Jump to content
Halc

Ballistic 'not-light' cone question

Recommended Posts

Posted (edited)

I am trying to figure out ballistic trajectories over cosmological distances.  All the literature seems to speak only of light, and shows worldlines only of comoving objects, not objects with motion relative to the comoving frame.

So suppose some early galaxy exists 1.7 billion years after the big bang, 12 billion years ago.  Some star emits light and neutrinos at that event. The light gets here today, so the galaxy at the time was something on the order of 4.7 GLY away (proper distance measured along line of constant cosmic time) from here (the place where our solar system will eventually be).  Due to expansion, the star is increasing its proper distance from us at something like 2.5c at the time, although the same galaxy is currently receding at more like 2.1c today.

My question is about the neutrino, which is not light, but rather a simplistic ballistic object. It is moving at nearly light speed relative to the star that emitted it. Does it get here? When?  If it doesn't get here close to today, how far does it get in the 12 billion years (cosmic time)? I don't know how to apply expansion of space into an integration of its movement, and special relativity is not help whatsoever. The light is easy because it always moves locally at c. Both light and neutrino increase their proper distance from 'here' at first since they're both well beyond the Hubble distance at the time but still well inside the particle horizon.

Does the speed matter?  Two such ballistic objects are ejected, one at .9999c and the other at .999999c.  Is there a major difference in when the two get here?  The neutrino has a lot of 9's.

 

Edited by Halc

Share this post


Link to post
Share on other sites

In supernova 1987A the arrival was almost simultaneous. Neutrinos actually arrived first (by about 3 hours), owing to the fact that photons had to scatter in getting out of the event, while the neutrinos were essentially unimpeded

https://en.wikipedia.org/wiki/Measurements_of_neutrino_speed#Supernova_1987A

 

For the difference between 0.99990c and 0.999999c you can still use d = vt

 

 

Share this post


Link to post
Share on other sites
Posted (edited)
28 minutes ago, swansont said:

In supernova 1987A the arrival was almost simultaneous.

I know that, but that supernova occurred barely outside our galaxy, hardly enough distance for cosmic expansion to play a role.

I'm asking about a hypothetical event 5 orders of magnitude further away than that, from a source with a recession speed greater than c.

Edited by Halc

Share this post


Link to post
Share on other sites
41 minutes ago, Halc said:

I know that, but that supernova occurred barely outside our galaxy, hardly enough distance for cosmic expansion to play a role.

I'm asking about a hypothetical event 5 orders of magnitude further away than that, from a source with a recession speed greater than c.

What role do you expect expansion to play on one that does not affect the other in the same way?

Share this post


Link to post
Share on other sites

Well, that's my question.  I'm not asserting anything here.

Suppose I launch a rock at half light speed.  The proper distance between it and some object 7 GLY away is constant, making it sort of stationary relative to it in a way. But over time, the Hubble 'constant' goes down, and that distant object is no longer stationary relative to our rock, so the rock starts gaining on it.  Given that logic, perhaps it will not only eventually get there, but it will pass by it half light speed. That would imply that expansion has no effect on peculiar velocity over time.  If so, then our neutrino gets here pretty much the same time as the light. Not sure if that logic holds water. 

Share this post


Link to post
Share on other sites
On 8/27/2020 at 12:49 PM, Halc said:

Suppose I launch a rock at half light speed.  The proper distance between it and some object 7 GLY away is constant, making it sort of stationary relative to it in a way. But over time, the Hubble 'constant' goes down, and that distant object is no longer stationary relative to our rock, so the rock starts gaining on it.  Given that logic, perhaps it will not only eventually get there, but it will pass by it half light speed.

I don't think that's right. I think the rock would have to lose energy to the expansion, which means it would arrive at a lower speed relative to its destination than its speed relative to the source when launched.

Suppose a photon and a neutrino were sent, with the same energy. Say they arrived roughly at the same time, and the photon was red-shifted to half its energy. Wouldn't the neutrino need to have roughly half the energy it was sent with?

In this example, the neutrino's speed is so close to c that it can lose half its energy to a decrease in speed and still be moving at a speed very close to c. In the case of a slower rock, I think it'd lose even more energy since it would take longer to travel.

Since the expansion wouldn't affect the rest mass, I suspect it's a fraction of only the kinetic energy (of the rock or neutrino or photon) that is lost, not total energy.

 

Share this post


Link to post
Share on other sites
On 8/27/2020 at 2:49 PM, Halc said:

Well, that's my question.  I'm not asserting anything here.

Suppose I launch a rock at half light speed.  The proper distance between it and some object 7 GLY away is constant, making it sort of stationary relative to it in a way. But over time, the Hubble 'constant' goes down, and that distant object is no longer stationary relative to our rock, so the rock starts gaining on it.  Given that logic, perhaps it will not only eventually get there, but it will pass by it half light speed. That would imply that expansion has no effect on peculiar velocity over time.  If so, then our neutrino gets here pretty much the same time as the light. Not sure if that logic holds water. 

That’s a different example, in several ways. A rock is not interchangeable with a neutrino.

4 hours ago, md65536 said:

I don't think that's right. I think the rock would have to lose energy to the expansion, which means it would arrive at a lower speed relative to its destination than its speed relative to the source when launched.

Suppose a photon and a neutrino were sent, with the same energy. Say they arrived roughly at the same time, and the photon was red-shifted to half its energy. Wouldn't the neutrino need to have roughly half the energy it was sent with?

In this example, the neutrino's speed is so close to c that it can lose half its energy to a decrease in speed and still be moving at a speed very close to c. In the case of a slower rock, I think it'd lose even more energy since it would take longer to travel.

Since the expansion wouldn't affect the rest mass, I suspect it's a fraction of only the kinetic energy (of the rock or neutrino or photon) that is lost, not total energy.

 

A photon losing half its energy does not imply a neutrino or a rock would. The wavelength relationship to energy is not the same for a photon and a massive particle.

Share this post


Link to post
Share on other sites
Posted (edited)
5 hours ago, md65536 said:

I don't think that's right. I think the rock would have to lose energy to the expansion, which means it would arrive at a lower speed relative to its destination than its speed relative to the source when launched.

That was sort of my thinking at the OP,  but I wasn't sure. Energy is conserved, so where does it go?  The 'expansion' doesn't seem like a form of energy that can receive it. Maybe all the slowing down stuff powers the dark energy, translated into acceleration of expansion.

Energy is conserved in an inertial frame, but not from one frame to another, so the rock has no kinetic energy in the frame in which it is stationary, but has KE in the frame of Earth from which it was launched.  Relative to the curved FLRW frame (also known as cosmological or comoving frame), the rock has been imparted with a sort of absolute KE which is derived from its peculiar velocity. The reply from @swansont thus made some sense in that light.  KE relative to an inertial frame makes sense only for a local object. What's the KE of some star receding from us at 2.3c?*  It isn't computable, but its KE based on its peculiar velocity does make sense.  This is why I was willing to accept that answer, even though it wasn't my initial guess.

*recession velocities are not measured in units that add relativistically, but rather add linearly, per the FLRW frame, so 2.3c doesn't have the usual meaning when computing something like KE.

Quote

Suppose a photon and a neutrino were sent, with the same energy. Say they arrived roughly at the same time, and the photon was red-shifted to half its energy. Wouldn't the neutrino need to have roughly half the energy it was sent with?

Good argument, but I've heard that the photon doesn't lose energy since relative to Earth it was always red-shifted to its observed measurement,.  But as I said, 'relative to Earth' loses meaning over these sorts of distances. The universe isn't Minkowskian.

The photon definitely does lose energy in the FLRW frame.  Its peculiar velocity must be c, so its peculiar energy (is there such a term??) must be lost to something in that frame. That or the curved frame does not have the property of conservation of energy.  How do the absolutists (which typically select this frame as the absolute one) deal with that? Violation of thermodynamic law has sunk an awful lot of theories.

Quote

In this example, the neutrino's speed is so close to c that it can lose half its energy to a decrease in speed and still be moving at a speed very close to c. In the case of a slower rock, I think it'd lose even more energy since it would take longer to travel.

I'm not going to disagree, but I'm getting conflicting answers. Is there an article or other page somewhere that discusses this? I didn't know how to frame a search on stack exchange.

Anyway, you seem to address one question in the OP: "Two such ballistic objects are ejected, one at .9999c and the other at .999999c.  Is there a major difference in when the two get here? "  Apparently the number of 9's makes a significant difference.  The speed is nearly identical, but the energy/mass is not.

Quote

Since the expansion wouldn't affect the rest mass, I suspect it's a fraction of only the kinetic energy (of the rock or neutrino or photon) that is lost, not total energy.

That's good, since it would be harder to swallow a view where the rock lost proper mass along the way, fading to dust...

50 minutes ago, swansont said:

That’s a different example, in several ways. A rock is not interchangeable with a neutrino.

Why not?  A neutrino seems to be a very small and typically fast rock. It has proper mass. It has an inertial frame in which it is stationary, all just like the rock. The photon has none of this.

Quote

A photon losing half its energy does not imply a neutrino or a rock would. The wavelength relationship to energy is not the same for a photon and a massive particle.

That would seem to need more justification than just an assertion. I thought it was a valid point, and cannot easily think my way around it, especially when it is expressed as 'peculiar energy' as I put it.

Edited by Halc

Share this post


Link to post
Share on other sites
4 hours ago, Halc said:

That was sort of my thinking at the OP,  but I wasn't sure. Energy is conserved, so where does it go?  The 'expansion' doesn't seem like a form of energy that can receive it. Maybe all the slowing down stuff powers the dark energy, translated into acceleration of expansion.

There's this: https://www.forbes.com/sites/startswithabang/2018/07/28/ask-ethan-where-does-the-energy-for-dark-energy-come-from/

Including:

Quote

When you have particles interacting in a static background of spacetime, energy is truly conserved. But when the space through which particles move is changing, the total energy of those particles is not conserved.

[...]

There was a paper written in 1992 by Carroll, Press, and Turner, which dealt with this exact issue. In it, they state:

    …the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically, one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρc^2.

Unfortunately I can't find anything about how expansion affects the energy of moving objects. I'd assumed it would be the same as with light.

Share this post


Link to post
Share on other sites
10 hours ago, Halc said:

 

Energy is conserved in an inertial frame, but not from one frame to another,  

But with expansion you don't have a single frame. 

 

Quote

 

Anyway, you seem to address one question in the OP: "Two such ballistic objects are ejected, one at .9999c and the other at .999999c.  Is there a major difference in when the two get here? "  Apparently the number of 9's makes a significant difference.  The speed is nearly identical, but the energy/mass is not.

What will make a difference is the length of travel, and if expansion is an issue, that the v of a massive object will diminish.

You should still be able to apply vt to the problem, but v will now be a variable.

A 10^-4 fractional difference in the speed means 10^-4 difference in arrival time over some distance. If that's a light year, then it's 10^-4 of a year, or 3.15 x 10^3 seconds. That's less than an hour. At 100,000 LY, it's about a year.

Quote

 

That's good, since it would be harder to swallow a view where the rock lost proper mass along the way, fading to dust...

Why not?  A neutrino seems to be a very small and typically fast rock. It has proper mass. It has an inertial frame in which it is stationary, all just like the rock. The photon has none of this.

The mass of the object matters , since the wavelength is h/p

Quote

That would seem to need more justification than just an assertion. I thought it was a valid point, and cannot easily think my way around it, especially when it is expressed as 'peculiar energy' as I put it.

Sorry, I though this was readily apparent. For a photon the wavelength shift is the is the same factor as the energy shift, since energy is hc/lambda - twice the wavelength is half the energy. But for a massive object, twice the wavelength is half the momentum, not half the energy.

 

Share this post


Link to post
Share on other sites
Posted (edited)
8 hours ago, swansont said:

What will make a difference is the length of travel, and if expansion is an issue, that the v of a massive object will diminish.

You should still be able to apply vt to the problem, but v will now be a variable.

A 10^-4 fractional difference in the speed means 10^-4 difference in arrival time over some distance. If that's a light year, then it's 10^-4 of a year, or 3.15 x 10^3 seconds. That's less than an hour. At 100,000 LY, it's about a year.

Time t will vary along with v, but also with d (length of travel), which itself should vary, and depends on time and expansion. You're giving an example where the differences in v and t are small, and difference in d is negligible? (Due to being nearby and/or negligible rate of expansion? Or assuming a fixed distance of expanded space, instead of being sent from a common source that is receding over time?)

However if we were talking about an object at the cosmological horizon, light from it would arrive after infinite time. For any speed less than c, there should be another nearer horizon, from which an object traveling at that speed would take an infinite time to arrive. For a speed like .999999c, if someone is in between that speed's horizon and the cosmological horizon, then light from them would arrive after finite time, but a rock at .999999c would never arrive.

So you can have "nearby" cases where expansion adds less than a second to the time between arrival of light and arrival of neutrinos, and the extreme distant case where the neutrinos take infinitely longer to arrive.

8 hours ago, swansont said:

For a photon the wavelength shift is the is the same factor as the energy shift, since energy is hc/lambda - twice the wavelength is half the energy. But for a massive object, twice the wavelength is half the momentum, not half the energy.

That would imply expansion does slow down an object, if the object is redshifted. If light from a distance source was redshifted to twice the wavelength by expansion, would a massive object launched from the source and traveling at near c (and arriving nearly at the same time as the light) also arrive with nearly twice the wavelength it was launched with?

Edited by md65536

Share this post


Link to post
Share on other sites
50 minutes ago, md65536 said:

Time t will vary along with v, but also with d (length of travel), which itself should vary, and depends on time and expansion. You're giving an example where the differences in v and t are small, and difference in d is negligible? (Due to being nearby and/or negligible rate of expansion?)

It’s not my example. I’m giving a response to the example.

 

50 minutes ago, md65536 said:

That would imply expansion does slow down an object, if the object is redshifted. If light from a distance source was redshifted to twice the wavelength by expansion, would a massive object launched from the source and traveling at near c (and arriving nearly at the same time as the light) also arrive with nearly twice the wavelength it was launched with?

That’s my understanding - everything gets redshifted. Cosmology says there is a cosmic neutrino background, with a temperature below 2K (neutrinos decoupled earlier than photons)

Share this post


Link to post
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.