Jump to content

Recording or perceiving the activity of an oncoming object


geordief

Recommended Posts

I have been reliably informed that ,if we could listen in directly to an object like a spacecraft that was approaching us at a relativistic speed then any ongoing conversations would be heard slowed down rather than  sped up as  my intuition have supposed. 

 

Since we also have the relativistic doppler  effect (ie in addition to time dilation) am I to think of a conversation that would be both high pitched and slowed down?

 

 

 

 

And what about the same situation but purely with sound?

 

If we had a  plane approaching  us at just below the speed of  sound  could we reflect a sonic beam against the cabin window and eavesdrop on the pilots conversation?

Would it be high pitched and simultaneously  slowed down?

Edited by geordief
Link to comment
Share on other sites

  • geordief changed the title to Recording or perceiving the activity of an oncoming object
1 minute ago, Lorentz Jr said:

What do you mean by "listen in directly"?

Well ,not by later recovering a recording of any conversation (that is obvious ,I suppose)

But hopefully  using whatever method comes closest to how we listen  or watch something in our own frame of reference (there is forcibly a delay )

 

So in my mind ,for the approaching spacecraft I imagined us sending a laser beam against perhaps  a  window and capturing a recording of the window's vibrations in the light that was reflected back to the emitter.

 

For the plane  travelling  just below the speed of sound I had a similar (same) method in mind

 

So ,as "directly "as possible (perhaps there are better methods?)

Link to comment
Share on other sites

15 minutes ago, geordief said:

Well ,not by later recovering a recording of any conversation (that is obvious ,I suppose)

But hopefully  using whatever method comes closest to how we listen  or watch something in our own frame of reference (there is forcibly a delay )

 

So in my mind ,for the approaching spacecraft I imagined us sending a laser beam against perhaps  a  window and capturing a recording of the window's vibrations in the light that was reflected back to the emitter.

 

For the plane  travelling  just below the speed of sound I had a similar (same) method in mind

 

So ,as "directly "as possible (perhaps there are better methods?)

In general I think the Doppler effect would speed up the conversation, just as it would raise the pitch of the speakers. Relativistic time dilation is another thing. 

 

Link to comment
Share on other sites

5 minutes ago, exchemist said:

In general I think the Doppler effect would speed up the conversation, just as it would raise the pitch of the speakers. Relativistic time dilation is another thing. 

 

Would it  (the time dilation)slow the speed of the conversation down without affecting the pitch? (Maybe it would affect the pitch too)

 

Would there be a particular relative speed where the doppler speeding effect would exactly counterbalance the relativistic time dilation slowing  effect?

Link to comment
Share on other sites

7 minutes ago, geordief said:

Would it  (the time dilation)slow the speed of the conversation down without affecting the pitch? (Maybe it would affect the pitch too)

 

Would there be a particular relative speed where the doppler speeding effect would exactly counterbalance the relativistic time dilation slowing  effect?

No. The pitch is determined by the rate at which successive peaks and troughs in the sound waves arrive. The starting and stopping of every syllable of the conversation will also arrive at the same rate, since both are determined by the speed of sound relative to the receiver.

Link to comment
Share on other sites

For the time dilation effect, let's consider a simple scenario. The ship approaches the observer with the speed 0.6 (c=1). Time dilation for this speed is 1.25.

The ship sends light pulses every second. When the distance is 10 light-seconds, a signal goes off. It reaches the observer 10 seconds later, on the observer's clock.

Next signal goes out in 1 s on the ship's clock. That is 1.25 s later on the observer's clock. By then the ship is 0.75 light-seconds closer. The signal reaches the observer 9.25 s later, on the observer's clock.

Thus, the observer receives the second signal 1.25+9.25 = 10.5 s after the first signal has been sent, i.e. 0.5 s after receiving the first signal. 

Looks like the "conversation" speeds up rather than slows down. 

Is my accounting correct?

Link to comment
Share on other sites

2 hours ago, geordief said:

Since we also have the relativistic doppler  effect (ie in addition to time dilation) am I to think of a conversation that would be both high pitched and slowed down?

Relativistic effects don't change the shape of the waveform. Whatever happens to the overall speed of the signal will also happen to the frequencies.

2 hours ago, geordief said:

I have been reliably informed that ,if we could listen in directly to an object like a spacecraft that was approaching us at a relativistic speed then any ongoing conversations would be heard slowed down rather than  sped up as  my intuition have supposed. 

The most "direct" way to listen in to the signal would be to have an array of listeners along the spaceship's path, each station receiving a short snippet as the spaceship passes.

When you're in front of the spaceship, the signal will seem faster according to the Doppler formula, as Genady showed.

When the line between you and the spaceship is perpendicular to the ship's velocity, the signal will seem time dilated by the usual gamma factor.

So there must be some angle [math]\theta[/math] where the signal comes in at normal speed.

If a time [math]dt_s [/math] passes in Earth's reference frame between two events on the spaceship, the interval in the spaceship will be

[math]\displaystyle{ dt' = \frac{dt_s }{ \gamma} } [/math]

If the ship travels a distance ds, from y = dy to y = 0 (so [math] dy = ds \sin \theta [/math]), during this time, and a receiver is a (relatively large) distance y (along the y axis) below the x axis, the time for the first light signal to reach the receiver will be

[math]\displaystyle{ t_{r1} = \frac{y + dy}{c} }[/math]

and the total time from when the first signal is sent to when the second signal is received will be

[math]\displaystyle{  t_{r2} = \frac{ds}{v} + \frac{y}{c} }[/math]

So the time interval for the receiver will be

[math]\displaystyle{   dt_r = \frac{ds}{v} - \frac{dy}{c} }[/math]

Now we set [math] dt' = dt_r [/math] to calculate [math]\theta[/math]:

[math] \displaystyle{ \frac{dt_s }{\gamma} = \frac{ds}{v} - \frac{dy}{c} }[/math]

[math] \displaystyle{ \frac{ds }{\gamma v }= \frac{ds}{v} - \frac{ds \sin \theta}{c}} [/math]

[math] c = \gamma c - \gamma v \sin\theta [/math]

[math] \displaystyle{  \sin\theta = \frac{c}{v} \left( 1 - \frac{1}{\gamma}\right) } [/math]

Edited by Lorentz Jr
Link to comment
Share on other sites

BTW, if my numbers are correct, they show a curious relativistic "optical illusion":

In 0.5 s, the observer "sees" the ship 0.75 light-seconds closer. It looks like the ship moves faster than light!

Link to comment
Share on other sites

So the premise in the(my)OP was wrong .A misunderstanding no doubt on my part  from a long time ago.

 

Moving on,if we have a spacecraft approaching a black hole  activities btw it and the event horizon will seem speeded up,won't they?

 

An observer in the spacecraft will  see all the objects which had disappeared into the BH ahead of him ,can record them and send back the recorded  pictures of all that happened just outside the event horizon from the inception of the BH.

 

It would need a very powerful slingshot to send back the holiday snaps (and it's own deductions if appropriate)

Link to comment
Share on other sites

2 hours ago, geordief said:

An observer in the spacecraft will  see all the objects which had disappeared into the BH ahead of him

I'm guessing here, but I don't think it will happen. Assuming they all free fall from far-far away, he will not catch up with anything ahead of him. All that stuff will be redshifted to undetectability, and he will not see it.

Link to comment
Share on other sites

14 minutes ago, Genady said:

I'm guessing here, but I don't think it will happen. Assuming they all free fall from far-far away, he will not catch up with anything ahead of him. All that stuff will be redshifted to undetectability, and he will not see it.

Hadn't considered  that.But some of that stuff will be moving in the opposite direction to him.Can he see that?

Some stuff will be orbiting  the BH;might he see that as it swings around in his direction?

Link to comment
Share on other sites

30 minutes ago, geordief said:

Some stuff will be orbiting  the BH;might he see that as it swings around in his direction?

Yes, he can see that. But that stuff is not very close to the EH. The minimal stable circular orbit for non-rotating BH is 3 times farther from the singularity than EH. Anything below that will fall in.

Link to comment
Share on other sites

6 hours ago, Genady said:

For the time dilation effect, let's consider a simple scenario. The ship approaches the observer with the speed 0.6 (c=1). Time dilation for this speed is 1.25.

The ship sends light pulses every second. [...] observer receives the second signal [...] 0.5 s after receiving the first signal. 

Looks like the "conversation" speeds up rather than slows down. 

Is my accounting correct?

Yes, that's correct. The relativistic Doppler effect includes delay of light and SR's time dilation. The Doppler factor period_received/period_sent = sqrt((1+beta)/(1-beta)) where beta*c is velocity, and for beta=-.6 the factor is equal to 0.5.

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

6 hours ago, Genady said:

BTW, if my numbers are correct, they show a curious relativistic "optical illusion":

In 0.5 s, the observer "sees" the ship 0.75 light-seconds closer. It looks like the ship moves faster than light!

Yes it appears as if it's moving faster than c, but not faster than light, because incoming light appears to arrive at the same time it appears that it was sent! To look faster than that (v < -c), you'd have to see it arrive before you saw it leave, see it in 2 places at once, etc.

The incoming (outgoing) object would also appear stretched (compressed) by the relativistic Doppler factor, so if something could approach the speed of light you could see it arrive just after it left, appearing stretched along its entire path, as well as bright and "very" blue.

2 hours ago, geordief said:

But some of that stuff will be moving in the opposite direction to him.Can he see that?

Sure, like if you consider an object falling into a black hole and sending out signals. A signal sent outward as the event horizon passes, becomes a part of the event horizon. Then if another observer falls in, it could see that signal as the event horizon passes it.

I think that if both observers fell in at similar speeds, a photon sent exactly at the event horizon's passing wouldn't be redshifted, because where would the energy go? (It's not like the horizon itself stretches out.) I may be wrong. It seems, if you were both far above a black hole, and one went in, the other would see the one in front slow and redshift to the point of appearing stopped and invisibly dark. Then if the other followed, they'd see the one in front redshift less and less until appearing normal speed at the event horizon, and I think they'd then see the one in front blueshift increasingly redshift again and eventually "end" as there's some point where any "outward"-directed light from after that point will reach the singularity before you can catch up to it. That is, if the interior of the black hole has Schwarzschild geometry. (I think it would be blueshifted because the light you could receive would be increasingly later in the first object's life, it would have to appear sped up???)

Edit: On second thought it seems that light would get "stretched" in opposite directions on either side of the event horizon, and it would immediately increasingly redshift as soon as the event horizon passed you. However, I'm trying to figure it out intuitively and could be way off.

 

This is related to a topic that came up before, and I found it useful to see the paths of light on a Kruskal–Szekeres diagram, when talking about what infalling observers might see.

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

Edit2: Okay looking at those diagrams again, the paths of light are 45-degree diagonal lines, now I remember what the diagrams showed: The second infalling observer could see images of the first observer from when they were "above" where the 2nd observer is now (higher r value). The 2nd observer would hit the singularity while still being able to "see" the first, but could only see the 1st's life up to some minimum r, whose value depends on how long the 2nd went in after the 1st. That implies the second observer would see the 1st redshifted inside the black hole.

Edited by md65536
Link to comment
Share on other sites

6 hours ago, geordief said:

An observer in the spacecraft will  see all the objects which had disappeared into the BH ahead of him ,can record them and send back the recorded  pictures of all that happened just outside the event horizon from the inception of the BH.

Mostly no. Any light inside the event horizon will "fall" toward the singularity, but not all at the same rate (light sent inward will get there faster than light sent outward). No light from inside can go outside to the spaceship. Or, light from the spaceship can't go to places that light going past it can't... if we can't see what's in the black hole then either the spaceship can't see it or we can't see the spaceship.

The event horizon passes over the ship at the speed of light, locally. It can "see" light that's part of the horizon, but it can't send signals faster than the light on the horizon.

However once the spaceship enters the horizon, it could see some light from objects from inside the BH that fell in before it. Basically, it can fall toward the singularity more quickly than light directed outward does. Or, everything including light is moving toward the singularity, but you can catch up to light that is aimed toward you.

Link to comment
Share on other sites

On 3/3/2023 at 4:51 AM, Genady said:

Just wanted to add that it is a well-known phenomenon in astrophysics, the "superluminal motion". 

Superluminal motion - Wikipedia

They use "apparently faster-than-light motion" to mean the same as "apparently faster than c", so my earlier distinction between the 2 was pointless.

I wish they'd distinguish though, because "faster than light" can mean "faster than light measured correctly (measured differently than the object)" or it can mean "can beat light in a race (measured the same way as the object)", and they mean only the former. Then people get in arguments and it turns out they're saying the same thing but meaning 2 different things. If an object can appear to move faster than c, then light can appear to move faster than c, and "light can appear to move faster than light" is confusing unless it's clear that it's implying 2 different measurements.

Link to comment
Share on other sites

22 minutes ago, md65536 said:

They use "apparently faster-than-light motion" to mean the same as "apparently faster than c", so my earlier distinction between the 2 was pointless.

I wish they'd distinguish though, because "faster than light" can mean "faster than light measured correctly (measured differently than the object)" or it can mean "can beat light in a race (measured the same way as the object)", and they mean only the former. Then people get in arguments and it turns out they're saying the same thing but meaning 2 different things. If an object can appear to move faster than c, then light can appear to move faster than c, and "light can appear to move faster than light" is confusing unless it's clear that it's implying 2 different measurements.

I agree with your clarification. In fact, there are many cases when people say "faster than light" while meaning "faster than c."

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.