Light speed and time

[Bart]

Can anyone explain for me the following case, please:

A rocket is moving at speed 0.5 c, from planet A to planet C, which are distant from each other by 4 light-days.

 

During the flight the rocket passes the planet B, located exactly halfway between planets A and C.

At the time of passing the planet B, from the rocket and from the planet B are simultaneously sent SMS messages to the planets A and C.

SMS sent from the planet B reaches the planets A and C at the same time, in two days.

Question, when the SMS sent from the rocket, will be received on the planet A and the planet C, respectively.

 

Thanks.

 

This is not a homework.

 

Edited September 28, 2011 by Bart

[imatfaal]

Same time - but the one to C will be blueshifted and to A will be redshifted

 

shift will be as follows

 

[math] \lambda_{observed} = \lambda_{signal}. \sqrt{\frac{1+v/c}{1-v/c}} [/math]

 

v is negative for closing gap and positive for opening

[Janus]

Can anyone explain for me the following case, please:

A rocket is moving at speed 0.5 c, from planet A to planet C, which are distant from each other by 4 light-days.

 

During the flight the rocket passes the planet B, located exactly halfway between planets A and C.

At the time of passing the planet B, from the rocket and from the planet B are simultaneously sent SMS messages to the planets A and C.

SMS sent from the planet B reaches the planets A and C at the same time, in two days.

Question, when the SMS sent from the rocket, will be received on the planet A and the planet C, respectively.

 

Thanks.

 

This is not a homework.

 

 

According to which frame?

 

According to the rest frame of planets A and C, the signals from both the Rocket and Planet B all arrive simultaneously at A and C.

According to the rest frame of the rocket, the signals sent from the both rocket and B arrive at planet C in 1.15 days and at planet A in 3.46 days. IOW, while the signals from both the rocket and planet B will arrive at each planet at the same time, the signals will not arrive at planet A at the same time as they do at planet C.

[imatfaal]

of course Janus is correct - and shows how dangerous it is to assume which frames are being used; I have given myself a slap on the wrist

[Bart]

According to the rest frame of the rocket, the signals sent from the both rocket and B arrive at planet C in 1.15 days and at planet A in 3.46 days. IOW, while the signals from both the rocket and planet B will arrive at each planet at the same time, the signals will not arrive at planet A at the same time as they do at planet C.

 

 

So how an astronaut in the rocket can understand that the signal sent at halfway between the planets A and C, comes to these planets at such different times, knowing that the planet B is exactly midway between the planets A and C and, that light travels at the same rate to the both planets?

 

Edited September 29, 2011 by Bart

[Janus]

 

 

 

So how an astronaut in the rocket can understand that the signal sent at halfway between the planets A and C, comes to these planets at such different times, knowing that the planet B is exactly midway between the planets A and C and, that light travels at the same rate to the both planets?

 

 

It's because the speed of light is constant that the astronaut will determine that the signals reach the planets at different times. The signals are emitted halfway between A and C. The signals travel away from the rocket at c in both directions. (after 1 sec, both the signals sent towards A and C will be 299,792,458 meters from the rocket.)

Planets A and C do not maintain a constant distance from the Rocket. After 1 sec planet A is 149,896,299 meters further away, and planet C is 149,896,229 meters closer. IOW, according to the astronaut's frame, planet A is running away from the signal and Planet C is rushing to meet it. By the time the signal and planet C meet, Planet C will be much closer to the rocket than it was when the signal was emitted, and by the time it reaches planet A, planet A will be much further away. Thus from the rocket's frame, the signal traveling to A travels further than the signal to A, and since both signals travel at the same speed, the signal reaches A after it reaches C.

[Bart]

Janus, thank you very much for your explanations and sorry yet, but still I don't well understand the calculation of 1,15 days to planet C and 3,46 to planet A? That's means for the astronaut in the rocket, who knows the real distance to the planets, that his messages sent to planet C run at speed almost 1,8 times greater than light speed c, and to planet A only 0,58 of c?

 

What is the physical difference between the case 1 and case 2, below?

 

Case 1.

 

A rocketis moving at speed 0.5 c, from planet A to planet C, which are distant from each other by 4 light-days.

 

During the flight the rocket passes the planet B, located exactly halfway between planets A and C.

At the time of passing the planet B, from the rocket and from the planet B are simultaneously sent SMS messages to the planets A and C.

SMS sent from the planet B reaches the planets A and C at the same time, in two days.

Question, when the SMS sent from the rocket, will be received on the planet A and the planet C, respectively.

 

 

Case 2.

 

A car is moving at constant speed 0.5 v, from town A to town C, which are distant from each other by 4 motorcycle-days

 

On the way the car passes the town B, located exactly halfway between town A and C.

At the time of passing the town B, from the car and from the town B are simultaneously sent messages to the towns A and C, by motorcycles with an arbitrary set constant speed v.

 

The same question, when the message sent from the car, will be received at the town A and the town C, respectively? (in the same time)

 

Edited September 29, 2011 by Bart

[Janus]

Janus, thank you very much for your explanations and sorry yet, but still I don't well understand the calculation of 1,15 days to planet C and 3,46 to planet A? That's means for the astronaut in the rocket, who knows the real distance to the planets, that his messages sent to planet C run at speed almost 1,8 times greater than light speed c, and to planet A only 0,58 of c?

 

according to the rocket, A and C are 1.732 light days from the rocket at the moment of emission, due to length contraction. Planet A is receding at 0.5c and planet C is approaching at 0.5c Given the speed of light is c, the time I gave are how log it takes for each planet to meet up with its respective signal. There is no such thing as the "real" distance between the planet's, only the distance as measured by each frame ( or conversely, the distance as measured in each frame is equally "real".)

 

And no, for the astronau,t the signals do not travel at anything but c. the closing speed between A and the signal is 0.5c and between C and the signal is 1.5 c, but closing speed does not reflect any real change in c

 

What is the physical difference between the case 1 and case 2, below?

 

Case 1.

 

A rocketis moving at speed 0.5 c, from planet A to planet C, which are distant from each other by 4 light-days.

 

During the flight the rocket passes the planet B, located exactly halfway between planets A and C.

At the time of passing the planet B, from the rocket and from the planet B are simultaneously sent SMS messages to the planets A and C.

SMS sent from the planet B reaches the planets A and C at the same time, in two days.

Question, when the SMS sent from the rocket, will be received on the planet A and the planet C, respectively.

 

 

Case 2.

 

A car is moving at constant speed 0.5 v, from town A to town C, which are distant from each other by 4 motorcycle-days

 

On the way the car passes the town B, located exactly halfway between town A and C.

At the time of passing the town B, from the car and from the town B are simultaneously sent messages to the towns A and C, by motorcycles with an arbitrary set constant speed v.

 

The same question, when the message sent from the car, will be received at the town A and the town C, respectively? (in the same time)

 

 

The difference is that "v" is not an invariant so the car will not measure the speeds of the motorcycles as being a constant with respect to himself. However, that does not mean that according to him, that the motorcycle heading for C will not get there first.

 

In this situation you have to use the addition of velocities theorem:

 

[math]w = \frac{u+v}{1+\frac{uv}{c^2}}[/math]

 

To get the velocity of the motorcycles with respect to the car as measured from the car frame.

 

Given that 0.5v is the relative velocity of the car to the road and v is the relative speed of the motorcycles, then:

 

The speed of the motorcycle heading toward A relative to the Car is:

 

[math]w = \frac{0.5v+v}{1+\frac{0.5v^2}{c^2}}[/math]

 

which is less than 1.5v

 

and the speed of the motorcycle heading towards C relative to the car is:

 

[math]w = \frac{v-0.5v}{1-\frac{0.5v^2}{c^2}}[/math]

 

which is more than 0.5v

 

 

If you take these speeds and the speed of the car relative to the roadway( 0.5v), you get the closing speeds between the motorcycles and the towns according to the car. The closing speed between motorcycle and A will be lower than that of between motorcycle and C. Given that A and C were equal distances from the car when the motorcycles left, a motorcycle will reach C before it reaches A, according to the Car.

 

Now it would be almost impossible to measure the difference in timing at the speeds at which cars and motorcycles really travel, so we would not tend to notice it.

 

For instance if v =200 kph then

 

[math]\frac{200 kph+100 kph}{1+ \frac{200kph(100kph)}{c^2}}[/math] = 299.999999999995 kph.

 

So close to 300 kph, that you would never have a chance of noticing it in everyday life.

 

But a v and u get closer to c the difference becomes more and more measurable.

Edited September 29, 2011 by Janus

[Bart]

Many, many thanks Janus for your reply.

 

Sorry again, but I need your more explanation. Case 2 was designed as an exact physical equivalent of the case 1, where the car represents the rocket and the motorcycles (instead of it the sound may be considered) represent the light carrying messages. Other media do not exist, so in my understanding they should not occur in the calculations for the case 2. Therefore I do not understand how and why in your calculations for the case2, there exists a third medium in the form of c?.

 

Edited September 30, 2011 by Bart

[Janus]

Many, many thanks Janus for your reply.

 

Sorry again, but I need your more explanation. Case 2 was designed as an exact physical equivalent of the case 1, where the car represents the rocket and the motorcycles (instead of it the sound may be considered) represent the light carrying messages. Other media do not exist, so in my understanding they should not occur in the calculations for the case 2. Therefore I do not understand how and why in your calculations for the case2, there exists a third medium in the form of c?.

 

 

First, I'm going to make a request. Just use the default formating for your posts. The way you are formatting them makes it near impossible to read when using the quote function. The formating does not show and all you get is the format tags. It ends up looking something like this:

 

[tag] word[endtag] [tag] a few words[endtag][tag]word[endtag][tag]word[endtag]...

 

with sentences and phrases broken up by formating tags. This makes it very difficult to respond to your post point by point.

 

You might think it looks nice in the post, but trust me, nobody really cares about what the posts look like, only their content.

 

 

As to your post. It doesn't matter whether light or anything else exists in your scenario or not. It is the speed "c" that is important, that and the fact that it is invariant ( it's just handy to use light in many examples because it happens to travel at c).

 

The fact that c is a finite and invariant speed has consequences that extend past things that travel at c. Removing light from the scenario does not change this. The addition of velocities formula I gave is an example. The "c" is there because it plays a fundamental role in how the universe works.

 

c plays the role infinite speed was thought to play before Relativity. Then, infinite speed was the invariant speed. (in fact, if you replace c with infinity in the velocity addition equation, it reduces to u+v.)

 

So to repeat, we use light in many examples because it is convenient to do so, not because light is required for the principles in the example to work.

[Bart]

First, I'm going to make a request. Just use the default formating for your posts. The way you are formatting them makes it near impossible to read when using the quote function. The formating does not show and all you get is the format tags. It ends up looking something like this:

 

[tag] word[endtag] [tag] a few words[endtag][tag]word[endtag][tag]word[endtag]...

 

with sentences and phrases broken up by formating tags. This makes it very difficult to respond to your post point by point.

 

You might think it looks nice in the post, but trust me, nobody really cares about what the posts look like, only their content.

 

 

As to your post. It doesn't matter whether light or anything else exists in your scenario or not. It is the speed "c" that is important, that and the fact that it is invariant ( it's just handy to use light in many examples because it happens to travel at c).

 

The fact that c is a finite and invariant speed has consequences that extend past things that travel at c. Removing light from the scenario does not change this. The addition of velocities formula I gave is an example. The "c" is there because it plays a fundamental role in how the universe works.

 

c plays the role infinite speed was thought to play before Relativity. Then, infinite speed was the invariant speed. (in fact, if you replace c with infinity in the velocity addition equation, it reduces to u+v.)

 

So to repeat, we use light in many examples because it is convenient to do so, not because light is required for the principles in the example to work.

 

Many thanks Janus for your last explanation, and sorry for my formating. I will be better from now.