Two Balls in Space: An Observer's Perspective

[Mustakafine]

Some friends and I were out at a pub last night and this question came up.

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

Does light propagate through space at the same speed regardless of the speed of it source relative to another object?

 

[Dickfore]

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

No, they would be traveling at a velocity:
$$\frac{\frac{c}{2} + \frac{c}{2}}{1 + \frac{1}{c^{2}} \, \frac{c}{2} \, \frac{c}{2}} = \frac{c}{1 + \frac{1}{4}} = \frac{4 \, c}{5} = 0.8 \, c$$
where c is the speed of light in vacuum.

 

[Mustakafine]

Speed of light = 299792458*m/s

Half of the speed of light = 149896229 m/s

I throw ball one at 149 896 229 m/s in one direction.

I throw ball two in exactly the opposite direction at 149 896 229 m/s.

Both balls are moving away from me at 149 896 229 m/s. An observer on ball one would see me moving away from him at 149 896 229 m/s. why would he not see ball 2 moving away at 149 896 229 m/s + 149 896 229 m/s which is my speed plus his which is the speed of light.

Two trains moving towards each other at 50 miles per hour each have a closing speed of 100 miles per hour. When they pass reach other the effective distance between them at any given time will be the sum of there two speeds.

 

[Ryan_m_b]

This is a valid question that everybody asks at some point. The problem is that adding the two speeds is technically wrong, it works for everyday life but in reality for moving observers instead of adding the speeds there's an equation that shows that you can never observe something moving at or faster than the speed of light. Look at what dickfore has shown you, this is the answer.

No matter how fast you are traveling you will always measure the speed of light to be the same. For the two observes they are experiencing time dilation, because of this they measure the closing speed as less than the speed of light.

 

[Dickfore]

The Law of composition of velocities (in the same direction in space) in Special Theory of Relativity is:
$$v_{CA} = \frac{v_{CB} + v_{BA}}{1 + \frac{v_{CB} \, v_{BA}}{c^{2}}}$$
This is quite different from what you had used based on Galilean transformation (non-relativistic kinematics).

 

[Mustakafine]

Okay let's try this from another angle.

Lets add observers C & D they are exactly 10 light seconds from the guy in the middle throwing the balls. The balls are thrown at the same time with observers A & B on board. They travel for exactly 20 seconds each and are caught by observers C & D. Because C & D and the thrower are stationary relative to each other they have not experienced any time dilation. Observers on the balls A & B did relative to each other. A saw B moving away at .8c and visa versa.

C & D observed 20 seconds of travel time and there is now a physical distance of 20 light seconds between the balls. A & B's watches are no longer in synch with C & D's and the throwers because of time dilation. However they still covered the 20 light seconds in 20 seconds to the stationary observers which means relative to themselves they were traveling at the speed of light as observed by others and measured in distance and time.

Can you explain in laymens terms please. Remember that this was a pub discussion. I would not know where to begin with the formulas above.

Many thanks for this guys. Very interesting way to pass time at work on a friday :)

 

[Dickfore]

How is this a laymen's discussion when it is posted in the 'Special and General Relativity' subforum? Check out the rules:

https://www.physicsforums.com/showpost.php?p=172576&postcount=1

 

[Mustakafine]

Is that the right link. It just leads to a page that says this.

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This forum is meant as a place to discuss the Theory of Relativity and is for the benefit of those who wish to learn about or expand their understanding of said theory. It is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories. All future posts of this nature shall either be deleted or moved by the discretion of the Mentors.

Thank you for your cooperation.

 

[Dickfore]

Okay let's try this from another angle.
Lets add observers C & D they are exactly 10 light seconds from the guy in the middle throwing the balls. The balls are thrown at the same time with observers A & B on board. They travel for exactly 20 seconds each and are caught by observers C & D. Because C & D and the thrower are stationary relative to each other they have not experienced any time dilation. Observers on the balls A & B did relative to each other. A saw B moving away at .8c and visa versa.

C & D observed 20 seconds of travel time and there is now a physical distance of 20 light seconds between the balls. A & B's watches are no longer in synch with C & D's and the throwers because of time dilation. However they still covered the 20 light seconds in 20 seconds to the stationary observers which means relative to themselves they were traveling at the speed of light as observed by others and measured in distance and time.

I don't have the slightest idea what you are trying to describe. Care to present a sketch of the situation you are describing?

 

[Ryan_m_b]

C & D observed 20 seconds of travel time and there is now a physical distance of 20 light seconds between the balls. A & B's watches are no longer in synch with C & D's and the throwers because of time dilation. However they still covered the 20 light seconds in 20 seconds to the stationary observers which means relative to themselves they were traveling at the speed of light as observed by others and measured in distance and time.

I don't really know what you are saying here, let me try to sketch it out.

C......AoB......D

Each full stop is half a light second. C and D are 10 light seconds away from o (thrower). A and B travel at half the speed of light. C, o and D observe A and B traveling at half the speed of light. A and B can measure their speeds by observing C, o and D. A will observe B receding from it at 80% of the speed of light and vis versa. What are you struggling with here?

EDIT: It occurs to me that the last thing I said may be wrong, A and B may observe the distance between them increasing faster than light (I'm not sure if that's true). If so what is your point?

 

[Mustakafine]

Good sketch ryan. Was doing one up in paint shop which would have been embarrassing to post to the wild.

It takes A 20 seconds to travel to C. And 20 seconds for B to travel to D. There are 20 light seconds of distance between C & D.

So after 20 seconds both a & b have arrived at C & D respectively so the total distance between them is 20 light seconds. So A & B relative to each other have traveled apart at the speed of light as observed by C & D but only observed each other move apart at 80% c.

Is that a bit clearer?

 

[Ryan_m_b]

Good sketch ryan. Was doing one up in paint shop which would have been embarrassing to post to the wild.

It takes A 20 seconds to travel to C. And 20 seconds for B to travel to D. There are 20 light seconds of distance between C & D.

So after 20 seconds both a & b have arrived at C & D respectively so the total distance between them is 20 light seconds. So A & B relative to each other have traveled apart at the speed of light as observed by C & D but only observed each other move apart at 80% c.

Is that a bit clearer?

Yes C and D would measure A and B moving away from each other as C but that doesn't mean anything FTL is actually occurring. A and B would not measure their receding speed as this.

 

[nitsuj]

No, they would be traveling at a velocity:
$$\frac{\frac{c}{2} + \frac{c}{2}}{1 + \frac{1}{c^{2}} \, \frac{c}{2} \, \frac{c}{2}} = \frac{c}{1 + \frac{1}{4}} = \frac{4 \, c}{5} = 0.8 \, c$$
where c is the speed of light in vacuum.

Is that simular to saying the space between the [STRIKE]balls[/STRIKE], objects is NOT increasing by 299 792 458 m/s from the frame of either of the objects.

But it is increasing by 299 792 458 m/s wrt the observer who is in the middle.

 

[FlexGunship]

Some friends and I were out at a pub last night and this question came up.

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

Does light propagate through space at the same speed regardless of the speed of it source relative to another object?

Speed of light = 299792458*m/s

Half of the speed of light = 149896229 m/s

I throw ball one at 149 896 229 m/s in one direction.

I throw ball two in exactly the opposite direction at 149 896 229 m/s.

Both balls are moving away from me at 149 896 229 m/s. An observer on ball one would see me moving away from him at 149 896 229 m/s. why would he not see ball 2 moving away at 149 896 229 m/s + 149 896 229 m/s which is my speed plus his which is the speed of light.

Two trains moving towards each other at 50 miles per hour each have a closing speed of 100 miles per hour. When they pass reach other the effective distance between them at any given time will be the sum of there two speeds.

Okay let's try this from another angle.

Lets add observers C & D they are exactly 10 light seconds from the guy in the middle throwing the balls. The balls are thrown at the same time with observers A & B on board. They travel for exactly 20 seconds each and are caught by observers C & D. Because C & D and the thrower are stationary relative to each other they have not experienced any time dilation. Observers on the balls A & B did relative to each other. A saw B moving away at .8c and visa versa.

C & D observed 20 seconds of travel time and there is now a physical distance of 20 light seconds between the balls. A & B's watches are no longer in synch with C & D's and the throwers because of time dilation. However they still covered the 20 light seconds in 20 seconds to the stationary observers which means relative to themselves they were traveling at the speed of light as observed by others and measured in distance and time.

Can you explain in laymens terms please. Remember that this was a pub discussion. I would not know where to begin with the formulas above.

Many thanks for this guys. Very interesting way to pass time at work on a friday :)

Okay, the problem is that Mustakafine has confused his high school physics with real physics. Dickfore and Ryan are both correct in that this is in the wrong forum, but let's at least settle this permanently:

Mustakafine is trying to apply the following math:

A central observer (O) witnesses two vehicles (A and B) travel away from his point of reference at 60 mph in opposite directions. After one hour's time, length AO is 60 miles and length OB is 60 miles. By back calculating the relative speed, he determines they were traveling 120mph away from each other, or 60mph+60mph.

THIS IS AN APPROXIMATION! It is not how the universe actually works. If you had an accurate enough stop watch, and a perfect enough measurement of velocity, you would find that the two cars were actually traveling away from each other at 119.9999999999998mph as given by the following formula:

$$v_{CA} = \frac{v_{CB} + v_{BA}}{1 + \frac{v_{CB} \, v_{BA}}{c^{2}}}$$

$$119.9999999999998 = \frac{60 + 60}{1 + \frac{60*60} {c^{2}}}$$

If you re-solve that problem with the very high speed you're talking about, you will find that they are NOT receding from each other at the speed of light.

EDIT: Try this out in Excel (I just did) and compare the results of the correct calculation and the approximation by addition. You can see that as you approach the speed of light, your error in the calculation gets higher and higher!

At one billionth the speed of light, Excel doesn't have enough decimal places to represent the error. Effectively zero. But at one millionth the speed of light (670 mph), the error between the correct value and the approximation of adding the two values is 1 x 10^-10%.

As you continue you will see that at a hundredth the speed of light (6.7 million mph) the error between your approximation by adding and the correct value is 0.01%. At a tenth the speed of light (67 million mph) the error is 1%. And at the speed of light (670 million mph) the difference between your approximation and the real value is 100%!

The error in your calculation for half the speed of light is 25%. This means that your answer (the speed of light) is 25% higher than the correct answer. 1.25X = Y where X is the correct answer and Y is your answer (the speed of light) then you should get Y/1.25 = 0.8c

Dickfore gave you the CORRECT answer originally.

 

[Mustakafine]

Thanks guys I get it now.

Didnt know this was the wrong forum for this type of question first day here and until we have another pub night trying to bash Einstein after a couple beers probably the last.

 

[FlexGunship]

...another pub night trying to bash Einstein after a couple beers...

In general, this has proved to be fruitless.

For everyone.

 

[Mustakafine]

God you physics guys are a bit uptight. The responders did not have to take any to answer this but it was a genuine question.

 

[FlexGunship]

God you physics guys are a bit uptight. The responders did not have to take any to answer this but it was a genuine question.

I think the reason you got a less than warm welcome is the fact that these answers are readily available to anyone with the inclination to look them up.

If you look back, both Ryan and Dickfore answered you correctly and with great compassion and willingness to explain. Dickfore gave you the formula to use, and Ryan gave you a nice explanation.

But that wasn't enough. You pressed on with additional posts. It would probably have been less frustrating if you weren't the seven millionth person to register an account on here for the sole purpose of belligerently rejecting Einstein, or fumblingly denouncing Gauss.

So, yes, this thread has proved frustrating for us "uptight physics guys."

 

[WannabeNewton]

God you physics guys are a bit uptight. The responders did not have to take any to answer this but it was a genuine question.

Don't worry about it. Your question is a perfectly valid question and most of the individuals here are not like what you, unfortunately, have experienced in this thread. Just remember that many of the vector space operations you learned usually apply to $\mathbb{R}^{n}$ and the minkowski space of SR has some different properties from $\mathbb{R}^{n}$ and in GR you have to use laws that do not have a preferred coordinate system.

 

[Passionflower]

A way to avoid all this 'confusion' is to use rapidities instead of speed.
Unlike speed rapidity is additive.

 

[harrylin]

Some friends and I were out at a pub last night and this question came up.

Although a pub discussion, here's my (and Einstein's) take on it - which seems to be slightly different from that of some others here.

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

Yes indeed, that is so by definition - although many people now use a different ("modern") definition of "relative speed". In Einstein's time (and still according to everyone in the eighties), the relative velocity equals the difference in velocities from your perspective.

As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

Each would see the other ball moving away at a speed less than that of light (as expressed by that equation that they threw at you. )

Does light propagate through space at the same speed regardless of the speed of it source relative to another object?

Yes, as far as we know the speed of light is independent of the speed of the source - and it has been verified to quite high precision. That is the main point of one of the "two postulates" of special relativity theory.Harald

 

[Dickfore]

A way to avoid all this 'confusion' is to use rapidities instead of speed.
Unlike speed rapidity is additive.

Right. Velocity is related to rapidity $\eta$ according to the expression:
$$\frac{v}{c} = \tanh{(\eta)}$$
Notice that the hyperbolic tangent has a modulus smaller than 1 for all real numbers.

 

[Mustakafine]

Flexgun – Oh I did try to look them up and got a whole whack of variant answers and lots of references to star trek.

seven millionth person to register an account on here for the sole purpose of belligerently rejecting Einstein, or fumblingly denouncing Gauss.

From what I can tell this is an open forum to discuss physics and related topics and questions. I came to this forum in good faith and intention to ask what I thought was a valid question. I did so politely and with nothing but good intent. Had I known before hand that I was the 7th million person to come and ask a stupid question about physics I would have never had posted on your elitist forum. Its almost like you think I am uneducated because someone chucked a formula at me and I did not understand it in layman’s terms.

This thread came about because of a conversation between my peers after learning about a test failure in a scram jet test flight, in a pub. At the table were all trained helicopter pilots in the British Army. All Commissioned Officers and all with university degrees, and all Sandhurst trained. Physics is what keeps our aircraft flying and we are all thankful for that. But if you came on a pilots forum, or in fact got in an aircraft with me the last thing I would do is mock you for asking a question that to me is basic knowledge but to you, regardless of your education, you simply do not know. I could for example hand over control of the aircraft and ask you to hover and give you a formula for why the aircraft can hover and expect you can actually from that formula hold a hover. But you would not be able to and we would smack into the ground in short order.

I could however with very basic instruction tell you what you need to do via the controls and with visual reference to hold a hover. Its as simple as riding a bike. Truly it is. Once your body figures it out the rest is physics.

The point to this post is very basic. I posted an honest question that I did not know the answer to. As Flexgun has stated my belligerent (which forms intent) and fumbling (which implies stupidity) interaction here is not wanted so I shall leave you guys to it. I have never come across any online community so hostile as this one.

 

[harrylin]

[..] The point to this post is very basic. I posted an honest question that I did not know the answer to. As Flexgun has stated my belligerent (which forms intent) and fumbling (which implies stupidity) interaction here is not wanted so I shall leave you guys to it. I have never come across any online community so hostile as this one.

Just have a look at a few other threads and you will see that you just had bad luck - most people here are friendly. Also, instead of ignoring the friendly ones and discussing with the hostile ones it's better to do the opposite!

And see the posting guidelines, such personal attacks are not allowed and you should report it:

hostile language will not be tolerated on Physics Forums. This includes [..] direct personal attacks or insults; snide remarks or phrases that appear to be an attempt to "put down" another member; and other indirect attacks on a member's character or motives.

Please treat all members with respect, even if you do not agree with them. If you feel that you have been attacked, and the moderators or mentors have not yet gotten around to doing something about it, please report it using the "Report" button. If you choose to post a response, address only the substantive content, constructively, and ignore any personal remarks.

-https://www.physicsforums.com/showthread.php?t=414380

So, welcome at Physicsforums

 

[ghwellsjr]

Some friends and I were out at a pub last night and this question came up.

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

Does light propagate through space at the same speed regardless of the speed of it source relative to another object?

You're right about the light from the moving balls being redshifted. In fact, you can tell the speed of an object just from the amount of redshift. The faster an object travels, the more the redshift. Redshift is caused by the increasing distance between two objects moving away from each other and thus the increasing time it takes for signals to travel between them. At 0.5c the redshift is so great that the balls would become invisible. But let's say the balls and the launcher have some sophisticated electronics in them so that they can emit a radio pulse at, say, once per second, and they have receivers in them to detect these pulses from the other objects and measure how far apart they are in time. Repetitive pulses like these are the same effect as redshift, which is a reduction in the frequency of light caused by increasing relative motion, but it is easier for us to talk about the time intervals between pulses which is the reciprocal of the frequency. So a redshift in frequency is the same as a lengthening of time intervals.

So let's start at the beginning and assume the launcher emits a pulse at the exact moment that it launches the two balls in opposite directions at 0.5c, one half the speed of light. We'll call this "Pulse0". First, let's focus on how one of the balls will detect the pulses from the launcher. Where will it be one second later when Pulse1 is emitted? It will be half way between Pulse0 and the launcher, correct? What about when Pulse2 is emitted from the launcher? Now the ball will still be half way between Pulse0 and the launcher but Pulse1 will have reached the ball, correct? So what will the electronics on the ball measure as the time interval between the first two pulses emitted by the launcher? Well, if it weren't for relativity, it would measure two seconds, but we know that moving clocks tick slower than normal when they are traveling at high speeds. At 0.5c, the timing circuitry in the ball will take 1.1547 seconds per tick. Now if you have two clocks ticking at different rates, the slower one will measure time intervals as shorter than the other clock. As a result, the ball measures the time interval between the pulses coming in from the launcher at 2 seconds divided by 1.1547 which is 1.73205 seconds.

We could repeat the same analysis for the other ball but since the geometry is symmetrical we can say right off the bat that both balls make the same measurement.

Now, what about the launcher's measurement of the "redshift" from the two balls? Let's assume the first pulse was emitted at launch time. The second pulse will be emitted 1.1547 seconds later because the timing circuitry in the balls is running slow by that amount. How far away from the launcher will each ball be when this happens? That's easy, 0.5 times 1.1547 or 0.57735 light seconds away. How long will it take for this pulse to travel back to the launcher? That's easy too, 0.57735 seconds. That means the total time will be 1.1547 plus 0.57735 or 1.73205 seconds.

So we see that the balls both measure the same "redshift" for the pulses coming from the launcher as the launcher measures of the "redshift" coming from the two balls. This illustrates the symmetrical relationships that are characteristic of Special Relativity. Two objects in relative motion emitting the same signal will be measured by each other identically. Note that we have done our analysis on just the first two pulses but we would get the same results if we continued for any number of other pulses.

Now let's see what each ball measures of the other ball's pulses. We already know when and where each ball is when it emits its second pulse but now we have to figure out where the other ball will be when it receives that second pulse because this determines the amount of "redshift" that each ball sees of the other ball. That was your question.

First, we can realize that we have already solved part of the problem as we know how long that second pulse has taken to get from the first ball to the launcher, 1.73205 seconds and we know where the second ball is at that time traveling at half the speed of light, 0.866025 light seconds from the launcher. All we have to do is figure out how long it will take for the pulse traveling at c to catch up to the second ball traveling at 0.5c. That's really pretty easy to do because we know that the pulse is traveling twice as fast as the ball so in another 1.73205 seconds the ball will have traveled another 0.866025 light seconds.

So the total time for the second pulse from the first ball to reach the second ball is 1.73205 plus 1.73205 or 3.4641 seconds. Now how long will this second ball measure this time to be? Since its clock is running slow by 1.1547 it is 3.4641 divided by 1.1546 or 3 seconds (rounded off). If we take the reciprocal of this, we get the conventional factor for redshift of 0.33333.

So instead of the redshift being zero (what you get at the speed of light), which is what you and your friends in the pub thought, it is only a factor of 1/3.

Now this is probably too much to share with your friends but maybe what will satisfy them is for you to point out to them that when you are considering redshifts between a "stationary" observer and a "moving" observer, the time dilation factor is applied just once but between two observers moving in opposite directions it is applied twice so the speed of light is never reached.

 

[pervect]

Some friends and I were out at a pub last night and this question came up.

If you were in space and you had two balls. You threw one ball in one direction at half the speed of light. You through the other ball in exactly the opposite direction then relative to each other they would be traveling the speed of light.

As the observer in the center I would be able to see light from both object but it would red shifted quite a bit. But what would an observer on each of the balls see of the opposite ball?

Does light propagate through space at the same speed regardless of the speed of it source relative to another object?

Going back to the original question, with just a little bit of logic you can work out what happens. The approach used is sometimes called k-calculus, but it doesn't involve any calculus.

You've already noticed that there is some doppler shift factor, k, that redshifts the light. Let's put you and the two balls all in the same spot at t=0.

So, if you emit a signal at time t, it will be received by both balls at time kt, where k is the doppler shift factor. If you think of the light wave as having some frequency, say 500 trillion cycles per second, then you can see that in one second, by your watch, you emit 500 trillion wavefronts. For you the spacing between wavefronts is 1/ 500 trillionth's of a second, for the receiver it's k times that. So the last wavefront of those 500 trillion oscillations in the one second interval will arrive at the ball in k seconds.

The other thing you need to know is that the situation is symmetric, so that if the ball emits a signal at time t by it's watch, it arrives at time kt by yours.

Therefore what happens is that a signal emitted at time t by one ball gets received by you at time kt. And you can retransmit that signal to the other ball, and find that it will be received at time k^2*t. The doppler shift factor k multiples.

You can use this to work out how velocities add, it's not the familiar formula you might expect.

The only mildly tricky thing is that the doppler shift factor k is the same for you to the ball as it is for the ball to you. This is true by physical experiment and by the assumptions of relativity, but isn't consistent with the idea that light has some velocity relative to "space". There isn't any way to determine such a velocity, people have tried and failed - see the Michelson Morley experiment where people tried to measure the Earth's velocity through "space".

 

[ghwellsjr]

Thanks for pointing this out, pervect, I did not know this, but as I look back at my example, it points this out. I had been focusing on the fact that the time dilation is "applied" twice but that is not a simple squared function.

Here are the added details: I had calculated the measured "redshift" time interval as 1.73205 seconds between the launcher and either one of the balls and between the two balls as 3 seconds which is 1.73205 squared. So again, thanks for bringing this to my attention. It might actually be a more cogent approach than mine. Well, it's definitely more cogent but you never know what approach might make more sense to another person. I really like that fact that there are so many different ways to come to the same conclusions in Special Relativity and they are all, of course, correct.

 

[harrylin]

[..] The only mildly tricky thing is that the doppler shift factor k is the same for you to the ball as it is for the ball to you. This is true by physical experiment and by the assumptions of relativity,

Yes of course.

but isn't consistent with the idea that light has some velocity relative to "space".

You surely mean it differently from the way you said it, for that idea happens to be the second postulate! As Einstein formulated it in his 1905 introduction: "another postulate [..] that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body".

There isn't any way to determine such a velocity, people have tried and failed - see the Michelson Morley experiment where people tried to measure the Earth's velocity through "space".

Yes, it's the same as with Newton's mechanics. Harald