Speed of Message Relative to a Space Station: Relativistic Addition

[Saibot]

Homework Statement

A spaceship leaves a space station at 0.5 c. It sends a message back to the station, which it measures to be moving at 0.7 c. What is the speed of the message as seen by the station?

Relevant Equations

𝑢=(𝑣+𝑢′)/(1+𝑣𝑢′/𝑐^2 )

u is the speed of the message relative to the station
v is the speed of the spaceship relative to the station
u’ is the speed of the message relative to the spaceship

u=(-0.5c+0.7c)/(1+((-0.5c)(0.7c))/c^2 )
=0.2c/0.65=0.308c

This just seems way too high, and I'm not sure if I'm doing it right.

Thanks a million in advance.

 

[vela]

Your answer is correct.

 

[Saibot]

Thanks vela :)

I'm trying to self-teach this; you have no idea how much that reassurance helps.

 

[phinds]

@Saibot, just to be sure you are completely clear about this, there are two things going on there. One is the speed of a carrier wave and the other is the speed of the message it is carrying.

Say you send out a 10 second burst of information on a radio frequency carrier wave. It leaves the source at c and arrives at the destination at c, because EM radiation is always seen as traveling at c. The MESSAGE, however is frequency shifted (the carrier wave is red shifted) such that the message that took 10 seconds leaving the source takes about 32.5 seconds to arrive (using the figures of your example --- 32.5 = 10/.308) and thus has the speed .308c

This would, of course, require the receiver to have some ability to downshift upshift the message by the appropriate amount, otherwise, in this case as an example, a 10 second normal voice transmission would sound like a VERY slow drawl.

 

[jbriggs444]

@Saibot, just to be sure you are completely clear about this, there are two things going on there. One is the speed of a carrier wave and the other is the speed of the message it is carrying.

None of that is relevant to the homework problem at hand. This is plain old vanilla relativistic velocity addition. There are no carrier waves. There are no radio messages. The message is sent by way of a hypothetical sub-light mechanism.

Maybe the mechanism is a very fast rocket. Maybe the mechanism is a very slow particle beam. We do not know. We do not care. The problem states that the message is sent at a speed relative to the space ship of 0.7 c. So it is written. So it shall be. One ought not normally argue with the givens of a problem.

Say you send out a 10 second burst of information on a radio frequency carrier wave. It leaves the source at c and arrives at the destination at c, because EM radiation is always seen as traveling at c. The MESSAGE, however is frequency shifted (the carrier wave is red shifted) such that the message that took 10 seconds leaving the source takes about 32.5 seconds to arrive (using the figures of your example --- 32.5 = 10/.308)

Wait a minute. The duration of the received message will be dilated according to relativistic Doppler. That's not velocity composition. That's an entirely different factor. Yes, a real time message will be received at a lower pitch and have a longer duration.

That has absolutely nothing whatsoever to do with the speed of transit of a hypothetical message being carried by means of a radio signal. Or by a modulated sub-light particle beam for that matter. In vacuum, a radio message proceeds at a velocity of $c$ regardless of relativistic doppler.

Nor does relativistic doppler do much to alter the audio characteristics of a USB flash stick carried in a relativistic rocket.

 

[Saibot]

Indeed, the message is sent via another rocket of some sort (I made the question up). It's still nice to see your thinking behind a carrier wave though!

Thanks to both of you for taking the time to explain.

 

[PeroK]

Indeed, the message is sent via another rocket of some sort (I made the question up). It's still nice to see your thinking behind a carrier wave though!

Thanks to both of you for taking the time to explain.

How could you double check this yourself? You could do it a long way with the Lorentz Transformation and hopefully get the right answer. But, that's a bit messy.

What I would do is use a spreadsheet and take a fixed $v = 0.5c$ and calculate $u$ for a whole range of $u'$. You could do as many as you want, but I suggest $-1.0c, -0.9c, -0.8c \dots 0.9c, 1.0c$.

That will give you a feel for how $u$ depends on $u'$ in this case.

Once you've done that, of course, you can vary $v$ as well.

 

[PeroK]

PS have you ever checked this out? In this scenario, using $c =1$ for simplicity, we have:
$$u = \frac{v + u'}{1 + vu'}$$$$u' = \frac{-v+u}{1-vu}$$What happens if you plug the expression for $u'$ into the first equation? Do you end up with $u$? You must, but is that worth checking out?