Speed of Light Always The Same?

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I know this has been beaten to death in this forum and other places -> The problem is that I have not come across an explanation which I can understand!!

I understand the most basic things of time dilation. And it makes sense to me.

However, I cannot grasp the concept of how light always travels at roughly the same speed in a vacuum.

If a spaceship travelling at 99% of the speed of light is racing after a photon of light, how is it that the photon is moving away from the spaceship at 100% of the speed of light...?
Shouldn't the spaceship be only a small distance from the photon which it is chasing?

It seems though that this is wrong. And that the photon should be moving away from the spaceship at c... meaning the photon would be extremely far away from the spaceship?!

I have read many explanations, but none do it for me >.<
Can someone explain this in a very simple manner?

excuse my stupidity on the matter..
 

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  • #2
HallsofIvy
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I know this has been beaten to death in this forum and other places -> The problem is that I have not come across an explanation which I can understand!!

I understand the most basic things of time dilation. And it makes sense to me.

However, I cannot grasp the concept of how light always travels at roughly the same speed in a vacuum.
Not "roughly" the same speed- exactly the same speed!

If a spaceship travelling at 99% of the speed of light is racing after a photon of light, how is it that the photon is moving away from the spaceship at 100% of the speed of light...?
Shouldn't the spaceship be only a small distance from the photon which it is chasing?
No, the photon moves away from the space ship at 100% the speed of light. That would be true in clasical physics. In classical physics, a stationary person would say that the photon is moving, relative to him, at 199% the speed of light because he would use the "speed addition formula" u+ v. But according to relativity (and verified by experiment) for very high speeds the correct formula is
[tex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/tex]
In your example u= c. That formula gives
[tex]\frac{v+ c}{1+ \frac{vc}{c^2}}= \frac{v+c} {\frac{c+v}{c}}= c[/tex]

That works out that way for any speed, not just .99c. That's why the speed of light, relative to anyone, no matter how fast, the spaceship that originates the light is moving relative to them, is c.

It seems though that this is wrong. And that the photon should be moving away from the spaceship at c... meaning the photon would be extremely far away from the spaceship?!

I have read many explanations, but none do it for me >.<
Can someone explain this in a very simple manner?

excuse my stupidity on the matter..
 
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  • #3
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Hmm sorry, I don't understand the mathematics.

I accept what you told me is true :) But it doesn't help me explaining how it is true... >.<

For example, I don't know why the spaceship would see the photon far away from them

But someone not moving and watching would see the photon close to the spaceship?

Or the other way around? hmm... I don't understand. Surely the photon is either close or far away from the spaceship... but not both?

What if we froze time and the spectator walks up to the photon... where would it be then?! close to the spaceship or far away?

(if you have a problem with freezing time, let's say we have a simulation of the universe in a computer. And we froze that instead.) :p
 
  • #4
JesseM
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Reposting some comments from another recent thread on this question:

Different observers measure the speed of anything using rulers and clocks which are at rest relative to themselves, and synchronized in their own frame using the Einstein clock synchronization convention. But the rulers of different observers don't measure distance the same way (each observer measures rulers that are moving relative to themselves to be shrunk by a factor of sqrt(1 - v^2/c^2) relative to their own ruler/clock system, the effect known as length contraction), the clocks of different observers don't measure time the same way (each observer measures clocks that are moving relative to themselves to have the time between ticks expanded by a factor of 1/sqrt(1 - v^2/c^2), the effect known as time dilation), and clocks that are synchronized in their own frame will be measured to be out-of-sync in other frames (clocks that are synchronized and have a distance x between them in their own rest frame will be out-of-sync by vx/c^2 in another frame where they are moving at speed v, due to what's called the relativity of simultaneity). So, it's because of all these differences that two observers can each think light is moving at c relative to themselves.

For a numerical example of how all these effects come together to ensure both observers measure a single light ray to move at c, see my post #6 on this thread.
 
  • #5
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you are confused for the same reason that all newbies get confused. 'relativity of simultaneity'. once you get the idea its not as hard as it sounds.
 
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  • #6
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So, it's because of all these differences that two observers can each think light is moving at c relative to themselves.
This wording is going to lead to cos-like discussion about what is "real" and "true". It's not just a case of each of the observers thinking that light is moving at c relative to themselves, light is moving at c relative to themselves.

The scenario raised here is almost precisely the one which got me interested in this area and I am not sure that someone who has approached from the OP's starting point will immediately appreciate the "relativity of simultaneity" approach.

The visualisation which I came up with to explain the phenomenon worked quite well for me, but months of effort have shown it is difficult to explain to others.

cheers,

neopolitan
 
  • #7
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the best advice I can give you is to learn to draw a spacetime diagram.
 
  • #9
JesseM
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This wording is going to lead to cos-like discussion about what is "real" and "true". It's not just a case of each of the observers thinking that light is moving at c relative to themselves, light is moving at c relative to themselves.
Well, it's only moving at c relative to themselves in a particular choice of coordinate system. Coordinate systems are human inventions, are they not? That's why I said "thinks", although I could equally well have said something like "measures using their own ruler/clock system constructed in the particular way defined by Einstein".
neopolitan said:
The scenario raised here is almost precisely the one which got me interested in this area and I am not sure that someone who has approached from the OP's starting point will immediately appreciate the "relativity of simultaneity" approach.
The relativity of simultaneity was only part of my answer, the more intuitive difference is that they measure length differently because of length contraction, and time differently because of time dilation (as well as because of the relativity of simultaneity if they are measuring the one-way speed of light, although you can dispense with simultaneity issues altogether if one of them is measuring the two-way speed of light and therefore needs only one clock), and each one defines "speed" in terms of their own distance/time measurements.
 
  • #10
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Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
 
  • #11
sylas
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Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
Not at two places at once... rather that for any fixed event (place and instant) the distances and times to other events will depend on the observer.

It's not that the photon is in two places at once, but that the single place-and-instant is at many different distances at once, depending on how you look at it.
 
  • #12
Fredrik
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you are confused for the same reason that all newbies get confused. 'relativity of simultaneity'. once you get the idea its not as hard as it sounds.
the best advice I can give you is to learn to draw a spacetime diagram.
I agree with these comments, but I would like to add that even though it's not as hard as it sounds (mathematically), it's even more counterintuitive than it sounds. (When you hear about it the first time, you aren't even aware of some of the more bizarre consequences of the invariance of the speed of light). That's why spacetime diagrams are so useful. The only way to develop some sort of intuition for these things is to starting thinking in terms of spacetime diagrams.
 
  • #13
Fredrik
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Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
A photon is the worst example you can pick, because photons aren't really localized in space. (That's a very difficult quantum mechanical result that I don't fully understand myself). The detection of a photon is a better example, because that's an event. It's something that happens at a specific time at a specific place. The set of all events is called "spacetime". All observers agree about what events are a part of spacetime. They just disagree about what coordinates to assign to the events. Each observer uses his/her own coordinate system, which is just a function that assigns four numbers to each event.
 
  • #14
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Well, it's only moving at c relative to themselves in a particular choice of coordinate system.
Can you demonstrate a coordinate system in which c is not moving at c relative to an observer? (Stress is on the relative and note I write c, not the figure which is a consequence of the standard metre and the standard second, so c in terms of fathoms per fortnight or furlongs per full moon is still c.)

cheers,

neopolitan
 
  • #15
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A photon is the worst example you can pick, because photons aren't really localized in space.
The OP specifically had a photon based scenario.

cheers,

neopolitan
 
  • #16
DaveC426913
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Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)


*numbers in this example are picked for their clarity of explanation, not for their accuracy
 
  • #17
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I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.

Here is my similar problem:

A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft.
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft.

In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.

Thanks for any help you can provide,
regards
 
  • #18
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I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)


*numbers in this example are picked for their clarity of explanation, not for their accuracy
due to relativity of simultaneity he will measure distances differently too. not just time (the length of an object is the distance between the front and back at one simultaneous moment)

this is why learning to draw spacetime diagrams is so important
 
  • #19
Fredrik
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The OP specifically had a photon based scenario.
I guess I'm just nitpicking the word "photon". The OP's question is fine if we interpret "photon" as "massless classical particle", which is probably what he had in mind anyway. It's a convenient way to think about light in SR. (The word "photon" is actually defined by a quantum field theory which says that photons aren't localized).
 
  • #20
JesseM
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Can you demonstrate a coordinate system in which c is not moving at c relative to an observer?
Do you mean a coordinate system in which light is not moving at c (or even at a constant speed) relative to the observer? If so you could pick something like Rindler coordinates which are commonly used for an observer experiencing constant proper acceleration (also discussed here). But really you have total freedom in defining non-inertial coordinate systems any way you like, for example if x,y,z,t represent the coordinates of an inertial coordinate system I could define a non-inertial coordinate system in terms of any transformation I can think of, a simple one would be:

x' = x - at^2
y' = y
z' = z
t' = t

But you could also do a crazy one like:

x' = x * sin(t/t0)
y' = y - t*c
z' = pi*z - 1008
t' = t + 3x/c + 7y/c + 0.5z/c
 
  • #21
JesseM
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I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)


*numbers in this example are picked for their clarity of explanation, not for their accuracy
But by picking fake numbers you've actually given a misleading explanation here--in reality you can't explain the fact that both observers measure the photon to move at c using only time dilation as you suggest, you also need to take into account length contraction and the relativity of simultaneity.
 
  • #22
JesseM
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I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.

Here is my similar problem:

A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft.
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft.

In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.

Thanks for any help you can provide,
regards
In the planet X rest frame:
Suppose as the spacecraft is leaving planet X, they synchronize clocks, so that the planet X clock reads t=0 and the spacecraft clock reads t'=0 at the moment it leaves planet X. Now suppose that planet X sends the signal 75 years later at t=75 on the planet X clock, and that at the same moment in this frame, the spacecraft is passing another planet, planet Y, which is at rest relative to planet X and 60 light years away in this frame (since the spacecraft was traveling at 0.8c in this frame, naturally 75 years after it left planet X it would be at a distance of 75*0.8=60 light-years from planet X). There is a clock at planet Y which is synchronized with the on at planet X in this frame, so it also reads t=75 when the spacecraft passes it. In this frame the spacecraft clock is slowed down by a factor of sqrt(1 - 0.8^2) = 0.6 due to time dilation, so it only reads t' = 75*0.6 = 45 when it passes planet Y.

The spacecraft starts out 60 light years away from planet X when the signal is sent at t=75, so after 300 years have passed in this frame the spacecraft will have traveled an additional 300*0.8c = 240 light years away, so 300 years later at t=375 years, it'll be 60 + 240 = 300 light-years away from planet X. And naturally if we assume the signal travels at c = 1 light year/year in this frame, the signal will also be 300 light-years away from planet X after 300 years have passed from the time it was sent, so t=375 years must be the time in the planet X frame that the signal reaches the craft. However, because of the time dilation factor, at t=375 the spacecraft clock only reads 375*0.6 = 225 years, so t'=225 years must be the time on the spacecraft's clock when the signal reaches it.

In the spacecraft frame:
All frames agree on local events, so it must be true in this frame too that the planet Y clock reads t=75 at the moment the spacecraft is passing it, at which time the spacecraft clock reads t'=45, as we figured out before. However, in this frame the planet X clock does not read t=75 at this moment, due to the relativity of simultaneity. The way the relativity of simultaneity works is that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at speed v parallel to the axis between them, the clock in the rear will show a time that's ahead of the clock at the front by vL/c^2. So, in the spacecraft's frame at the moment it's passing planet Y, the planet X clock must be behind by (0.8c)*(60 light-years)/c^2 = 48 years, so the planet X clock reads t=75-48=27 years at the moment the spacecraft is passing planet Y in the spacecraft frame. Since the planet X clock is slowed down by a factor of 0.6 in this frame, we know the planet X clock will take an additional 48/0.6 = 80 years to tick forward to t=75, i.e. the planet X clock will read 75 at t'=45+80=125 in the spacecraft frame. And again, all frames agree on local events, so it must be true in this frame as well that the signal was sent when the planet X clock read t=75, at t'=125 in the spacecraft frame.

Meanwhile, because of length contraction, if the distance between planet X and planet Y was 60 light years in their own rest frame, then in the spacecraft rest frame the distance is shrunk by a factor of sqrt(1 - 0.8^2) = 0.6, so the distance between planet X and planet Y is only 60*0.6 = 36 light-years in the spacecraft rest frame. So at t'=45 when the spacecraft was next to planet Y, planet X was 36 light-years away; and since planet X is moving at 0.8c away from the spacecarft in this frame, that means 80 years later at t'=125 when planet X was sending the signal in this frame, planet X must have been an additional 80*0.8=64 light-years away, for a total of 64+36=100 light-years away from the spacecraft at the moment the signal was sent at t'=125. So, under the assumption that light travels at c in this frame too, the signal will reach the spacecraft 100 years later at t'=225.

By the way, another way to approach this problem would be to use the Lorentz transformation, which says that if an arbitrary event has coordinates x,t in one frame, then in a second frame moving at v along the first frame's x-axis (and with the two frames having a common origin, so x=0,t=0 in the first frame coincides with x'=0,t'=0 in the second), this same event will have coordinates:

x'=gamma*(x - vt)
t' =gamma*(t - vx/c^2)

with gamma = 1/sqrt(1 - v^2/c^2)

In this problem v=0.8c so gamma=1/0.6=1.666..., which means for example that if the event of the signal being sent had coordinates (x=0, t=75) in the planet X frame, in the spacecraft frame the event of the signal being sent has coordinates:

x'=1.666...*(0 - 0.8*75) = -100
t' =1.666...*(75 - 0.8*0) = 125

And if the event of the signal reaching the spacecraft happened at (x=300, t=375) in the planet X frame, in the spacecraft frame this event has coordinates:

x'=1.666...*(300 - 0.8*375) = 0
t'=1.666...*(375 - 0.8*300) = 225

So, you can see that this agrees with the previous calculations, and it's a little simpler to figure out although if you use the Lorentz transform you don't get to think about the different contributions of time dilation, length contraction and the relativity of simultaneity.
 
  • #23
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read einsten's "special theory of relativity",trust me ,all your doubts will vanish.
 
  • #24
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Thanks JesseM, this really helped in understanding it all, and the semi-complex nature of special relativity. Definitely printing that out for keeps in helping my understanding.

regards
 
  • #25
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Do you mean a coordinate system in which light is not moving at c (or even at a constant speed) relative to the observer? If so you could pick something like Rindler coordinates which are commonly used for an observer experiencing constant proper acceleration (also discussed here). But really you have total freedom in defining non-inertial coordinate systems any way you like, for example if x,y,z,t represent the coordinates of an inertial coordinate system I could define a non-inertial coordinate system in terms of any transformation I can think of, a simple one would be:

x' = x - at^2
y' = y
z' = z
t' = t

But you could also do a crazy one like:

x' = x * sin(t/t0)
y' = y - t*c
z' = pi*z - 1008
t' = t + 3x/c + 7y/c + 0.5z/c
Yes, the first c should have been "it" from your quote and "it" was light.

As for your coordinate system, you just jumped out of SR and the question was posed in the context of SR (look at the OP's original question).

Let me rephrase, can you demonstrate a coordinate system in which light is not moving at c relative to an inertial observer?

It doesn't make sense to have a non-inertial coordinate system if the observer is inertial, but I think you will find that even if a non-inertial coordinate system is used then, in that non-inertial coodinate system, the speed of light relative to the observer will be c. So, the choice of inertial coordinate system is not the culprit in your example, but rather non-inertia.

And not being inertial is not a human invention.

cheers,

neopolitan
 

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