the head for you. Did I? If not, please point out where I missed your concern.
 
84
1
George many thanks. I always asumed that the speed of light was measured in one direction, if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant.This I think is not much different to measuring something as going there and back.
Do the effects of length and time dilation mean that two people could measure the same thing and get different answers or two people measure different things and get the same answer.
How can just movinf faster affect your view of the universe?
 
3,871
88
[..] if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant. [...]
Regretfully that is wrong. In SR, light propagation is modeled like waves and not like bullets. Please make sure to understand that first; correct understanding comes step by step.

For example, you could first read up on MMX which the foregoing lecture actually discusses (and I agree with that sequence of explanation):
http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html
(warning: the section with "The only possible conclusion" is plain wrong, as mentioned in post #17 with a link to a discussion in which this is elaborated; but you can simply skip that section).

You can also find a different discussion here:
http://en.wikipedia.org/wiki/Michelson–Morley_experiment

Try if you can derive those equations, based on Newton's mechanics and Maxwell's hypothesis of "a constant speed of light": It is assumed that the speed of light in vacuum is everywhere c in all directions, independent of the speed of the light emitter (and as you see, although the required math is very basic, even Michelson made a blooper when he first did so! :tongue2:).
 
Last edited:

ghwellsjr

Science Advisor
Gold Member
5,120
146
George many thanks. I always asumed that the speed of light was measured in one direction, if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant.This I think is not much different to measuring something as going there and back.
As harrylin pointed out, the ballistic model of light is incorrect but beyond that, I'm not sure what you are describing with respect to a gun firing bullets. Wouldn't you need multiple guns because if we think of the light source as being a gun, in effect, we really need a second and third gun at each of the final detectors to simulate sending the message back to the clocks, correct?
Do the effects of length and time dilation mean that two people could measure the same thing and get different answers or two people measure different things and get the same answer.
I'm not sure what you are asking here. If we go back to the linked diagram, there are two people involved, one on the ground and one on the spaceship. Do you consider them to be measuring the same thing or different things?
How can just movinf faster affect your view of the universe?
The obvious answer is that the faster you move, the faster your view of the universe appears to be moving past you. But I don't think that is what you are looking for. If not, could you please rephrase your question?
 
84
1
It just seems to get more complicated.
Going back to the spaceship and forgeting about actually measuring the speed of light. We will keep the stationary light emitter and the one at the back of the ship and the ship moving at 0.5c. We will however put the observers on treadmills synchronized at rest at say 5 km/hr. As the emitter on the ship passes the one on the ground each emits a blip of light. What will I see with regards to the 2 blips of light and the different observers. I am on the stationary treadmill.
Will the blips of light remain moving together at a constant speed?, the stationary treadmill will still be going at 5 km/hr, what will I see happen to the treadmill on the ship.
Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c.
There is also the question now about the difference between waves and photons.
Obviously waves are waves, although I dont see how they can propagate through nothing, are photons then like bullets? if so how do waves and photons with the same energy relate to each other.
Finally the MM experiment seems to assume a moving aether, would it make any difference if it was a stationary field just producing resistance so as to cause a wave to form, as waves seem to form due to the resistance of the medium or am I way out here.
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
20,620
4,352
It just seems to get more complicated.
Going back to the spaceship and forgeting about actually measuring the speed of light. We will keep the stationary light emitter and the one at the back of the ship and the ship moving at 0.5c. We will however put the observers on treadmills synchronized at rest at say 5 km/hr. As the emitter on the ship passes the one on the ground each emits a blip of light. What will I see with regards to the 2 blips of light and the different observers. I am on the stationary treadmill.
What do you mean by a "blip" of light? A single pulse or flash? A single photon? A constantly updating blip like on a radar?

Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c.
Are you saying you are trying to figure out an actual way of measuring this, or that you are attempting to understand how such a thing is possible?
There is also the question now about the difference between waves and photons.
Obviously waves are waves, although I dont see how they can propagate through nothing, are photons then like bullets? if so how do waves and photons with the same energy relate to each other.
Light is an EM wave. The field itself carries the energy and the vector of the electric and magnetic fields of the wave are what are oscillating. The wave interacts in quanta called photons. They are NOT little bullets.
 
Re: Speed of light

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave
 
3,871
88
It just seems to get more complicated.
That is normal for physics; and it's hopeless without doing the necessary exercises.
[..]Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c. [..] Finally the MM experiment seems to assume a moving aether, would it make any difference if it was a stationary field just producing resistance so as to cause a wave to form, as waves seem to form due to the resistance of the medium or am I way out here.
Instead, MMX assumes a moving interferometer in a light medium. So, if you understand waves, please present your MMX calculation, and we can take it from there.
It is like becoming familiar with multiplication in order to next understand exponents. But if you don't understand waves, then it's necessary to go back even one more step...
 
3,871
88
Re: Speed of light

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave
You only "got" 1/3 of the reasons:
- clock synchronization
- time dilation
- length contraction

See: https://www.physicsforums.com/showthread.php?t=620279
All effects together are presented in the equation of post #3 (with "relative speed", he just means "speed").

Your question is the topic of that thread, and you can still discuss more there.
 
Last edited:
You only "got" 1/3 of the reasons:
- clock synchronization
- time dilation
- length contraction

See: https://www.physicsforums.com/showthread.php?t=620279
All effects together are presented in the equation of post #3 (with "relative speed", he just means "speed").

Your question is the topic of that thread, and you can still discuss more there.
Thanks. I just posted over there.
 
84
1
The spaceship example post 42 for blip read pulse if you like.

Light has particle, wave duality what is a photon then and how does it relate to the wave/particle nature. Waves and particles have energy so why do we also need photons.

All I was trying to work out at the outset was how can the speed of light be constant and independant of the motion of the emitter.
I am standing still, an emitter travelling at 0.5c emits a pulse of light as it passes me, the light moves away from me in all directions at c, relative to the emitter though the forward direction of the pulse would be 0.5c, and backwards 1.5c and sideways c. So if the emitter could measure the speed of light moving away from it it should measure the forward speed as 0.5c moving away and backwards 1.5c moving away, the speed of light remains constant but the measured speed is dependant on the motion of whoever is doing the measuring.
What I got from the answer in post 2 was that both I and the emitter would measure the speed in all directions as c.
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
20,620
4,352
The spaceship example post 42 for blip read pulse if you like.
Alright. Also, your example there is...hard to understand. You have an emitter sending out pulses of light. You have a spaceship traveling at 0.5c with an emitter at the back of it. Then you have 2 treadmills that are synchronized, but one is stationary with respect to the ship? That doesn't make any sense to me.

Light has particle, wave duality what is a photon then and how does it relate to the wave/particle nature. Waves and particles have energy so why do we also need photons.
An EM wave of a specific frequency will only interact in little chunks or packets we call photons. Each photon carries with it part of the energy of the wave, where the amount of energy is E=hf, with h being Planks constant, and f being the frequency of the EM wave. The wave can only interact in this manner.

All I was trying to work out at the outset was how can the speed of light be constant and independant of the motion of the emitter.
I am standing still, an emitter travelling at 0.5c emits a pulse of light as it passes me, the light moves away from me in all directions at c, relative to the emitter though the forward direction of the pulse would be 0.5c, and backwards 1.5c and sideways c. So if the emitter could measure the speed of light moving away from it it should measure the forward speed as 0.5c moving away and backwards 1.5c moving away, the speed of light remains constant but the measured speed is dependant on the motion of whoever is doing the measuring.
What I got from the answer in post 2 was that both I and the emitter would measure the speed in all directions as c.
Yes, both you and the emitter would measure the light to move at c. From the emitters point of view the light in all directions moves away at c. To reconcile this, you have to account for things like doppler shift, time dilation, etc. If the emitter is moving away from you then while the light still travels at c from it to you, it loses part of its energy from your frame of reference, while it retains it from the emitters frame.

Honestly your best bet to understand all this is probably to buy a book on the basics of relativity. Without getting into at least a little bit of the basic math this is a very difficult thing to understand properly. You can easily find some at any bookstore or online.
 
3,871
88
Re: Speed of light

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave
Hi Dave,

Now that I better understand your question about "accounting for" the invariance of the speed of light, and giving in to the confusion between "constant" and "invariant", I post my answer about how "the speed of light" can be invariant here. :tongue2:

(but how can the title in your post be just "speed of light"? :confused:)

Neither time dilation alone, nor together with just length contraction, can give the right answer on such one-way speed of light questions. I stressed that in post #3 of this thread.

The Lorentz transformations and the resulting equation in post 3 in the other thread can be thought of as being built up from a combination of effects, two of which by nature and one man-made:
- length contraction by factor γ
- time dilation by factor γ
- clock synchronization as if the measurement system is in absolute rest

BTW, that is roughly how these transformations were originally derived, with a bit of trial and error. So, here it is worked out from the point of view of "rest" system S for your co-moving system S'.

1. Only assuming time dilation.
0.75c -> γ ≈ 1.51
Result: you would measure the light to speed away from you at 0.25c*1.51≈0.38c
Not OK, contrary to your thinking here above.

2. Adding length contraction.
Result: Now you would measure the light to speed away from you at approximately 0.38c*1.51≈0.57c
Still not OK, contrary to your thinking in a more recent post.

Note that there is an error from slightly out of relative sync clocks according to S which I did not account for here; I prepared an example for post #10 with different speed, but nobody was interested.

3. You now make a "local" clock synchronization for system S', such that the one-way speed equals the (average) two-way speed.
Thus we first calculate the physical effects for light in the other direction:
1.75c*1.51≈2.65c
42.65c*1.51≈4c (= speed of light coming at you, as approximately measured by you)

Next we calculate the total time Δt' in S' over a length L' of that system:
L'/0.57c + L'/4c = 2L'/c
(I put the equal sign because the result is exact). 2L'/c is just the time that it would take if the light propagated at speed c over L' and back. The "two-way" speed according to S' is thus c.

With clock synchronization you simply make the one-way speed according to S' equal to this two-way speed. See section 1 of: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Note that the "experience" that he speaks of at the end of that section relates to the average two-way speed of light in the equation there, because the one-way speed is largely a matter of convention ("definition").

Next you will "measure" with your system S' in both directions a speed of light equal to c. :smile:
 
Last edited:

Related Threads for: Speed of light constant

  • Posted
Replies
7
Views
3K
  • Posted
Replies
20
Views
2K
  • Posted
Replies
6
Views
2K
  • Posted
Replies
4
Views
2K
Replies
20
Views
1K
  • Posted
Replies
5
Views
2K
  • Posted
Replies
5
Views
2K
  • Posted
Replies
10
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top