the speed of light is constant in all IFOR (1 Viewer)

[kony]

i need some clarification please.

this is somewhat contradictory.

so we have the train moving at 0.8c and theres an observer outside. inside the train, a passenger observes a light wave moving 1metre in 1/c seconds.

outside, the distance observed is shorter. i.e. the light moves 1*sqrt(1-0.8^2) = 0.6m.

but it also takes more time, as observed by the outside observer (time dilation).
time becomes [1/c]/[sqrt(1-0.8^2)] = 1/0.6c

the velocity of light, therefore, becomes (0.6)/(1/0.6c) = 0.36c!!


i think i got one of the length contraction and time dilations mixed around (it should cause the whole sqrt(1-0.8^2) thing to cancel out)

 

[Forbidden.]

Speed of light is constant except when refracted from one medium to another.
There's general relativity or special relativity, but one corresponds to saying light is not constant while another (definitely an Einsteinian one) says light is always constant.

 

[kony]

1/c is the time it takes light to move 1m.

since distance/velocity = time.

distance = 1m
velocity = c

time: 1/c

 

[airie]

inside the train, a passenger observes a light wave moving 1metre in 1/c seconds.

I take it that you meant that both the 1m distance and 1/c sec time are measured in the frame of reference of the train?

Then yes, the distance measured by someone outside the train is shorter; however, the contracted length is 1m, as measured in the train, so the proper length measured by someone outside the train would be l₀ = l_v/rt(1-v²/c²) = 1/0.6 m. And as you pointed out, the dilated time observed by someone outside the train would be t_v = t₀/rt(1-v²/c²) = 1/0.6c seconds. Thus, the speed of light measured by someone outside the train would be distance/time = l_v/rt(1-v²/c²) / [t₀/rt(1-v²/c²)], which does equal c.

Your apparent problem arises from your confusion between proper length and contracted length. Proper length is measured by someone outside the train, while contracted length is measured in the train, not the other way around

 

[kony]

Then yes, the distance measured by someone outside the train is shorter; however, the contracted length is 1m, as measured in the train, so the proper length measured by someone outside the train would be l₀ = l_v/rt(1-v²/c²) = 1/0.6 m.

i disagree.

we define *proper length* to be the length measured in the frame of reference where the thing you're measuring is stationary, so the proper length in this case would be inside the train.

hence, in all other frames of reference, the length is shorter, which goes back to my original point.

airie, isn't the formula for length contraction l₀ = l_v * sqrt(1-v²/c²), i.e. multiply, not divide.

 

[airie]

we define *proper length* to be the length measured in the frame of reference where the thing you're measuring is stationary,

Exactly; When you take the measurement of the length, this distance is taken in the moving train. Thus this is the contracted length that you measure.

airie, isn't the formula for length contraction l₀ = l_v * sqrt(1-v²/c²), i.e. multiply, not divide.

No; l_v = l₀ * sqrt(1-v²/c²). Length contraction, remember. Therefore l₀ = l_v/sqrt(1-v²/c²).

 

[Forbidden.]

Exactly; When you take the measurement of the length, this distance is taken in the moving train. Thus this is the contracted length that you measure.


No; l_v = l₀ * sqrt(1-v²/c²). Length contraction, remember. Therefore l₀ = l_v/sqrt(1-v²/c²).

It can get pretty annoying when you have to calculate length contraction or time & mass dilation from a different frame of reference.

But I'm still not sure when to reverse l_v and l₀ or for t or m for that matter.

 

[kony]

the distance measured by someone outside the train is shorter; however, the contracted length is 1m, as measured in the train, so the proper length measured by someone outside the train would be l₀ = l_v/rt(1-v²/c²) = 1/0.6 m.

hey airie, in the first sentence you say that to the outside observer, the length of that the light moves is shorter, but in the last bit you say that it is 1/0.6 m, which is clearly longer than 1m.

anyway, once more for clarification:

a passenger holds up a metre-ruler in the train in the direction of the motion of the train. to the passenger, the ruler is exactly one metre long, since he is not in relative motion to the ruler.

now, the guy standing outside on the platform. he sees the train move relative to him, so everything in the train, including the train, becomes contracted.

therefore, the length of the ruler is contracted to 0.6m, if the train is moving at 0.8c.

the contracted length is then 0.6m, the proper length is 1m.



i know something's wrong here, since if this was right, c wouldn't be constant. but airie, what you've said so far is a bit unclear. please elaborate, thanks.

 

[airie]

hey airie, in the first sentence you say that to the outside observer, the length of that the light moves is shorter, but in the last bit you say that it is 1/0.6 m, which is clearly longer than 1m.

No no no you totally misunderstood me x.x

Alright, sorry if I caused you any confusion. But when I said that "the distance measured by someone outside the train is shorter" I meant that, a length measured in the train would be contracted to a person outside the train, say A (but this is NOT observed by the person inside the train, say B, since they are stationary in relation), while A does not observe length contraction in their own frame. Therefore, observed by A, a same distance would be measured shorter in the train, and longer outside the train.

(Is that clear enough? ... Or am I just plain confusing? >.< )

Of course, you could also take A's frame of reference to be the moving one and B's the stationary, since motion is relative. Then B will measure the proper length, and B sees A measure the contracted length. But since A's frame is the moving one, time is dilated in A's frame, so for a same time interval, B observes their own time to be normal, but A's time to be of a smaller value (since A's time is observed by B to be slower than B's). And then you get the same light speed in both frames.

(Perhaps you didn't keep in mind that time dilation, length contraction and mass dilation in one particular frame is only observed by someone OUTSIDE the frame, in another frame in relative constant motion? Or that, you can't operate on quantities measured in different frames of reference together? As in, you can't divide B's measurement of length by the time that B observes A to measure etc...Do you get what I'm saying though? I can be very inarticulate at times )

 

[kony]

ok i think i know what you mean now. thanks.

the only thing is this:


say you are on earth observing a spaceship travelling past you. so to you, the time on the spaceship is dilated. but what does that mean?

i.e. does that mean that for every 1 second you time on earth, more than 1 second pass on the spaceship according to you?

or does that mean that for every 1 second you time on earth, less than 1 second pass on the spaceship according to you?

 

[xiao1985]

i believe it's the latter

you age faster (experience mroe time) whilst standing still.

 

[kony]

yes, of course. i can't believe i asked such a noob question.

it is obviously the latter, since all those "evidence" for time dilation and stuff, like the one where they put one atomic clock on the rocket and fly it around for a bit. at the same time an identical one is on earth. the one on that went in the rocket is the one that showed a shorter time elapse.

even though this evidence is dodgy, since none of the frames of references are inertial. (including the Earth, which has a constantly changing velocity)

 

[xiao1985]

lol true that...

for the purpose of the calculation and thought experiments though, they are considered inertial...

 

[airie]

the only thing is this:


say you are on earth observing a spaceship travelling past you. so to you, the time on the spaceship is dilated. but what does that mean?

i.e. does that mean that for every 1 second you time on earth, more than 1 second pass on the spaceship according to you?

or does that mean that for every 1 second you time on earth, less than 1 second pass on the spaceship according to you?

time is dilated in A's frame, so for a same time interval, B observes their own time to be normal, but A's time to be of a smaller value (since A's time is observed by B to be slower than B's)

I would think that this question was already answered in my explanation

yes, of course. i can't believe i asked such a noob question.

it is obviously the latter, since all those "evidence" for time dilation and stuff, like the one where they put one atomic clock on the rocket and fly it around for a bit. at the same time an identical one is on earth. the one on that went in the rocket is the one that showed a shorter time elapse.

even though this evidence is dodgy, since none of the frames of references are inertial. (including the Earth, which has a constantly changing velocity)

lol. You just gave your own explanation there

But yeah, like xiao said, Earth can be considered an inertial frame cos you're not looking at its motion relative to the sun or anything, and since the rocket frame was moving relative to Earth and it was the one that underwent acceleration, time dilation is observed in it when it is brought back to Earth.

So I take that there aren't any problems now