Check your time dilation intuition

[Genady]

Imagine a uniformly cooling white dwarf of uniform temperature. A cosmic ray is flying through it. Will the ray observe it cooling faster, slower, or with the same rate as a rate observed by an external observer which is at rest relative to the dwarf?

[TheVat]

Well, the cosmic ray's clock is moving more slowly relative to an observer on, say, a nearby planet or space station.  So if the ray were passing through an object of negligible mass, then it would observe that's object's clocks as moving faster, therefore cooling faster.

But that white dwarf's mass is not negligible, so its clock is also moving slower to an observer on the nearby planet.  (even on Earth, with it's small mass, the center of the Earth has aged 2.5 years less than its surface)  

So it would depend on the competing effects of gravitational time dilation and relative velocity time dilation.  Generally, given that cosmic rays (like protons) move very close to c, I would think that would be the stronger effect and the proton would "see" the star as cooling faster.  

[Genady]

42 minutes ago, TheVat said:

Well, the cosmic ray's clock is moving more slowly relative to an observer on, say, a nearby planet or space station.  So if the ray were passing through an object of negligible mass, then it would observe that's object's clocks as moving faster, therefore cooling faster.

But that white dwarf's mass is not negligible, so its clock is also moving slower to an observer on the nearby planet.  (even on Earth, with it's small mass, the center of the Earth has aged 2.5 years less than its surface)  

So it would depend on the competing effects of gravitational time dilation and relative velocity time dilation.  Generally, given that cosmic rays (like protons) move very close to c, I would think that would be the stronger effect and the proton would "see" the star as cooling faster.  

Hmm ... I don't say if the answer is right or wrong, but I think that comparing the ray's clock with the object's or the dwarf's clock is confusing. The question compares the ray's observation with the external observer's observation, and I think that only their clocks should be compared.

For definiteness, less say that the external observer is an astronomer on Earth, i.e., very far from the dwarf.

[TheVat]

 

Then we have something like the muon paradox, don't we?  To the muon, time is moving slower on Earth as the Earth speeds towards it at near-c, while to the Earth observers the muon's clock is slower and so it is not decaying as fast as expected.  

So the cosmic ray observes the WD as speeding towards it near-c, and the WD aging more slowly, therefore cooling more slowly.  But the rest frame of the observer sees the CR/muon as aging more slowly.  So you can't really reconcile those observations.

And, afterthought, there is an almost instantaneous velocity change, because of the vector reversal for the CR muon.  One nanosecond, the white dwarf is approaching the muon, but the next it is receding.   And there is also a Doppler effect on the WD spectrum that the muon can observe.  (edit:  I should use proton CR because it will last longer than a muon, right?)

So, back to your OP question:  the perspective that matters is the muon, which will see the star cooling more slowly due to its relative velocity.  To an earth observer, which shares a rest frame with the dwarf, the dwarf will be cooling normally because they (observer and star) are co-moving.  

Is the instantaneous velocity change important here?  Also, would a proton make it through the packed neutrons core of a WD?  Would it lose a lot of its velocity if it did?  

[Genady]

Let's check each effect in turn. Doppler. No effect since the CR can measure the temperature "by touch" or in the direction perpendicular to its velocity.

Velocity reversal. I don't see any effect. It does not affect the temperature reading. The velocity does not reverse relative to the astronomer. Anyway, time dilation depends on speed and does not depend on velocity.

30 minutes ago, TheVat said:

would a proton make it through the packed neutrons core of a WD?

If it does not, we compare the observations it does before its demise.

 

31 minutes ago, TheVat said:

Would it lose a lot of its velocity if it did?  

Perhaps, but it will only weaken the effect.

[exchemist]

3 hours ago, Genady said:

Imagine a uniformly cooling white dwarf of uniform temperature. A cosmic ray is flying through it. Will the ray observe it cooling faster, slower, or with the same rate as a rate observed by an external observer which is at rest relative to the dwarf?

Purely from the SR viewpoint, I presume that, in the frame of reference of the cosmic ray, it is the dwarf that is moving fast relative to it, so it should see it exhibit time dilation, i.e. cool slower than an observer in the frame of reference of the dwarf. 

However, this thing also has a big gravitational field, so I presume GR may also come into it. But I don't have a good feel for how GR works- especially from the frame of reference of the entity affected by the field it - so I'll need to wait for the right answer to be revealed, I suppose.

[TheVat]

Okay,  then the significant observation is the proton watching the center of dwarf approach at (I checked) .996 c.  Using Lorentz,

 γ = √(1 - v²/c²) 

the dwarf is now seen by proton as a pancake that is shortened to .089 its rest frame diameter.  The proton finds its path is less than a tenth the distance of a rest frame.   And all clock-like functions in the dwarf are perceived as slower, happening at 8.93 percent of the dwarf's proper time.  

The proton is essentially free-falling as it moves in, so would GR be significant?

I mean it is already at 99.6 c, so how much would the gravity pull of the star accelerate it?  

Edited September 5, 2023 by TheVat
forgot equation

[Genady]

The only way to avoid confusion of what slows relative to what is to look at things which are frame independent. The fundamental frame independent things are events. 

So, let's take two events. Event 1: the proton takes a temperature measure. Event 2: 1 ns later on its clock, the proton takes another temperature measure. The change in temperature between the two events is 10 degrees (on some scale). The rate of cooling observed by the proton is 10 degrees/ns.

For the astronomer on Earth, the proton's clock moves slow. While it has advanced 1 ns, the astronomer's clock has advanced 2 ns. Thus, the astronomer observes the same cooling of 10 degrees in 2 ns, i.e., 5 degrees/ns. 

So, the proton observes faster cooling than the astronomer.

A gravitational time dilation adds to the effect. It does not depend on speed, but on the strength of gravity at time and place of the event. Both events occur in the gravitational field of the star, which perhaps is stronger than the gravitational field experienced by the astronomer. This makes the proton's clock run even slower. Let's say, while the proton's clock has advanced 1 ns, the astronomer's one has advanced 2.5 ns. Thus, the astronomer observes the cooling rate of 4 degrees/ns compared to the proton's 10 degrees/ns.

[md65536]

4 hours ago, TheVat said:

Well, the cosmic ray's clock is moving more slowly relative to an observer on, say, a nearby planet or space station.  So if the ray were passing through an object of negligible mass, then it would observe that's object's clocks as moving faster, therefore cooling faster.

That's not how SR works. The observations are reciprocal, not absolute. If the planet observes the cosmic ray's clock ticking slowly relative to its own due to relative motion, the cosmic ray observes the planet's clock also ticking slowly relative to its own.

[TheVat]

3 hours ago, md65536 said:

That's not how SR works. The observations are reciprocal, not absolute. If the planet observes the cosmic ray's clock ticking slowly relative to its own due to relative motion, the cosmic ray observes the planet's clock also ticking slowly relative to its own.

Of course, and I corrected in my subsequent post.  I actually meant to say the correct way, I was typing all this distracted and couldn't use the edit in time.  If you read my two later posts hopefully it is clear that I understand the reciprocal nature of observations from the FoRs.  

We had a wildfire nearby, so I'm checking other posts to see if other errors happened.   Sorry for confusion.  

[iNow]

1 hour ago, TheVat said:

I'm checking other posts to see if other errors happened.   Sorry for confusion.  

Completely unacceptable. How dare you be human, or worse still, only as good as ChatGPT!

[TheVat]

5 hours ago, Genady said:

The only way to avoid confusion of what slows relative to what is to look at things which are frame independent. The fundamental frame independent things are events. 

So, let's take two events. Event 1: the proton takes a temperature measure. Event 2: 1 ns later on its clock, the proton takes another temperature measure. The change in temperature between the two events is 10 degrees (on some scale). The rate of cooling observed by the proton is 10 degrees/ns.

For the astronomer on Earth, the proton's clock moves slow. While it has advanced 1 ns, the astronomer's clock has advanced 2 ns. Thus, the astronomer observes the same cooling of 10 degrees in 2 ns, i.e., 5 degrees/ns. 

So, the proton observes faster cooling than the astronomer.

 

This is sort of clear, but the measurements by the proton-man are still frame dependent.  In his one nanosecond, he travels farther in the star interior which is Lorenz contracted.   And temperature is a measure of average kinetic energy.  Something seems off, but...as @iNow observes I am afflicted with human fallibility.  😀

Great thread, btw, as many of yours are.  

[Genady]

30 minutes ago, TheVat said:

temperature is a measure of average kinetic energy

It's average kinetic energy of a random motion of molecules. We cannot boil water by running around a kettle, unfortunately. Temperature measurements themselves are not affected.

I want to point to the origin of this OP. It started with this exercise from MTW:

My explanation above is my partial answer to the question at the end. I've added the conditions of uniform temperature in the OP to make the second term on the right vanish. OTOH, I've added in my explanation the gravitational time dilation effect, which is not considered in the exercise.

Thank you for the last remark. 

[Markus Hanke]

4 hours ago, Genady said:

We cannot boil water by running around a kettle

Very nicely put  +1

[TheVat]

10 hours ago, Genady said:

It's average kinetic energy of a random motion of molecules. We cannot boil water by running around a kettle, unfortunately. Temperature measurements themselves are not affected.

LoL running around kettle.  If you have a temperature probe which is plunging through the kettle at relativistic speed, wouldn't it be impacted by molecules with an average velocity that is higher?  Moving through the star, your proton-man would get higher temperature measurements, as atoms whacked into the probe at near-c.  (i am being whimsical, because I know that a proton taking measures is purely a gedankenexperiment)  (and we still get your correct answer  which is faster cooling because there are two readings - we can just as easily get this from Lorentz contraction, where in one nanosecond by Proton's clock, it has traveled farther through the star, and more time with more cooling has passed for the star and observers in its rest frame)

[Genady]

The probe would read higher energy, but that reading will not be a temperature. To get the temperature from that reading, the peculiar motion will need to be subtracted.

Like we subtract the peculiar motion of Earth to get the isotropic reading of CMB.

[Janus]

Three words: Relativity of simultaneity.

For the cosmic ray, the white dwarf would not be of uniform temperature.

In the astronomers frame, as the ray enters one side of the star, the opposite side will be, at that moment, at the same temperature.

But for the cosmic ray, the events of the opposite sides being the same temp would not be simultaneous. The exit point would already be cooler than the entry point at the moment of initial contact.  The star cools at a slower rate as a whole, but since the exit point had a "head start" the Temp there will be much lower when the cosmic ray exits than you would get by just taking the entry point temp an subtracting the rate of cooling times the time it took to cross the star.

 

[Markus Hanke]

There’s also the matter of the star’s shape to consider - in the frame of the particle, the star isn’t spherical, but a flattened disc along the direction u. Thus both the time and the space parts of the gradient must be considered, and treated in the same way (for covariance) - therefore (2.37) is indeed reasonable so far as its general form is concerned.

[Genady]

After considering the last two comments, by @Janus and @Markus Hanke, I've realized that my attempt

On 9/5/2023 at 11:07 PM, Genady said:

to make the second term on the right vanish

has failed. The spatial distance in the rest frame between the two measurement events has to be accounted for as well.

Thank you.